# Can I put a resistor in parallel with LED to limit current consumption?

I want to power one 3V LED with a 3V source. Let's say I'm using a 10 Ohm limiting resistor, and the LED eats 20mA of current. Using a CR1225 with 50mAh capacity that will last for about 2.5h.

What if I connected another 10 Ohm in parallel with the diode + limiting resistor? The voltage across would still be 3V - enough for the LED - but the current going through the LED would be halved. What about the total power consumed by the circuit?

I don't mind if the LED gets dimmer, I'm ok running 10mA or less through it. All I want is to increase the amount of time it emits light.

Would that work?

• A 3V LED stands a good chance of not lighting up at all on a 3V source. Post a data sheet for your LED so we know what we're dealing with. Dec 21, 2017 at 13:52
• "but the current going through the LED would be halved" not correct. According to battery manufacturers, CR2032 (CR1225 have the same technology) has an output resistance of about 10ohms. With only LED and resistor the current will be around let's say 3mA (source this datasheet), while with the 10 ohms in parallel the voltage from the battery will be 1.5V (so current in LED = roughly 0). And you are lucky if it runs 10 mins Dec 22, 2017 at 14:02
• @frarugi87 where do you see 10ohms in that data sheet? Dec 22, 2017 at 14:13
• @Trevor you have to guess it from the graphs. For instance, for the last one, when you increase the output current from 0 to 100mA the voltage drops from 3 to 2V; a 1V drop at 100mA is equivalent to a 10 ohm output resistance. This value "seems" consistent with the other measurements (0.7V at 50mA - 14 ohm, 0.25V at 20mA - 12.5 ohm). This is just an order of magnitude, not a precise value Dec 22, 2017 at 15:18
• @frarugi87 ah.. thanks.. I thought I was going blind.. or at my age..blinder. ;) Dec 22, 2017 at 15:42

If you add the resistor across the battery, so it's in parallel with your existing circuit it will draw an additional 300 mA (I = V / R), assuming the battery is able to supply it. This is not what you want.

You could reduce the current draw by adding the additional resistor in series with your current circuit.

• @pipe, I have edited my answer, thanks for your comment. Dec 21, 2017 at 13:57
• If I added another resistor in series with the diode - and decrease the current draw - would that make the battery last longer? It seems to me that the additional resistor would use some power as well. Dec 21, 2017 at 14:19
• @Alichino Yes, if you add another resistor in series it will make the battery last longer. Your 50mAh battery can supply 50 mA for 1 hour, or 25 mA for 2 hours, or 12.5 mA for 4 hours etc. Reducing the current draw will increase its life. Dec 21, 2017 at 14:21
• @Alichino you need to read up on how resistance and power are related. More resistance = less current = less power. Think of your battery like a dam letting water out through a pipe... if you restrict the pipe more (add resistance) the water flows out of the dam more slowly... Dec 21, 2017 at 16:03

If you add another resistor in parallel with the LED and its limiting resistor it will not change the current going through the LED, just draw more current from the battery. That MIGHT drop the battery voltage a little bit which would reduce the LED current, but that really is a secondary effect.

You effectively have two independent circuits attached to the battery. As such, the current taken from the battery is increased by the 300mA that will go through the 10R shunt, for a total of 320mA. Not for long from a 50mAh battery though.. less than 10 minutes.

If you want it to last longer, increase the 10R until you cannot accept the dimmer brightness any more.

BTW: Running a 3V LED from a 3V supply is generally a bad idea.

No, that won't work.

The battery will be empty sooner.

Maybe you're (wrongly) taking the 50 mAh from the battery into account. A battery does not provide current, it provides a voltage. The current only flows when a load is connected.

The load (LED + series resistor) then determines the current.

The CR1225's 50mAh indicates how much current it can deliver for how much time, it is not the maximum current that you should be loading it with. A Fresh high quality CR1225 might deliver 100 mA if your load has a low enough resistance. That will deplete that poor CR1225 within an hour though.

Placing a resistor in parallel is generally a bad idea. It is like making your car go slower by breaking while not releasing the gas/accelerator.

To make the LED last long:

Use a RED LED as there require the lowest voltage meaning the LED will still light up as the battery depletes.

Use a LED with a high efficiency. Some LEDs are quite bright at a low current of only 1 mA. That will make the battery last for a few days.

• RE: A battery does not provide current, it provides a voltage. // That is wrong. A battery is supplying both the voltage and the current. That is important because the voltage that a battery can supply depends on the current draw (and how much the battery has discharged overall).
– MaxW
Dec 21, 2017 at 21:10
• @MaxW Then explain why the voltage is printed on the battery but almost never the current. Yes a battery does supply a current but is does not force a current. It forces a voltage and when a load is connected a current can flow. The 1st order behavior of a battery is a voltage source. Dec 21, 2017 at 21:38
• +1 for answering a yes/no question with a leading yes/no answer. Dec 21, 2017 at 22:41
• @Bimpelrekkie - My car battery is dead. I connected 8 hearing aid batteries in series each of which supplies 1.5 volts. My voltmeter says 12 v for my homemade battery pack. Why won't my battery pack start my car?
– MaxW
Dec 22, 2017 at 0:39
• @MaxW Your voltmeter does not say 12 V when you are actually starring the car. You're treating it as the same situation which it is not. The voltage won't be 12 V due to the internal resistance of the batteries and you know it. But since you think batteries supply current you can just connect the hearing aid batteries in parallel until enough current flows to start the car. Dec 22, 2017 at 6:51