# CPF University: EE Course (Lessons)



## Darell (Jun 29, 2004)

Mr. Al welcomes everybody!


Here are the links to the initial lessons. The following posts in this thread will contain each new lesson as it is created.
Course Overview 
Lesson 1: Math Tools
Lesson 2: REPLACEMENT, ORDER OF OPERATIONS, OPERATOR PRECEDENCE 

This thread is only for the chapters to be posted - *NOT for discussion*. Please post here for discussion of the course. I will leave this thread unlocked so that Mr. Al can post here, and chapters can be cut/pasted here if they start in another thread. again *NO DISCUSSION* in this thread - only the educational material. The next two posts will be the first two chapters that I will copy/past up here. If Mr. Al still wants to put the material inline in his EE thread, then anybody can (and should) toss a copy up here too. Thanks.


----------



## Darell (Jun 29, 2004)

*Re: CPF University: EE Course (LESSON-1)*

Serial number: EE-Ch0001-p101 

Chapter 1: Math Tools 

Like many other crafts, this one requires some tools. 
These tools are mostly intellectual ones. 

----------------------------------------------------------- 
Part 101: Symbols, terms, elementary operations, equations 
----------------------------------------------------------- 


Symbols 

Addition (+) 

This simply means that we add one quantity to something else. 
Examples: 
1+1 
3+5 
2+9 

Notice we can change the order of the operations: 
2+9 
9+2 

and we would get the same answer. 



Subtraction (-) 

This simply means we subtract one quantity from another. 

3-1 
5-2 
9-1 
1-9 

Notice here that we cant change the order of the operations. 
9-1 
is not the same as 
1-9 

Subtraction is the inverse operation of addition. If we add 
3+2 
we get '5' 
and if we subtract one of the numbers we used for the addition 
from the answer '5' we get the remaining number: 
5-2 gives us 3, the other number used in the addition, and 
5-3 gives us 2. 


Terms 

Anything separated by an addition sign or a subtraction sign is 
called a 'term'. 
Examples: 
1+3-2 
has three terms 
3+4+5+6 
has four terms 
8-3-2+1.33 
has four terms 


More Symbols 

Equals (=) 
The equals sign is probably the most important of all. This sometimes 
tells us that out of a maybe infinite number of possibilities, only one works. 
Examples: 
1+1=2 
1+2=3 
1.4+1.5=2.9 
3+5=4+4 

Notice that in each case above everything to the 'left' of the 
equals sign is equal to everything to the 'right' of it. 
This is the way the equals sign works, and it's very useful for 
finding answers to some very complex questions. 

Not equals (!=) or (<>) 
This sign is used to show when something is not equal to something else. 
There are two forms in common useage as shown. 
4!=5 simply means four is 'not equal' to five. 
4<>5 is another way to show this. 

Since the symbol "!" is used to show factorial sometimes, it's better 
to use '<>' when the context includes calculating factorials. 

There is also the equals sign with a question mark, which is used 
to show that we have an equation that we dont know is proved to be 
true yet. 
Example: 
4+5?=9 
Of course once we prove this, we remove the question mark: 
4+5=9 
unless of course it proves to be false. 



Multiplication (*) 

Basic multiplication is just adding the same number over again for 
a number of times. Thus, 5*4 means to add 5 four times, or to add 
4 five times. 
Examples: 
1*2 
3*7 
8*9 
9*8 

Notice it doesnt matter what order we do the multiplication in, 
we get the same answer: 
2*3 
is the same as 
3*2 


Division (/) 

Division is the inverse operation of multiplication. 
Examples: 
3/2 
5/4 
4/2 

Notice that the order is important again. 
4/2 
is not the same as 
2/4 


Terms and equations 
When we have an equation, we simply have a set of terms on 
the left and a set of terms on the right. 
Examples: 
1+2=3+0 
4*5+3=3*7+2/1 

One important point here is that TERMS are separated by addition 
or subtraction signs, NOT by multiplication or division signs. 
This last equation 
4*5+3=3*7+2/1 
thus has only two terms on the left and two on the right. 
This is because 
4*5 
is a single term 
and 
3*7 and 2/1 
are also single terms. 


Ok, now here are some self testing questions. 
If you cant answer any of these or have problems, just yell! 

1. How many terms on the left of: 3+4*2-1=5*2 
2. How many terms on the right of: 3+4*2-1=5*2 
3. When we multiply 3*2 we get 6. Write at least one way 
to show how the inverse operation works. 
4. When we add 3+9 and get 12, and then take 12 and 
subtract 9 we get 3, which gives us the first number in the 
addition back again (3). Why does this happen?


----------



## Darell (Jun 29, 2004)

*Re: CPF University: EE Course (Lesson - 2)*

Serial: EE-Ch0001-p102-{A939C0B7-558E-4687-9271-AB92E4935FD0}

We need to establish some ground rules when evaluating
numerical expressions. These rules have long been set
almost since the dawn of man's desire to understand
the world numerically.


REPLACEMENT, ORDER OF OPERATIONS, OPERATOR PRECEDENCE

When calculating an expression, we proceed from left to right
once the precedence of all the operators is the same.
To calculate 1+2+3 we start with 1+2, and to calculate
2*3*4 we start with 2*3.

When we have to calculate a multi-term expression we cant do
the whole thing at once, we have to do two terms at a time
or evaluate a product first.

We can do:
1+3
right away without any problem because there are only two terms,
but when we go to do:
4+2+7
we have to do two terms first, then continue to the next term
something like this:
4+2 gives us 6, then 6+7 gives us 13.

Notice we had to do two terms first, then proceed to the next term
in order to calculate the whole expression. We cant add all three
numbers all at the same time really.

This becomes even more apparent when we try to add many terms:
1+2+3+4+5+6-7+9

We have to start somewhere, and the rule is we have to start at the
leftmost term and proceed to the right (once we reduce the expression
to all additions and subtractions or the operators have all the
same precedence).
If you want to keep track of intermediate results the following helps:

We start with the full expression:
1+2+3+4+5+6-7+9

Then, we start with the left most terms,
1+2
We do the operation, then replace the two terms with the
result. Since 1+2 equals 3 we replace these two terms with
the number 3, and we get:
3+3+4+5+6-7+9

Next we do '3+3' and get 6, and replace '3+3' with '6':
6+4+5+6-7+9

Next, 6+4 and replace that:
10+5+6-7+9

Likewise 10+5:
15+6-7+9

15+6:
21-7+9

21-7:
14+9

Now we are down to our last two terms, so we do the operation and
we are done...

14+9 equals 23:
23

And we have the result!
1+2+3+4+5+6-7+9=23

In each case i simply copied the remaining expression and pasted it
to the next line, then replaced the left two terms with the result
from doing the operation between those two terms.
Notice each time we do this we reduce the expression by one term.
Eventually we get all the terms combined into one single term,
which is of course the answer.
It's very important to do the left two terms before anything else,
because we MUST proceed from left to right in order to evaluate
expressions when all the operators are of the same precedence.

Now so far we've only done expressions with addition and subtraction.
When we do addition and subtraction with multiplication and division,
we have to take into account the precedence of the operators.
Precedence simply means operators with higher precedence get
calculated first. Of course we have to know the precedence of
operators in order to know which ones are higher, so here's a list:

addition 1
subtraction 1
multiplication 2
division 2

Notice some are higher then others.

Now let us see how this affects our calculations...

Calculate 1+2*3=? (the question mark indicates a number we dont know yet)

Because multiplication is higher than addition, we have to start with
that...

We start with the original expression:
1+2*3=?
calculate 2*3 and replace it with it's result, 6:
1+6=?
Now we do the last two terms and get 7, so we have the answer:
1+2*3=7

Notice that if we do the '1+2' first we would get the wrong answer:
1+2=3, and 3*3=9, and '9' is not the correct answer...
1+2*3<>9
This should show how important operator precedence is.


Self test exercises

1. compute the answer to: 3+4*5+7-2*2 = ?
2. compute the answer to: 3+4*3*8 = ?
3. compute the answer to: 2+3*5+1 = ?
4. compute the answer to: 8-3*2+3*2*3 = ?
5. compute the answer to: 8-3*2+3*2+3 = ?
6. compute the answer to: ? = 8/2+3*2+3*2-3


----------



## MrAl (Jun 30, 2004)

*Re: CPF University: EE Course (Lesson - 3)*

Good morning!

Answers to Ch0001-p102 (yesterdays) self test:

3+4*5+7-2*2=26
3+4*3*8=99
2+3*5+1=18
8-3*2+3*2*3=20
8-3*2+3*2+3=11
13=8/2+3*2+3*2-3


Serial: EE-Ch0001-p103-{A939C0B7-558E-4687-9271-AB92E4935FD0}

We can change the order of operations by introducing notation
which shows which items should be calculated first.


PARENTHESES

There are two symbols we use:
open parenthesis:
(
close parenthesis:
)

One set of parentheses consists of one open and one close.
This set is used to enclose a number of terms in an expression
like this:
(3+1)

What this means is that we are to calculate whatever is inside
the set of parentheses before doing anything else.

Example:

2+(3+1)=?

The parentheses here tells us to add '3+1' before adding the '2'.
Without the parentheses we would have had simply:
2+3+1=?
and we would have added the '2+3' first, but since the parentheses
enclose the '3+1' we have to do that first.
Evaluating 2+(3+1)=?
first we add 3+1 to get 4, so we end up with the reduced expression:
2+4=?
Now we have no parentheses left and there is only one operator so
we proceed from left to right, adding 2+4 to get 6, so
2+(3+1)=6
and we have found the answer.

Note that if we had simply added from left to right ignoring the
parentheses we would get the same answer:
2+3+1=6
but this is a coincidence and is not normally the case.

Take for example:
2*(3+1)=?
All we did here was replace the first (+) operator with a multiplication (*).
Now lets do it, starting of course with the parentheses:
3+1=4, reduce expression to:
2*4=?
Two times 4 equals 8 so we have our answer,
2*(3+1)=8

Note had we ignored the parentheses in this case we would have gotten
the WRONG answer:
2*3+1=7
which is not the same as:
2*(3+1)=8

Using our 'not equals' method of expressing this, we could
write:
2*3+1 <> 2*(3+1)

Note this last equation has two expressions, one on each side.
Quite often we will run into things like this, where we have to
calculate BOTH sides of the equation in order to make progress.

Prove
1+2=4-1

Here we simply calculate both sides and see if they are the same or not...
3=3
so they are the same in this case so the equation is true.



To recap,

When calculating an expression follow this order:
1.
Evaluate anything inside a set of parentheses
2.
Evaluate highest precedence operators, then
evaluate next lowest precedence operators, following
this order until only additions and subtractions remain.
3.
Complete the evaluation by combining the remaining terms


BTW, the words 'parentheses' and 'parenthesis' are often
abbreviated by simply 'parens'.


Let's take a minute to do an expression...

2*3*(4+2)-(7+2)*3=?

We start by doing anything inside parens. Since there are two sets,
we start with the set to the left (as usual, we always start from the left).
We find (4+2) so we do that... we get 6 so we rewrite the expression
replacing 4+2 with 6...
2*3*(6)-(7+2)*3=?
Now since there is only one thing inside the parens we can drop them, and
the result it this:
2*3*6-(7+2)*3=?

We have another set of parens, so we do that next...
(7+2) evaluates to (9) so we substitute (9) for
that:
2*3*6-(9)*3=?
Again we have a single thing inside of a set of parens so
we can drop this set too, and we end up with this:
2*3*6-9*3=?
Notice now we have no parens left! We have thus succeeded in
removing all the parens from the equation.

Next, we look for the highest precedence operators, which are
multiplies (*).
We have two terms, so we start with the term on the left, 2*3*6
and evaluate that
2*3*6
starting on the left,
2*3=6
replacing '2*3' with the 6 gives us
6*6
so we end up with 36.
Now back to the equation, 
2*3*6-9*3=?
we found that
2*3*6=36
so we replace that and end up with
36-9*3=?
Now we have to do the '9*3' so we calc that out and it equals 27 so
we replace that:
36-27=?

Now we are down to two terms, so we finish up:
36 minus 27 equals 9, so we have our answer...

2*3*(4+2)-(7+2)*3=9

Not too bad huh?

If we take each equation one step at a time we can
always find an answer.



NESTED PARENTHESES

Here we take a quick look at what to do when we have more
then one set of parens when one of the sets is inside of
another set.

Evaluate:
2*(2*(3+1)+3)=?

Note that here we have two sets as before, but now one set
is 'inside' the other set. What to do?
When this occurs, the rule is to perform the calculations
from the 'inner-most' set to the 'outer-most' set.
If it's not apparent or looks confusing, start by eliminating
all the numbers and operators one at a time until you end up
with only parens. Let's do this with the above equation...
If we eliminate all the stuff that is NOT a paren, we end up
with this:
(( ))
We can immediately notice that there is one inside set and
one outside set. When we look back at the original equation
we note that the inside set belongs around the '3+1' so
we have to do that first.

Thus, in evaluating 
2*(2*(3+1)+3)=?
we start by calculating 3+1. We get 4, so we replace that in the
original equation:
2*(2*(4)+3)=?
Since we end up with a single number (4) inside a set of parens
we can eliminate those parens to get:
2*(2*4+3)=?
Now we're down to one set of parens, so we would proceed as
mentioned earlier...
2*4=8 so we replace that and get...
2*(8+3)=?
8+3=11 so we relace that and get...
2*(11)=?
remove parens and get...
2*11=?
The final multiplication equals 22, so the answer to the original
equation is:
22
or written out completely:
2*(2*(3+1)+3)=22



Exercises

1. Evaluate: 2*3+4=?
2. Evaluate: ?=3*(2+1)
3. Evaluate: ?=(9-3)*4+7*(4+2)/2
4. Evaluate: ?=(5*(2+3)+4)*3+(4+1)*2
5. Evaluate: ?=(((((6+1)))))

Answers next time...

Take care,
Al


----------



## MrAl (Jul 1, 2004)

*Re: CPF University: EE Course (Lesson 4)*

Hi again!


Answers to last self test:

1. 2*3+4=10
2. 9=3*(2+1)
3. 45=(9-3)*4+7*(4+2)/2
4. 97=(5*(2+3)+4)*3+(4+1)*2
5. 7=(((((6+1)))))

If you didnt get #5 or it was confusing, you should take another
more careful look at parentheses.


Serial: EE-Ch0001-p104-{A939C0B7-558E-4687-9271-AB92E4935FD0}

Today we'll look at something fairly simple. We are going to
replace the question mark we have been using with a letter of
the alphabet. We can use any letter we want to use.

Recall the last self test looked like this:

1. Evaluate: 2*3+4=?
2. Evaluate: ?=3*(2+1)
3. Evaluate: ?=(9-3)*4+7*(4+2)/2
4. Evaluate: ?=(5*(2+3)+4)*3+(4+1)*2
5. Evaluate: ?=(((((6+1)))))

All we are going to do now is replace the question mark "?" with
the lower case alpha character "y" (the lower case letter 'y' in
the english alphabet) for each problem...

1. Evaluate: 2*3+4=y
2. Evaluate: y=3*(2+1)
3. Evaluate: y=(9-3)*4+7*(4+2)/2
4. Evaluate: y=(5*(2+3)+4)*3+(4+1)*2
5. Evaluate: y=(((((6+1)))))

Notice we havent changed anything at all really, just what we are
using to represent a quantity that has not been determined yet.

Since we have already done these problems we know what the 'y' will
be for each one, so if we were to replace the y with the correct
value again we would (again) end up with the same set we had before:

1. 2*3+4=10
2. 9=3*(2+1)
3. 45=(9-3)*4+7*(4+2)/2
4. 97=(5*(2+3)+4)*3+(4+1)*2
5. 7=(((((6+1)))))


Simple right?

Let's look at #2 a little closer for a minute...

Evaluate: y=3*(2+1)

Notice that the 'y' is on the left side, and the numerical part is
on the right side. This is the way most formulas are presented.
The 'y' represents the 'unknown' quantity in the equation--the
value we are seeking. The stuff on the right tells us how to go
about finding this unknown quantity.
The answer to this might be something very useful... from a resistor
value for our flashlight, to the voltage required to run a particular
type of LED. Being able to understand how to find these unknowns
can be of great benefit when we want to find something useful in
electronics. Now knowing how to find these values means we have to
ask someone else and hope we get the right answer. If we want to
know a whole range of 'what if' senarios we would have to ask
a whole bunch of questions in some cases. It's much more rewarding
to be able to calculate the unknowns and come up with the answers
and then test them in our own circuits! If you're one of the people
who like doing this, then you're certainly in the right place 

Notice we used the letter 'y' in all these. We could have
used any letter, but usually the letter is chosen to have
some relationship (if any) to what we are seeking.

We could have used lower case 'a':
Evaluate: a=3*(2+1)

We could have used lower case 'b':
Evaluate: b=3*(2+1)

We could have used upper case 'A':
Evaluate: A=3*(2+1)

or we could have chosen a more descriptive name made up of
several easily read characters that represents the actual
thing who's value we are after:

perhaps a resistor value:
resistor=3*(2+1)

or the voltage present at a particular place in a circuit:
volts=3*(2+1)

The only thing changed is the 'name' of the variable.

Some conventional names for various electrical quantities are:

v for volts
e for volts
i for amps
r for resistance

and for various parts on schematics:

R for resistor
B for battery
C for capacitor
L for inductor (not I, which is usually for current)
D for diode
S for switch
T for transformer
Q for transistor (not T, which is for transformer)


Self test

Solve the following for the unknown variable

1. v=3*(4+2)
2. x=7+9*2
3. p=4+(8+2)*(2+3)
4. MyResistor=25*0.2+2
5. MyResistor=25*0.4+2

Identify the unknown variable in the following

6. y=3+4
7. x=9*8+3+5
8. 7*x+3=8*3

Notice in #8 above the 'x' is not alone, but has other
terms with it on the same side. Something to ponder
is how we would get the answer to this problem which
would of course give us the value of x, which could
be something we really want to know.


Take care,
Al


----------



## MrAl (Jul 2, 2004)

*Re: CPF University: EE Course (Lesson 5)*

Good morning!

Answers to yesterdays self test problems:

1. v=3*(4+2), v=18
2. x=7+9*2, x=25
3. p=4+(8+2)*(2+3), p=54
4. MyResistor=25*0.2+2, MyResistor=7
5. MyResistor=25*0.4+2, MyResistor=12

6. y=3+4 the "y"
7. x=9*8+3+5 the "x"
8. 7*x+3=8*3 the "x"

Pondering problem #8...

To solve #8 we would first get the unknown variable
"x" to one side of the equation so it's alone. Then we would
be able to calculate the answer just as we have done before.

This will be the subject of our discussion today. We'll find
out how to solve for a single variable in an equation like #8.


Serial: EE-Ch0001-p105-{A939C0B7-558E-4687-9271-AB92E4935FD0}

Find the value of the unknown variable x in:
7*x+3=8*3

Since x isnt alone, we want to find a way to get it alone on
one side of the equals sign so it appears like this: "x=".

Since there are other partial calculations we can do without
any problem first thing we do is simplify. Since 8*3 can be
calculated immediately we do that first and replace it with
it's result. Since 8*3=24 we reduce the equation to:

7*x+3=24

Now it's a tiny bit simpler then it was before, so we are that
much closer to solving it for 'x'.

The next thing we will notice is that there are two terms on
the left and one of them is a single number. The next thing
we do is very useful for solving these kinds of equations:
since we found a +3 on the left side and we want to get x
alone we add a minus 3 (-3) to BOTH sides of the equation.
This 'cancels out' the 3 on the left side which will leave
the term '7*x' on the left side, which is closer to having
x alone on one side.

Adding -3 to both sides is the same as subtracting 3 from
both sides, so we do that now, and equation then looks like this: 
7*x+3-3=24-3
where you can see 3 is being subtracted from both sides.
Now we notice we have numbers we can calculate right away,
the 3-3 and the 24-3, so we do that right now and end up with:
7*x+0=21

Since 0 doesnt add anything to 7*x we can drop that and we end
up with a simpler equation:
7*x=21

Recall that when we had a +3 to get rid of on the left side
we performed the 'inverse operation' to get rid of it.
The inverse operation of adding a number is to subtract a number,
and since it must be that same value to get rid of it we used a 3.
Now we have
7*x=21
and x still isnt alone, but we notice that x is MULTIPLIED by 7.
Now if we could get rid of that 7 we would have x alone on one side
and be in a position to calculate the value of x, so we think of a
way to get rid of that 7. Since x is multiplied by 7 we use the
inverse operation (division) to get rid of the 7. We must
divide BOTH sides of the equation to maintain equality between the
two sides, so we simply divide both sides by 7. This would look like
this:
7*x/7=21/7
Regrouping so the numbers appear together:
x*7/7=21/7

We can notice now that 7/7=1 and 21/7=3, so we replace those
with 1 and 3:
x*1=3

Finally, since 1 times any number equals that number, x*1 equals x
so we get rid of the 1 next and end up with:
x=3

Since there is nothing else to calculate we are done.

Here we found the value of x by performing a number of operations
on the equation. The operations we performed were always the
inverse operation of whatever we wanted to get rid of in order to
get x alone on one side. This is the way many single variable
equations work.

Important points:
1. Whatever you do to one side of an equation you must do to the other side
2. To get rid of something on one side of an equation you perform the inverse
operation on both sides.

Self test problems:

Find the value of the unknown variable in each of these

1. x*2=8+2.2
2. 9=45*x
3. (x+2)*5=20
4. (R+2)*3+9=7*6
5. 8*x-2=27+3

Note it doesnt matter which side the unknown variable is on, as usual.


Next time we'll do something a little simpler which turns out to be
very useful... we'll take a look at formulas.


See ya next time,
Al


----------



## MrAl (Jul 5, 2004)

*Re: CPF University: EE Course (Lesson 6)*

Hello again!

I hope your fourth of July went off with a bang, but now
it's back to work! 


Answers for last self test:

1. x*2=8+2.2, x=5.1
2. 9=45*x, x=0.2
3. (x+2)*5=20, x=2
4. (R+2)*3+9=7*6, R=9
5. 8*x-2=27+3, x=4


Formulas are very handy when you have them. Formulas
allow you to quickly determine the value of your resistor or preform
a test to see if your voltage is too high or too low, to name a few.
Next we are going to take a look at formulas and how to use them,
and even run into a few real life circuit formulas that might
be encountered when using a Nichia or Luxeon LED.


Serial: EE-Ch0001-p106-{A939C0B7-558E-4687-9271-AB92E4935FD0}


One day you find a circuit you want to use and the designer has
provided a formula to calculate the required series resistor value
for you to use in series with your LED in order to use this circuit,
and this value changes with the voltage of the battery you wish to
use with the circuit.

Here's a typical formula:
R=(V-4)*50

Notice that there are two unknown variables in this formula.
Usually the unknown variables are identified in some way like this:

R is the resistor value, in ohms
V is the voltage, in volts

This way you know what variable is being used for what.

Now we have to be able to calculate the resistor value from
this formula, and looking at the formula we notice that we
also have V in the right side, so this means we have to
know beforehand what the value of the voltage will be.
Since we know what battery we want to use, we know the 
voltage already, so we simply 'plug' that value into the
formula, then calculate the value of the remaining unknown
variable "R".

Lets say we start with a battery voltage of 9 volts:

Recalling the formula and writing it down again...
R=(V-4)*50

Knowing the voltage V is 9 volts we rewrite the equation
and replace the V with 9 like this:

R=(9-4)*50

Simple right?

Now we simply perform the indicated operations and we come out
with the value of the required resistor:

R=(5)*50
R=250

Pretty simple right?

Now we know we would have to get a 250 ohm resistor to make the
circuit work with a 9 volt battery.

What about if we wanted to use a 10 volt battery?

Recall the formula and write it down:
R=(V-4)*50
Replace the V with the 10:
R=(10-4)*50
Calculate it out:
R=(6)*50
R=300

So we see that increasing the battery voltage to 10 volts
requires a resistor value of 300 ohms.

Now lets take a quick look at what happens if we have two 
quantities to replace before we do the calculation.

Suppose we have the same situation as before, except the designer
included current in the formula as well as voltage?

Let's say we have the new formula:

R=(V-4)*50+100*I

Now we have to know both the voltage and the current in order
to calculate the value of the resistor, R.

For example, let's say we have again a 9 volt battery and
we want to pump 0.02 amps through the circuit. We know
both the voltage (9) and the current (0.02) so we simply
replace those two variables with these values and calc
the result...

recall the formula:
R=(V-4)*50+100*I

plug in the values for V and I, making sure not to mix them up:
R=(9-4)*50+100*0.02

calculate it out:
R=(5)*50+2
R=250+2
R=252 

So the final value of R comes out to 252 ohms.


Formulas are one of the most important uses of math in electronics.
You'll be able to do really a lot of things just knowing how
to use formulas alone, even if you dont know anything else about
the circuits they are used with.



Self test
(some of these exercises are actual real life circuit examples)

Find the required value of resistor with the given quantities for
the unknowns on the right side of the equations

1. R=i/5+400, where i=0.35
2. R=(v+2)/4+90, where v=2
3. R=(v-3)/i, where v=9 and i=0.02

Find the required voltage for the given value of resistor:
4. V=0.1*R, R=100
5. V=i*R+3.5, where i=0.3 and R=1

Extra Credit 
6. Extra credit if you can spot the formulas that would probably
be used with your typical Nichia or Luxeon 1 watt LED 


Take care,
Al


----------



## MrAl (Jul 6, 2004)

*Re: CPF University: EE Course (Lesson 7)*

Hi again, and welcome to another part of the EE Course!


Answers for last self test...
the answer is enclosed in brackets like this: [answer]

Find the required value of resistor with the given quantities for
the unknowns on the right side of the equations

1. R=i/5+400, where i=0.35 [R=400.07]
2. R=(v+2)/4+90, where v=2 [R=91]
3. R=(v-3)/i, where v=9 and i=0.02 [R=300]

Find the required voltage for the given value of resistor:
4. V=0.1*R, R=100 [V=10]
5. V=i*R+3.5, where i=0.3 and R=1 [V=3.8]


Serial: EE-Ch0001-p107-{A939C0B7-558E-4687-9271-AB92E4935FD0}

Today we are going to take a look at one of the most important formulas
in the history of electronics--Ohm's Law. The basic idea behind Ohm's
Law is that it shows the relationship between current and voltage in
an impedance. Since we will be starting with Ohm's Law for direct current
circuits, we'll be looking at the formula as it applies to resistances.

Before we do that though, we will establish what a voltage and current is.



A voltage is a 'potential' to cause current to flow though a resistance.
If you have voltage and you connect a resistance across it, a current will
flow though the resistance and the level can be determined by Ohm's Law.

The most important thing about a voltage is that it is observed in nature to
occur (and can be measured) ACROSS an element (such as a resistor).
The most important thing about a current is that it is observed in nature to
occur as a flow of tiny particles THROUGH an element. The sum of the tiny
particles makes up the total current flow.

This cant be stressed enough. Try to remember this relationship:

Voltage <-> ACROSS
Current <-> THROUGH

The voltage exists as a potential to force current to flow though
an object. The current flows as a result of the object reacting
to the voltage potential. The voltage potential is usually 
there first, then once the object reacts current begins to flow
though that object.

The Voltage and Current shown above differ from the objects they
act upon in that they are energy suppliers, while the objects are
generally energy absorbers. Although some objects can also supply
power, for now we will be looking at objects called resistors,
which always absorb energy.

Resistors are the most common circuit elements (or objects) so we
will start using them first, then move on to other types.
The resistor will therefore be the first element on our circuit
element list:

Circuit elements:
Resistor, absorbs energy, two terminals

Note we add the number of terminals for reference.

Let's take a quick look at the whole list of things:

Voltage <-> ACROSS, Supplies energy
Current <-> THROUGH, Supplies energy
Circuit elements:
Resistor, Absorbs energy, Two terminals

Next, we need to know what units are used to measure these quantities.

Voltage is of course measured in 'volts', and it's symbol is usually V or E.
Current is measured in 'amperes' usually abbreviated 'amps', and it's symbol is I.
Resistance is measured in 'ohms', which is the ratio of volts to amps, or V/I.

Voltage is measured with a voltmeter, current with an ammeter, and resistance
with an ohmmeter, although you usually get all three functions in a typical
modern digital voltmeter.

It's interesting to think about what voltage and current really is, but it's
easier to think of voltage, current, and resistance all at the same time
instead of trying to understand each one separately. When we do this we
find a relationship that will help understanding each one a little better.
This relationship happens to be Ohm's Law, and can be expressed in a simple
little equation like this:

V=R*I

From our previous work, we can see that this is a formula that requires
knowing both R and I in order to solve for V. If we knew R and I we could
plug them in and get the answer pretty quick.

Let's say we have a resistor who's value is written on the side and it
says "100 ohms". Let's also say we happen to know the current and it's 
0.35 amps. Now all we have to do is plug and calculate:

V=R*I
V=100*0.35
so
V=35 volts

Pretty nice huh? 

But suppose we didnt know the current, but we knew both R and V?

First, we would rearrange the formula a little...

starting with the original Ohm's Law formula:
V=R*I
we want to first solve for I because we know R and V...

We see that I would be alone if the R was on the other side, so
we divide both sides by R:

V/R=R*I/R

and we end up with:
V/R=I

Turning this around we get:
I=V/R

(Note it doesnt matter which side is which)

and we have our formula for I. 
Now if given the voltage and resistance we could easily calculate
the current, I.

Ok, what if we knew both I and V and didnt know R?

Starting with the original Ohm's Law formula:
V=R*I

we want R alone on one side, so we divide by I:
V/I=R*I/I

Since the I/I=1 we end up with:

V/I=R
or
R=V/I

Now if given the voltage and current we would be able to 
find the value of the resistor, even if it wasnt marked!

If we plug in V=10 volts and I=1 amps, we would quickly find R=10 ohms.

The last formula we found was
R=V/I

and we've gone over a quick example knowing the V and I, but now
lets take a quick look at the physical aspect of this formula:
what it says about voltage and current and resistance in nature.

The resistance R is equal to the voltage V divided by the current I.
Since we know the voltage is ACROSS the object and the current is
THROUGH the object, we know the resistor has a voltage across it
and a current though it at the same time. If there is a voltage
across it, then there is also a current flowing through it.

Looking at the units involved, we see that on the left side we
have 'ohms' for the resistor, and on the right we have
'volts' divided by 'amps'.
Another way of stating this relationship is that
"1 ohm equals 1 volt per amp".

This is easily verified by looking at a few values of V and I
with a constant R of 1 ohm in R=V/I.


Conclusion For Today

I know we talked about a lot of things today, but if you get anything
out of this part of the course i ask that you find a way to remember this
about current and voltage...

Voltage <-> ACROSS

Current <-> THROUGH

and that these two supply energy to our passive circuit elements.



Note also, if you will, that when you go to measure these two
quantities you measure voltage ACROSS an object, and you 
measure current THROUGH an object.


Self test questions

1. What is the most important thing about the behaviour of a voltage?
2. What is the most important thing about the behaviour of a current?
3. The voltage across a resistor of 100 ohms is measured to be 25 volts.
What is the current in amps?
4. We measured 10 volts across a resistor and 1 amp though it. What is
it's value in ohms?
5. A 200 ohm resistor has 0.1 amp flowing though it. What would we
measure across it: voltage or current, and what would the value and units be?
6. A Nichia LED is wired in series with a 50 ohm resistor and the series
combination is wired to a battery. The resistor's voltage drop is 1 volt.
What is the current flowing though the resistor, in amps?
Hint: Voltage drop is the voltage measured across something.

Good luck with today's self test!

Take care,
Al


----------



## MrAl (Jul 8, 2004)

*Re: CPF University: EE Course (Lesson 8)*

Hi again, and welcome to another part of the EE Course!


Answers for last self test...
the answer for each is enclosed in brackets like this: [answer]

1. What is the most important thing about the behaviour of a voltage?
[The most important thing about a voltage is that it always appears across 
a circuit element]
2. What is the most important thing about the behaviour of a current?
[The most important thing about a current is that it always flows through
a circuit element]
3. The voltage across a resistor of 100 ohms is measured to be 25 volts.
What is the current in amps?
[Since i=v/r, i=25/100, so i=0.25 amps]
4. We measured 10 volts across a resistor and 1 amp though it. What is
it's value in ohms?
[R=V/I, so R=10/1 which equals 10 ohms]
5. A 200 ohm resistor has 0.1 amp flowing though it. What would we
measure across it: voltage or current, and what would the value and units be?
The first part of the answer:
We've been asked for an 'across' variable, and voltage is the across variable,
so it would be [voltage], and the value is determined using Ohm's Law:
[V=I*R, so V=0.1*200, so V=20 volts.]
6. A Nichia LED is wired in series with a 50 ohm resistor and the series
combination is wired to a battery. The resistor's voltage drop is 1 volt.
What is the current flowing though the resistor, in amps?
Hint: Voltage drop is the voltage measured across something.
We know the voltage drop across the resistor and the value in ohms of
the resistor, so we dont care what else is wired in the circuit really,
we still use Ohm's Law:
[I=V/R, so I=1/50, so we get I=0.02 amps for the current]



Serial: EE-Ch0001-p108-{A939C0B7-558E-4687-9271-AB92E4935FD0}

Today we are going to talk about powers and roots, because these are
also very important in circuit analysis. Being able to take a number
'up to a power' will enable us to look at the rest of Ohm's Law so
we can determine power in a resistor using a number of methods.
Of course taking a 'power' in math is not the same as power dissipation
in an electrical circuit, although we do need to know both.
After today's discussion we should be ready to move to some actual
real life circuits!

To begin with, i want to stress again that taking a power in math
is not the same as finding the power dissipation in say a resistor,
but we will be using the math 'power' to find the power dissipation in
a resistor.

We've covered addition, subtraction, multiplication and division,
and their respective symbols (+), (-), (*), (/),
and now we look at the symbol for power:
(^)
Enclosed in the parens above is the symbol for taking a power in math.
It requires two numbers just like multiplication.
Some examples:
2^3
means to "take 2 up to the power of 3"
and
4^2
means to "take 4 up to the power of 2"
which is also called "finding the square of"
so we could also say that
4^2
is "to find the square of 4"

Note that when the power is exactly 2 we can also call it
the 'square' as we did above.

The meaning of something like 4^2 is very simple, almost as
simple as understanding multiplication from knowing how addition works.
When we multiplied something, it was like adding one of the numbers
the number of times as indicated by the second number:
4*2 is the same as 4+4, which both of course equals 8.

Well, taking the power is the same except we use multiplication:
4^2 is the same as 4*4.

See how simple that is?

The SECOND number (the 2 above) is the number of times we have to
multiply the FIRST number by itself to get the answer. Thus,
4^2=4*4 which of course equals 16, so
4^2=16 .

If we had 3 for the second number, as in this:
4^3
then all we do is multiply 4 times itself again, only this
time we have to do it THREE times, because the second number is 3 this time:
4^3=4*4*4 which equals 64.

Pretty simple.

3 in the last case above is called the 'Power'.

It gets a little tricky when you have a decimal in the power like this:
4^3.1

Note that you cant multiply 4 times itself 3.1 times, so you have to use
a calculator to do this one. Usually the manual that comes with the
calculator tells you how to take powers, so you'll have to consult
the manual that came with your calculator. Most calculators use
the symbol "x^y" to take a power.

Using the calculator that comes with windows, i found the 
value of 4^3.1 to be 73.516694719810240435112124593787, but
even though there are lots of decimal places there, it's still
only an approximate answer to taking 4 up to 3.1, but it's 
usually close enough to get by with.

The Inverse of Power

Just as addtion and multiplication had inverses, so does Power.
The inverse of Power is "Root", or taking a root.
The symbol for taking a root has to be done using graphics,
so for now we'll call it by it's functional equivalent:
Root(x,y)
which will mean to take the y root of x.

As you probably guessed, we usually need the calculator for
this one too. A lot of calculators use the 'inverse' key
to get the root. You press the inverse key and then the
'power' key (x^y) and you get the root.

There are some roots that are easy to spot. For example,
root(4,2)
means to take the second root of 4, and we know
2*2 equals four, so the second root of 4 is 2.

root(16,2)
we know 4*4 equals 16, and that's multiplying 4 times
itself two times, so we know the second root of 16 is 4.

Went we find the second root of a number it's also
called 'finding the square root' of a number. When
the root is exactly 2 this is called the square root.
Since quite a few problems involve roots that are not
whole numbers, we end up using the calculator most of
the time anyway 

With multiplication, we could reverse the two numbers and
get the same answer:
5*2=2*5

but with powers and roots, we CAN NOT reverse the two numbers:
5^2<>2^5

To see why, do 5^2 and then do 2^5 and compare the answers to
find that they are NOT the same. This means the order must
be left alone when doing powers and roots too.


Powers, Roots, and Equations

As you might have guessed again, powers are used in equations
just like all the other operations we covered. A typical
equation might look like this:

y=4^2

For this, we simply grab the calculator and punch in 4, x^y, 2, =
(or something similar) and we get the answer, 16, so 
y=4^2
y=16

which wasnt difficult at all.

What about when we have something like this:

40=10*I^2

First, we have to establish an operator precedence for powers and roots,
which turns out to be one level higher then multiplication.

Let's look at the precedence table now...

Add and subtract 1
multiply and divide 2
powers and roots 3

This means we must do powers and roots before any other operation.


So how do we solve for I in the previous equation then...
40=10*I^2

Since we have to do powers first, this is the same as:
40=10*(I^2)

We see that we have to get the 'I' alone eventually, and (I^2) is a 
product, so we divide both sides by 10 first:
40/10=10*(I^2)/10
regroup:
40/10=10/10*(I^2)
do the math we can do:
4=1*(I^2)
drop the 1:
4=(I^2)
drop the parens:
4=I^2

and now what? The 'I' still isnt alone.
Well, what did we do before? We used the 'inverse' operation,
so we do that here also, except now we have to use the inverse
of taking a power, which is taking a root. Of course what
we do to one side we have to do to the other, so we have to
take the 2 root (square root) of both sides:

root(4,2)=root(I^2,2)

We see that we can immediately do root(4,2) and get
the left side simpler, because root(4.2)=2, so we do that
and end up with:
2=root(I^2,2)
but we still have a root and a power, what next?

We have a special case here, where the 'Power' is the same
as the 'Root'. In other words, we are taking the square root
of the square, so since the power 2 is exactly equal to the
root 2 they cancel out, and we get root(I^2,2)=I . Nice?

We can therefore rewrite the equation:
2=I

and turning it around...

I=2

so we have our long sought answer, I=2 amps.

Looking back at our original equation,
40=10*I^2
we found that in this equation I has to equal 2 amps.



Self test problems:
(hint: use a standard calculator to help)

1. y=5^2, find y
2. x=6^3, find x
3. v=10*i^2, with i=3 amps, find v
4. v=10*i^2, with v=10 volts, find i
5. v+2=20*i^2, with v=8 volts, find i.
6. y=root(81,2), find y
7. v=2*i^3+4*i^2+8*i+1, find v when i=0.01
8. y=3.4^3.14, find the value of the unknown variable

Note #7 is called a "power series in i" because it contains a series of terms
made up of powers of a variable 'i'.

In addition to this part, here is some more 
information to look at:
http://mral.peu.net/index.php?page=ChapterOne

The last question for today is this:

In the above Figure 1c, the resistor on the far right,
what is the value of this resistor in ohms?


Good luck!

Take care,
Al


----------



## MrAl (Jul 10, 2004)

*Re: CPF University: EE Course (Lesson 9)*

Hi again, and welcome to another part of the EE Course!


Answers for last self test...
the answer for each is enclosed in brackets like this: [answer]

1. y=5^2, find y ... [y=25]
2. x=6^3, find x ... [y=216]
3. v=10*i^2, with i=3 amps, find v ... [v=90 volts]
4. v=10*i^2, with v=10 volts, find i ... [i=1 amp]
5. v+2=20*i^2, with v=8 volts, find i ... [i=0.70710678 amps approx]
6. y=root(81,2), find y ... [y=9]
7. v=2*i^3+4*i^2+8*i+1, find v with i=0.01 ... [v=1.080402 volts]
8. y=3.4^3.14 ... [y=46.6491569 approx]


Serial: EE-Ch0001-p109-{A939C0B7-558E-4687-9271-AB92E4935FD0}


There are certain relationships among powers and roots that makes it
possible to solve a number of equations knowing these relationships.
Today we'll take a quick look at a few more relationships involving
powers and roots.

Each one of the following represents a particular kind of relationship.

[1] Order of operations:

y=a^n^m

means to take n^m first and then take 'a' up to the result of n^m.
Using parens this looks like this:
y=a^(n^m)

Note that with powers we have to work from right to left.
This is just the opposite of when we have a series of additions
or multiplications.

[2] Inversion of an exponent results in taking the root:
a^(1/n) is the same as root(a,n) which is taking the nth root of a.
This means we could substitute an inverted exponent for a root:
root(4,2)=4^(1/2)

[3] Taking a root of a power when the root and the power are the same
(root(a,n))^n)=a
This can also be done like this:
(a^(1/n))^n is the same as a^(n/n)
and since n/n=1 this is the same as a^1
which equals a.

[4] For a^(n/m) the 'n' is a power and the 'm' is a root.

[5] Addition of exponents is the same as multiplying the base with exponents:
a^(n+m) = a^n * a^m

[6] Subtraction of exponents is the same as dividing the base with exponents:
a^(n-m) = a^n / a^m

[7] Multiplication of exponents is the same as taking sucessive powers:
a^(n*m) = (a^n)^m
Note this is NOT the same as a^n^m which equals a^(n^m).

[8] The root of a power is the same as a power of a root:
(a^(1/n))^m = (a^m)^(1/n)

[9] A product taken up to a power is the same as taking each factor up to the power:
(a*b*c)^n = a^n * b^n * c^n

[10] A division taken up to a power is the same as taking the numerator up to the power
and dividing by the denominator raised to the power:
(a/b)^n = (a^n)/(b^n)


We'll be covering these again when an equation comes up during the analysis of
a circuit, but knowing them will help find the solution faster.

Now we are ready to cover the second part of Ohm's Law, sometimes called
"Joules Law".

The basic equation is this:

P=V*I

where
P is power in watts
V is voltage in volts (across variable)
I is current in amps (through variable)

Note that in the above we have to know V and I in order to 
calculate the power. Also, if we knew the power and didnt know
either voltage or current, we could solve for either by
rearranging the equation to solve for the unknown. This works
pretty well as usual, but there is one more thing to take note of...

Recall that Ohm's Law looks like this:

V=I*R

Well, this is pretty nice, because we can calculate V or I if
we knew the other two variables, and one more thing...

If we look at both Joule's Law and Ohm's Law together:

P=V*I
V=I*R

we see they both have V and I in them.

What this means is that if we solve one for say V, and insert this into
the other equation for it's value of V, we will have eliminated the
other equation's variable 'V' which gives us another form.

Let's try it...

First we take 
V=I*R 

which is already solved for V, namely, V is equal to I*R, so
we now take the other equation (Joule's) 

P=V*I

and insert what we found from the first equation (Ohm's)
in place of the variable we solved for, 'V', and we get

P=R*I*I

which means we can now find power knowing only the resistance and
the current!

This is just a glimpse the power of math as applied to circuits.

Of course we could simplify the result a little by noting that
I*I
is just the square of I, so we end up with:

P=R*I^2


Self test problems
(hint: use the relationships above when needed)

1. y=2^(a+b), find an alternate form
2. Y=3^(a-b), find an alternate form
3. y=a^(-2), find y with a=3
4. Find the power in a resistor knowing I=1 amp and R=10 ohms
5. Find the form of the combination of Joule's and Ohm's Laws that allows you
to find the power knowing only voltage and resistance.
Hint: do the same as we did in finding P=R*I^2 .


Good luck!

Take care,
Al


----------



## MrAl (Jul 11, 2004)

*Re: CPF University: EE Course (Lesson 10)*

Hello again,

Here's the next part in this Course.
I hope you find it interesting as we get into our first circuits
this time!


Answers to last self test, answers to each in brackets: [answer]

1. y=2^(a+b), find an alternate form [y = 2^a * 2^b]
2. Y=3^(a-b), find an alternate form [Y = (3^a)/(3^b)]
3. y=a^(-2), find y with a=3 [y = 1/(a^2) = 1/9 ]
4. Find the power in a resistor knowing I=1 amp and R=10 ohms [P=R*I^2, so P=10]
5. Find the form of the combination of Joule's and Ohm's Laws that allows you
to find the power knowing only voltage and resistance.
Hint: do the same as we did in finding P=R*I^2 .
[For this problem, you solve Ohm's Law for 'I' and then insert that result
into Joule's Law where 'I' appears, which then reduces to P=(V^2)/R.
This allows you to find the power knowing only the voltage and resistance.]


Serial: EE-Ch0001-p110-{A939C0B7-558E-4687-9271-AB92E4935FD0}


For the following discussions refer to this web page:
http://mral.peu.net/index.php?page=ChapterOne
You'll have to scroll down the page to get to the drawings. (Edit: Darell included them here - hope I got the right ones!)

IDEAL CIRCUIT ELEMENTS

Figure 2:





Figure 2a "wire":
A wire (or conductor), but this is a special kind of wire.
It has zero resistance so it doesnt drop any voltage.
The voltage measure at point 'A' is the same as point 'B'
no matter how much current is flowing through it.
We will use this kind of wire to connect our circuit 
elements.

Figure 2a "Battery":
You dont have to look any farther...this is the perfect
battery. By 'perfect' i mean just that! It's always
fully charged and it can deliver any current you want,
it's voltage is always that which is marked next to it,
and it's got zero series resistance. Oh yeah, and it
NEVER runs down! Nice, huh? That's the 'ideal' battery
for ya.

Figure 2a "Voltage Source":
This is pretty much the same as the perfect battery above,
except it's drawn a little different. It's always got
the exact voltage that is marked next to it and it never loads
down, just like the perfect battery.

Figure 2b Left Side:

This is more like the 'real' battery you find in stores.
This has a series resistance (marked 'Rs'). As the current
increases, the series resistor drops more of the voltage.
The internal 'perfect' battery B1 still acts as the ideal 
battery, but we cant access it's terminals from outside the
gray box. Everything inside the gray box is internal to
the battery, so we can only get at the terminals marked
with the little circles.

Figure 2b Right Side:
This is the same as the 'real' battery except it uses
the voltage source to replace the battery.


As you can see from the above and from the diagrams, we always
use ideal circuit elements to represent real life parts such
as batteries. Even the resistors shown on the schematics
are not real life resistors, but are 'ideal' in that they
have certain qualities that make things easier to deal with.
When we find that an ideal element is too far from reality
to be worth using, we add more ideal elements to it until
it becomes something close to what you would find in real life.
The 'real life' battery above is a good example, where we
add a resistor in series with a perfect battery to approximate
the real life battery, which has a series resistance.
Later, we'll find we have to do the same thing with the
capacitor and inductor because they too have series resistances
that usually have a significant impact on the outcome of
the analysis of a power supply.


BEGINNING TECHNIQUES OF CIRCUIT ANALYSIS

To be able to analyse a circuit we have to follow certain
rules, especially with the polarity of elements in our
circuit. We begin by marking the polarity of all the
elements in our circuit after drawing the schematic.
Also, there are certain details about a schematic drawn
out on paper (or monitor) that aid greatly in simplifying
the whole process.

Fig 3:




Fig 3a:
Here we have a battery connected to a resistor. As you can
see, one lead of the battery connects to one lead of the
resistor, and the other lead to the other lead of the resistor.
The voltage arrows are drawn for both the battery and the
resistor, as well as one small arrowhead for the current.
The tip of the voltage arrow is positive, while the tail negative.
The arrowhead for the current is simply the direction of
current flow.
You'll notice that because the battery (+) terminal is connected
to the resistor upper lead, that resistor terminal is marked (+)
also. This is because it's a direct connection to the (+)
terminal of the battery.
Another thing you'll notice is that the current arrow POINTS
TOWARD the resistor, and if you moved that little arrowhead
along the wire until it touched the resistor itself you
would call that end of the resistor 'POSITIVE' and mark it
(+) in red. Since the other end must therefore be negative, we
could mark the other end (-) as shown in red also. Since the
voltage arrow has head positive and tail negative, the voltage
arrow for this resistor is drawn as shown (right next to R1).

The rule here is simple:
When the current arrowhead points toward the resistor, mark
that end of it positive (+).
When the current arrowhead points away from the resistor,
mark that end of it negative (-).
You then know where to draw the voltage arrow.

Also note that the current flows THROUGH the resistor,
and comes out the other end. If we drew it coming out
of the resistor (near the bottom of the resistor)
it would be pointing away from the resistor.

The battery also has a voltage arrow for it drawn.
Note that if you were to follow the battery's voltage
arrow to the resistors voltage arrow they would
be pointing toward each other.

Fig 3b:
Here we have the series resistance for the battery shown also.
There isnt much difference between these two circuits, except
3b has two resistors in series with the battery, while 3a only
has one. What this means is that there will be THREE voltage
arrows because we now have three elements. We havent changed
the way we draw the arrows, we still follow the same rules,
it's just that now we have more elements so we have to do it
more then once or twice.
You'll note that there are three current arrows drawn in 3b too,
in red. This doesnt mean there are three currents however,
because the SAME current flows through all three elements.
The voltage arrows, on the other hand, are different because
each element responds differently to the same current.
In reality, we only need one arrowhead for current.
Note also that both voltage arrows for the resistors point
in opposite direction to the voltage arrow for the battery.
This is again because if you follow the rules for drawing the
voltage arrows, the current arrow points toward the bottom of
Rs and toward the top of R1, so the voltage arrows are drawn
accordingly.

Fig 3c:
Here we have the same as in figure 3b, except again the
battery is replaced with it's equivalent voltage
source and resistor Rs is now R1 and R1 is now R2, plus
the resistors are lined up on the right side.
The effect of drawing it differently results in the 
arrow for R1 being inverted from fig 3b, but that's ok
because we dont care about that, we only care that we
followed the rules for drawing the arrows, which are
based on the direction of current flow, not the orientation
of the parts or whatever else.
Also note that the source voltage arrow is marked 'Vx'
and the voltage across R1 is marked 'V1' and the voltage
across R2 is 'V2'. This is because we eventually want
to know ALL of these voltages in order to understand
the circuit better. We call them something so we have
a variable to work with later.

Fig 3d:
Notice that this is the same as 3c, except it doesnt have
any arrows drawn but rather has large black dots drawn
inbetween each and every element in the circuit.
These dots correspond to points in the circuit, and are
called "Nodes".
Notice that we only need one node whereever two elements
join, but in more complex circuits where more then two
elements join we still only need one node per group.
Another way to state this is if we were to solder several
parts together at the same point, that would be called 
a node. We only need one node per point, and we dont need
two nodes say along a perfect conductor (wire).
The idea is to draw as few nodes as possible, yet still
get every connection.
You'll also note that for all three nodes, each one has
a number next to it. This is a simple procedure of
just numbering the nodes. For this circuit in 3d i started
from the number zero (0) as shown, because that point
might be the ground for the circuit, however it's possible to
start with the number 1 as well.
As a side note, sometimes the circuit in 3d is called a
'voltage divider' because the voltage at node #2 is a
division of the source voltage V. Later we will find 
that there is an exact relationship between the voltage
V and the voltage at node #2 and the value of the two
resistors. In other words, knowing V, R1, and R2, we
can calculate the voltage at node #2.
The node voltages are referenced to node #0, in that
the voltage at a given node is measured from that
node to node zero (0).

Sometimes it's convenient to label voltages in the circuit
using the nodes as subscripts. Since a voltage requires
two points for it's specification, we need to show two
nodes for each voltage.
For example, the source voltage (battery) in 3d can be called
v[1,0], or even v[10] because it's measured from node 1 to
node 0. Similarly, the voltage ACROSS R1 can be called
v[1,2], because it's measured from node 1 to node 2.
In the literature, this is often noted with actual subscripts
so these voltages look like this:
v10,v12, etc, except of course the numbers are written below
the line of print as subscripts usually are.


Next time we'll find out how to solve for the voltage
at node #2 in fig 3d. We'll also probably be able to get to
our first LED circuit!
I dont know about you, but i can hardly wait 


Self test problem:

In figure 3c, we can 'lump' resistors R1 and R2 into one simple
resistor by adding their values. If R1 is
10 ohms and R2 is 20 ohms and the battery (Vx) is 30 volts,
how much current flows as marked by the arrowhead?
Hint: add the two resistances to form a single resistor, then
redraw the circuit with one resistor, then use Ohm's Law V=I*R
to find the answer.


See ya next time!

Take care,
Al


----------



## MrAl (Jul 12, 2004)

*Re: CPF University: EE Course (Lesson 11)*

Hello again!

Thanks for taking part in the course.
We didnt get to an LED circuit this time, im sorry, because
we had to cover two important laws first which will aid in
finding out much information about a simple circuit like
an LED and a series resistor and a battery. We should be
able to get there next time however.
Today's circuits are 'almost' an LED circuit, and we'll see
why next time.


Answer to last self test:

In figure 3c, we can 'lump' resistors R1 and R2 into one simple
resistor by adding their values. If R1 is
10 ohms and R2 is 20 ohms and the battery (Vx) is 30 volts,
how much current flows as marked by the arrowhead?
Hint: add the two resistances to form a single resistor, then
redraw the circuit with one resistor, then use Ohm's Law V=I*R
to find the answer.

Answer:
[The current will be 1 amp, because 30 volts divided by the total
resistance of 30 ohms equals 1 amp]


Serial: EE-Ch0001-p111-{A939C0B7-558E-4687-9271-AB92E4935FD0}

Today we'll talk about Kirchhoff's Laws. These laws help
us to analyse circuits because they are applicable to any
circuit we can find.


Refer to: 
http://mral.peu.net/index.php?page=ChapterOne






Kirchhoffs's Voltage Law

Our first Law for network analysis is Kirchhoff's voltage law.
It's really quite simple and it applies to all circuits
no matter how complicated.

Fig 4a shows a voltage V in series with two resistors
R1 and R2. The voltage arrows (polarities) are already drawn.
What we do next is to sum the voltages around the path which
the current takes (current shown by the arrowhead as usual).

Kirchhoff's voltage law states this:

"The sum of the voltages around a closed path equals zero".

What this means is if we follow the current path along the
conductors through each component we will obtain a sum
of all the voltage drops equal to zero.

Taking our circuit in Fig 4a again, we can start anywhere, so
we will start at the BOTTOM of the source marked '30v'.
The rule is, if you first encounter the tail of an arrow you ADD
that voltage to the total sum, and if you first encounter the TIP
(or head) of an arrow you SUBTRACT that voltage.
In other words:
TAIL ---> ADD
HEAD <--- SUBTRACT

Let's try this with Fig 4a ...

Starting at the bottom of the source labeled "V" we 
first encounter the tail of V. Since V=30v (30 volts)
we encountered a tail of a 30v arrow so we add that to
the sum. Doing this gives us 30v to start with, so we
write that down:
30

Next, we follow the current arrow up though the source and
around until we encounter R1. We have now encountered an
arrowhead, who's arrow is marked "10v". Since we got an
arrowhead now we subtract that 10 from our sum so far.
Our sum so far was 30 so we subtract 10:
30-10=20

Next, we follow the direction of current again which takes
us through R1 and then we encounter another resistor and
another arrowhead of it's voltage arrow, so we again subtract,
only this time it's 20v so we subtract 20 from our sum so far.
Our sum before this was 20, so we subtract the 20 from the
current arrow and we get 0 (zero).

Thus, the sum of the voltage drops in a series circuit is
equal to zero.


Kirchhoff has another law, which will also help us.
The next law is called
Kirchhoff's current law.

Refer to Fig 4b where we see a large 'node'. This node
has three currents flowing into it and two currents flowing
out of it. Current arrowheads show direction as usual, and the
value of each current is shown by each number next to each
arrowhead.

Kirchhoff's current law states this:

"The sum of the currents entering a node equals the sum of
the currents leaving the node"

Let's try this with Fig 4b...

Since we have three currents entering the node, we write them
down as a sum on one side of an equation:

1+4+5

Note all that's done here is to sum all the currents entering
the node, as this is part of the voltage law above.

Next, we sum the currents leaving the node:

6+4

Now we equate the currents entering and the currents leaving:

1+4+5=4+6

because that's what the law states.

Now we simply add up the two sides and see if they are equal:

10=10

and they do come out to be the same, as the law states.


Summary:

Kirchhoff's voltage and current laws apply to any circuit we can find
so they turn out to be extremely important in circuit analysis.
Thanks to Mr. Kirchhoff for his amazing find!


Self test problems:
(refer to figure 4)
These are very important questions!

1. In fig 4a, say we had V=40v and R1's voltage 20v and R2's voltage
20v, what is the sum of the voltage drops around the closed current
path?
2. In fig 4b, say we have the three currents 2,4,5 entering the node
and 6 is again one of the currents leaving the node, what is the
last current leaving the node?
Hint: follow Kirchhoff's current law calling the unknown current 'i'
then write the equation as we did above and solve for 'i'.
3. In fig 4c, we have two voltage sources (probably batteries)
but it doesnt make any difference at all, as we still follow
the same rules and laws.
Let's say we have V2=1 volt, V3=1 volt, and V4=1 volt.
What is the value of V1?
4. In fig 4d, say we have i1=1 amp, i2=2 amps, and i3=3 amps.
What is the value of i4? (how much current)
5. In fig 4d, say we have i1=1 amp, i2=3 amps, and i4=7 amps.
What is the value of i3?


Take care, and good luck,
MrAl


----------



## MrAl (Jul 13, 2004)

*Re: CPF University: EE Course (Lesson 12)*

Hello and Welcome!

Here's another part of the Course.
I hope you find it interesting, as we now have our first LED circuit
included here this time.


Answers for last time:

1. In fig 4a, say we had V=40v and R1's voltage 20v and R2's voltage
20v, what is the sum of the voltage drops around the closed current
path?
[The sum is zero no matter what the case, and to prove it all we
have to do is follow the current loop around in the direction of
the current arrow and use the rules we learned. This would give
us: 40-20-20=0 as expected.]

2. In fig 4b, say we have the three currents 2,4,5 entering the node
and 6 is again one of the currents leaving the node, what is the
last current leaving the node?
Hint: follow Kirchhoff's current law calling the unknown current 'i'
then write the equation as we did above and solve for 'i'.
[Following KCL, we have 2+4+5=6+i, or 11=6+i, or 11-6=i, so i=5]
3. In fig 4c, we have two voltage sources (probably batteries)
but it doesnt make any difference at all, as we still follow
the same rules and laws.
Let's say we have V2=1 volt, V3=1 volt, and V4=1 volt.
What is the value of V1?
[Using KVL, we have V1-V2-V3-V4=0, so V1-1-1-1=0, or V1=3]

4. In fig 4d, say we have i1=1 amp, i2=2 amps, and i3=3 amps.
What is the value of i4? (how much current)
[Using KCL again, we have i1+i2+i3=i4, so 1+2+3=i4, so i4=6 amps]
5. In fig 4d, say we have i1=1 amp, i2=3 amps, and i4=7 amps.
What is the value if i3?
[Again using KCL, we have i1+i2+i3=i4 (same as before), so 1+3+i3=7, so
4+i3=7, or i3=7-4, which comes out to i3=3 amps. Checking, 1+3+3=7 or 7=7
so it's correct.]



Serial: EE-Ch0001-p112-{A939C0B7-558E-4687-9271-AB92E4935FD0}

Today we'll get into quite a few new things, and finally we get
to our first LED circuit. I dont know about you, but im going
to celebrate tonight! 

First we'll look at some units of current so we can write currents
in different forms a bit easier and faster, then we'll look at
series and parallel circuits and the main features about them,
and then we'll finally get to our first LED circuit.



Lets take a quick look at some units of current

Units of current:

milliamps = ma
1 ma = 1 amp / 1000

microamps = ua
1 ua = 1 amp / 1000000

nanoamps = na
1 na = 1 amp / 1000000000

So if we write:
20ma
what we mean in amps is:
0.020 amps








We now come to the basic two circuit types, and learn some
valuable information about each one:

Series circuit:
Current is the same though every element (Figs 5a, 5b, 5d, 5e).
Voltages across each element is generally different, but could be
the same level by chance.

Parallel circuit:
Voltage is the same across every element (Fig 5c only).
Current through each element is generally different, but could be
the same level by chance.


Refer to figure 5

Fig 5a:
Here we have a voltage source (battery) in series with two resistors.
We know the supply voltage is 5v because it's marked, and we know
the value of R2 also because it's marked. What we want to find is
the value of R1.
To find R1, we would like to use Ohm's Law because it tells us
the resistance if we know the voltage and current, but wait, we
only know the voltage is 1.4v so far, so before we can find R1
we have to find the current in the 'loop' (marked I1).
How can we find this current?

Well, we know two things:
1. In a series circuit, the current is the SAME through every element
2. Ohm's Law

Starting with #1 above, since the current is the same through every element
it must be the same though R1 as it is through R2, so if we knew the
current through R2 we would right away know the current through R1.
Once we know the current through R1 we could then find it's resistance,
because we already know the voltage is 1.4 volts (marked).

So let's start by finding the current through R2...

Using Ohm's Law, I=V/R, so I=3.6/180, which comes out to 20ma.

Since we now know the current through R2, we also know the
current through R1 is the same so it's 20ma also.

Now that we know R1's current, and since we know it's voltage, we
can finally find it's resistance again from Ohm's Law:
R=V/I, so R1=1.4/0.020, which comes out to 70 ohms.

Fig 5a redrawn with the resistor labeled with the resistance we just
found is shown in Fig 5b, and we are done.


Fig 5d:
Here we have another series circuit, but the voltages for the
two resistors are not shown. How can we find these voltages?
We dont know the current either, so that makes it a little
more difficult. We cant use Ohm's Law (yet) because we dont
know two quantities we can plug into one of his forms.

In this case, we have to recognize that we cant do anything 
else except 'Lump' the two resistors into one. Since we know
both their values, this should be easy. Once we know
the combined resistance, we can find the current using Ohm's
Law, and then we'll be able to get farther with this circuit.

This brings us to two rules about combining (or 'lumping')
resistors:

1. Resistors in series add...that is, RT=R1+R2+R3...etc.
where R is the total resistance.
2. Resistors in parallel add the reciprocals, then take
the reciprocal of the result..that is, 1/(1/R1+1/R2+1/R3...)

We need the rule for series resistors for this problem, so
we add the two resistors to obtain the total:
R1+R2=70+180
and so
70+180=250 ohms total resistance

Now knowing the total R, we can redraw the circuit with one
resistor and use Ohm's Law knowing the V and the R:
I=V/R, so I=5/250, which comes out to 0.020, or 20ma.

Now that we know the current through both R's, we go back
to the original circuit (Fig 5d) and apply the current through
R1 to get it's voltage (using Ohm's again) and to R2 to get
it's voltage also.
Doing this we use V=R/I for both:
V=R1/0.020, so V=70/0.020, which comes out to 1.4 volts, so we
know R1's voltage drop now, then

V=R2/0.020, so V=180/0.020, which comes out to 3.6 volts, so
now we know R2's voltage as well.

Knowing all the currents and voltages, we have completed the
analysis and we're done.


Next, we go to Fig 5e, where we have our first LED circuit
of the Course !!

Here in Fig 5e we have a voltage source (battery), one resistor,
and one white Nichia type LED. We know the current through
all the elements of the circuit because it's marked, but if
it wasnt marked we could have looked on the graph to the right
(Fig 5f) and noticed that 3.6 volts corresponds to 20ma for
that LED (typical graph shown).
Before we proceed, notice that this is almost the same circuit
as figure 5d except R2 is replaced by an LED.

To show how similar this is to Fig 5d let's calculate the static
resistance of the LED (the static resistance is it's resistance at
a given operating point, in this case 3.6 volts and 20ma).
Since we know both it's voltage and it's current at this operating
point, we can easily find it's resistance using Ohm's Law:
R=V/I, so R=3.6/0.020, which comes out to 180 ohms.
Before we proceed here, i'd like you to note that this resistance
is the static resistance of the LED, and it only applies when
the led is operated at very close to if not exactly 20ma.
If we were to change this, this resistance would also change.
As long as we maintain 20ma through the LED we can use this
value however, and that is what we intend to do here.
More about that at a later date too...

Now compare again with Fig 5d and note that the resistor R2 is
also 180 ohms. This means our two circuits (Figs 5d and 5e)
should come out to be similar.

Ok, so now we know the LED's static resistance, but how do we find
the value of R1 which gives us this magic operating point of 20ma?

Well, we were able to find R1 in the previous Fig 5d, and this circuit
isnt any different once we establish the operating point (20ma), so
why not proceed using the same basic technique?

Since we already know the voltage of the LED (3.6v) we dont have
to go through all that trouble. All we have to do this time is
apply Kirchhoff's Voltage Law!

The sum of the voltage drops in a closed path equals zero, so
let's sum the voltage drops calling R1's voltage VR1 for now...

Starting from the bottom of the source, we have a tail, so
we get 
+5
to start, then
we hit the head of R1, so we SUBTRACT VR1:
5-VR1
next, we hit the head of the voltage of the LED, so we subtract 3.6:
5-VR1-3.6
and equate them all to zero:
0=5-VR1-3.6
Carrying out the easy operation, 5-3.6 first we end up with:
0=1.4-VR1
Adding VR1 to both sides we end up with:
VR1=1.4
and we have the voltage across R1.

Now that we have the current through R1 and the voltage across it,
we can calc it's resistance with Ohm's Law again:
R1=V/I, so R1=1.4/0.020, which comes out to 70 ohms as expected.

Finally, note that if you replace the LED with it's static
resistance you end up with the same circuit! Ha ha!
So it turns out that an LED circuit is (almost) nothing
more then a circuit with resistors!

See how important resistors are in electrical circuits?


Questions for testing your knowledge:

1. In figure 5e, replace the LED with one
that only has a 3.4 volt drop across it while
20ma is flowing through it.
Find the required value of R1, and find
the LED's static resistance at this operating point.

2. Again in figure 5e, replace the battery with
a 100 volt source instead of a 5 volt one.
Using the pictured value of voltage for the LED
of 3.6 volts and again with 20ma flowing, 
calculate the required value of R1 and also
calculate the power dissipated in it using 
Joules Law P=V*I .


Good luck, and see ya next time!

Take care,
MrAl


----------



## MrAl (Jul 15, 2004)

*Re: CPF University: EE Course (Lesson 13)*

Hello again and welcome to another part of the EE Course.

Answers for last time:

1. In figure 5e, replace the LED with one
that only has a 3.4 volt drop across it while
20ma is flowing through it.
Find the required value of R1, and find
the LED's static resistance at this operating point.

Answer:
Since the LED has 20ma flowing and it has 3.4 volts
across it we easily calc the static resistance by
using Ohm's Law:
R=V/I, so R=3.4/0.020, which comes out to 170 ohms.
The required value of R1 to maintain 20ma through the
LED is calculated by using KVL (Kirchhoff's Voltage Law)
around the current loop, solving for the voltage drop,
then the resistance of R1:
5-V1-3.4=0
doing the math,
1.6-V1=0
or
V1=1.6 volts (voltage across R1)
Now we know both V and I for R1 so we can calc R1:
R1=V/I, so R1=1.6/0.020, which comes out to 80 ohms.


2. Again in figure 5e, replace the battery with
a 100 volt source instead of a 5 volt one.
Using the pictured value of voltage for the LED
of 3.6 volts and again with 20ma flowing, 
calculate the required value of R1 and also
calculate the power dissipated in it using 
Joules Law P=V*I .

Answer:
The required value of R1 is calculated again using KVL
after replacing the source with 100v instead of 5v:
100-V1-3.6=0 (using KVL)
so
V1=96.4 volts
Now that we know the voltage across R1 we can calc it's
value:
R1=V1/I1, so R1=96.4/0.020, which comes out to 4820 ohms.
To calc the power, we use
P=I*R
so
P=0.020*96.4
which comes out to 
1.928 watts (the unit of power is the watt)
which is almost 2 watts.
In practice, we would not use a 2 watt resistor however,
but rather double this value and use that, so we would have
to purchase a 4 or 5 watt resistor.
We would also have to take into account temperature rise,
because this resistor will heat up and could melt something
if it's mounted too close to the resistor.
We'll look at temperature rise at a later time.
This question showed the importance of calculating the
power in a resistor. If we didnt calculate it and used
a 1/2 watt resistor, it would have heated up and burned
out or maybe even started a fire.


Today's part of the course:

Serial: EE-Ch0001-p113-{A939C0B7-558E-4687-9271-AB92E4935FD0}

Today we are going to look at another way of looking at the
LED in a series circuit, and also look at a parallel circuit.

Refer to Figure 6 for the following discussion...







In fig 6a, we have our series LED circuit again, with all the
voltages and currents shown as usual with arrows. We talked
about how the LED voltage is constant at the operating point
of 20ma so we could replace it with a resistor, and we calculated
the value of resistance for that resistor to be 180 ohms.
Fig 6b has this circuit drawn.
Next, we will look at replacing the LED with a voltage source.
In fig 6c the circuit is redrawn replacing the LED with another 
voltage source. Notice the current is still the same, and of
course the voltage across the 'LED' is still the same, 3.6v,
so everything about the circuit is still the same, even though
we have replaced the LED with a voltage source. If we didnt know
the value of R1 we could still easily find it by using KVL and
solving for the voltage across R1 and then finally the value
of R1 as we have done before. It's interesting that because at 
a given operating point we can replace the LED with either a
resistor or a voltage source and still find the value of R1.
Replacing the LED with the voltage source is usually simpler
though. In either case, the LED absorbs power, and the 5v 
source delivers power.
Note that when we replace the LED with another type of element,
we are assuming the operating point will be obtained once the
circuit is turned on. If the operating point changes, the
assumption changes so the circuit will behave differently.



Parallel Circuits

We now turn our attention for a little while to parallel
circuits.


In fig 6d, we find a circuit with two resistors and one
voltage source. The two resistors are said to be in parallel
because the voltage across them is the same and they are connected
together at both ends. All the currents and voltages are marked with arrows
as usual. Note that V1 is the same as V2 because the two resistors
are in parallel, and they are both in parallel with the source.
Now let's say we know the voltage Vx is 4 volts, and we know the
resistor R1 is 12 ohms and the resistor R2 is 6 ohms.
How do we find the current flowing out of the battery, and how
do we find both currents I2 and I3 for both resistors?
Since R1 is in 'parallel' with R2 we can find the lumped
value (total resistance) again and replace it with one
single resistor to find I1.
To find the values of I2 and I3 we can use Ohm's Law.

To start, we find I1 by lumping R1 and R2, so we use the formula
for parallel resistors:
R=1/(1/R1+1/R2)
This gives us a single resistor we can substitute for R1 and R2
so we will have a simpler circuit in which we can easily find I1.
R1=12 and R2=6 so we have:
R=1/(1/12+1/6)
R=1/(1/4)
R=4
So we get R=4 ohms.
Knowing the voltage is 4 volts, we get the current I1 easy:
I=V/R, I1=4/4, so I1=1 amp.

Now to calculate the current through each resistor we simply
divide the voltage by each resistor in turn:
I2=4/12 which is 1/3 amp, and
I3=4/6 which is 2/3 amp.

Now we can check by adding I2 to I3 and see if we get I1, because 
using KCL (Kirchhoff's Current Law) the sum of currents entering a node
equals the sum of currents leaving a node, and the node is the junction
of the tops of R1 and R2 and the (+) terminal of the source.
Adding,
I2+I3=1
so it checks out.

We thus found two methods of finding the current I1:
1. Lumping the two resistors and solving the resulting circuit
2. Calculating the individual currents and then using KCL


When we start doing more complex circuits we will find these facts
about parallel circuits just as handy as those things about series
circuits. All these methods help to find answers to much more
complex circuits because they usually have a combination of
series and parallel circuits within them.


Self test questions:

1. Replacing the LED in fig 6a with a voltage source we get the circuit
in fig 6c. Find the value of R1 required with a source voltage of 6 volts.

2. In Fig 6c, is it possible to use a 3v source (for the 5v one) and
still maintain the operating point of 20ma?

3. In fig 6d, say we have Vx=6 volts and R1=10 ohms and R2=10 ohms.
What is I1, I2, and I3?

4. In fig 6d, say we have Vx=3 volts and R1=3 ohms and R2=12 ohms.
Calculate I1, I2 and I3.

See ya next time!

Take care,
MrAl


----------



## MrAl (Jul 17, 2004)

*Re: CPF University: EE Course (Lesson 14)*

Hello again,

Here's the next part in the Course.


Answers for last time's self test:


1. Replacing the LED in fig 6a with a voltage source we get the circuit
in fig 6c. Find the value of R1 required with a source voltage of 6 volts.
answer:
Using KVL
6-V1-3.6=0
V1=2.4
at 20ma,
R1=V1/I, R1=2.4/0.020
which equals 120 ohms.

2. In Fig 6c, is it possible to use a 3v source (for the 5v one) and
still maintain the operating point of 20ma?
answer:
No, because 3v is less than the LED's operating point of 3.6v.
The LED might still light dimly, because it would operate at
a different point, but we're only concerned with getting it to
light with 20ma, not with less current.

3. In fig 6d, say we have Vx=6 volts and R1=10 ohms and R2=10 ohms.
What is I1, I2, and I3?
answer:
Using Ohm's Law, I2=Vx/R1, so I2=6/10 or 0.6 amps, while I3=Vx/10, also 0.6 amps,
so I2=0.6 and I3=0.6, and since I1=I2+I3 (using KCL) I1=0.6+0.6=1.2 amps.
Thus, I1=1.2 amps, I2=0.6 amps, and I3=0.6 amps.

4. In fig 6d, say we have Vx=3 volts and R1=3 ohms and R2=12 ohms.
Calculate I1, I2 and I3.
answer:
Using OL (Ohms's Law) we have
I2=Vx/R1, so I2=3/3 which equals 1 amp, and 
I3=Vx/R2, so I3=3/12, which equals 1/4 or 0.25 amps.
Using KCL, I1=I2+I3 so I1=1+0.25, which equals 1.25 amps.
Thus, I1=1.25 amps, I2=1 amp, and I3=0.25 amps.


Today's part of the course:

Serial: EE-Ch0001-p114-{A939C0B7-558E-4687-9271-AB92E4935FD0}


Today we'll talk a little about series and parallel circuits
when they happen to be present in the same circuit. This
occurs so much that it's very important to be able to 
recognize these 'subcircuits' within a larger circuit.
We start with the now somewhat familiar series LED circuit.

Refer to fig 7...







In fig 7a, we find our series LED circuit again. The currents
and voltages have all been labeled already.
Looking now at fig 7b we find almost the same circuit, except
we have two resistors R1 and R2 in the place where R1 alone
was in 7a. So what's the difference...the only difference is now
we have two resistors in parallel instead of just a single
resistor.
What we have to do before we can solve for the current I3 is
we have to combine the values of R1 and R2 to form a single
resistor, and then we end up with the same circuit as fig 7a.
This has been illustrated in fig 7c.
Note in 7c the two currents are shown simply as the addition
of currents I1 and I2 from fig 7b. I1 and I2 add to make I3.
Ok, so we have to find I1 and I2 and then sum to get I3, that
will give us the total current flowing through the LED.

Let's see what happens with R1=105 ohms and R2=210 ohms...

To get fig 7c from 7b all we have to do is combine the resistors
into one lumped value, and then we'll know the equivalent resistance
as well as the total current, so we start by using the formula
for parallel resistors:
R=1/(1/R1+1/R2+1/R3+...)
so
R=1/(1/R1+1/R2)
so
R=1/(1/105+1/210)
and using a calculator, we get
R=70 ohms.

It's a simple matter now to see that this is the same circuit only it
uses two resistors instead of one, so the current is still 20ma.
It's good to note however, that if we didnt have the correct value
(70 ohms) on hand perhaps we could find two values to use in parallel
to use as a replacement for one single resistor. Another simpler
example would be R1=140 and R2=140, which comes out to 70 ohms also.

Ok, so now that we found the single resistor R, we could solve for the
current in fig 7d using KVL:
5-V1-3.6=0
so again the voltage drop across R is 1.4 volts,
and knowing it's value to be 70 ohms we again get 20ma,
so I3=20ma again.

Another way to solve this would be to notice that the LED's operating
voltage of 3.6 volts is already given, and since we are using the Nichia
type LED the current through it must be 20ma, so we would know this
without calculating anything. There is a catch however, and that's what
we are going to talk about next.


Component Parameter Variations

In the previous parts of the course we were assuming the Nichia LED
voltage was 3.6 volts at 20ma because we looked at the graph supplied
by the manufacturer of the part and found the 'typical' value of voltage
for a given current. In reality, however, there is usually a 'spread' 
of voltages associated with a given current for a part like this, we
next we'll try to take that into account to get some real world values
of current.

It's ok to estimate the voltage of the LED to be 3.6 volts at 20ma, but
one has to remember that the actual voltage at 20ma might range from
maybe 3.4 volts to 3.8 volts, so it's good to look at the circuit
again and make sure it will still work if we happen to get an LED that
is working at the far end of the spec. As it turns out, this isnt
that hard to do.

Looking again at 7d, we find all the voltages and currents are labeled.
The LED voltage is labeled "3.6v" but we are going to change that first
to 3.4 volts and then to 3.8 volts to be sure we cover the whole range
of possible LED voltages. We'll use 70 ohms again for R.

We will use "vLED" as an abbreviation for the LED voltage.

First we change vLED to be 3.4 volts and analyze the circuit:
using KVL, 5-V1-3.4=0,
so V1=1.6 volts . Note this is a little higher then before.
Now using OL (Ohm's Law):
I=V/R, so I=1.6/70, which is approx 0.022857 amps (using a calculator).
Since I1+I2=0.022857 amps and I3=I1+I2, I3 equals 0.022857 amps also,
so the current through the LED is now almost 23ma. 
Since the Nichia LED is run as high as 30ma for some applications,
23ma shouldnt pose a problem if used in normal ambient temperatures.
Imax for the LED would therefore be 23ma and would be considered safe
for the LED.

Next we change vLED to be 3.8 volts and analyze the circuit a second time:
using KVL, 5-V1-3.8=0,
so V1=1.2 volts . Note this is a lower then before.
Now using OL (Ohm's Law):
I=V/R, so I=1.2/70, which is approx 0.01714 amps (using a calculator).
Since I1+I2=0.01714 amps and I3=I1+I2, I3 equals 0.01714 amps also,
so the current through the LED is now reduced to about 17ma. 
Since the Nichia LED brightness at 17ma isnt that much lower than at 20ma,
17ma shouldnt pose a problem.

In the above analysis, we found that a variation in the manufacturing process
of an LED could affect our circuit performance, but in this particular
circuit we found it wouldnt affect it that much so the circuit should
work as expected for a long time.

Now let's take a quick look at the same circuit only with
less input voltage.

Let's say we use a battery that is only 4 volts, so the battery 'source'
voltage in fig 7d goes down from 5v to only 4 volts.
Let's also say we design the circuit to work at this 4 volts so we 
still get 20ma at 4 volts battery.

To start, we have to calc a new resistor to provide 20ma at 4v battery:
using KVL, 4-V1-3.6=0, so V1=0.4 volts. This is significantly lower
then before when the battery was 5v.
Now calc the new resistor R value using OL, R=V1/I, so R=0.4/0.020
which equals 20 ohms, so R=20 ohms for a battery of 4 volts.

Next, we need to calculate the variation in LED current with the
possible manufacturing range spread for the LED of 3.4 to 3.8 volts
again. To do this, we again replace the LED voltage with 3.4v and
then 3.8v in turn and calculate the results again.
Doing this first for 3.4 volts:
4-V1-3.4=0, so V1=0.6 volts, which using OL gives I3=V/R, so I3=30ma,
then with vLED=3.8 volts, we get:
4-V1-3.8=0, so V1=0.2 volts, which makes I3=10ma.

The above two calculations show that 

Imax=30ma
and
Imin=10ma

for the new circuit with the 4v battery.

What this means is that we might find either 30ma flowing or
10ma flowing through this LED. That means either we'll be
operating at fairly high brightness or we'll only be 
operating a minimal half brightness, for any possible LED.

This would probably not be something we would want to see,
because we might end up with a flashlight that is too dim.
Thus, we would have to conclude that this probably isnt
a good circuit to use for a flashlight or other lighting purpose.

We can also conclude that by lowering the input battery voltage
and readjusting R to provide 20ma at the maximum battery voltage
caused a problem. In general, the lower the battery voltage
the more this becomes a problem, so that if you buy say 10 LED/s
and some of them read lower voltage then others, you can expect
a spread in brightness instead of having each LED light to 
the same level.

One way around this problem is to use a constant current supply
instead of just a battery, but we'll have to save that solution
for a later date as there are other things we have to cover
before that in order to understand all of the principles involved.

Self Test Questions

1. In fig 7b, with R1=70 and R2=70 and the LED voltage of 3.6v find
the battery voltage that will provide 20ma to the LED.
Hint: use KVL and solve for the battery voltage.

2. In fig 7b, with an LED voltage of 3v and battery of 5v find two resistors
that are the same value that will provide 20ma to the LED.
Hint: a shortcut for finding the resistance of two resistors of equal value
in parallel is to take half the resistance of one resistor.

3. In fig 7b again, if battery=5v and LED=3 volts and R1=300 ohms, find the 
value of R2 that will provide 20ma to the LED.


Take care,
MrAl


----------



## MrAl (Jul 19, 2004)

*Re: CPF University: EE Course (15)*

Hello again and welcome to Chapter 2 of the course!


Answers for last self test...

1. In fig 7b, with R1=70 and R2=70 and the LED voltage of 3.6v find
the battery voltage that will provide 20ma to the LED.
Hint: use KVL and solve for the battery voltage.
Answer:
Since R1 in parallel with R2 is 35 ohms, and 35*0.02 is 0.7v, we need
a supply voltage that is 0.7 volts greater then the LED voltage, so
3.6+0.7=4.3 volts.

2. In fig 7b, with an LED voltage of 3v and battery of 5v find two resistors
that are the same value that will provide 20ma to the LED.
Hint: a shortcut for finding the resistance of two resistors of equal value
in parallel is to take half the resistance of one resistor.
Answer:
Using KVL to find the voltage across the resistors we get 2v, and Ohm's Law
2/0.02=100 ohms total, then knowing two resistors in parallel form a
resistor who's value is half, we multiply 100 times 2 and get 200 ohms,
so it would take two 200 ohm resistors in parallel to get 20ma through
the LED.

3. In fig 7b again, if battery=5v and LED=3 volts and R1=300 ohms, find the 
value of R2 that will provide 20ma to the LED.
Answer:
Since battery V=5 and LED voltage is again 3 volts, we again have 2v across
the two parallel resistors, and from problem #2 we found we needed 100 ohms
total. Knowing one value is 300 ohms we can find the
other value by solving the equation for parallel resistors making R2 the
unknown.
Recalling the formula for two parallel resistors:
R=1/(1/R1+1/R2)
and plug in the value for the total R=100 and for R1=300 we get:
100=1/(1/300+1/R2)
To solve for R2 we multiply both sides by "(1/300+1/R2)" and we get:
100(1/300+1/R2)=1
Now we distribute the '100' to both terms inside the parens:
100/300+100/R2=1
simplify:
1/3+100/R2=1
multiply both sides by '3' we get:
1+300/R2=3
subtract 1 from both sides we get:
300/R2=2
multiply both sides by R2 we get:
300=2*R2
divide both sides by 2 we get:
150=R2
so this means the other resistor would have to be 150 ohms.
As a side note, we could have done it this way...
Recall the formula:
R=1/(1/R1+1/R2)
first take the reciprocal of both sides and we end up with:
1/R=1/R1+1/R2
now subtract 1/R1 from both sides:
1/R-1/R1=1/R2
then take the reciprocal of both sides again:
1/(1/R-1/R1)=R2
and we end up with a neat little formula for R2 knowing R total and R1 !
R2=1/(1/R-1/R1)


For today:

Serial: EE-Ch0002-p101-{A939C0B7-558E-4687-9271-AB92E4935FD0}


Today we are going to talk about more general series and parallel circuits,
and look at a circuit with both series and parallel subcircuits contained
within it. We'll reduce the circuit to a simpler form by solving the
lumped values for the series and/or parallel combinations we can find,
then solve for the total current.
Lastly, we'll take a very brief look at a circuit that cant be solved
using lumped values to catch a glimpse of what's coming up in the course.
Not to worry, however, since it wont be long when we can solve this one
almost as easy as all the other circuits.






In fig 8a, note all the elements are connected end to end, while if you compare it
to fig 8b, all the elements are connected together at BOTH ends. Fig 8a is a
series circuit while 8b is a parallel circuit. 

Some more facts about series and parallel circuits:

Voltages add in a series circuit.
Currents add in a parallel circuit.

Current is the same through every element in a series circuit.
Voltage is the same across every element in a parallel circuit.

Note also in 8a that the voltage of the source is divided up
among the resistors, as each one has a part of V across it,
but in 8b each resistor has a part of the current.

To solve the circuit of fig 8a we would first add all the resistor values,
obtaining one resistor and then apply Ohm's Law.
To solve the circuit of fig 8b we would first add all the reciprocals of
each resistor and then take the reciprocal obtaining one resistor and then
apply Ohm's Law.
In each case we reduced the circuit to a simpler form and then applied 
Ohm's Law.

Next we turn to fig 8c. Here we have an unusual combination of resistors,
some in parallel and some in series. Note R2 and R3 are in parallel and
so are R4 and R5, but R6 is in series with those two sets, and R1 is
in parallel with the series set. To find the lumped resistor value
we would simply apply what we know about series and parallel resistors
to find the lumped value.

Noting R2 and R3 are in parallel, we apply the formula for parallel
resistors and call the resultant resistor value R23:
R23=1/(1/R2+1/R3)
Now if we wanted to, we could redraw the circuit to show a single
resistor in place of R2 and R3 who's value would be R23.

Next, we do the same thing with R4 and R5, calling their lumped
value R45:
R45=1/(1/R4+1/R5)
Replacing R4 and R5 with the single R45 would mean we have three
resistors in series on the right hand side of the circuit, R23, R45,
and R6, so we simply add their values to get a single resistor we'll
call R23456:
R23456=R23+R45+R6
Replacing R23, R45, and R6 with the single resistor R23456 gives us
a circuit with only two resistors in parallel, R1 and R23456, so
we use the parallel resistor formula again and call the final resistor
value RL:
RL=1/(1/R1+1/R23456)

and now we have our final resistor value, RL, so we can apply
Ohm's Law to get the current flow out of the source V1.


Let's try a numerical example next...

Let's make R1=3, R2=2, R3=2, R4=1.5, R5=3, and R6=1 ohms,
and a source voltage V1=3 volts.

We start again by finding the parallel combo of R2 and R3:
R23=1/(1/2+1/2)=1/1=1, so R23=1 ohm
Then R4 and R5 in parallel:
R45=1/(1/1.5+1/3)=1 ohm (might want to use a calculator)
so R45=1 ohm
Now we need R6 in series with those two, so
R23456=R23+R45+R6 which is 1+1+1=3 ohms, so R23456=3 ohms,
then we need R1 in parallel with R23456, so
RL=1/(1/3+1/3)=1.5 ohms, so the final resistor is 1.5 ohms.
Knowing the source voltage is 3 volts, we use Ohm's Law
to find the current from the source I, so
I=V/R, so I=V1/1.5 which equals 2 amps, so we're done for now.

We found the lumped resistor for this circuit, then applied
Ohm's Law and we were able to find the total current flowing
out of the source. This is probably the most important
current to know, although there are other things
we wish to know about this circuit that we will cover next time.

Next, we look briefly at the circuit in Fig 8d, where we find
another combination of resistors connected to a source voltage.
Take a quick look and see if you can find parallel and series
combinations of resistors that will reduce this circuit to
one resistor and one source, but dont look too long, because
it's NOT POSSIBLE with this circuit! This is one of those
circuits where there is NO combination of series or parallel
resistors that will reduce the circuit to one resistor with
one source. In spite of this, series and parallel circuits
are extremely important because the method of lumping circuit
elements to reduce circuit complexity is still applicable to
many circuits. We'll look at the circuit in fig 8d another time
very soon, but for now we'll try to solve a few circuits like
fig 8c, remembering although EVERY circuit can not always
be reduced to series/parallel combos this way, it's still a
very powerful method.


Self Test Questions:

1. In fig 8c, with the following values for the resistors:
R1=80
R2=50
R3=50
R4=75
R5=37.5
R6=30
and
V1=4 volts
Find the total current flowing through V1.

2. In fig 8c, with the following values for the resistors:
R1=320
R2=200
R3=200
R4=300
R5=150
R6=120
and
V1=8 volts
Find the total current flowing through V1.


Good luck, and i'll see ya next time!

Take care,
MrAl


----------



## MrAl (Jul 21, 2004)

*Re: CPF University: EE Course (16)*

Hello again and welcome to another part of the course!


Answers for last time

1. In fig 8c, with the following values for the resistors:
R1=80
R2=50
R3=50
R4=75
R5=37.5
R6=30
and
V1=4 volts
Find the total current flowing through V1.
Answer:
The total lumped resistance is 40 ohms, so I=4/40=0.1 amps.

2. In fig 8c, with the following values for the resistors:
R1=320
R2=200
R3=200
R4=300
R5=150
R6=120
and
V1=8 volts
Find the total current flowing through V1.
Answer:
The total resistance is 160, so I=8/160=0.05 amps (50ma).


For today:

Serial: EE-Ch0002-p102-{A939C0B7-558E-4687-9271-AB92E4935FD0}

Today we'll talk about what's known as voltage and current division.
This is where you have a circuit with at least two resistors where
either the voltage (series) or the current (parallel) divides into
two voltages (series) or currents (parallel). Because either the
resulting voltages or currents are related to the value of the
resistors, we can easily come up with a formula to calculate the
effect of the division either in the series or in the parallel case.







Voltage and Current Division

In fig 9a we have the same series/parallel circuit from last time
when we found I1. Next, we would like to find I2. 
One way to do this would be to divide V1 by R1 to get I3, then
using KCL we could get I2 by subtracting:
I2=I1-I3
That would be fairly easy, but say we didnt know V1 and we were only
given I1. How would we get I2 knowing only the current I1 and
the resistance of all the resistors?
Fig 9b shows the same circuit with some of the R's lumped.
Finally, Fig 9c shows the value R26 as the lumped value of
the right hand side of the original circuit of fig 9a, resistors
R2, R3, R3, R5, and R6.
Here (fig 9c), we are going to assume we know by I1 and the value of R1
and R26 but we dont know the value of V1. By using a concept
known as 'current division' we can find I2 and I3.
The first thing we do is take the sum of R1 and R26, 
and use it as a denominator. Next, we take the resistor that is
NOT in the path of the current we want to find and use it in the
numerator. In other words, if we want to calculate I2 then instead
of taking R26 in the numerator we take R1, and if we want
to calc I3 we take R26 in the numerator. After that we
multiply by the total (known) current I1.

In formula form:

I2=I1*R1/(R1+R26)
I3=I1*R26/(R1+R26)

Note both formulas have the sum of the two resistors
in the denominator, and have the resistor who's current
we DON'T want to calculate (simply the other resistor
of the two) in the numerator, and then multiply that by
the total current I1. This is called current division.
For more than two resistors it gets a bit more complicated,
but it's very handy because two resistances like this are
fairly common.

There is also a rule for 'voltage division', which 
involves two resistors in series with a source voltage.
In the case of voltage division, we still take the sum of
the two resistors in the denominator, but then to find the
voltage across a given resistor we take the value of that
very resistor in the numerator, rather then the opposite
resistor's value.

Refer to fig 9d:
The formula for voltage division looks like this:

V1=V*R1/(R1+R2)
V2=V*R2/(R1+R2)

where
V is the voltage source
R1 is one resistor
R2 is the other resistor
V1 is voltage across R1
V2 is voltage across R2

Note also that in fig 9d the point labled V2 is a voltage referenced to ground.

Self Test Questions

1. In fig 9c, if R1=10 and R26=10 and I1=1 amp, what is I2?
2. In fig 9c, if R1=10 and R26=30 and I1=1 amp, what is I3?
3. In fig 9d, if R1=100 and R2=100 and V=10 volts, what is V2?
4. In fig 9d, if R1=100 and R2=300 and V=4 volts, what is V2 and what is V1?


Good luck, and see ya next time!

Take care,
MrAl


----------



## MrAl (Jul 23, 2004)

*Re: CPF University: EE Course (17)*

Hello again and welcome!


This time the answers (and some hints) will be provided with the
self test questions so you can check your work immediately instead
of having to wait for the next part.

Answers for last time

1. In fig 9c, if R1=10 and R26=10 and I1=1 amp, what is I2?
ans: I2=I1*R1/(R1+R26)=0.5 amps
2. In fig 9c, if R1=10 and R26=30 and I1=1 amp, what is I3?
ans: I3=I1*R26/(R1+R26)=0.75 amps
3. In fig 9d, if R1=100 and R2=100 and V=10 volts, what is V2?
ans: V2=V*R2/(R1+R2)=5 volts
4. In fig 9d, if R1=100 and R2=300 and V=4 volts, what is V2 and what is V1?
answers:
V2=V*R2/(R1+R2)=3 volts
V1=V*R1/(R1+R2)=1 volt


For today:

Serial: EE-Ch0002-p103-{A939C0B7-558E-4687-9271-AB92E4935FD0}

Today we'll have to talk a little about simultaneous equations. We are
used to solving one equation and extending that to doing more then one
equation at the same time wont be too much harder. We'll find that
using a calculator will be needed most of the time for equations of
maybe 3 or greater, but it ends up being so simple because all
we end up doing is copying the coefficients of the equations to
the calculator in a specific format and pressing a few buttons to
get the answer even to fairly complex systems of equations. If for
some reason you find this section difficult, hold on 'till the end
where the calculator again makes everything fairly simple. If you are
willing to pay attention you'll find you dont have to do much real
work to get to the answers.

Before we start, i'd just like to mention that sometimes we use a shortcut
notation for multiplication. A term such as 2*a can be written as simply
2a
where the 'a' is written right next to the '2'.
This means the same as "2*a".

Also, since there are two major calculator types being used for the
calculations there will be two entries for finding answers: one for 
the TI89 and one for the HP49G. You only have to refer to the line
that contains the calculator you are using. For example:

TI89: (2+3)^(-1) <enter>
HP49G: INV(2+3) <enter>

means if you are using the TI89 you enter (2+3) and then take it up to 
the (-1) power, but if you are using an HP49G then you use the function
called "INV" built in, which is available on the 'CAT' menu. Both these
entries come out to the same answer, but the HP is a little different then
the TI so their entries have to be done a little differently.


Simultaneous Equations

We've already looked at equations of the form:

5=2a+b (as noted above, this is the same as 5=2*a+b)

where we can solve for either a or b, but there also exists
a form of equation that actually contains more then one equation
to be solved as a 'system' of equations. That is, all the
equations are taken to belong to the same problem and are
written just for the purpose of solving one single problem.
That problem involves more then one variable and we'll need 
to find the values for all of the variables.

Take the following system of equations for example:

5=2a+b
8=2a+4b

This is a system in two unknowns, a and b, and the equations 
are usually just written one after the other just as shown, one
equation on one line and the next equation of the set on the next line.
The difference here is that these two equations form a system for
solving for BOTH variables a and b, while previously we had only solved
for one variable, either a or b, using only one equation.

We first look at adding and subtracting whole equations...

Lets say we want to add the first equation to the second one above:
[8=2a+4b]+[5=2a+b]=?

In order to get the answer to this, we need to realize that when
adding equations we have to keep the individual terms separate,
and all we have to do is add the 'like' terms to each other to
form the result. The result will also be an equation.
'Like' terms are those terms which appear in both equations
that have the same variable:

2a is on one side and the other so they are 'like'
4b is 'like' b because they both have 'b'
8 is 'like' 5 because they both have no variable

Thus, when we do the adding we only add the like terms:
2a+2a=4a
4b+b=5b
8+5=13

and these new terms form our resulting equation:
13=4a+5b

Note all we did here is add the individual terms separately, 
then form a new equation. Nothing strange there.

The whole thing written out looks like this:
[8=2a+4b]+[5=2a+b]=[13=4a+5b]

Now the whole idea is to find out what a AND b both are, and
adding the two didnt seem to help, because we ended up with
an equation which STILL has both variables a and b in it:
13=4a+5b

Let's try subtracting the two:
[8=2a+4b]-[5=2a+b]=?

[8=2a+4b]-[5=2a+b]=[3=3b]

Now we've gotten somewhere! We got rid of the terms with 'a' in them
so now we can find out what b is equal to.
Taking our resulting equation and dividing both sides by 3 we get
1=b

Now all we need to do is substitute 1 in for b in one of the equations
to solve for a, so say we chose 5=2a+b ...
knowing b=1 now we can substitute and solve for a:
5=2a+b
5=2a+1
4=2a
so
a=2

We therefore have solved for both a and b using the two simultaneous equations
and we're done.


Now lets say we have a system of three equations in three unknowns:

8=2a+b+c
17=2a+4b+3c
12=4a+b+c

We begin by noting that the first and third equations both have (b+c) in them,
so subtracting we'll end up with an equation in 'a' alone so we can solve for a:


[12=4a+b+c]-[8=2a+b+c]=[4=2a], a=2

We next take two of the equations (one of which wasnt used yet)
and replace their 'a' with what we found for a, 2, and proceed to
solve for one of the other variables b or c. We have to work
with two equations at the same time as we did above for the system
of two equations:

17=2a+4b+3c
8=2a+b+c

17=4+4b+3c
8=4+b+c

13=4b+3c
4=b+c

Now we can get rid of the c variables in both equations
if we multiply the second one by 3 and then subtract...
We end up with these two equations:

13=4b+3c
12=3b+3c

Subtracting the second from the first, we get rid of the '3c' term in
both equations so we can get the value for b:

1=b

Now all that's left to do is substitute the values we found for a and b
into one of the original equations and solve for c. Thus, we solved
for a, b, and c given three equations in the three unknowns.

It's good to know that we wont always have to do this by hand, as
there are other methods that use matrix algebra to solve for the
unknowns, and some calculators can handle this.

For example, the system of two we did above can be written more compactly...

5=2a+b
8=2a+4b

can also be written like this:

[[2,1][2,4]] for the right hand side and
[[5][8]] for the left hand side.

Note a matrix is used for the side that contains the variables and
a vector for the other side.

Now to solve for the variables a and b we simply enter this data
into the calculator calling [[2,1][2,4]] a matrix and calling
[[5][8]] a vector (or column matrix).
We then have two choices...
We can either use the 'simult' function on our calculator, or
we can do the following:
Take the inverse of the matrix and multiply this by the vector, which
gives us the values of a and b.

To take the inverse of the matrix we might just enter it like this:

TI89: [[2,1][2,4]]^(-1) <press enter>
HP49G: INV([[2,1][2,4]]) <press enter>

which will give us another matrix, which we then multiply by
the vector to get the values of both a and b:
TI89: [[2,1][2,4]]^(-1) * [[5][8]] <press enter>
HP49G: INV([[2,1][2,4]]) * [[5][8]] <press enter>

we get: [[2][1]]

(note on the HP49G calculator [[2][1]] may appear vertically)

As you can see, the calculator makes life much easier 

Let's look now at our three equation system we did above...

8=2a+b+c
17=2a+4b+3c
12=4a+b+c

This can also be entered into a calculator and solved by forming the matrix:

[[2,1,1][2,4,3][4,1,1]]

and the vector:

[[8][17][12]]

Note all we do is enter the coefficients on the right hand side of the
equations into a matrix and enter the ones from the left hand side
of the equations into a vector to get the proper forms. That's almost
too easy 
Then, to get the answer again we use the 'simult' function or take the
inverse of the matrix and multiply the result by the vector:

TI89: [[2,1,1],[2,4,3],[4,1,1]]^(-1) * [[8][17][12]]
HP49G: INV([[2,1,1],[2,4,3],[4,1,1]]) * [[8][17][12]]

After pressing enter we get the values of a, b, and c in a vector
which looks like this:
[[2][1][3]]
which are the values for a, b, and c respectively.


Note that the form used for the calculators are very specific,
in that you must order the coefficients correctly and use
the brackets '[' and ']' exactly as shown. Each equation 
forms a new 'row' in the matrix which is entered in
brackets like this: [1,2,3]
while a new 'column' is formed in a new set of brackets.
For example, the first equation right side makes a row:
[2,1,1] (note the 1's come from having an implied '1' as coefficient)
the second row from the second equation:
[2,4,3]
and the third from the third equation:
[4,1,1]

Put them all on the same line and you get:
[2,1,1][2,4,3][4,1,1]
then add more brackets around the whole thing:
[[2,1,1][2,4,3][4,1,1]]
and you now have the proper form for the matrix.
The 'vector' is formed in the same way, only using the
left side of the equations for the numbers:
[8]
[17]
[12]
put all together:
[8][17][12]
then add brackets around:
[[8][17][12]]
and now you have the proper form for the vector.

All you do now is take the matrix up to the power -1 and
multiply the result times the vector to get the answer.

The above works the same on the TI89 and HP49G, but
with the HP49G you must use the built in function "INV"
as we have been doing above.

The general form for the TI89 is:
Matrix^(-1) * Vector

while for the HP49G it is:
INV(Matrix)*Vector

On the TI89 you can look up the function used to
solve simultaneous systems and use that if you wish,
while the HP49G uses a editing screen. You'll have to
look these up the the manuals if you wish to use them
instead of taking the inverse and multiplying as we did above.
Note that the TI85 might accept another form for the vector.

As a final note, once you get the values of all your 
variables it's a good idea to plug them into one or more
of the equations to check to see that they are correct.


Self Test Questions
(hints and answers below...look at hints first if you have to)

1. Solve the following system of equations for a,b, and c:
a+2b+3c=7
6a+5b+4c=14
7a+2b+2c=21

2. Solve the following system of equations for a,b, and c:
4a+2b+3c=203
6a+5b+4c=406
7a+2b+2c=609


Hints for questions only if you need them:
1. TI89: [[1,2,3][6,5,4][7,2,2]]^(-1) * [[7][14][21]]
HP49G: INV([[1,2,3][6,5,4][7,2,2]])) * [[7][14][21]]

2. TI89: [[4,2,3][6,5,4][7,2,2]]^(-1) * [[203][406][609]]
HP49G: INV([[4,2,3][6,5,4][7,2,2]]) * [[203][406][609]]

Answers (use to check your work):

1. The system
a+2b+3c=7
6a+5b+4c=14
7a+2b+2c=21
can be put into matrix form for the calculator as:
TI89: [[1,2,3][6,5,4][7,2,2]]^(-1) * [[7][14][21]]
HP49G: INV([[1,2,3][6,5,4][7,2,2]]) * [[7][14][21]]

which after hitting enter comes out with:
[[3][-4][4]]
so
a=3
b=-4
c=4

2.
4a+2b+3c=203
6a+5b+4c=406
7a+2b+2c=609
TI89: [[4,2,3][6,5,4][7,2,2]]^(-1) * [[203][406][609]]
HP49G: INV([[4,2,3][6,5,4][7,2,2]]) * [[203][406][609]]

which comes out to:
[[105][28][-91]]
so
a=105
b=28
c=-91

See you next time!

Take care,
MrAl


----------



## MrAl (Jul 26, 2004)

*CPF University: EE Course (18)*

Hello and welcome again to another part of the course!


Serial: EE-Ch0002-p104-{A939C0B7-558E-4687-9271-AB92E4935FD0}

Today we'll be looking at a major circuit analysis technique
so that we will be able to analyze a wide class of circuits.
Although we are still dealing with mostly resistors as circuit
elements, it wont be long before we extend what we do here
to other simple circuit elements such as capacitors and inductors.






In fig 10a we have a general resistor circuit with a single voltage source,
and this circuit cannot be analyzed as easily as previous circuits we looked
at because there are no combinations of parallel or series resistors to
lump. R1 and R3 are 'sort of' in parallel, but there's R5 between them
so they arent both connected at both ends, they are only connected at
one end. Same with R2 and R4, and R5 is somewhere
in the middle of everything. There's no series or parallel resistors
so we have to do something else for this kind of circuit.

Here we get into our first major circuit analysis method called
'Nodal Analysis'. Nodal analysis is probably the best form of
network analysis because it promises to solve any general type
of circuit whereas some other types of analysis cant be applied.
This alone means it's very good to know.

Nodal Analysis is so named because it pays most attention to what's
happening at the nodes of the circuit. in fig 10b, we see the same
circuit with the nodes highlighted with red circles and they are also
numbered from 0 (zero) to 3. A node is formed whenever two or more
components are connected together. There are four nodes total for
this circuit.

The first step therefore is to number the nodes.

The second step is to assign voltage variables to each node, as in
fig 10c. Here, each node has been assigned a variable except since
node 0 has been called zero voltage (the reference node) that node
has been assigned a value of zero (0) volts.

The third step (fig 10d) is a little harder. For each node draw three
current arrows pointing either toward the node (current entering the node)
or pointing away from the node (current leaving the node). It makes
sense to have at least one pointing toward the node and one pointing
away from the node for each node, although there could be more of each too.

The fourth step (also fig 10d) is to draw the voltage arrows across each 
element. The voltage for each arrow is the difference between the two nodes
it bridges across. For example, the voltage across R1 is the difference
between the voltage v1 and the voltage v2, so the difference is v1-v2. R1's
voltage arrow is thus labeled "v1-v2". Since the first voltage (v1) is
assumed to be more positive than the second (v2) the arrow gets drawn with
it's head pointing toward v1 and it's tail toward v2.

Now that we've got everything labeled, the next step (step five) is to
write the equations using everything we know and everything we assumed.
In doing this, we sort of pretend that we know v2 and v3 even though
we really dont (yet). Once done, we will know all the node voltages
and then can readily compute any current we wish in any part of the
circuit.
When we write the equations, we simply write one for each node.
To make it a little easier, we start at the first node and work our
way around to each node one after the other. We use KCL for each node,
which means we simply sum all the currents entering that node and all
the currents leaving that node, then go to the next node. We use
Ohm's Law to get each current, knowing both the voltage and resistance
for each component.

Before we start writing the equations, it is helpful to realize that
two of the voltage arrows are labeled with a node voltage minus zero (0)
so that arrow's voltage is simply the node voltage indicated; we can drop
the zero.
The voltage v1 is assumed to be known, so we'll make it 10 volts.
Also, we call a current entering the node positive and a current
leaving the node negative. 


Starting with node 2, we have one current entering and two leaving:

i1=(v1-v2)/R1=v1/R1-v2/R1
i2=-(v2-v3)/R5=-v2/R5+v3/R5
i3=-v2/R2=-v2/R2

Summing all three currents to zero:
0=i1+i2+i3
so
0=v1/R1-v2/R1-v2/R5+v3/R5-v2/R2

Rearranging terms so that they appear in order v1,v2, etc, and like
terms are next to each other:
0=v1/R1-v2/R1-v2/R5-v2/R2+v3/R5

Thus, our node #2 equation is:
0=v1/R1-v2/R1-v2/R5-v2/R2+v3/R5



Next, we do node #3:
i1=(v1-v3)/R3
i2=(v2-v3)/R5
i3=-v3/R4

Summing to zero:
0=i1+i2+i3
0=v1/R3-v3/R3+v2/R5-v3/R5-v3/R4

So our node #3 eq is:
0=v1/R3+v2/R5-v3/R5-v3/R3-v3/R4



Next, we recall both node eqations:

node 2:
0=v1/R1-v2/R1-v2/R5-v2/R2+v3/R5
node 3:
0=v1/R3+v2/R5-v3/R5-v3/R3-v3/R4

Here we have two equations that form a system:
0=v1/R1-v2/R1-v2/R5-v2/R2+v3/R5
0=v1/R3+v2/R5-v3/R5-v3/R3-v3/R4

Since we know the value of v1=10 volts, we replace that:
0=10/R1-v2/R1-v2/R5-v2/R2+v3/R5
0=10/R3+v2/R5-v3/R5-v3/R3-v3/R4

Since doing that gives us constants, we move them to the 
other side of the equations by subtracting:
-10/R1= -v2/R1-v2/R5-v2/R2+v3/R5
-10/R3= +v2/R5-v3/R5-v3/R3-v3/R4

Next, we group like terms of variables v2 and v3 and lump coefficients:
-10(1/R1)= -v2(1/R1+1/R5+1/R2)+v3(1/R5)
-10(1/R3)= v2(1/R5)-v3(1/R5+1/R3+1/R4)

Since we assume we know all the values of all the resistors,
we now have a system of two equations in two unknowns, where the
unknowns are v2 and v3.

This is nothing new for us and we solve it the same way we did with
previous systems, by putting the coefficients into a matrix and
a vector by punching in the numbers into the calculator in the
correct order. Before we do that though, let's assign some
values to the resistors:
R1=5
R2=10
R3=10
R4=5
R5=10

Knowing all the resistor values we can now simplify the system of
equations by replacing each R in turn with it's value:

-10(1/5)= -v2(1/5+1/10+1/10)+v3(1/10)
-10(1/10)= v2(1/10)-v3(1/10+1/10+1/5)

Doing the math, we end up with:
-2= -v2(0.4)+v3(0.1)
-1= v2(0.1)-v3(0.4)

Now any term with a variable that has a negative sign gets its
sign moved to the constant:
-2= v2(-0.4)+v3(0.1)
-1= v2(0.1) +v3(-0.4)

Now we simply form the matrix and the vector as before:
[[-0.4,0.1][0.1,-0.4]] for the matrix
[[-2][-1]] for the vector
then do the math on the calculator:
TI89: [[-0.4,0.1][0.1,-0.4]]^(-1) * [[-2][-1]] <press enter>
HP49G: INV([[-0.4,0.1][0.1,-0.4]]) * [[-2][-1]] <press enter>


and the answer comes out to:
[[6][4]]
so
v2=6
and
v3=4

and we're done.

Since we now know all the 'node' voltages we can calculate
any current in the circuit as well using Ohm's Law.
For the current through say R5 we simply find the voltage
across it (v2-v3) and divide by it's resistance:
iR5=(v2-v3)/R5
so
iR5=(6-4)/10=2/10=0.2 amps

so the current through R5 is 200ma.

Note above we used the notation "iR5" to indicate the
current through R5. Usually this is written "i" with a 
subscript "R5" but here we cant use subscripts so we just
write it "iR5". If we wanted to indicate the voltage
across R5 we could write "vR5" using this same convention.
Sometimes it's written v(R5) or v[R5] also, or even v_R5.
As long as we recognize the convention used we'll know what
the meaning is.


That's Nodal Analysis!



Self test problems:

1. In fig 10a, let the source voltage be 10 volts and let the
value of the resistors be as follows:
R1=5
R2=10
R3=10
R4=5
R5=20
Using the method outlined above, find the current through R5.

2. In fig 10a, let the source voltage again be 10 volts and let the
value of the resistors be as follows:
R1=20
R2=10
R3=10
R4=20
R5=40
Find the current through R5.


Good luck!

Take care,
Al


ANSWERS (You shouldnt look here until you've tried the exercises)

1. In fig 10a with the source voltage equal to 10 volts and resistors:
R1=5
R2=10
R3=10
R4=5
R5=20

we end up with two equations:
-10(1/5)= -v2(1/5+1/20+1/10)+v3(1/20)
-10(1/10)= v2(1/20)-v3(1/20+1/10+1/5)
simplified:
-2= v2(-0.35)+v3(0.05)
-1= v2(0.05)+v3(-0.35)
which can be entered into the calculator as:
TI89: [[-0.35,0.05][0.05,-0.35]]^(-1) * [[-2][-1]] <enter>
HP49G: INV([[-0.35,0.05][0.05,-0.35]]) * [[-2][-1]] <enter>

which gives:
[[6.25][3.75]]
so
v2=6.25 volts and
v3=3.75 volts
Since the voltage across R5 is v2-v3, the current is:
iR5=(v2-v3)/R5=2.5/20=0.125 amps



2. In fig 10a, with the source voltage again be 10 volts the
value of the resistors:
R1=20
R2=10
R3=10
R4=20
R5=40

we end up with the two equations:

-0.5= v2(-0.175)+v3(0.025)
-1 = v2(0.025) +v3(-0.175)

which can be entered into the calc as:
TI89: [[-0.175,0.025][0.025][-0.175]]^(-1) * [[-0.5][-1]]
HP49G: INV([[-0.175,0.025][0.025][-0.175]]) * [[-0.5][-1]]

which comes out to:
[[3.75][6.25]]
Since again we want the current through R5 we first find
the voltage across it then divide by it's resistance:
iR5=(v2-v3)/R5
so
iR5=(3.75-6.25)/40
= -2.5/40
= -0.0625 amps
Note this time it comes out negative, because we assumed the
current flow was to the right. It's not a big deal however,
it's still 62.5ma flowing.


----------



## MrAl (Jul 29, 2004)

*Re: CPF University: EE Course (19)*

Hello again!

Todays course section is 'sort of' optional. That is, we will
probably be using Nodal Analysis most of the time, but the
method given here today is still good to know in case you want
to use it or if someone else talks about a circuit using this
method you'll be able to follow the conversation better.

Serial: EE-Ch0002-p105-{A939C0B7-558E-4687-9271-AB92E4935FD0}








In fig 11a we again have a circuit in which there are no series or
parallel combinations of resistors that will simplify the circuit.
We can easily analyze this circuit using nodal analysis by calling
the node at the bottom zero (0) and the node between R1 and R2
we can call simply 'v'. Then, using KCL at node 'v' we have:
iR1+iR2=iR3
where
iR1=(V1-v)/R1
and
iR2=(V2-v)/R2
and 
iR3=v/R3

Putting these together we get:
(V1-v)/R1+(V2-v)/R2=v/R3

Knowing V1,V2,R1,R2, and R3 we could then easily solve for v:

V1/R1-v/R1+V2/R2-v/R2=v/R3
-v/R1-v/R2=v/R3-V2/R2-V1/R1
v/R1+v/R2+v/R3=V2/R2+V1/R1
v*(1/R1+1/R2+1/R3)=V1/R1+V2/R2
v=(V1/R1+V2/R2)/(1/R1+1/R2+1/R3)


Now we want to look at another method called "Mesh Analysis".
With Mesh Analysis we solve for 'mesh' currents and once
we know all the mesh currents we can solve for any current
or voltage in the circuit. Sometimes a circuit is easier
to analyze using Mesh Analysis so we're going to take a minute
to look at this.

In fig 11b we see the same circuit drawn with the addition of
two looping current arrows (red). The inside of each loop arrow
is labeled with a current, I1 or I2, one for each loop.
The main idea here is that these mesh currents are presumed to
be flowing through each element they pass, so I1 flows through
V1, R1, and R3, while I2 flows through R3, R2, and V2.
Take note that either current flows through some elements,
while BOTH currents flow through R3. I1 flows through
R3 from top to bottom, while I2 flows through R3 from bottom
to top. This of course means the two currents are fictitious,
but nevertheless this is still a very powerful method.

The rule for drawing the current loop arrows is fairly
simple -- one current loop for each looping set of components
in the circuit. This procedure is complete when every element
of the circuit has at least one current loop arrow passing by it.

One more requirement is that all the mesh currents must
be drawn in the same clock direction...that is, either all
clockwise or all counter-clockwise. In fig 11b they are
all clockwise so this condition is satisfied.

Ok, so after we've drawn the mesh current arrows we next
draw some voltage polarity arrows as we've done before.
This has been done and is shown in fig 11c. Note the
direction of the arrows are opposite to the mesh current
flow.

Next we write out the equations for each current loop,
taking note of the voltage polarity arrows directions
and component values.

Before we start writing equations, let's assign some
real world component values and values to the two
voltage sources which could be simply two batteries...

V1=10v, V2=10v, R1=10, R2=10, R3=5 ohms

The following rule applies when following the current arrows:
When we encounter first a voltage arrow tail we call that
voltage positive and when we encounter first a head we call
that voltage negative.

As we encounter each element we also use Ohm's Law to find the
voltage. This simply means we multiply the current times it's
resistance as usual. Since there may be more then one current
'loop' flowing through a given element, we must include each
currents contribution to the voltage across that element.

Ok, starting with I1 at the bottom of the circuit and following
it's direction around it's loop:

+V1 -I1*R1 -I1*R3 +I2*R3 = 0 
(note for this equation R3 has both I1 and I2 flowing, and we encounter two
arrows for R3: one for I1 and one for I2, so each is included in the equation)

and now for I2 doing the same things:

-I2*R3 +I1*R3 -I2*R2 -V2 = 0

We have thus completed the two required equations, and now we 
recall them both to form a system of equations in two unknowns,
I1 and I2:

+V1 -I1*R1 -I1*R3 +I2*R3 = 0
-I2*R3 +I1*R3 -I2*R2 -V2 = 0


Now we'll just rearrange them so we have one constant on the right
and the variables I1 and I2 on the left:

-I1*R1 -I1*R3 +I2*R3 = -V1 
-I2*R3 +I1*R3 -I2*R2 = V2


After multiplying the top eq by (-1) we end up with:

I1*R1 +I1*R3 -I2*R3 = V1
-I2*R3 +I1*R3 -I2*R2 = V2

Group like terms and reorder:

I1*(R1+R3) -I2*(R3) = V1
I1*(R3) -I2*(R3+R2) = V2

Now let's plug in the real world values:

I1*(10+5) -I2*(5) = 10
I1*(5) -I2*(5+10) = 10

Do the math:

I1*(15) -I2*(5) = 10
I1*(5) -I2*(15) = 10

Move the (-) signs:

I1*(15) +I2*(-5) = 10
I1*(5) +I2*(-15) = 10

and then write this out in a form suitable for the calculator:

TI89: [[15,-5][5,-15]]^(-1) * [[10][10]]
HP49G: INV([[15,-5][5,-15]]) * [[10][10]]

and we get:
[0.5][-0.5]]
so
I1=0.5 amps and
I2=-0.5 amps

Now we know both mesh currents so let's calculate the total
current through R3. Since I1 flows from top to bottom it's 
influence on the total current is positive, while I2 flows in
the opposite direction so it's influence is negative. Since
I2 flows opposite to I1 we have to multiply the current I2 by
minus 1 (-1) and add it to I1 to get total current through R3:
iR3=I1-I2
iR3=0.5-(-0.5)
iR3=1 amp


Last for today is fig 11d. This is an example of a non planar circuit
which can't be analyzed using mesh analysis. Note there is no way
to draw the circuit without at least one crossing current path
(the crossing current path is shown by the little loop in the
wire crossing the path to the resistor on the right most side
of the drawing).


Good luck and see ya next time,
MrAl



Self Test Question

1. For this exercise, pick three values of resistors and two voltage
sources, use them for the elements of fig 11c, then analysze the
circuit using mesh analysis to find the value of the current
flowing down through R3. Using the same values, analyze the
circuit using nodal analysis to check your answers. You should
of course get the same current either way.
A good set of values is:
V1=3v, V2=11v, R1=10 ohms, R2=5 ohms, R3=5 ohms
(Scroll down for the solution using these values)





Solution:
In fig 11c with V1=3, V2=11, R1=10, R2=5, and R3=5,
the current flowing down through R3 is 1 amp.


----------



## MrAl (Aug 2, 2004)

*Re: CPF University: EE Course (20)*

Hello again!

Today we are going to talk about a shortcut method of analyzing
a circuit with more then one source in it. This method is so 
handy we cant go without it!

Serial: EE-Ch0002-p106-{A939C0B7-558E-4687-9271-AB92E4935FD0}









In fig 12a we again have a circuit in which there are no series or
parallel combinations of resistors that will simplify the circuit.
This is the same circuit we used last time, except the node at the
top is labeled "V" and the node on the bottom is zero.
We can easily analyze this circuit using nodal analysis or mesh 
analysis like we did last time, but this time we will try a very
different method. This new method might be called:
"Kill all sources" but officially it's called the
"Superposition Theorem".

The idea is to 'kill' all sources in the circuit and then one at
a time remove the 'kills' and analyze each node as we would any
other circuit. The nice thing about this method is that for
any given circuit the resulting set of circuits that remain to
be analyzed are often GREATLY simplified. This means we can usually
get fast answers even with circuits with many sources in them
because as each source is killed the circuit gets more and more
simple.

What it means to 'kill' a source is very simple:
For a voltage source it gets short circuited, and for
a current source (which we havent covered yet anyway) it
gets open circuited. To 'unkill' the source you simply
put it back the way it was originally.

Now lets look at an actual example...

If you did the mesh analysis on this circuit last time the first
thing you will note when you follow this analysis is how much
simpler it is and how much faster we get the answer, except
maybe for the drawing of the pictures 

In fig 12a we'll set the values as follows:
V1=3v, V2=11v, R1=10 ohms, R2=5 ohms, R3=5 ohms

Now to start the analysis first we would 'kill' both sources,
V1 and V2 by short circuiting both of them. Then, we would
'unkill' say V1 first and start analyzing. The shorted V2
circuit is shown in the top of fig 12b and the resulting circuit
is shown right below it in 12b also. The thing most notable
now is that two of the resistors R2 and R3 now appear in 
parallel, whereas before we shorted V2 this wasnt true.
Also, i've called the top node Va (the bottom is still zero).
The reason for the new name is because Va is a new node, more
or less, and will not be the same voltage as V in fig 12a because
we've changed the circuit by shorting V2.
Ok, now we must find out the voltage of Va in the bottom of fig 12b
so we start by simply finding the value of R2 and R3 in parallel:
Rp=1/(1/R2+1/R3)=1/(1/5+1/5)=1/0.4)=2.5
so we replace R2 and R3 with a 2.5 ohm resistor. This leads
to a circuit with R1 in series with a 2.5 ohm resistor driven
by voltage source V1 so we use voltage division:
Va=V1*2.5/(10+2.5), note R1=10
Since V1 is 3 volts this means
Va=3*2.5/12.5 which equals 0.6 volts, so
Va=0.6 volts and we have the answer to the value of the 
node that comes from the contribution from source V1.

Now we have to do the same thing with the other source as shown
in fig 12c.

R1 in parallel with R3 is:
Rp=1/(1/R1+1/r3)=1/(1/10+1/5)=1/(0.1+0.2)=10/3

Next we can again use voltage division with values 10/3 and R2
so we get:
Vb=V2*(10/3)/(5+10/3)
so
Vb=11*10/(15+10)=11*10/25
so
Vb=4.4 volts.

Now we have the two answers, Va and Vb.
To get the final answer of what V is, we simply add Va and Vb:

V=Va+Vb

and we get 

V=5 volts

Since we know R3=5 ohms we can get the circuit using Ohm's Law:

iR3=5/5=1 amp

Which agrees with the analysis from last time.

It's very interesting that by changing the circuit a little we
could reduce the circuit itself into two smaller circuits
(fig 12d) that could both be analyzed easier then the original
circuit and so provided us with another means of finding the answer.
Pretty cool if you ask me!!

With all good things there are some drawbacks, and this method
is no exception. In order to use this method on a given circuit
he first requirement is that ALL elements of the circuit are
what's known as "linear" elements. A linear element is defined
pretty intensely, but let's try to attach some form of definition
to what a linear element is.

A linear element has a linear voltage current relationship, so
that a multiplication by it's current of a factor K results
in a multiplication of it's voltage by the same factor K.

Now i dont know how much good that will do you at this point, so
let's look at a simple example in another field...

You drive two strong nails into the wall approximately on the same
level line, spacing them about 12 inches apart. You stretch a
rubber band across both nails so it stretches across horizontally.
Now you begin hanging weights in the middle of the band, so that
for each weight you measure a deflection of the horizontal 
band in the center which now forms a vee shape.
Let's say the first weight of 1 ounce causes a deflection of
1 inch. Now you hang a second weight of 1 ounce and it too
causes another defection of 1 inch so the total now is 2 inches.
You hang a third weight of 1 ounce and another 1 inch results so
now it deflects by 3 inches. 
In each case we hung the same weight on the band the same result
occurred: one additional inch of deflection.
The rubber band has thus responded linearly, in that a change
of input of a certain degree led to a change in output of
the same amount as with the original input amount.
On the other hand, if we keep hanging weights on it eventually
it breaks and so it 'stretches' far more for that last one ounce of weight.
This second kind of response would make it 'non linear', which is
simply put, "not linear".

Soon we'll encounter the LED again to see just why this is called
a nonlinear device. Things will make much more sense then too,
and why it's important to know a nonlinear device from a linear one.
For now, for the purposes of what we are doing at this point it will
probably be good enough to know that at least a resistor is a
linear element. When we come to other elements, we'll look a little
more at this too. It's interesting however, to note that in most
analysis the capacitor and inductor are considered linear too, so
we'll find this method is of use later as well.


Thanks for joining in today!

Take care,
MrAl

Last minute note:
Please dont try to short out your voltage sources
in circuits that you or someone else has built up.
This is a theoretical method, and if you want to
try it in practice you need to first open circuit
each voltage source (like a battery) before you
short the circuit where the battery originally
connected. Please remember this!

Today's Self Test Question

Take the circuit of fig 12a and make the following changes:
V1=11v, V2=3v, R1=5 ohms, R2=10 ohms, R3=5 ohms

Find the voltage at the node labeled "V". 
If you see a shortcut, that's very good, but you should
try this method at least once anyway.
Scroll down for the answer.





V=5 volts.
All we did was swap V1 and V2, and R1 and R2, from the previous circuit
so the analysis comes out the same! If you didnt notice this it's ok
and you should try this method anyway.


----------



## MrAl (Aug 9, 2004)

*Re: CPF University: EE Course (21)*

Hello again!

Today we are going to talk about using an LED again, only this time we will
take the nonlinearity of the LED into account and see why this makes a
difference.

Serial: EE-Ch0003-p101-{A939C0B7-558E-4687-9271-AB92E4935FD0}






In figure 13a we see an LED wired in series with a resistor (R1) and a
set of four alkaline AA cells which makes up the battery.
In figure 13b the physical battery has been replaced by it's equivalent
real life circuit approximation: a voltage source in series with a 
small value resistor. The small value resistor is labeled "Rs" and
it's made equal to the series resistance of each battery times four,
which comes out to 0.9 ohms. The symbol right after the "0.9" is
the symbol for ohms, and saves us from writing out "ohms" each time.
All the voltage and current arrows have been assigned and labeled.
The voltage for the LED is labeled "V2".

We can analyze this circuit as we've done other circuits by using
Kirchoffs Voltage Law (KVL) around the current loop.
Starting at the bottom node (zero volts ground) and working
around the circuit in the direction of current flow:
0=6-V3-V1-V2
Now above we have one known source (6 volts) and three unknown,
and we also dont know what I is yet but we know what Rs and R1 are
so let's change this equation into a form with current as the variable,
while we let the voltage of the LED be some function of current as well.
We'll note that the LED's voltage is a function by the notation:
"vfLED(i)"
where
vLED=vfLED(i)
which simply means that the voltage across the LED is a function
of the current through it, and the function is called "vfLED".
Now what this means is we 'sort of' know the voltage across
the LED, and it's simply "vfLED(i)".
What this REALLY means is that we REALLY know the voltage 
across the LED if we know the current though the LED.


Ok, now back to the equation:
0=6-V3-V1-V2

We can use Ohm's Law for V3 and V1 because:
V3=I*Rs
and
V1=I*R1
and we know both Rs and R1, so we end up with:

0=6-I*Rs-I*R1-V2

We use the functional notation as discussed for the LED to replace V2
and we now end up with:
0=6-I*Rs-I*R1-vfLED(i)

Now we simply replace the 'i' with "I" and we have and equation
who's only variable we dont know yet is "I":

0=6-I*Rs-I*R1-vfLED(I)

Now let's rearrange so the constant voltage source of 6 volts is
on one side and the other stuff is on the other side:

I*Rs+I*R1+vfLED(I)=6

And we're almost done.

All we need now is a 'model' of the LED voltage as current varies.
A model is simply a mathematical representation of a circuit element
that relates one quantity to another. In the case of the LED, we 
need a model that takes the current i as an input parameter and
gives us the voltage v across it.

For the purpose of this discussion and because we are not pulsing
the LED, the following model for the Nichia LED will be used:

vLED=(4.3-7.5*i)*i^(0.0625)+10*i {0.001<=i<=0.100}

Note three things about this model:
1. The right side variable is the current i
2. The left side is the voltage across the LED
3. There is a limitation on the current for which the model is accurate

What this means is that we can input the current and immediately
get the voltage across it. In real life, this would take a 
few seconds to stabilize, but we can still use it assuming
our circuit will temperature stabilize in say five seconds.

Now that we have a model for the LED, we can rewrite the equation replacing
V2 with this new function and we get:

I*Rs+I*R1+(4.3-7.5*i)*i^(0.0625)+10*i=6

Now let's make all the lower case i's upper case:

I*Rs+I*R1+(4.3-7.5*I)*I^(0.0625)+10*I=6

And we finally have our equation!

All that's left to do now is solve this equation for I and
we'll have the answer to what we need. We start by replacing
every resistor we know by it's actual value, so we replace
Rs and R1 and we end up with:

I*0.9+I*22+(4.3-7.5*I)*I^(0.0625)+10*I=6

Ok, now we've done all we can do so far, but we dont seem to
have a way to solve for I yet.
Unfortunately, it's very hard to solve for I directly but
luckily we have a few simple methods.

To solve for I we can either try values for I (starting at
maybe 0.020 amps, calculating the left side of the equation,
then comparing to "6" to see if we should increase or decrease
I to get closer to the 6) or, we can use a numerical solver.
A numerical solver does all the work for us. Once we enter
the equation we tell it to solve for I and it takes a few
seconds and then shows the answer. The TI89 and the
HP49G both have this feature and it's not hard to use.

If you'd like to try solving by hand, here's a typical
way to go about it...

Start with the equation we found:
I*0.9+I*22+(4.3-7.5*I)*I^(0.0625)+10*I=6

and separate the left side from the right side:
I*0.9+I*22+(4.3-7.5*I)*I^(0.0625)+10*I
6

Now guess a value for I and use it in:
I*0.9+I*22+(4.3-7.5*I)*I^(0.0625)+10*I

Let's say we guess 0.030 amps for I, we insert for each I and get:
(0.030)*0.9+(0.030)*22+(4.3-7.5*(0.030))*(0.030)^(0.0625)+10*(0.030)
which when computed comes out close to:
4.26

Now we compare this answer to the right hand side, which was '6'.
Noting that 4.26 is lower then 6, we increase I a little, to say
0.035 amps (sometimes you have to decrease the variable however,
it depends on the equation--it's trial and error).

Ok, so doing the same with 0.035 amps we get approximately:
4.4258

This is closer to 6 then the result we got using 0.030, so we
assume we need to increase again. Let's try 50ma this time.

Doing the same with 0.050 amps we get about:
4.9

This is even closer to 6 then before, so we increase again to 70ma
and get approximately:
5.5

Now we're getting closer to 6 so we increase a little to 75ma and
get:
5.65

Then to 80ma and get:
5.79

Then to 85ma and get:
5.93

We're now pretty close to 6.00 so we increase a little more to 90ma
and get:
6.08

Now we are over 6.00 so we decrease just a small amount to 88ma. We get:
6.022
which is pretty darn close to 6.00, but lets decrease a tiny bit more to
87ma and we get:
5.994
Now we have one result that is slightly below 6.00 and one result that
is slightly above 6.00 so we know the current is between 87ma and 88ma
and that's usually accurate enough so we're done.

Alternately, we could have subtracted the 6 from the equation and tried
to get a value that is close to zero instead.


Thanks for joining in today!

Take care,
MrAl



Today's Self Test Question

In the circuit of fig 13b replace the 0.9 ohm resistor with 1 ohms
and the 22 ohm resistor with 50 ohms. Find the current through the
LED.
(scroll down for the answer after you try this)




Answer:
The current through the LED is between 44ma and 45ma if you do it by hand.
If you use a numerical solver it's very close to 44.84ma .


----------



## MrAl (Aug 16, 2004)

*Re: CPF University: EE Course (22)*

Hello again and welcome to another part of the Course!

Today we are going to calculate the resistor values when paralleling
LED's. After that we will be in a much better position to analyze
what happens when the LED's are paralleled directly instead of using
a separate resistor for each LED which we'll do next time.

Note that the results we find here are applicable to either Nichia
type or Luxeon type LED's, except for the Luxeon type LED's we
would need a different LED model equation. In fact, these results
apply to any type of LED as long as you have the right model 
equation.

Serial: EE-Ch0003-p102-{A939C0B7-558E-4687-9271-AB92E4935FD0}






In fig 14a we have a circuit in which there are two LED's and each one has
it's own series dropping resistor. In fig 14b we have the same circuit
with the battery replaced by it's real world approximation of a source
in series with a small resistor Rs. The value of Rs is shown as 0.9 ohms.
All the voltage arrows have been drawn also, as well as the current
arrows which show how the current Is splits into two separate currents.

Now, let's say we want to pump 30ma through each LED, and to start with
we'll assume the LED characteristics are exactly the same which means the
LED's are matched. This means Is is twice either LED current I1 or I2
because Is splits into I1 and I2, and I1=I2.

We dont yet know what V is so we'll use a nodal equation at that node.
Is=I1+I2

As usual, we pretend we know the voltage of each LED, which will be
vLED1 for LED1 and vLED2 for LED2, but since we assume they are the
same for this analysis we also define vLED=vLED1=vLED2. This gives
us a single LED voltage to work with, vLED, and we know it represents
either LED voltage in the circuit.

Having the voltage of each LED and using the variable "V" for the top node
we then know both currents I1 and I2:
I1=(V-vLED)/R1
I2=(V-vLED)/R2

Since both LED voltages are the same we know both R1 and R2 will be
the same because we want equal current flowing in each LED, so we 
define another resistor and call it "R" so that:
R=R1=R2
In this way we know R represents either resistor value.

Also, since we want equal current flowing we know that I2=I1,
so this makes it possible to write an equation for Is using only I1:
Is=2*I1

Using V and Vb we can write an equation for Is:
Is=(Vb-V)/Rs

Using V and vLED and R we can write an equation for I1:
I1=(V-vLED)/R

and knowing Is=2*I1 we can write:
Is=2*(V-vLED)/R

We thus end up with two simultaneous equations in variables V and vLED:
Is=(Vb-V)/Rs
Is=2*(V-vLED)/R

Now wouldnt it be nice if we could eliminate one variable?
Since V appears in both equations, we can eliminate that variable.
First, expand both equations so there are no divisions...
Rs*Is=(Vb-V)
R*Is=2*(V-vLED)

Now eliminate parentheses:
Rs*Is=Vb-V
R*Is=2*V-2*vLED

Now we can get both equations to contain a lone V variable by dividing
both sides of the bottom equation by 2, and we end up with:
Rs*Is=Vb-V
R*Is/2=V-vLED

Now the top eq has a "-V" and the bottom eq has a "V" so
if we add them the V disappears:

TopEq+BottomEq=>

Rs*Is+R*Is/2=Vb-vLED

Now we have an equation we can solve for R to find out
what resistance to use to obtain a given current through
each LED:

Rs*Is+R*Is/2=Vb-vLED

and solving for R we get:

R=2*(Vb-vLED-Rs*Is)/Is

The only thing left to do is substitute the voltage for an LED
with it's model (from last time). Repeating the model we
will use for the LED voltage:

vLED=(4.3-7.5*i)*i^(0.0625)+10*i

Again, this relates the voltage across the LED (vLED) to the
current through the LED (i). Thus, knowing the current i we
also then know the voltage across it.

Now all we need to do is figure out what i should be and we
can then solve for vLED and insert into the equation for R
and come up with a suitable value for both resistors R1 and R2.

Since half the current of Is flows though each LED i will
equal one half of Is, or 
i=Is/2

but we already know we want to drive each LED at 30ma, so
we simply insert 0.030 into the equation for vLED and compute
the value of the voltage across either LED:

vLED=(4.3-7.5*(0.030))*(0.030)^(0.0625)+10*(0.030)

Now we have all numbers on the right side so we simply calculate
the value of vLED (using a calculator), and it comes out to:

(4.3-7.5*(0.030))*(0.030)^(0.0625)+10*(0.030)=3.573

approximated to four significant figures.

Now that we know the voltage across an LED, we can use the formula
for R to get the value of each resistor:

R=2*(Vb-vLED-Rs*Is)/Is

replacing vLED with 3.573 we get:

R=2*(Vb-3.573-Rs*Is)/Is

We know Is is equal to twice any LED current so that comes out to:
Is=0.060

and Rs and Vb are already shown on the schematic in fig 14b so:

R=2*(6-3.573-0.9*0.060)/0.060

All that's left to do now is to calculate the value, which comes out to:

79.1 ohms.

We can use a standard value of either 75 ohms or 82 ohms.

Ok, so we've calculated the value of both resistors R1 and R2 in fig 14b,
so what about the circuit in fig 14c ?

Since we are assuming the LED's are matched (next time we'll drop this
restriction and see how bad the circuit can end up) the circuit of
fig 14c can be constructed from the circuit of fig 14b by simply
placing a short between the tops of the two LED's. In doing so,
the two resistors R1 and R2 end up in parallel, so we can replace
those two with a single resistor which is equal to the parallel
combination of the two. Recalling the formula for two parallel
resistors:
Rp=1/(1/R1+1/R2)

and since we found R1=R1=79.1 ohms, the value of R3 (fig 14c)
would be:
R3=1/(1/79.1+1/79.1)
or simply half the resistance:
R3=(79.1)/2=39.55 ohms.

The nearest standard value is 39 ohms, so we could make
R3 in fig 14c equal to 39 ohms. Of course we'd also
have to calculate the power in R3 to make sure we had
the correct size resistor.

Next time we'll look at how the circuit of fig 14c can
quickly become a problem if the LED's arent matched
well enough to share current equally.



Thanks for joining in again today!

Take care,
MrAl


Self Test Questions:

Using the circuit of fig 14b, let's say we want 40ma flowing
in each LED. Calculate both resistors R1 and R2 (approximating
the voltage across an LED to four significant figures as above)
approximating each resistor to three significant figures.

Using the circuit of fig 14c, calculate the resistor R3 with
40ma flowing through each LED assuming the LED's are well
matched.

(Scroll down for answers after you try these)






ANSWERS:

An LED voltage approximated to four figures is 3.671 volts.
Resistors R1 and R2 would be about 56.4 ohms each.
In fig 14c, R3 would be half R1 or R2, which is about 28.2 ohms.

In an actual circuit we would use a close standard value resistor
and then do the analysis again to make sure the current is
close enough to what we want it to be using the actual chosen
standard value.


----------



## MrAl (Aug 26, 2004)

*Re: CPF University: EE Course (23)*

Hello again and welcome to another part of the Course!

Today we are going to analyze what can happen when the LED's are paralleled
directly instead of using a separate resistor for each LED.

Note again that the results we find here are applicable to either Nichia
type or Luxeon type LED's, except for the Luxeon type LED's we
would need a different LED model equation. These results actually
apply to any type of LED as long as you have the right model 
equation.

Serial: EE-Ch0003-p103-{A939C0B7-558E-4687-9271-AB92E4935FD0}






In fig 15a we have a circuit in which there are two LED's and both are
driven from the same resistor R3.

To begin, since R3 is in series with Rs we can lump them into one
single resistor which we'll call R. In this way we have two LED's
driven from a 6 volt battery in series with a resistor R.
Recalling from last time we found if we wanted the current through 
each resistor to be 30ma we would need a resistor value of about
39 ohms, so we'll make R equal to that plus the cells internal resistance
of 0.9 ohms. This means R=39.9 ohms.

We dont yet know what either LED voltage is so we'll use a KCL equation at
the node where Is splits into I1 and I2:
Is=I1+I2

We also pretend we know the voltage across one LED and call it vLED and
this allows us to write:
0=Vb-Is*Rs-Is*R3-vLED
and since we lumped Rs and R3 into one resistor R we can rewrite this:
0=Vb-Is*R-vLED

Since we the two LED's are in parallel the voltage across them must
be the same even though the currents are I1 and I2, so we can write:
vLED1(I1)=vLED2(I2)

Solving 0=Vb-Is*R-vLED for Is we get:
Is=(Vb-vLED)/R
which expanded looks like this:
Is=Vb/R-vLED/R

Since either LED voltage can be used to represent the vLED term,
we'll use vLED1(I1) and this gives us:
Is=Vb/R-vLED1(I1)/R

and recalling
Is=I1+I2

these two taken together gives us a system of two equations:
Is=Vb/R-vLED1(I1)/R
Is=I1+I2

One way to solve these is to simply realize that they both equal Is
so we eliminate Is by simply setting the two right sides equal to
each other:
Vb/R-vLED1(I1)/R=I1+I2

Recalling that the voltage across both LED's are the same:
vLED1(I1)=vLED2(I2)

we can take these last two equations together as a system:
Vb/R-vLED1(I1)/R=I1+I2
vLED1(I1)=vLED2(I2)

and now we're getting close to the solution.

Since the second equation has both I1 and I2 in it, if we solve
the first equation for I2 and insert the result into the second
equation we can obtain a single equation of voltages that only
depend on one current, I1.

Solving the first eq for I2 gives us:
I2=Vb/R-vLED1(I1)/R-I1

and inserting this result into the second equation vLED1(I1)=vLED2(I2)
gives us:
vLED1(I1)=vLED2(Vb/R-vLED1(I1)/R-I1)

which is an equation that depends only on one unknown variable, I1 !

There's a slightly easier way assuming we use a calculator, and that
is to define I2 as a 'function' (or program) and then simply try to
solve 
vLED1(I1)=vLED2(I2)
numerically.

This works out very well except you have to know how to enter a 
function into the calculator. The HP49G and the TI89 work 
differently in this respect, so depending on what calc you use
you'll have to enter the function in a different way. Also,
the TI89 can't handle upper case letters (it converts them all
to lower case) so you'll have to use all lower case letters
with the TI89. The TI85 doesnt have that problem.

Recalling the model we've been using for vLED:
vLED=(4.3-7.5*i)*i^(0.0625)+10*i

we first need to create two new models for LED's where one model
has a forward voltage a little less then average and one has
a forward voltage a little more then average. To do this, we
simply change the inner term 4.3 to 4.1 for one model and 4.5 for
the other. Note that all we do here is add or subtract 0.2 for
a new model.
Doing this gives us the two new models:
vL1=(4.1-7.5*i)*i^(0.0625)+10*i
vL2=(4.5-7.5*i)*i^(0.0625)+10*i

and that's about the easiest way to handle that task.

Now we'll simply assign vL1 to LED1 and vL2 to LED2.
We're using L1 and L2 to represent the LED's just to make it a little
faster to enter into the calculator, but normally we wont do this because
L's are usually used to represent inductors, such as L1, L2, L3, etc.

Now to enter these functions into the calculators...

TI89... (if you're using the HP49G scroll down to that section)

First we'll make sure the variable I1 is 'clear' (ie not used).
You can enter
I1 enter
to see if it's been used.
If it displays a number it's used and has to be cleared with the
operator DelVar by doing:
DelVar I1 enter

You want to make sure the variable i has been cleared also.

Now enter the equation for vL1 first:
(4.1-7.5*i)*i^(0.0625)+10*i STO> vL1(i)
where "STO>" represents the 'store' operation by pressing the "STO>" key.
Note the calculator will convert vL1 to display in lower case as
"vl1", so you'll end up with lower case but it will still work.
This is a big problem with the TI89 but at least it works.

Now to enter the second model equation you can recall the one just
entered by using the up arrow key and highlighting it, then pressing
enter. You'll get the same equation on the first line and then you
can simply change the vL1 to vL2 and the 4.1 to 4.5 to enter it
because those are the only differences in the two models.
After changing those two characters and hitting enter you'll then
have both models entered into the calculator.

Now you'll need to enter the equation for I2 into the calc:
Vb/R-vL1(I1)/R-I1 STO> I2(I1) enter
and that will do it.

Now enter the values for Vb and R:
6 STO> Vb enter
39.9 STO> R enter

and of course it converts Vb to vb and R to r, but other than that
it works ok.

Now we have both models and the equation for I2 as well as the values
for Vb and R so all that's left to do is go into the APPS/Numerical Solver
and solve for I1.

Choosing APPS and then 9 enter, we enter the other equation:
vL1(I1)=vL2(I2(I1))

Note I2 is a function of I1 so it's entered as I2(I1) and not just I2.

Now we hit the down arrow key and enter in a guess for i1 (changes it to lower case)
of 0.040 and hit F2 for 'Solve'...some time later we get the answer:
i1=0.0450007

Now we know i1 is about 45ma, so we go to the 'home' screen and enter in:

i2(i1) enter
and we get the answer for i2, which is about
0.0164, or 16.4ma.

Thus, we've solved for BOTH currents I1 and I2 and we found out that they
are very different!

I1=45ma
I2=16.4ma

and since Is=I1+I2 this means the total current
flowing is:
Is=61.4ma

Now although the total current (which we would have adjusted for in the circuit)
is very close to the target value of 60ma, we've found that that doesnt mean
that there is 30ma flowing in both LED's...in fact, it's no indication at all!
One LED is running very high, while the other LED is running fairly low.
This not only means one is going to be brighter, it's also going to
burn out sooner.




Now using the HP49G...

For the following discussion, the symbol "-->" will be used to indicate
hitting the red arrow key and then the zero (0) key, which generates
an arrow pointing to the right on the HP screen. Also, "STO" will indicate
the store operation using the STO> key.
Also, when you hit the red arrow and the plus (+) key you get symbols
that look like this on the screen:
<<
>>
which indicates you are now about to enter a program or function.

First we need to enter the two model equations, vL1 and vL2:
<< --> I
'(4.1-7.5*I)*I^(0.0625)+10*I'
>> STO vL1
enter

The above is generated by first hitting red arrow key then plus key,
then using the red arrow key and zero key to enter "-->" then
ALPHA I, then red arrow key then period (.) key (new line) then red arrow key then single quote
then (4.1-7.5*I)*I^(0.0625)+10*I
then move cursor down with down arrow key then STO vL1 enter.
If everything was done right, it should look like that above.

Now we need to enter the second model equation for vL2:
<< --> I
'(4.5-7.5*I)*I^(0.0625)+10*I'
>> STO vL2
enter

Now the equation for I2:
<< --> I1
'Vb/R-vL1(I1)/R-I1'
>> STO I2
enter

Note the format of entering these equations is the same:
first the symbols << >>, then -->, then the variable name I or I1,
then the equation enclosed in single quotes, then after the >>
comes the STO and then the function name vL1, vL2, or I2.
Also acceptable is:
<< --> I1 'Vb/R-vL1(I1)/R-I1' >> STO I2
but i dont think it would fit all on one line anyway.

Lastly, we enter the numerical solver on the HP by red arrow key
then NUM SLV, then 1 (or enter) then enter the equation for the
LED voltages:
vL1(I1)=vL2(I2(I1))
then press the down arrow key to highlight I1 and press 
F6 for 'SOLVE'. After a little while the answer comes up:
4.50007380532E-2 which is the same as:
0.0450007 approximately, which is close to 45ma.

Now that we know I1, we want I2 so we press the ON key 
(to get back to the main screen) then enter:

I2(I1) enter

and we get approximately: 1.641324E-2 which is
about 16.4ma.

Thus, we've found the following information about the currents:
I1=45ma
I2=16.4ma
so the total current Is is 
61.4ma

Notice that even though the total current is close to the target
current of 60ma the individual currents through the LED's are
not the same (we wanted 30ma and 30ma) but are rather very 
different. One LED is carrying 45ma and the other only 16ma
meaning not only will one be brighter than the other, but
one will burn out long before the other.


Conclusion

We did an analysis of two LED's in parallel and found that
one LED might carry much more current than the other. If
there were more LED's in parallel and/or the target total
current level raised to say 80ma the situation could have been
even worse.


Thanks for joining in again today!

Take care,
MrAl


Self Test Question:

Set the resistor R3 equal to 26.1 ohms (fig 15a) and find the
current through both LED's using the two modified models used
here today. Scroll down for the answer when you're ready.








ANSWER:

To get the answer once we have all the equations entered in
all we have to do is change R to 27 ohms (26.1+0.9) and then
go back into the numerical solver and solve for I1 again.
Once we get I1 again we only have to do I2(I1) again to
get the new value of I2. Once we have both I1 and I2 we
just add them to get the total current Is. The answers
here are rounded:

I1=61.5ma
I2=24.8ma
Is=86.3ma

Note again the current ratio I1/I2 is almost 3 to 1 meaning
I1 is almost three times I2!


----------



## MrAl (Nov 11, 2004)

*Re: CPF University: EE Course (Links)*

Hello again,

Here are some links associated with this course.

PDF's of this course (courtesy of korpx of CPF) can be found here:
http://www.shell.linux.se/raven/diverse/cpf_mral_eecourse/list.html

All the drawings used for this course can be found here:
http://mral.peu.net/index.php
(click on 'index')
You'll also find various other things like power supplies
and procedures and test set-ups used for testing various 
electronic components like transistors and diodes at
that same location.

[Updated link 09/2005]
Here's the link to the HP49 calculator:
http://www.emu-france.com/?page=fichiers&idFile=839
Follow these instructions to download it...
On the "Menu" (on the left side) find and click on:
"Emulateurs"
Then under "Calculatrices (5)" find and click on:
"Hewlett-Packard"
Scroll down and under "Liste complete" you'll find:
"HP49 Emulator 1.18"
Click on that, then click on:
"Télécharger HP49 Emulator 1.18" to download it.

After unzipping, run "YUser.exe" to run the calculator.


Take care for now,
MrAl


----------

