# Help on wiring LEDs and 2032 batteries in series/parallel



## bfinleyui (Jan 15, 2013)

So I'm not an EE type of guy, but i'm slowly learning.

After doing the research, i thought I had this figured out...

I have 6 LEDs, each at 3v, with full brightness at 20mah.

I stacked 6 2032 batteries, each putting out 3v, for a total of 18v. They supposedly put out 220mah as well. Being that I thought I was drawing a total of 120mah, I had *assumed* that these would then drive at full brightness.

Unfortunately, that doesn't seem to be the case. What am I don't wrong here?

I know if I run the batteries in parallel, it would up the mah, keeping it at 3v, but that's obviously not going to work so well trying to push 6 LEDs. 

What am I missing about the mah math there? Thanks.


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## Burgess (Jan 15, 2013)

Don't confuse *mah* with *ma* .


Milliamps (ma) is a unit of Current.


MilliampHours (mah) is a unit of Capacity.


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## bfinleyui (Jan 15, 2013)

That's exactly my issue, thank you. I would ninja edit to make it clearer, but I'll leave my stupidity here for someone else to hopefully find and benefit from.

In that vein, then, I need to find a power source that'll give me 3v and 20mA to 6 LEDs. Does it exist, or am I going to be stuck wiring each to its own coin cell? I need them to run for about 6-8 hours (wedding reception).

Three NiMH would get me 3.6v and 1200mAh, conservatively. I would then use a resistor to bring the 3.6v down to 3, and then wire the LEDs in paralell after the resistor, correct? This should give me, assuming 20mA per LED and 6 LEDs, 10 hours of life, correct? Or am I still reading this wrong?

Edit: Still reading it wrong, apparently.

I would put the resistors (33ohms, in this case) in line between the negative lead of each LED and the shared negative back to the battery. Is that correct? That's according to this site


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## Gregozedobe (Jan 16, 2013)

Are you sure your LEDs need exactly 3.00V ? Many white LEDs have a vF of 3.2-3.4V and that can make a big difference. Also your battery voltage will drop over 6-8 hours unless you use largish batteries, so if you require constant brightness you will need bigger batteries or some kind of driver to control voltage.


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## argleargle (Jan 16, 2013)

bfinleyui said:


> That's exactly my issue, thank you. I would ninja edit to make it clearer, but I'll leave my stupidity here for someone else to hopefully find and benefit from.



Respect for that. Discussion has to start somehow. I've done it too.

Anyway, I'm probably about to say something wrong or stupid, too.  Ohm's law states that it isn't necessarily the power source causing the 20mA to happen. V/R = I, that is to say divide the Voltage of your battery by total resistance of the circuit and the result is what the current will be. This isn't taking into account that all batteries have an internal resistance that can screw up your math under certain conditions, but it's a pretty good starting point. Try a multimeter to take resistance values of the LEDs you've got? Resistance values that are wired in series are simply added up with addition.

mAh (milliAmphours) is basically the capacity rating of the battery. If you're looking at runtime, then mAh of the battery is what you should be considering. Is the size/space limited for your batteries? Why not 2xDcell batteries in parallel and don't worry about it? By this, I mean groups of 2 d cells in series, with the "packs of 2" wired in parallel?

Another idea: Get an adjustable DC power supply, run extension cords and hide it somewhere. Simply crank the voltage up until you're satisfied with the brightness and call it a day?


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## bfinleyui (Jan 16, 2013)

OK, so if I'm reading this right, and I think I am...

I could take a pair of 1.5v AA batteries, which each contain, conservatively, about 1800mAh, wire them in series, to give myself a 3.0v, 1800mAh power source.

Then, I would wire, in parallel, as many LEDs as I want, it would just drain the battery quicker.

For example, if I had 6 3v LEDs, running at 20mA, in parallel, they'd be using about 120mAh, giving me 15 hours of time.

I know that's an oversimplification of the current degradation curve, and that i'm probably more likely to get somewhere in the neighborhood of 10 hours, which is more than enough for my purposes.

My question, why is a resistor really necessary on the back side of these LEDs? Almost everyone tells me I'll need *something* on the back side, but nobody really seems to clarify why... all the calculators I come up with tell me a I don't need one...

Thanks for everyone's help.

Edit:Would it be better to put another 1.5v battery in that stack, and put a 100ohm behind each LED? That works out to driving each LED at 18mA, according to the calculators. I get maximum *consistent* brightness, without having to be the guy who looks like a moron buying 1ohm resistors?


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## argleargle (Jan 16, 2013)

bfinleyui said:


> OK, so if I'm reading this right, and I think I am...
> I could take a pair of 1.5v AA batteries, which each contain, conservatively, about 1800mAh, wire them in series, to give myself a 3.0v, 1800mAh power source.
> Then, I would wire, in parallel, as many LEDs as I want, it would just drain the battery quicker.
> For example, if I had 6 3v LEDs, running at 20mA, in parallel, they'd be using about 120mAh, giving me 15 hours of time.
> ...



[email protected] each x 2 series, yeah. You've got that. (As far as I can tell.)

As far as the limiting resistor... If you wire a bunch of leds in series, they each add together and act as their own limiting resistor. It depends on what voltage you're throwing into them and their internal resistances as to whether or not you'd even need a limiting resistor. The problem with series wiring of emitters is that if one fries, they all go dark.

If wired in parallel, then you could place the limiting resistor in SERIES with that. Think of the parallel led array as one object... a black box, if you will. Then it'd be (wired in series) battery -> limiting resistor -> parallel array object -> back to the battery. For parallel wiring, you'd probably need a limit resistor as there will be a voltage drop across the resistor and across your parallel array object.

So... series led = no limit resistor depending on your input voltage versus the combined resistance.
Parallel led = a limit resistor most likely.

Extra resistance in the circuit keeps the leds from frying. I used to make little handheld 5mm led lights about 10 years ago, 2x5mms in parallel with 100ohm resistor in series with that using 3AA batts in series (hopefully I'm remembering the numbers correctly.) They seemed to run forever (or 14 hours before dimming much or something, IIRC.) There's nothing wrong with one resistor in series with each led and the led/resistor series unit wired together in parallel, I've read a couple of guides that suggest just that to prevent a casacade failure if one led burns out. In parallel, as one led burns out, the felt current increases on each of the remaining leds... hastening their failure, if you've only got the one limit resistor in series with the parallel array.

I don't have a testbed or the stats on your leds, but 3 leds in series with 2 arrays of this in parallel might not need a limit resistor due to your combined resistance if 6 leds in series is too dim. The current will, of course, be divided between the 2 arrays since this is what current is supposed to do when it hits a parallel junction.

Hope I'm making sense and that I'm correct. As always, if anyone would like to jump in here and talk... please do! I'm all ears.


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## bfinleyui (Jan 16, 2013)

Makes a lot of sense. I'm going to give it a shot when I get home, on the breadboard with:

3xAA, split out to 6 LEDs, each with a 100ohm resistor after them, then back to the negative terminal on the battery stack. 

That should do it, thank you for your help. 

Edit: like so:


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## AnAppleSnail (Jan 16, 2013)

bfinleyui said:


> Makes a lot of sense. I'm going to give it a shot when I get home, on the breadboard with:
> 
> 3xAA, split out to 6 LEDs, each with a 100ohm resistor after them, then back to the negative terminal on the battery stack.
> 
> ...



That's a correct diagram. Technically speaking, you can really get away with using a single 100 ohm resistor. It cuts your parts count and, as long as your single resistor is well within its dissipation specs, won't harm the resistor or your reliability much. 

Assume Vbatt = Vnom = 4.5
Assume Vled = Vf = 3.2v

V=Vled + Vres
4.5v = 3.2v + Vres
Vres = 1.3
Vres = IR
I = 20 mA * 6 = 0.12A
Vres = 0.12 * R
1.3 = 0.12*R
R=10 ohms.

Is my math right? I seem to have a different resistor value than you do.


You split a power lead by connecting wires together. If I'm forced to do it freehand with solder, I wrap ALL the wires together at a junction, then solder them. This is where it helps to have a big soldering iron next to your little one.

I get decent brightness with quality AA and LEDs with 2 AAs in series connected to any number of LEDs in parallel. However, there is a short time at high brightness and a long time at low brightness. As the available voltage at the battery drops, the current to the LEDs drops, so that 2 AAs and one LED will be meaningfully (In pitch dark) lit for months.


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## bfinleyui (Jan 16, 2013)

My resistor values were just a product of futzing with numbers in the calculator to get to a radioshack-friendly resistor value.

I was using 3v as a general number for my LEDs. The data sheet says 2.9-3.1v on the LED, so I picked 3v, when combined with a 100ohm resistor and a 4.5 input, gave me 18mA on the LEDs.

The brightness difference between 20 and 18mA doesn't seem noticeable, to me, so i just ran with that math, plugging my number in, i get...

Assume Vbatt = Vnom = 4.5
Assume Vled = Vf = 3.0v

V=Vled + Vres
4.5v = 3.0v + Vres
Vres = 1.5
Vres = IR
I = 18 mA * 6 = 0.108A
Vres = 0.108 * R
1.5 = 0.108*R
R=13.888 ohms.


I think I did that math right...


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## argleargle (Jan 16, 2013)

Haven't done this stuff in forever. I think you're not supposed to substitute numbers if you're invoking more than one equation until you've constructed the full equation... or something like that.

Let's just work it a different way. We'll use one-shot equations and stated values only. We *WONT* use a derived number to plug into another equation and we should be good.

V=IR, therefore R = V/I, where I is desired current of .018A and V is 4.5. R=(4.5)/.018a = 250 ohms to run one led only only off of 3aa batts in series with one resistor.

Ok, seems to make sense. Current divides in parallel, since all branches of the parallel are the same we can state that total current is desired current*6. Therefore, R = (4.5)/(.018a*6) = 41 ohms total resistance of limit resistor in series plus the resistance of the parallel array of leds! 1/Rled = 1/Rled1 +1/Rled2 ... +1/Rled6. ....so we need the resistance value of the leds you're using to finish this answer.

We can't answer this question with straight Ohm's law without the impedance value of the leds. If you're not sure, maybe go wire all the leds in parallel and check the resistance of the parallel array with your multimeter?

Hmm. I'm really shaking out the cobwebs here. I could be wrong and know for a fact I've partied too hard during class. Anyone else want to jump in?


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## argleargle (Jan 16, 2013)

Oh, and check the resistance of the battery as well. (Resistance of battery + Resistance of LED parallel array + Resistance of limit resistor) all in series should equal 41 ohms to run each led at 18ma.

Therefore Rlimit = (Vbatt)/Iled) - Rbatt -Rarray = 41 - Rbatt -Rarray.

I'm sure that a 100ohm resistor is too big. The math says the leds will be dim. Don't forget that you can string two or more resistors in series if you don't have the correct resistor value. Note that this isn't a resistor next to each (in series with each) led in the parallel array. This is to use just one limiting resistor.

Bfinley, the answer may very well BE 13 ohms. Thinking about this is fun. There's usually more than one way to view a problem and if it's within 5% or so of the "actual" value, good enough.

I doubt there'll be a FIRE or anything....


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## bfinleyui (Jan 16, 2013)

Microtivity doesn't list the impedance on their website or the paperwork, so I guess it's going to be a "try it until it stops blowing up" situation 
As far as multimeter? You honestly think a guy who's on here trying to figure out how to turn a light on with a battery has a multimeter? Someday, surely, but I'm just crawling right now...


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## argleargle (Jan 16, 2013)

A meter is like 15bux at Radioshack, alternately you might pick one up for 5 bucks used from a dropout EE student. I bought one that way once. 

"try it until is stops blowing up," I laughed. Good attitude. I'm of a similar mind myself.


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## bfinleyui (Jan 16, 2013)

The lower resistor, the... (insert some number in the 10-41 range because we're not quite sure), that's only the case if i do an array, correct? If I'm going with a resistor in-line with each LED in the array, 100ohm is the correct resistor, yeah?


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## argleargle (Jan 16, 2013)

bfinleyui said:


> The lower resistor, the... (insert some number in the 10-41 range because we're not quite sure), that's only the case if i do an array, correct? If I'm going with a resistor in-line with each LED in the array, 100ohm is the correct resistor, yeah?



Yeah, somewhere 10-41 ohm for the single limiting resistor... too high will be dim output, but will be "safe." This will be your easiest to play with. Try a 20ohm and then a 10ohm. This will be cheapest in components, but theoretically more vulnerable to failure. The "Economy" or "Bean Counter Method," because you don't have to re-design the circuit based on wandering values of cheap components spit off the assembly line by the millions.

Still need the led impedance if you want a limit resistor in series on each branch of a parallel led setup, formula will be 1/Rarray= 1/(Rled1+Rlim) + 1/(Rled2+Rlim) + ... +1/(Rled6+Rlim). Some say that this is the most "bulletproof" way to wire it. The "Engineer's Method."

My vote? Try 2 strings of 3 leds in series. Run the 2 strings in parallel and see if it needs a resistor at all. Since 6 in series was too dim, this should brighten things up. (3 in series) in parallel with a total of 2 strings of these, failing that 2 leds in series with 3 groups of these in parallel. Still vulnerable to failure, but we skipped out on using discrete resistors alltogether. This should be easy since you've got a breadboard, as you said. This cuts down on trips to RatShack for yet another variety pack of resistors. Utilize internal resistance of your components. This method will vary your design based on specific values of the individual components that you have at the time. This one is the Path of MacGuyver.


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## bfinleyui (Jan 16, 2013)

The 6 in series were with 2032s that wouldn't give me enough mA to get things bright enough, i'll give it a shot with two 3-led series in an array, to see if it's bright enough to utilize the 2032s in some way. otherwise I'll just buy a big lot of AAs and do it that way. The 2032's don't seem like they'll work for anything more than a 1-to-1. I'm sure I'll find a use for the 100 2032's I bought on ebay for $15.


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## AnAppleSnail (Jan 16, 2013)

LEDs do not have impedance, exactly. They're one of the circuit components where Ohm's law does not apply in a useful way. That is, you could solve for an R at any given V*I, but it will be different at any other V*I. What we get with LEDs is an approximate forward voltage at some current, and with really nice LEDs a graph showing the general trend of forward voltage to current. This graph is exponential to some power, with a small change in voltage leading to a large change in current.

So an LED is a magic box whose current throughput is a function of voltage. Check the Cree datasheets for a great visual. I'm on a phone and can't easily get to them. This lets us substitute Vled into a loop law equation, where the voltage drop in the circuit plus the battery voltage is zero.

Vled + Vres + Vbatt = 0
Vled is known at a given current.
Vres is calculated at that same current and the voltage difference.

Of course, my mistake was forgetting that you have several 100 ohm resistors in PARALLEL, which equates to 1/(n/100) ohms, where n is the number of LEDs in parallel.


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## argleargle (Jan 16, 2013)

So we're clear, that's 6 led in series, no resistor added, with a stack of 6 x (3v 2032) =18 v?

I think I've noticed a problem...
http://www.eemb.com/Li-MnO2_battery.html
2032, 210 mah capacity each. If they're stacked, that's 210mah and all you get that way. You must parallel another pack to get reasonable runtime. Also, the maximum discharge current is rated at *THREE* milliamps! (with peak pulse current still only 15mah) We were intending to run each led at 18 milliamps, correct? Again, having fat parallel stacks in your battery pack will help with this. 

I think your original experiment didn't work because the voltage sagged as the cells exceeded maximum discharge rate because all leds and batteries were in series.

I also think this is quite doable once you get your battery stacks sorted out. Parallel the leds and parallel the battery pack.

*Did we get it? *:thumbsup:Muahahahaha!:devil:


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## bfinleyui (Jan 16, 2013)

argleargle said:


> So we're clear, that's 6 led in series, no resistor added, with a stack of 6 x (3v 2032) =18 v?
> 
> I think I've noticed a problem...
> http://www.eemb.com/Li-MnO2_battery.html
> ...



We all understand, yes. 2032's cant put out enough current to drive 6 LEDs at anything resembling real brightness. Max discharge current somewhere listed it around 25mA. 

Rather than parallel 4 stacks of 6 2032's, I'll stick with 3 AAs. Especially since after I figure this out I have to make 25 of em (oye)

I'll report back in a few hours with pictures and (hopefully) a blind spot from staring at a half-dozen 18mA LEDs)


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## argleargle (Jan 16, 2013)

Best of luck, let us know, and I look forward to your beamshots.


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## argleargle (Jan 16, 2013)

AnAppleSnail said:


> LEDs do not have impedance, exactly. They're one of the circuit components where Ohm's law does not apply in a useful way. That is, you could solve for an R at any given V*I, but it will be different at any other V*I. What we get with LEDs is an approximate forward voltage at some current, and with really nice LEDs a graph showing the general trend of forward voltage to current. This graph is exponential to some power, with a small change in voltage leading to a large change in current.
> 
> So an LED is a magic box whose current throughput is a function of voltage. Check the Cree datasheets for a great visual. I'm on a phone and can't easily get to them. This lets us substitute Vled into a loop law equation, where the voltage drop in the circuit plus the battery voltage is zero.
> 
> ...



Yeah, forgot that an LED is a "somewhat" reactive component like a cap or coil. You're completely right, of course. Good catch, thanks.


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## bfinleyui (Jan 16, 2013)

Heyyyoooo it worked!

Ended up jamming 10 LEDs, each with a 100ohm resistor on there, and suddenly... 

Onto the pictures...
















Thanks to everyone for their help!

Going to give it a shot with just the single limiting resistor now. Off to sift through the math and figure out what that'd be. Thanks!!

Edit: Guess I should point out, these will be part of the centerpieces at my wedding reception. No open candles at our reception venue = gettin tricky with LEDs


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## bfinleyui (Jan 16, 2013)

And I just wired it up with 30ohm worth of resistors after the array of 10 LEDs (meaning they run a little lower than 18mA, but i couldn't see a diff between 20 and 30), and it works great. Might do an experiment to see just how many resistors I can put in there before brightness is affected, but it would appear that will be the easiest way to go.

Battery pack to a bunch, that splits to 10 LEDs, wired to a bunch, 30ohms after that resistor and back to the battery pack.

Now to figure out the hell to lay this out on a 5-inch circle. Oye.

Thanks again for everyone's help. I learned a lot (including that I'll never be an EE)

Next up: trying to figure the best way to make (reliable) battery packs. Electrical tape is not holding the regular jumpers very well.


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## argleargle (Jan 16, 2013)

Well kick ***! Love the paper lanterns. Glad you got it sorted! Looks great!

Suggestion on the battery packs: Ratshack sells 4aa battery boxes complete with a switch. You can install 3 cells and jump the gap with a piece of wire with no soldering. It's what I used to make my little 2x 5mm lights with.

Im breathing a sigh of relief now that the single resistor solution was between 10 and 41


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## bfinleyui (Jan 17, 2013)

So, I'm back with some more questions.

After the centerpieces went so well, I got interested in doing my own up-lighting. Being that these aren't going to be on the tables, but along the walls, I scrounged a 5V wall wart (5.0v, 1.3A).

Did the math, and 13 3v blue LEDs in parallel running at 20mA on a 5v needed to get rid of 19 ohms, so i put a pair of 10ohm resistors on the negative side, and plugged it in to test out. 

The first time, I had it wired wrong, I think, and burned a few LEDs. 

So I reconfigured the breadboard, so along the right side, along the full-length positive lead, i had 13 jumpers bringing the positive over to individual rows. On that row, I put the positive lead of the LED. I put the negative on its own row, and then jumpered the negative over to the full-height negative lead on the left side of the board. On that same full-length negative lead, I put two 10ohm resistors. I connect the power source, plug it in, and they light up great. But then, the LEDs quickly get blazing hot. Like burn-your-finger hot.

So where'd I screw up, my wiring or my math?

Did I accidentally put them in series somehow? Is that not the proper place to put my resistors?

Edit: After using a different calculator specifically for LEDs in parallel, it brought up a whole other possibility (probability)... How many watts do I need to dissipate in that resistor? I just grabbed 1/4w resistors. Should I split these up into another configuration? Get 1W resistors? Or have I found the reason that when you get to this many LEDs, I should be using a resistor on each rung of the parallel array?


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## AnAppleSnail (Jan 18, 2013)

The wall wart does it. Cheap ones climb in voltage to reach rated load.


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## bfinleyui (Jan 18, 2013)

AnAppleSnail said:


> The wall wart does it. Cheap ones climb in voltage to reach rated load.



So what are my options?


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## AnAppleSnail (Jan 18, 2013)

bfinleyui said:


> So what are my options?



Get a fixed-voltage supply, use more LED (My favorite option), or fudge your resistor values until it works out right. The cheap wall-warts (Say, 12v) tend to climb about 1/5th their voltage before they settle in. If you blow $15 on a multimeter, you can check the 'open circuit' voltage of your "5v" wall wart and see up to 7v usually. This lets you plan on using real numbers, not stated guesses. Good luck!


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## bfinleyui (Jan 18, 2013)

Let's say it's 7v on the open circuit.

Will that 7v stay constant? I'm seeing some places where it says that the more mA i pull, the lower the voltage is? So do I calculate my resistors based on the voltage under load, or the voltage when it's just the multimeter?


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## argleargle (Jan 18, 2013)

I think it's a safe bet that cheap power supplies will not remain constant, even under what should be a constant load due to heat.

Assuming the power supply is actually throwing 7 volts under load, Power (watts) is given by P = V I. Current adds in parallel, such that 20 ma * 13 = 260ma, assuming for a second that this is what they're running.

P = 7 * .26 a = 1.82 watts of power through the parallel array. with respect to 1/4 watt resistors, I think you nailed it on your above edit, bfinley. See if you can get some higher wattage rated resistors. You could also try a parallel array of larger value resistors in series with the parallel array of the led since current splits among the branches of a parallel array.

The actual numbers will be different and don't count on the power supply to keep spitting out the same voltage cold versus warm... or even a constant level of output. Once you're talking about some kind of mangled waveform out of a cheap dc power supply, the math gets far crazier. Treating it as a straight level dc is probably good enough for an approximation. Don't forget that cheap resistors can vary somewhat in actual value from one piece to the next one from the same package.

Didn't goof on my setup or thinking, did I?


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## bfinleyui (Jan 18, 2013)

Is there a way to normalize the output from the wall wart, before it gets to the LEDs? Something that says "only give me 5v or less" or something, so that if anything, i just get dimmer, not blindingly hot?


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## bfinleyui (Jan 18, 2013)

argleargle said:


> I think it's a safe bet that cheap power supplies will not remain constant, even under what should be a constant load due to heat.
> 
> Assuming the power supply is actually throwing 7 volts under load, Power (watts) is given by P = V I. Current adds in parallel, such that 20 ma * 13 = 260ma, assuming for a second that this is what they're running.
> 
> ...




Hmmm. Any place I can get reliable, but inexpensive, wall warts? 

How far can I let it vary before I worry? I have friends who are willing to give me all their old cell phone chargers, and I'd like to chop them up and use them to drive these uplights. If I keep it cool (by dissipating the watts over more resistors, rather than letting one get really really hot), the variability should be reduced, correct?

Also, I'm headed to the hardware store for a multimeter today 


Edit: or would all my problems go away if I just nut up and solder a 100ohm resistor behind each LED in the array?


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## Lynx_Arc (Jan 18, 2013)

I have found calculating resistances for LEDs to be the hard method for figuring it out. I use the variable resistor and dual meter method. I have a variable power supply or a set of batteries and I use a variable resistor starting at allowing 0v to the LEDs and cranking it up slowly with an ammeter inline and a voltmeter on the power supply if variable. This way you can crank up the power to a specific current level to the LEDs and then disconnect the power and measure the resistance. Like others have said LEDs are not a resistive load they vary in resistance with the voltage supplied to them 
One other thing to realize is that most generic LEDs don't last as long as claimed at the rated current drive levels (~20ma) such that if you are going to put a lot of hours of use on them you may need to reduce the drive current down considerably. LEDs tend to be more efficient at lower drive levels also. If you are wanting a more constant light output you may consider a linear regulator too which has an advantage that is can be set up to keep the output the same regardless of if you decide to use 4 batteries instead of 3 or even a wall wart.


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## argleargle (Jan 18, 2013)

bfinleyui said:


> Hmmm. Any place I can get reliable, but inexpensive, wall warts?
> How far can I let it vary before I worry? I have friends who are willing to give me all their old cell phone chargers, and I'd like to chop them up and use them to drive these uplights. If I keep it cool (by dissipating the watts over more resistors, rather than letting one get really really hot), the variability should be reduced, correct?
> Also, I'm headed to the hardware store for a multimeter today
> Edit: or would all my problems go away if I just nut up and solder a 100ohm resistor behind each LED in the array?



No reason you can't use the cheap wall wart you have, you'll just have to work around its problems. Keep in mind that a lot of numbers in a circuit can vary up to 5%-20% from the "calculated correct value" and still work, but is it overdriven or underdriven? In real-world electronics, there is often no "one correct value." ...more of an "acceptable range without fire."

If you choose to dissapate heat over more resistors, this will work but their resistance values will need to start shooting up as you add more branches to a parallel array of resistors that's in series with the battery and the parallel array of led.

If you choose to use lower-watt resistors due to price (a common design trend) and use more of them by having one in series with each rung of the led parallel array, you might come out cheaper at the end of the day and it'll still work. I've seen designs where it's one resistor in series and then 3 led in parallel, with multiple instances of this wiring type all in parallel.

...and you're right about the really hot resistors. This can cause a change in resistor value, perhaps even a permanent one. I recall a laser pointer mod that involved burning up a resistor with a soldering iron.

An old electrical engineer was telling me about his experiences designing tv sets. He said he hated it because they'd design the perfect circuit, then management would make them start taking stuff out until it quit working and put the last thing back in. Apparently if you're making 10 million of them, the cost of .19 cents per $500 product makes or breaks a production run.


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## argleargle (Jan 18, 2013)

bfinleyui said:


> Is there a way to normalize the output from the wall wart, before it gets to the LEDs? Something that says "only give me 5v or less" or something, so that if anything, i just get dimmer, not blindingly hot?



Yeah, that's a called a DC to DC power supply. It goes by many names: driver, regulator, power filter, power conditioner and so on.

http://en.wikipedia.org/wiki/Voltage_regulator


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## bfinleyui (Jan 18, 2013)

Alright, scrounged up a pile of 100ohm resistors and put them behind each LED, they've been running for an hour at 20ohm and they're not getting hot. Tested the wall wart and with a completely open circuit, it didn't push past 5.3, so that looks stable enough for my needs.

Just ordered 300 100ohm resistors (for the uplights, 13leds x 10 units) and 100 10ohm resistors (for the centerpieces, 2 or 3 resistors x 24 units), and while there, I found they're the supplier for the 3xAA Battery boxes I'd ordered from another site, at half the price. 

The one thing I can't seem to find, and I hope I haven't screwed it up, is whether the battery holders are series or parallel. I'm assuming series, because, why wouldn't it be? But is there any way I can confirm for sure? I tried to find parallel holders, and i can't find any, so I assume they don't really exist...

I'm sure I'll be back when I'm elbow deep in smoking resistors, but for now, i think I'm fairly set, thank you all, again, a lot.


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## argleargle (Jan 18, 2013)

bfinleyui said:


> The one thing I can't seem to find, and I hope I haven't screwed it up, is whether the battery holders are series or parallel. I'm assuming series, because, why wouldn't it be? But is there any way I can confirm for sure? I tried to find parallel holders, and i can't find any, so I assume they don't really exist...



That's easy, with your new $15 multimeter, test the output voltage. The Ratshack battery holders with switch and leads are series packs. I like them because they are easily reconfigured on the inside or can be wired in gangs any way you want, assuming the humble AA or AAA cell can give you what you want.

DON'T forget!! You can FIX this if you don't like the way it's configured. The packs are full of little metal tabs that you can pull and add jumper wires. Make it *BE* your way, don't rely on it being the way you want it. Series adds voltage, parallel adds current.

...and whatever you do, post pics and stuff about what you build. Please? I asked nicely.


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## argleargle (Jan 18, 2013)

Lynx_Arc said:


> I have found calculating resistances for LEDs to be the hard method for figuring it out. I use the variable resistor and dual meter method. I have a variable power supply or a set of batteries and I use a variable resistor starting at allowing 0v to the LEDs and cranking it up slowly with an ammeter inline and a voltmeter on the power supply if variable. This way you can crank up the power to a specific current level to the LEDs and then disconnect the power and measure the resistance. Like others have said LEDs are not a resistive load they vary in resistance with the voltage supplied to them
> One other thing to realize is that most generic LEDs don't last as long as claimed at the rated current drive levels (~20ma) such that if you are going to put a lot of hours of use on them you may need to reduce the drive current down considerably. LEDs tend to be more efficient at lower drive levels also. If you are wanting a more constant light output you may consider a linear regulator too which has an advantage that is can be set up to keep the output the same regardless of if you decide to use 4 batteries instead of 3 or even a wall wart.



I really liked your post and it highlights something I said earlier in this thread about how there is always more than one way to approach a problem. I posted about how the easiest way was to get a variable DC power supply and crank the voltage until you're happy and call it a day.  That's all well and good until you plug it into a cheap power supply with +25% expected and labeled voltage. Either that or your variable power supply isn't correctly calibrated. Yeah, it'd go wrong then.

I like your method, too. The only theoretical problem I can see with it is that testing a circuit with any testing device changes what you're testing, even by a small amount. The circuit will be theoretically different for using that ammeter versus without it. That ammeter changes the resistance or reactivity of the circuit in some way, even if it is just a little bit of added latency to a waveform or logic pulse if working with a digital or ac circuit.

I said I liked your method, dammit! Don't put words in my mouth.  I think you're cool! Say some more stuff.


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## bfinleyui (Jan 19, 2013)

argleargle said:


> That's easy, with your new $15 multimeter, test the output voltage. The Ratshack battery holders with switch and leads are series packs. I like them because they are easily reconfigured on the inside or can be wired in gangs any way you want, assuming the humble AA or AAA cell can give you what you want.
> 
> DON'T forget!! You can FIX this if you don't like the way it's configured. The packs are full of little metal tabs that you can pull and add jumper wires. Make it *BE* your way, don't rely on it being the way you want it. Series adds voltage, parallel adds current.
> 
> ...and whatever you do, post pics and stuff about what you build. Please? I asked nicely.



Most definitely. There will be PLENTY of pictures, of the amazing products I make, of the horrible failures and injuries along the way, and what will surely be my favorite: a gallery of the worst solders from this project. If my calculations are correct, there will be thousands of them.

On that topic, for parallel, do I have to have a separate connection for each anode or cathode, or could i give all my LEDs a wire, smudge all the positive power connectors together, solder them into a clear section of the + rail, and do the same for the negative? any issues with a 'clump' of soldered wires? The one I prototyped took the arc reactor method, with about a 1/4 inch of stripped wire every 3 inches, and solder to that. and it became a nightmare.

My favorite part of this whole experience has been the rise in self assuredness when discussing these things with friends. Hell, this morning I bought a 16x2 LCD panel, and intend to wire it up with the daughter board from adafruit to display information from inside my beer fermentation chamber.

Thanks guys, this has been an amazing learning experience, and I will definitely lurk and post in the forums here as I learn more.


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## Lynx_Arc (Jan 19, 2013)

argleargle said:


> I like your method, too. The only theoretical problem I can see with it is that testing a circuit with any testing device changes what you're testing, even by a small amount. The circuit will be theoretically different for using that ammeter versus without it. That ammeter changes the resistance or reactivity of the circuit in some way, even if it is just a little bit of added latency to a waveform or logic pulse if working with a digital or ac circuit.


If you use the 10 amp scale the difference is negligible. Once you get above 100ma you can usually read the 10A scale well enough to get close to the right current. Trying to get it exact isn't needed as on a 20ma load if you are 5% off that is only 1ma difference which won't make a difference on LEDs.


> I said I liked your method, dammit! Don't put words in my mouth.  I think you're cool! Say some more stuff.


My method is the easiest and allows you to also check actual light output such that you can crank it down some and see if it is a lot of difference. My advice is if you use this method slowly crank up for no light with dark adapted eyes till you can see well enough to read the meter and then crank slowly till you get to about 5, 10, 15, and 20ma per LED. You may find that 10 or 15 ma levels gives you enough light output and runtime will be longer at those rates plus the LEDs will last longer.


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## argleargle (Jan 19, 2013)

bfinleyui said:


> and what will surely be my favorite: a gallery of the worst solders from this project. If my calculations are correct, there will be thousands of them.



Ha! An Air Force radio tech I used to know always said "The bigger the glob, the better the job." He was joking, of course. It was his way of admitting that he wasn't so great at soldering. Note I didn't say soldiering. That's different.



bfinleyui said:


> On that topic, for parallel, do I have to have a separate connection for each anode or cathode, or could i give all my LEDs a wire, smudge all the positive power connectors together, solder them into a clear section of the + rail, and do the same for the negative? any issues with a 'clump' of soldered wires? The one I prototyped took the arc reactor method, with about a 1/4 inch of stripped wire every 3 inches, and solder to that. and it became a nightmare.



With regular aa and aaa battery supplies, you don't have to worry about arc jumping. Remember, for a spark to jump less than half an inch, it's got to be over 2000 volts or so. The tendency to arc increases by exponentially as the distance is halved. That is to say, even a little bit of separation makes it even more remotely unlikely to happen. We're not throwing around hundreds of volts at hundreds of amps here. Even an automotive ignition spark is unlikely to jump 3/4 of an inch. It's basically unheard of.



bfinleyui said:


> My favorite part of this whole experience has been the rise in self assuredness when discussing these things with friends. Hell, this morning I bought a 16x2 LCD panel, and intend to wire it up with the daughter board from adafruit to display information from inside my beer fermentation chamber.



This sounds damn awesome. Adafruit, eh? You sure it's not Raspberry Pi that you're fermenting there? 

Take care, hope to talk to you some more, don't be a stranger! ...and remember if it's even got a single LED on it, it's fair game here if you explain it properly... with beamshots!


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## bfinleyui (Jan 19, 2013)

Raspberry Pi is currently monitoring temperature, sending a reading every minute to the database, then plotting that information. In the future, it'll be a little more intricate. I have a solid state relay ready to control a refridgerator on/off to keep it at the proper temperature, some small servos that could be used to dispense hop additions, etc.

All this is leading up to a June 29th wedding. Rehearsal dinner and hotel welcome baskets have homebrewed beer and cider, and then the lighting decorations at the reception are going to be all DIY LED projects. Scrounging plastic containers that could be converted to look somewhat like uplights. (round pasta take-out containers seem to be the leader)


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## argleargle (Jan 19, 2013)

Ultra cheap white wax-paper cone shaped cups make a dandy diffuser for a single led.  Sounds like we're on the same frequency.

Remember, federal law states that you can make 100 gallons of beer or light wine per adult in your house per household. If you need to rent a room in another state, just let me know. :thumbsup:


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## bfinleyui (Jan 19, 2013)

You gonna tell the state? Thought not.


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## argleargle (Jan 19, 2013)

bfinleyui said:


> You gonna tell the state? Thought not.



Sure. In no particular order, the States of: Confusion, Inebriation, Drowsiness, Disorder, Panic, Disarray, Aggravation, Doubt, Elation, and Boredom.

:hahaha::tinfoil: <--- That's my tinfoil hat, see?


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## AnAppleSnail (Jan 19, 2013)

With the runtimes we have these days, the last step will still be illumination.


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