resistance and current in coin cell led lights

i have a few questions. very sorry if these are obvious or stupid questions, i'm just missing some of the finer details about powering leds. i've done a lot of research to no avail and would really appreciate any advice on this:

white leds want 3.2-3.6 volts, but button cell keychain lights (like the photon clones) use no resistor and 2 batteries. so the leds are powered by 6 volts. isn't this likely to shorten the life of the led by a lot?

also, it seems that while testing my button cells voltage while powering an led, the voltage is around 2 volts, not 3. when not under load, the batteries test at about 3 volts. is this typical of all batteries? also, is this why the button cell lights dont kill leds quickly (4 actual volts powering the led)?

I'm misunderstanding something about voltage vs. current. with a circuit containing no resistor (photon clone), what will dictate the current? also, when an led rated at 3.7 volts is being under-powered by 2 volts, what dictates current then? i thought the resistor sets the current thru limiting, am i wrong?

do batteries have a maximum current output? if so does it differ with battery types? with no resistor will they output maximum current? the array wizard says i can use a 2 ohm resistor to feed 6 volts into a 3.7 volt led at 1000ma. that makes me think having no resistor at all would deliver even more current, so obviously i'm missing something here. also, are these calculations meant to be done with the batteries assumed (3v) voltage, or with the 2v the battery produces under load?

does anyone make a disc shaped resistor to insert between 2016 button cells in a photon clone? otherwise hard to fit one in there.

thank you in advance to anyone with advice!!

Great questions! I am looking forward to the responses!

Two common misconceptions can lead you astray when dealing with problems like this. The first is that a battery is a fixed voltage source. For some limited number of applications, you can sometimes get away with approximating it as one, but not in this case. The second is that an LED acts something like either a resistor or a constant current load. Both of those are very inaccurate views of what an LED is like. Once you have a better idea of the real nature of the battery and LED, all you need is Ohm's Law to see what's going on.

Let's start with the battery. A first approximation of a battery is a fixed voltage source in series with a resistor, the battery's internal resistance. Take a look at the data sheet for a CR2016 coin cell at http://www.panasonic.com/industrial/battery/oem/images/pdf/Panasonic_Lithium_CR2016_CR2025.pdf. Only one of the graphs gives a clue as to the internal resistance, the second one down on the left. Notice (at 20 degrees C) that the open circuit voltage is about 3.0 volts, and that at about 1.5 mA, the voltage has dropped 0.2 volt to 2.8 volts. So at 1.5 mA and the 50% discharge point, the internal resistance is (3.0 - 2.8) / .0015 = 133 ohms. In reality, the internal resistance isn't constant with load, and it increases as the battery discharges. But this gives you an idea of the ballpark value we're dealing with here.

Next consider the LED. The forward voltage of an LED is relatively constant, and changes little with large differences in current. It varies from type to type and individual LED to LED, but it's in the neighborhood of 3 volts or so at typical currents, so I'll use that for illustration here.

Now suppose that you put two of those (50% discharged) CR2016 cells in series and short circuit the output. You'd get about 6 / 266 ~ 23 mA. That's it -- it's absolutely all you can get from the battery. Let's say you connected them to an LED which has a forward voltage of 3 volts. You'd get a current of (6 - 3) / 266 ~ 11 mA. Although the current into the LED is half the short circuit current, that's going from a complete short circuit to a moderate load. So the internal resistance tends to make the battery produce a relatively constant current as the load voltage or resistance varies -- just the opposite of a constant voltage source.

When the battery is fresh, the voltage is a little higher and the internal resistance lower. So the LED draws more current. As the LED current increases, its forward voltage increases, but not by enough to have much of an impact on the current sourced by the battery with its internal resistance. In fact, LED heating causes its forward voltage to drop some. The battery's open circuit voltage drops as the battery discharges, and that plus the internal resistance end up providing an equilibrium current to the LED.

As the battery voltage declines and internal resistance increases with discharge, the LED current decreases.

So that's why you don't need an external resistor for these lights -- the battery's internal resistance is adequate to limit the current. Of course, the efficiency is only around 50% -- half the battery's energy is dissipated as heat within the battery. But using a low-resistance battery of the same voltage and external resistor wouldn't do any better -- it would just move the loss to the external resistor.

c_c

moon lander said:

Good questions moon lander, with involved answers, I will go one at a time;

i have a few questions. very sorry if these are obvious or stupid questions, i'm just missing some of the finer details about powering leds. i've done a lot of research to no avail and would really appreciate any advice on this:
​

white leds want 3.2-3.6 volts, but button cell keychain lights (like the photon clones) use no resistor and 2 batteries. so the leds are powered by 6 volts. isn't this likely to shorten the life of the led by a lot?
​

The current is the critical factor, which results from the applied voltage (like pressure). If you applied 6 volts with a power supply that can maintain the 6 volts regardless of the current that would result (constant voltage supply), yes you would burn the leds out very quickley (seconds).
Many lights have voltage or current regulators, not just resistors to set the proper current for the LED.

also, it seems that while testing my button cells voltage while powering an led, the voltage is around 2 volts, not 3. when not under load, the batteries test at about 3 volts. is this typical of all batteries? also, is this why the button cell lights dont kill leds quickly (4 actual volts powering the led)?
​

Yes, all batteries "sag" in voltage. The more current you draw, the more the voltage drops (instantly). This is due to internal resistance of the battery. Some cells have high internal resistance, like alkaline cells, some have very low like NiCad. Lithium cells can be made with both high or low internal resistance. Generaly, cells designed for slow discharge have high resistance and cells designed for fast discharge have low resistance. Why design with high resistance, because they typically can store more energy.
Energizer has some very good "curves" for there cells showning dischare at different current and the resulting voltage sag. The internal resistance of a 2032 coin cell is about 2 ohms. Very high, they are designed to discharge at only a few mA. See pdf with curves;
http://data.energizer.com/PDFs/cr2032.pdf

I'm misunderstanding something about voltage vs. current. with a circuit containing no resistor (photon clone), what will dictate the current? also, when an led rated at 3.7 volts is being under-powered by 2 volts, what dictates current then? i thought the resistor sets the current thru limiting, am i wrong?
​

The current is being limited by the resistance of everything in the circuit including the batteries, the LED, any resistor, and the wire (assume no voltage or current regulator present). The resistance of the LED Varies with the voltage applied, yes that is correct. Also the coin cells have about 2 ohms resistance on there own! So, if you want to calculate the current being drawn, you need a curve of the LED resistance verses voltage (available from the LED vendor) and the resistance of the batteries and any other resistance in the circuit (wire, connectors etc.)

do batteries have a maximum current output? if so does it differ with battery types? with no resistor will they output maximum current? the array wizard says i can use a 2 ohm resistor to feed 6 volts into a 3.7 volt led at 1000ma. that makes me think having no resistor at all would deliver even more current, so obviously i'm missing something here. also, are these calculations meant to be done with the batteries assumed (3v) voltage, or with the 2v the battery produces under load?
​

Batteries do have a current limit. Different for different cells. You can explode some batteries if discharged to quickely. Some batteries, like alkaline will generally not explode, but due to the high internal resistance, they will burn most of there energy up in heat inside the battery and the voltage will sag very low and you will not get proportionaly more current out. Batteries are very complex. You can get much info at differnet site, but energizer.com tech info has lots fo PDF documents on all kinds of cells type showing capacity, voltage, discharge curves etc.
I don't know anything about the "arry wizard". But, you must know what type of battery you are using to determine the voltage sag and effective voltage applied to the LED.

does anyone make a disc shaped resistor to insert between 2016 button cells in a photon clone? otherwise hard to fit one in there.
​

I have seen very large disc resistors, but not small ones. You could check at digikey.com, they have just about everything.

thank you in advance to anyone with advice!!​

Click to expand...

No problem. Please see answers above. I have spent several years on this type of stuff. Lots of info from vendors you may want to check out.
-BPH

Actually the Lithium coin cells measure in excess of 3v each new perhaps a pair would measure approx 6.4v new in series. The reason they do not fry the LED quickly is the batteries themselves under the current load sag in voltage due to their internal resistance under such load. Essentially it is like putting a seperate resistor inline with the LED to limit current. I have measured perhaps 60ma or so off new batteries to a 5mm white LED but this doesn't last long nor do most people run coin cells lights more than a few minutes at a time.

What has been mentioned so far, so I won't rehash things. One point worthy of noting is this is one of the reasno why the coin cell lights that run off of single 2032 cells deliver much longer battery life. One is that those batteries have significantly more enregy content than 2016s. Also, one 2032 battery has about a quarter of the resistance of two 2016 batteries in series. But, since the votlage of the 2032 is a closer match to the ~3 forward volts of red LEDs, this means that it won't cook the LED even without a resistor -- and at the same time that not nearly as much energy will be wasted heating up the battery due to resistance.

This also means that it's not really possible though, as was suggested in another thread, to extend the battery life of a coincell light by dropping in two 2032 batteries (short of adding in an external resistor... unfortunately a regulator circuit coudln't really be made to fit inside a coincell light) -- the lower internal resistance of these batteries would cause them not to sag as muhc, and thus to kill the 5mm LEDs. I tested two 2032 batteries with a Cree LED and found they drove it at 300mA, compared to 70mA with two 2016s!

wow thanks all for the info, its all coming together now... i just tested an led hooked up to 2 cr2032s and it drew 100ma and began to drop immediately, about 2ma/second.