LEDs - Expanding Battery Capacity

Greetings! We want to expand the existing battery capacity on two small LED lights to run for a longer time on C or D batteries for a children's project. I feel sure we probably will need resistors to protect the LEDs with the larger batteries.

I am not sure how to compute the resistor size since I can't find the exact specs on the LEDs (they're encased in the plastic lights.) I know Ohm's Law, I'm just not sure how to apply it here. I also have a multimeter, a soldering iron and can follow instructions :naughty:

To make things easier, we thought we'd stick to the same, existing voltages:

LED light #1 runs on 3 LR44's => ~ 4.5 volts REPLACE WITH (3) C or D batteries in series => ~ 4.5 volts

LED light #2 runs on 1 CR2032 => ~ 3.0 volts REPLACE WITH (2) C or D batteries in series => ~ 3.0 volts

What's the best way to approach this? I've searched and Google'd it but I don't see anything like this. I would hate to blow up a child's favorite light.

Many, MANY THANKS!!!!!

This depends on the LEDs as some have a forward voltage drop of ~3.2v and others around 2v. What are the color of the LEDs? Since they are running on small button cells it is safe to assume they are low current 20ma type LEDs. To calculate the resistor is straight forward. Take battery voltage - LED forward voltage divided by the current. For example, if the first LED has a Vf of 3.2v, then: (4.5-3.2)/.02=65 Ohms (68 ohms is closest common resistor value). To make sure the resistor power rating is correct: (battery V - LED Vf) times current=power in watts, therefore: (4.5-3.2)*.02= 26 milliwatts, so even an 1/8w resistor will work.

Two very useful pages for calculating such things:
http://led.linear1.org/1led.wiz
http://led.linear1.org/led.wiz

The site listed is very imformative but in this case I'm not sure it is going to be much help to the OP without knowing the forward voltage and current rating of the lights in question. JohnR66 has given a pretty good estimate above.

Would it be possible to replace the LED with one of known values without wrecking the light? In that case you could pop into any electronics components store and pick up a couple of 5mm White replacements.

I believe you are right to think you need resistors when changing to a larger battery. Both the LR44 and the CR2032 have significant internal resistance that can protect the LED from overcurrent. C and D cells have much lower internal resistance, and can easily overcurrent the small LEDs you are talking about.

The problem is that the voltages in the circuit are not necessarily anything like the battery voltages, because of the internal resistance. It would be much better if you could measure the voltage across the LED while running with the recommended batteries (at full charge). That would give us a much better basis on which to calculate resistor values.

JohnR66 said:

This depends on the LEDs as some have a forward voltage drop of ~3.2v and others around 2v. What are the color of the LEDs? Since they are running on small button cells it is safe to assume they are low current 20ma type LEDs. To calculate the resistor is straight forward. Take battery voltage - LED forward voltage divided by the current. For example, if the first LED has a Vf of 3.2v, then: (4.5-3.2)/.02=65 Ohms (68 ohms is closest common resistor value). To make sure the resistor power rating is correct: (battery V - LED Vf) times current=power in watts, therefore: (4.5-3.2)*.02= 26 milliwatts, so even an 1/8w resistor will work.

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Many, MANY THANKS for your response! The LEDs are all white. I really appreciate your giving good estimates to guide me and the sample numbers. I will work through your suggestions and formulas carefully and post the results here to make sure I understand it all correctly.

I want to try this with several different battery sizes, C, D and AA and will also experiment with the breadboard. I have some of my own LED lights that I'd love to run on Cs or Ds. I have to believe that others have gotten stuck with fully drained CR2032's and wondered how they could run their LED light with what they had in their electronics parts box.

Is it true that I can shine a bright light on the LED and measure the voltage across the battery clip to estimate the LED's forward voltage drop, or is that just silly Internet hockum?

Again, many, MANY THANKS for your reply!

Only the cheaper LED lights rely on the internal resistance of the battery to limit current. Are the LEDs connected via a PCB of some sort? If so it's likely resistors are integrated with the circuit, and you can upgrade the batteries without a problem. In fact many LED lights have a resistor replacing the LED pin, or integrated within the die itself.

If the LED light doesn't have a PCB, or you need to remove it, you don't need to calculate the forward current empirically. Just use the datasheet of an LED of similar size (3mm, 5mm, SMD, etc) and colour and use one of the LED resistor calculator website posted above and Bob's your father's brother.

For the 3 LR44 light, yes, use: Voltage supplied - Voltage required / Amps = Ohms. For the 2032 light, I wouldn't bother with a resistor at all. A 2032 is a lithium cell and doesn't have nearly the voltage sag of 3 alkaline or silver oxide LR44 cells. 3 volts from 2 Cs or Ds may be about right. If that light has a PCB, it may be a buck (lowers the voltage) or a boost (increases the voltage) circuit or it may have a resistor (lowers the voltage). If you add any resistor to a light with a buck circuit or with a resistor, it will be dimmer. If you lower the voltage below the operating voltage, it won't work at all. If you add a resistor to a boost circuit, you might get the same brightness but you'll make the light less efficient and the batteries won't last as long. The newer l.e.d.s may have a forward voltage of 3.2 volts. If it is an older l.e.d. like a Nichia CS, it could be around 3.6 volts. The 3 LR44 light could uses a 3.6 volt l.e.d. while the 2032 light could use a 3.2 volt l.e.d. I can't really provide further advice without more information. Good luck.

Hooked on Fenix said:

For the 3 LR44 light, yes, use: Voltage supplied - Voltage required / Amps = Ohms. For the 2032 light, I wouldn't bother with a resistor at all. A 2032 is a lithium cell and doesn't have nearly the voltage sag of 3 alkaline or silver oxide LR44 cells. 3 volts from 2 Cs or Ds may be about right. If that light has a PCB, it may be a buck (lowers the voltage) or a boost (increases the voltage) circuit or it may have a resistor (lowers the voltage). If you add any resistor to a light with a buck circuit or with a resistor, it will be dimmer. If you lower the voltage below the operating voltage, it won't work at all. If you add a resistor to a boost circuit, you might get the same brightness but you'll make the light less efficient and the batteries won't last as long. The newer l.e.d.s may have a forward voltage of 3.2 volts. If it is an older l.e.d. like a Nichia CS, it could be around 3.6 volts. The 3 LR44 light could uses a 3.6 volt l.e.d. while the 2032 light could use a 3.2 volt l.e.d. I can't really provide further advice without more information. Good luck.

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I agree, but for slightly different reasons. For the LR44 light, absolutely you need a resistor, as 3 C cells will be able to provide well over an amp into the LED, which will fry it for sure. The LR44 cells have so much internal resistance that they can't possibly damage the LED.

For the 2032 light, you probably don't need a resistor. But the reason isn't that the resistance of a 2032 is lower than that of two alkaline C cells in series (it isn't). The reason is that the voltage of either a 2032 or two alkaline cells probably isn't high enough to overdrive the LED. Very few (if any) white LEDs can be overdriven at 3.0V. And no 2032 cell and few alkaline cells (2 in series) can provide a lot of current at voltages over 3.0V

The internal resistance of a brand new, high quality CR2032 cell is in the range of 5-10 ohms. The internal resistance of a brand new, high quality alkaline C cell is under 1 ohm, so two in series are under two ohms. D cells are even lower. The reason alkaline cells have a (well deserved) bad rep for voltage droop isn't because of initial internal resistance. It's because during discharge, the voltage drops a lot. Two alkaline cells (in series) or one lithium cell start out at around 3V and would be considered dead (by the manufacturer) when the voltage drops to 1.6-1.8V. The difference is that at 90% discharge, the lithium cell is probably at 2.3V, while the alkalines are more likely at 1.9V or less. At 80% discharge, the difference is even greater.

When considering whether the batteries will burn out your LEDs, you need to be thinking about brand new, high quality cells, which are the most likely to burn out your LEDs. A poor quality or partly discharged cell is not worth thinking about for this situation.

The last batch of Cree 5mm LEDs I bought had a typical Vf of 2.95 volts at 20ma and I hand picked a few from the batch that had a Vf of 2.85v @ 20ma. The reason (in part) for this is because Cree uses a relatively large die in their 5mm LEDs. The die I believe they are using in the white, blue and green 5mm LED is the TR260 which is a 50ma part. Continuous current is limited because of the inability of the 5mm package to dissipate much heat. So, the chip is under driven at 20ma.