Need help with resistor question.

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Does using a resistor make the batteries last longer or is the energy just wasted. I guess I'm not sure it it keeps the voltage from comming out of the battery or just keeps the voltage away from the LED. Example: Using three 1.5 volt N batteries in series. I usually run the LED at 4.5 volts with no resistor. If I use a resistor to get the volts down to 4.0 would the batteries last longer? Thanks for any help on this.

 

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Brad,
The resistor cuts down on the current flow so it would lenthen the battery life. However, pushing current through a resistance causes heat, which is wasted energy. On most small circuits, this waste is minute and inconsequential. Using dropping resistors is only practical in very low power circuits, because you can lose nearly half the energy in the resistor. This is why more sophisticated power circuits have evolved. You can calculate how much power the resistor is using by multiplying the voltage dropped across it by the current flow in amperes to get watts.

 

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Thanks Chet,

So If I use two small 12 volt batteries in my 2 AAA minimag light with a 1K resistor I should be OK. I figured 24 volts - 4 volts for the light would be 20 volts devided by .02 mA would be 1K. According to your post, I should take voltage dropped (which is 20) times .02 mA = .4 watts
I don't know if using .4 watts is good or bad but I'll give it a try and see how it goes. Thanks.

 

[Mike]

Brad,

Using 2 12V batteries in series is a total waste of the 2nd battery if you are only lighting one LED. All the energy of the 2nd battery will just be dissapated by the resister. If you want to make the LED light longer using 2 12V batteries than you need to have the batteries in parrallel.

 

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Mike's right-
The thing is, the less voltage you need to drop across the resistor, the better the efficiency. As you can see from your calculations, you are wasting 0.4 watts to make heat and only using 0.08 watts(4 x 0.02) to make light. Only one sixth of the energy is used by the LED!
Chet

 

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Thanks Mike and Chet. I think I get it now.
I converted the Minimag last night. It's running at 50 mA. Nice and bright but I see I don't need the second battery if it's just being wasted. What I really wanted was longer light without changing the batteries.
I took out the resistor on the Solitaire with one 12 volt battery and put just the spring back in. I measured it at almost 60 mA. Hasn't burned out yet : ) I'll take the second battery out of the minimag tonight and just use a longer spring and smaller resistor. Thanks much. One thing about the minimag is you can leave the reflector in and focus the light.

 

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You know the funny thing is that despite all the warnings about excessive voltage and current, static electricity, and generally being delicate, I've found LEDs to be amazingly robust and able to take abuse. Maybe they're just making them better now. I know you can fry them, but I think one can tell by just being a little careful and taking ordinary precautions like feeling them during testing. It sounds like the internal resistance of your battery is what is limiting the current to 50 mils. It's nice not to have to use resistors.
Chet