# REALLY Simple Guide To Figuring Out Voltage In Maglite



## LuxLuthor (Jan 15, 2008)

I get a number of questions in PM's from people trying to figure out how to use and select the proper bulb and battery combination voltage for their particular Maglite setup. I remember not too long ago how confusing all of this was, so let me do my best to give a few SIMPLE pointers. _*PLEASE, NO EXPERT, DETAILED INPUT OR COMMENTARY from all of those electrical engineers who sleep on superconductors!*_
 OK....let's use a practical example. I want to use 2 of AW's C Li-Ion cells in a 2C size STOCK Maglite to light up a WA 1111 6V bulb, using this 1D size light I bought a while ago from FiveMega. I put the bipin bulb into a gold bi-pin bulb holder I also got from FiveMega, and put that into the stock Maglite bulb holder, and tighten down the retaining ring. Next, I look at my bulb testing thread here, and click on the 1111 bulb test data here.

 I know that each of the AW C Li-Ions when charged will have about 4.15V, and two of them in series gives 8.3V. When I look down the left "Applied Volts" column, I notice that when 8.3V is applied, the bulb flashes (dies). I am now worried that my brilliant idea of using 2 x Li-Ions at 8.3V (fully charged) will kill my bulb, and I wonder if I should use a different setup. :thinking:

 However, what needs to be taken into account is the amount of resistance (to the flow of electrical current) that is related to the various parts of the flashlight, and which effectively decreases the voltage actually seen at the bulb. There is a basic law you may have heard of called Ohms Law which we will use to understand this. Ohm's Law says:

*V = I x R* or *Voltage *(Volts) = *Current *(Amps) x *Resistance *(Ohms)

Now to see the effect of the flashlight parts having resistance and lowering the amount of voltage delivered to the bulb, I need to get some data in 3 measurements.

 1) With my DMM (digital multi-meter), I measure the voltage of my two C Li-Ions outside of the light, which I know are not fully charged. Touching each together, like they would be inside the light. *I get a reading of 7.94 Volts. *This is known as "Vbat" (battery voltage)

 2) With light switch off, I put the Li-Ions in the light. I then put the tailcap in position, with spring touching back of Li-Ion, but do not push down and screw it on. Now I turn on my DMM with it set to measure Amps. I turn on the switch for the light while it is resting on a brick...but with the tailcap off, there is not a complete (closed) circuit, so light stays off.
Now I take my black (Neg) DMM probe and touch it to tailcap edge where there is bare aluminum just above threads. I simultaneously touch my red (Pos) DMM probe to the back edge of flashlight tube, and my light comes on because I have now closed the circuit. *I read the current as 3.52 Amps.* When I take away my DMM probes, the circuit is now "open" and the light turns off.​



​3) I now turn off the light switch, screw on the tailcap, and screw off the head so I can see the bulb in the bipin holder. I use the DMM's Pincher leads, and clip them onto each leg of the bipin inserted into the gold bipin holder.​


​Now I turn on the light, and get a DMM *reading of 6.65 Volts*. This lower Voltage reading is known as Vbulb.​


​
So now we have readings of

*Vbat = 7.94V*
*Amps = 3.52A*
*Vbulb = 6.65V*

The *difference between Vbat & Vbulb is 1.29V *(7.94 - 6.65), which is the voltage drop because of resistance in the light (including bulb). Now we use Ohms Law to see how much resistance is in my setup, since we know the current, and how much the voltage dropped.
1.29V = 3.52A x ??? Ohms so then solving for Ohms:

1.29V / 3.52A = *0.366 Ohms* ​
There are 1,000 milli-Ohms (mOhms) in 1 Ohm, we could also say that *I have 366 mOhms of resistance in my setup.* In effect, this resistance is both "protecting" my bulb by lowering the voltage delivered, but is also decreasing my lumen output.

Now I know that when using *fully charged Li-Ions at 8.3V*, I would only see about 7V at the bulb, so it is not going to flash. *Good News!* But now if I want to have around 7.5V delivered to the bulb for better lumen output, then I need to do some things to reduce the resistance down to about 220 mOhms in my light setup.
8.3 Vbat - 7.5 Vbulb = 0.8 V 

This smaller drop in voltage would both protect my bulb, and give me better lumen output...so I want to try and reduce my resistance to: 0.8V / 3.6A = *222 mOhms*​How to reduce your resistance is another topic.


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## sed6 (Jan 15, 2008)

:thumbsup: Another great post Lux! So much of the info you put out is truely useful for us newish flashaholics. I have more of your posts bookmarked than anyone else I can think of. Thanks Lux!


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## littlejohnle (Jan 15, 2008)

Wow, great write up. Lots of very useful info. 
So thats how you measure the current!!

Another thread that may have to be stickyed.


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## LuxLuthor (Jan 15, 2008)

Thanks guys. Be careful if you are measuring current in a higher level (like 9-12Amps), because with the initial spike of voltage/current when first turning on your light you can blow the 10A fuse in the DMM. Been there, done that. Measuring higher current is done with a "shunt" like you see in pix of my destructive bulb testing setup, but is a more complex scenario.


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## cernobila (Jan 15, 2008)

Lux,

Now I know what is going on.......after all this time on here I finally made a significant step forward in my knowledge in this area.

now, how can I tweak my light to give me the ideal resistance.......I am ready for the next step......YEAH.


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## cat (Jan 15, 2008)

Thanks, LuxLuthor. 

I noticed that the unfused positive jack in my DMM says 10A. What happens if it gets more than 10A ? The DMM becomes the fuse? 
I was wondering whether that meant I couldn't use it to measure vbulb on a 646** setup. 

I thought it might mean that I need a Fluke multimeter.  A few months ago, I thought M6's were way too expensive. Now I've spent more on MAG stuff, and I think I need a bench power supply.  And it seems like I need a high-end RC charger to charge torch batteries! 

That reminds me, I was thinking how I was going to buy Fluke test leads that that cost 3 times more than my DMM. :duh2:


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## rizky_p (Jan 15, 2008)

Hi Lux, thanks for helping a newbie like me..

One question, how do you measure the VBatt? is it like measuring forward votage on LED?

thanks


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## js (Jan 15, 2008)

cat said:


> Thanks, LuxLuthor.
> 
> I noticed that the unfused positive jack in my DMM says 10A. What happens if it gets more than 10A ? The DMM becomes the fuse?
> I was wondering whether that meant I couldn't use it to measure vbulb on a 646** setup.
> ...



DMM's 10 Amp current measuring lead position IS fused, and will blow if you put significantly more than 10 Amps through it. I blew the 10A fuse on my Fluke early in my hotwire career by moving the lead for measuring current, changing the dial to measure current, THEN I decided I wanted to know the starting, unloaded, hot voltage of the pack, so I dialed up DC Volts and put the leads across the battery contacts. POOF! Instant blown fuse--because while I did change the DIAL from DC current to DC volts, I did not change the LEAD POSITION!

*So, be careful to ensure that you have the right lead position for the right dial position and right current range.* If you change the dial to measure current you must switch the leads. If you change it again, you must switch back.

Lux,

Thanks for taking the time and effort (and pictures) to make this thread!


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## mdocod (Jan 15, 2008)

hello js,
There are actually quite a few truly unfused multi-meters out there that will turn their internals into a big gooey fuse if you aren't careful. And there are quite a few who's low current measurement setting is the one that is fused while the high current side is unfused. (go figure).... 

My old Mastech is a perfect example of this, the current measuring "plug" is rated "20A Unfused" on the front, and in fact, I have measured as high as 32 amps with it.. (probably not good for it)... I have searched it's internals through and through for a large 20A+ fuse for that channel and it does not exist, there is a small fuse on the "200mA fused" plug. I even had friend who has a similar multi-meter make a "mistake" (plugged into current metering tried to check for line voltage, POOF) with his recently, and ended up with a destroyed multi-meter. Luckily they are only about $10-20... In some cases, it's easier to just find a replacement cheap multi-meter than it is to find a fuse that will fit some of them! LOL...


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## LuxLuthor (Jan 15, 2008)

*Guys, with respect please keep this topic on a simple level. Thanks*

*Cat *- Like a fuse in your car or any other product, when you pump too much voltage and/or current through it, the fuse will "blow" and either break the circuit, or end up not being able to test the function (like current in this case). 

Fuses are designed to protect your DMM. You don't have to use a Fluke or Fluke probes, that's just what I had on hand to demonstrate this topic. You can attach wires to the bipin legs and measure voltage of wires (just don't use too thin of wires, and don't let them touch each other when wrapped around bipin legs).

*rizky_p* - You just line up batteries and measure voltage as shown in pix...red (+) probe goes to (+) end of battery. These are two Li-Ion Pila brand batteries.


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## sb_pete (Jan 15, 2008)

Thanks Lux.
I gotta get me some o' them pincher leads, any suggestions?

Thanks again
-Pete


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## LuxLuthor (Jan 15, 2008)

Search for multimeter leads. Fluke brand run about $28

There are various types and brands on this link.


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## rizky_p (Jan 16, 2008)

LuxLuthor said:


> *
> 
> rizky_p - You just line up batteries and measure voltage as shown in pix...red (+) probe goes to (+) end of battery. These are two Li-Ion Pila brand batteries.
> 
> ...


*

isn't that supposed to be the VBat? not the Vbulb?

thanks*


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## LuxLuthor (Jan 16, 2008)

rizky_p said:


> Hi Lux, thanks for helping a newbie like me..
> 
> One question, *how do you measure the VBatt*? is it like measuring forward votage on LED?
> 
> thanks



That is Vbat which was what you asked. I showed measuring Vbulb in first post with the pincher leads.


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## js (Jan 16, 2008)

Lux,

Sorry. You're right. I'll edit my post down to the simple core of it, which was simply this:

*ALWAYS ENSURE THAT THE DMM LEADS ARE IN THE CORRECT HOLES TO MEASURE THE PARAMETER YOUR DIAL IS SET TO MEASURE*.

If you switch from voltage to current measuring, you must move the red lead. If you switch back, you must move the red lead yet again.

This is simple. This is important. This is why I posted.


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## cat (Jan 16, 2008)

LuxLuthor said:


> *Cat *- Like a fuse in your car or any other product, when you pump too much voltage and/or current through it, the fuse will "blow" and either break the circuit, or end up not being able to test the function (like current in this case).
> 
> Fuses are designed to protect your DMM.
> You don't have to use a Fluke or Fluke probes, that's just what I had on hand to demonstrate this topic. You can attach wires to the bipin legs and measure voltage of wires (just don't use too thin of wires, and don't let them touch each other when wrapped around bipin legs).



This DMM (I think it's quite typical of a cheap DMM) has 3 jacks one above the other on the righthand side: the black COM one, the "normal" red positive one, and then the red one marked 10A UNFUSED. 

The background to my question: 
A while back (and it seems like a long time ago because of the steep learning curve) I read something about measuring vbulb by poking the DMM probes into the socket and having to have thin enough probes. Then, recently, I saw Petrev's photo showing it being done with some well-shielded crocodile clips clipped to wires running connected to an unmounted Kiu socket. 

When I noticed the other day that my DMM said 10A unfused, I wondered if it was capable of what I'll want to do when all my parts arrive: measure the vbatt with bulbs that will be drawing a bit more than 10A. 

So,...DMM test lead clips on wires from the bi-pin bulb, DMM set to measure voltage, red test lead in the normal and the bulb is a 64458 that might be drawing more than 10A... will that not be too much for the DMM to handle?

My local electronics component shop has only got test leads with those tiny crocodile clips with...unsuitable plastic sheaths. So, look at online catalogs...and the most suitable seem to be the hook probes (similar to your pincher probes.)


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## cat (Jan 16, 2008)

js said:


> *ALWAYS ENSURE THAT THE DMM LEADS ARE IN THE CORRECT HOLES TO MEASURE THE PARAMETER YOUR DIAL IS SET TO MEASURE*.
> 
> If you switch from voltage to current measuring, you must move the red lead.



...From the "normal" /default jack to the one that says 10A UNFUSED? 

And (see my previous post), if I am measuring voltage (or trying to) and the vbatt is 21.something V and the bulb is a 64458, will the current draw of slightly more than 10A be a problem?


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## cat (Jan 16, 2008)

Lux, are we limited to Voltage here? (I have a newbass question about measuring resistance.) :tinfoil:


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## js (Jan 16, 2008)

Another simple but VERY important warning:

HALOGEN LAMPS CAN EXPLODE! Wear safety glasses if you are working with bare, powered up lamps. DO NOT TOUCH THE GLASS WITH YOUR FINGERS, either, as the oil will cause an explosive failure when the glass heats up. If you do handle a lamp, clean it with isopropyl alcohol and let dry before powering it up.


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## LuxLuthor (Jan 16, 2008)

cat said:


> This DMM (I think it's quite typical of a cheap DMM) has 3 jacks one above the other on the righthand side: the black COM one, the "normal" red positive one, and then the red one marked 10A UNFUSED.
> 
> The background to my question:
> A while back (and it seems like a long time ago because of the steep learning curve) I read something about measuring vbulb by poking the DMM probes into the socket and having to have thin enough probes. Then, recently, I saw Petrev's photo showing it being done with some well-shielded crocodile clips clipped to wires running connected to an unmounted Kiu socket.
> ...



If you are using a DMM that does not have a fuse when measuring up to 10Amps (i.e. is "unfused"), then *yes, subjecting it to higher than 10A current will damage your DMM. Don't do that 64458 current measurement.* This brand has unfused 10A like you describe. Note paragraph titled *Current Measurement - 10A DC *under *"Measurement For Managers."

Fire away on resistance measurement question.
*


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## Icebreak (Jan 16, 2008)

LuxLuther -

Great guide. Good of you to help folks out.


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## Kremer (Jan 16, 2008)

Lux,
To be accurate in that internal resistance calculation don't you have to measure the voltage of the batteries under load, and not open circuit? If the cells don't sag at all then there's no problem, but if they do then I don't think that calculation holds because your Vbatt is open circuit and Vbulb is under load, you really need the Vbatt to be taken while under load to only capture the internal resistance of the light.
~Dougk


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## cat (Jan 16, 2008)

LuxLuthor said:


> If you are using a DMM that does not have a fuse when measuring up to 10Amps (i.e. is "unfused"), then *yes, subjecting it to higher than 10A current will damage your DMM. Don't do that 64458 current measurement.* This brand has unfused 10A like you describe. Note paragraph titled *Current Measurement - 10A DC *under *"Measurement For Managers."
> 
> Fire away on resistance measurement question.
> *



ok. When I saw it, I just had a rough idea of 10A being the sort of current draw we could be dealing with, but when I wrote that, I went to the Destructive Incan Bulb Tests thread and checked the charts for the bulbs I'll be using, and the only one that's over 10A is the 64458. 
I bought the DMM for motorbike wiring and it's ok for that, but now I'm getting a better understanding of why some DMM cost 50 times more. 

My resistance question is probably another reason...
I just went back to that link there and I think I found out why I couldn't measure resistors. The DMM only measures from 200 Ohms!  But...it seemed to be ok when I measured the resistance of a 10A blade fuse, at 0.4 Ohms. 
So, this DMM has some limitations, for what I want it to do. :laughing: ...I can see myself going to look at the Fluke specs again.

[edit:] I suppose all I really need to do is measure the voltage at the Kiu socket, without the bulb, to see what the bulb's going to get at startup.


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## LuxLuthor (Jan 17, 2008)

Kremer said:


> Lux,
> To be accurate in that internal resistance calculation don't you have to measure the voltage of the batteries under load, and not open circuit? If the cells don't sag at all then there's no problem, but if they do then I don't think that calculation holds because your Vbatt is open circuit and Vbulb is under load, you really need the Vbatt to be taken while under load to only capture the internal resistance of the light.
> ~Dougk



Doug, unless I am mistaken, that is not my understanding of what we mean by Vbat. If I put those batteries into my light and measured the voltage in the bipin holes (with no bulb), I get the same 7.94 volts because the circuit is not closed (with a load).

While testing the condition of NiMH/NiCad cells, that really should be done under a closed circuit with a resistor ("under load") to verify their performance. This is not the case with Lithium Ion cells, as their voltage tested just like I did gives a reliable indicator of their charge capacity.

However my post was demonstrating how much resistance related drop in voltage occurs from the entire flashlight & bulb setup when this set of batteries turns on the lamp. The Vbat you suggest "under load" would depend entirely upon what resistor induced load you used.


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## Kremer (Jan 17, 2008)

LuxLuthor said:


> However my post was demonstrating how much resistance related drop in voltage occurs from the entire flashlight & bulb setup when this set of batteries turns on the lamp. The Vbat you suggest "under load" would depend entirely upon what resistor induced load you used.



Yes, the entire light, *with* that set of batteries, because that method includes the internal resistance of the cells themselves also. That's not a bad thing to include because it is present in the light while operating. 
~Dougk


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## TKO (Jan 17, 2008)

Lux,

The first column on your Destructive Incan Bulb Tests spreadsheet is "*Applied* Volts". This would be *Vbulb*?


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## mr.squatch (Jan 17, 2008)

Sorry if I missed it, but is the voltage at the bulb different with a running bulb in it than if you were to check the voltage by pulling out the bulb and touching the two contact points with a dmm? 

Reason being, we have a 6d rop hi that is currently running on 6d alkalines. At 9v it should flash the bulb, but I figured since they were alkaline, they wouldn't put out all the power. Not to mention resistance or battery drain. I didn't think it was as bright as it should be, so I decided to dmm it and see. I was expecting to get 7v or so. I pulled the bulb, touched one to the center and one to the outer tube (+ and -) and clicked the switch. It read 8.92v which was really a surprise. It should flash the bulb at that voltage, and that's after running it for an hour or so on the batteries. Also at that voltage it should be way brighter. Thus my wondering if there is still a voltage drop to consider when the bulb is in and running. 

Thanks for the explaination lux good writeup

g


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## cernobila (Jan 17, 2008)

mr.squatch said:


> Sorry if I missed it, but is the voltage at the bulb different with a running bulb in it than if you were to check the voltage by pulling out the bulb and touching the two contact points with a dmm?
> 
> Reason being, we have a 6d rop hi that is currently running on 6d alkalines. At 9v it should flash the bulb, but I figured since they were alkaline, they wouldn't put out all the power. Not to mention resistance or battery drain. I didn't think it was as bright as it should be, so I decided to dmm it and see. I was expecting to get 7v or so. I pulled the bulb, touched one to the center and one to the outer tube (+ and -) and clicked the switch. It read 8.92v which was really a surprise. It should flash the bulb at that voltage, and that's after running it for an hour or so on the batteries. Also at that voltage it should be way brighter. Thus my wondering if there is still a voltage drop to consider when the bulb is in and running.
> 
> ...



I am no expert here but after LuxLuthor's instruction here, I measured both the voltage at the bi-pin holder without the bulb, which was 8.00V (2x C Li-ion cells in a 2x C Maglite) then I inserted the WA1111 bulb and measured at the pins with the light "on" and it read 6.9V, so I have lost 1.1V due to resistance in this set-up. So, if yours is 8.92V without the bulb then it most likely is running on about 7.7-7.8V while the bulb is "on". I guess that every bulb will give slightly different resistance level.


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## LuxLuthor (Jan 17, 2008)

TKO said:


> Lux,
> 
> The first column on your Destructive Incan Bulb Tests spreadsheet is "*Applied* Volts". This would be *Vbulb*?



Yes...and in my testing setup, I used very thick wire so I had as little resistance in my testbed setup as I could get.


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## LuxLuthor (Jan 17, 2008)

*mr.squatch, *if I understood what you did, it sounds like were still testing Vbat with your reading of 8.92V. You got that reading with no current flowing through your circuit yet.

Remember that a voltmeter is testing the POTENTIAL electrical driving force energy (measured in volts) that is *waiting *to move current through a circuit from the built up positive charged end of the battery to the negatively charged end of the battery. (Electrons actually move away from the negative end towards the positive end--reverse direction of current) However, when the DMM takes its measurement, for practical purposes, it detects this difference in charge potential with no current moving to get the reading. 

Think of measuring volts (with a DMM) like you would think of measuring water pressure. You could hook up a water pressure meter to the end of your garden hose that would "feel" how much water pressure is pushing on the meter (usually measured in "psi" - 'pounds per square inch'), without any water actually squirting out...which is like measuring Vbat.

That analogy is similar to a DMM measuring the potential electrical force waiting to pump electrons through a circuit. However, when you put the bulb in the socket and turn on the light, *now *it is like you are taking a voltage "pressure" reading of the side of the garden hose after opening the nozzle valve as the "water is gushing out of the hose." The "voltage driving pressure force" will drop to some degree. 

Said another way....you can feel how hard the garden hose gets in your hand when you close your spraying nozzle (it has a high pressure). Then compare to how soft the hose gets when you open the sprayer (it now has a lower pressure). That is similar to Vbat versus Vbulb....but you have to open the garden hose sprayer and let the water run (turn on the light with bulb in socket) to see the effect of using the light. 

In this analogy, the moving water in the garden hose is like the current of a circuit. A resistor is like a water valve that limits how much current can flow through it. The higher (stronger) the resistor (like a smaller valve size opening), the less current (water) can flow. If you lower the resistor (open the water valve wider), more current (more water) can flow.

LOL! I hope that was what you meant.


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## mr.squatch (Jan 17, 2008)

Good analogy, explained sufficiently  Thanks buddy. I'll have to do some more testing tomorrow. I was kinda nervous about running 6d's on the rop bulb but so far it's taken everything we've thrown at it. i'm thinking resistance fixes at this point are only going to hurt us. 

g


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## LuxLuthor (Jan 17, 2008)

To get the resistance lowering fixes optimzed for your setup, you almost need to be willing to blow a couple of bulbs, or find out from someone using the exact same setup to share their experience. That is why we hope for a regulated driver from AW.

A regulated driver will take a higher voltage from extra batteries (more batteries with higher voltage means more run time), and regulates the Vbat down to optimum Vbulb that you want it set to.


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## Patriot (Jan 17, 2008)

Lux does it again! Thanks a bunch for you informative and helpful threads Lux. I learn something every time.


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## drew2001 (Jan 18, 2008)

... again thanks for bringing light to the subject.


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## LuxLuthor (Jan 18, 2008)

Drew, my comments in this post are not intended to be confrontational, but rather trying to explain what it means to keep a simple topic--simple. 

I don't disagree with your purist view, but for the vast majority of flashaholics there needs to be a SIMPLE, relatively easy agreed upon manner of determining Vbat with and without the load of the flashlight, and your comments take this thread into an unnecessarily complex direction. Beginners don't want this level of nuanced precision. Most of them never heard of Ohms Law, and don't have a DMM.

This is a SIMPLE guide. My example voltage drop included all parts of the light at that snapshot moment in the life of the bulb & battery's charge/condition/age. *People use their batteries to power their lights, not a bench power supply, which 99% of the CPF members don't have, nor do they know what it is or care about it.* 

This discussion on Vbat is presenting the generally agreed upon manner of how people test the voltage drop with the light on and off. Taking your issues and testing requirements into account would leave the *beginner *unable to figure anything out, even on a simple level.

As I said in the beginning we are concerned with understanding the concept of voltage drop under load, and why a set of batteries that seems to have a voltage that looks like it will flash their bulb does not do that. There is a reason I just used the garden hose analogy.


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## drew2001 (Jan 18, 2008)

.


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## chuck614 (Jan 18, 2008)

Lux,
Could you give us newbies a similar explanation of current? How is amperage computed for cells wired in series and parallel? If I know my voltage is right for a bulb of a stated wattage, how do I determine the mAh rating needed to power that bulb? How does the "C" rating of a cell effect my choice in batteries? Thanks.


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## LuxLuthor (Jan 18, 2008)

chuck614 said:


> Lux,
> Could you give us newbies a similar explanation of current? How is amperage computed for cells wired in series and parallel? If I know my voltage is right for a bulb of a stated wattage, how do I determine the mAh rating needed to power that bulb? How does the "C" rating of a cell effect my choice in batteries? Thanks.



Great question. I'm gonna answer it in two posts. This post gives the SIMPLE concept; the next post will deal with the reality of what we actual see, and how to figure out what current you need for your bulb.

The simple answer is no matter how many cells you have *in series* (each of which increases the total voltage), the *mAh rating* (or "capacity") on the battery label *will stay the same*. 

In contrast, putting *2 cells in parallel* (called "2p") *will double the mAh* of the cell label; 3 cells in parallel (3p) will triple the pack capacity. 

The "C" (capacity) rating is literally the same as the battery label mAh rating. However, if you have 2 cells in parallel, that pack will now have a "C" rating that is also double that of the individual cell label.

Let's look at an Eneloop 2,000 mAh rated NiMH AA cell. That means that the cell has enough stored electrical energy to *theoretically *deliver 2,000 mA (= 2A) for an hour. Think of it like how many gallons of gasoline you have in your car tank. All NiMH cells are designated to have 1.2 Volts, so in this example the 1.2V 2,000mAh descriptions are called the cell's "_*nominal*_" values.

If you have 3 cells in series ("3s"), you would have 3.6V but still 2,000mAh capacity in your pack. Put 3 cells in parallel ("3p"), and now you have 1.2V, but 6,000mAh capacity in this pack.

If you have a light that needs 6 Volts, then you need 5 cells in series (5s), but if you want a higher capacity to make it run longer, or to put out more total current, then you need 10 cells in 2p5s (or 5s2p), where each of the 5 pairs will have their Negative ends connected together, and their Positive ends connected together like this pix. (_thumbnail_)



 



This *2p5s pack has 6V & 4,000mAh capacity*. However, as you might suspect, it is not as simple as the 2000mAh rating would imply. Not all batteries are the same quality. Some ratings are totally misrepresented (they have a much lower capacity than their label says). In any case, you are never going to be able to get the total number of mAh energy capacity stated on the label to be delivered.


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## LuxLuthor (Jan 18, 2008)

OK, now let's deal with the reality of your question.

A single 2,000mAh Eneloop theoretically can deliver 2 Amps for an hour....or deliver 1A for 2 hours. It can also deliver 4 Amps for 1/2 hour, or 8 Amps for 1/4 hour (15 minutes).

So a 2p pack with 4,000mAh provides 4 Amps for an hour, or 1 Amp for 4 hours. It can also provide 8 Amps for 1/2 hour, or 6 Amps for 2/3 hour (40 mins). 

Let's assume your light bulb is the Osram 64430 bulb that I tested in this report. You can see from the report that this bulb is going to need between 5.8 to 8 Amps of supplied current depending on the voltage we choose to make the tungsten filament glow brightly. Let's pick 6A that we want for our discussion here to keep the math simple.

In our example, this pack can provide 4A/6A x 1 hour (0.67 x 60 mins) which is 40 minutes. But does it really provide a full 40 minutes?

------------ Make sure you understand that part before reading on --------------------

In the real world, batteries do not provide the full amount of their mAh rating--some of their stated capacity remains in the battery. A better quality cell will provide closer to that mAh number on their label than a lower quality brand. In addition, some higher quality batteries are able to put out a much higher Amp rate than others at a given voltage. So how do you actually know what the performance of your battery will be vs. what the label says? We need to find someone who independently tests the battery.

It just so happens that we have a member here at CPF, Silverfox, who has done this very thing, and posted his results in the battery section. This thread lists the NiMH cells that Silverfox has tested. Basically, he charges up a battery, and then uses this CBA-II device to discharge it at a particular Amp rate. Then he charges it again, and selects another discharge Amp rate...and again....and again..... 

The CBA-II program displays the voltage of the cell and how long the battery runs at each of his Amp rates. This is called "Battery Discharge Testing" and Silverfox has done this at various Amp rates for each of the battery brands you see on his thread!!! (He has also done this for other types of batteries, such as Lithium cells in other pinned threads in the battery section).

Let's take a peek at his results for the Eneloop AA cell. If this is too complex, or beyond your level of interest, either stop here, or come back to it later...but there is no other way to figure this part out other than looking at a battery's discharge performance data. If you are like I was the first time I looked at one of these graphs, I had no idea what was going on....so let me explain it below.







OK, the Volts of the battery are on the left. The time the battery discharged its' capacity is on the bottom. Each curve line is color coded, and is the results of discharging a fully charged battery at a specific Amp load rate. He actually tested this battery 9 different times, at different Amp rates !!! The top curve is draining the battery at 0.5 Amps, and the bottom line is draining it at 10 Amps. The box summarizes the results.

Now let's look for the 6A rate we need for our Osram light bulb. It is the 3rd curve from the bottom, and it blends together with the 4A rate. If you look in the box, the 6A line says that it lasted 1.753 Amp-Hours, or 17.5 minutes. _ (The math is 1/6 x 1.753 Ah x 60 min/h = 17.5 min).

_So if a single Eneloop (2000mAh) can deliver 6 Amps for 17.5mins, then a 2p (4000mAh) pack will deliver 6 Amps twice as long for 35 mins !!! Note however that it will not provide the full 40 minutes that I mentioned at the top of this post.

If you looked carefully at the left Volts side of this graph, you also noticed that the 6 Amp curve spent most of the time BELOW the nominal 1.2 Volts for NiMH cells...probably averaging about 1.15 Volts (higher with fresh charge, and lower as battery drains)....so at 6 Amps, our 5s2p pack is really only going to be delivering an average of 5 x 1.15V = 5.75 Volts.

Another benefit of looking at this curve is that we better add another Eneloop in series to make it 6s (or two if we want 6s2p), because the cell starts dropping its voltage when more Amps are demanded of it. So with another cell (or two cells for 2p if you want more run time capacity), our voltage is a better match for our bulb at 6 x 1.15V = 6.9 Volts with our 6 Amps of current.

I think I better stop here, as I know this getting more complex. Understanding this post however, is really valuable to this hobby!


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## NA8 (Jan 19, 2008)

A website for even more fun.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmlaw.html#c2


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## chuck614 (Jan 19, 2008)

Thanks Lux! This is a great help. If we can't grasp the fundamentals, we'll never be able to grow in the hobby. This is a good foundation for all newbs to grow on.


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## rizky_p (Jan 19, 2008)

LuxLuthor said:


> That is Vbat which was what you asked. I showed measuring Vbulb in first post with the pincher leads.



what i meant was VBulb. 

thanks.


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## Anglepoise (Feb 15, 2008)

Thanks for explaining that applied voltage in the bulb tests was the measured voltage with the bulb on. Now I know why certain bulbs I was afraid might 'flash' did not.
Really appreciate the time and effort to helpthose of us new to incandescence.


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