Calculating loss using resistor to balance

[acourvil]

I've seen a number of threads related to use if resistors to obtain the appropriate current for a given LED, and to balance current when using parallel circuits. I know there is power lost by doing this; could someone tell me how to calculate what is lost?

For example, in a sample circuit using a 4.5V source and multiple 5mm LEDs, with the LEDs having a Vf of ~3.3. If I use a 75ohm resistor, I get the desired current of approximately 20mA in the circuit; what am I losing to the resistor?

 

[Illum]

IIRC, the rate of energy loss or the power dissipation, P, in the resistor can be calculated from P = IV

 

[TMorita]

IIRC, the rate of energy loss or the power dissipation, P, in the resistor can be calculated from P = IV

Well, P=IE, but E=IR, so P=(I^2)R.

This way, you don't need to measure E across the resistor.

Toshi

 

[acourvil]

O.k., got it. The "loss" is simply the power consumed by the load created by the resistor.

So in the example above, P=(.0020)^2*75, or .0003 watts?

 

[lctorana]

Three points to note:

1 You slipped an extra zero in. It's actually 0.02^2 * 75 = .03W, or 30mW

2 For a 1.2v drop at 20ma, you need about 60 ohms rather than 75.
And if you use 3 rechargeable cells, you will get more conistent output with an even lower value resistor.

3. Also, you mentioned "multiple 5mm LEDs", so that 30mW per resistor.

Either you will use many 60 ohm resistors, or you can parallel all the LEDs and use a single resistor of much lower value.

Have fun!

 

[acourvil]

Three points to note:

1 You slipped an extra zero in. It's actually 0.02^2 * 75 = .03W, or 30mW

Damn, I never was good at keeping track of zeroes

2 For a 1.2v drop at 20ma, you need about 60 ohms rather than 75.
And if you use 3 rechargeable cells, you will get more conistent output with an even lower value resistor.

I was using round numbers. The actual voltage was a bit higher, and the 75 ohm resistor actually resulted in a 22mA current. But I appreciate the detail.

3. Also, you mentioned "multiple 5mm LEDs", so that 30mW per resistor.

Either you will use many 60 ohm resistors, or you can parallel all the LEDs and use a single resistor of much lower value.

Yes, I need to see how well matched the LEDs are. If they are well matched, I may simply use a potentiometer to dial in the total desired current.

Thanks for the pointers on this. I see that what I was having trouble understanding was actually pretty simple, but I just wasn't getting my head around it.