## Calculating the LED resistor

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### Calculating the LED resistor

I've got a torch with 24 LEDs, the torch has three AAA-cells with 4.5 V. There was only one resistor for all LEDs, 0.015 Ohm. I would like to use the LED-array for a mechanical TV, the LED-driver has 12 V maximum. If I calculate a normal LED with 3.2 V and 20 mA, I would need for each LED a 440 Ohm resistor. Now it is only possible to use one resistor. How is it calculated? Simply 1/R, so about 2.3 Ohm? It would fit, as for 4.5 V I would need 68 Ohm resistors for each LED, or 1/68 = 0.015, as I would take only one resistor. DrZarkov
I think I've had a cranial implosion.

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### Re: Calculating the LED resistor

I've tried it the other way: If I calculate the resistor like for a normal row of LEDs, I would have the factors: LED: V=3.2, 20 mA for one LED, for 24 LEDS = 480 mA, Input V= 12 V. That results in a resistor of 18 Ohm. Sounds more logical to me. I'll give it a try with 18 Ohm. DrZarkov
I think I've had a cranial implosion.

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### Re: Calculating the LED resistor

Volker, I can't think of a resistor of 0.015 ohm. That is the resistance of a PCB trace.

I think that I know which torch you are thinking of. I have too such a torch, with 24 LEDs at the front and 5 at the top. I just opened that torch and observed that the resistor is 1.5 ohm, brown-green-gold. For the 5 LED cluster the resistor is 10 ohm, brown-black-black.

In the torch there is only 4 volt, so the LEDs are all in parallel. I measured the voltage over the LEDs to be 3.2 volts, while the battery voltage was 4.1 volt. So the voltage over the resistor was 0.9 volt and the current will have been 0.6 A. That is 25 mA per LED.

But if you have 12 volt available switch 3 LEDs in series. Then you get 8 series in parallel and the current goes down to 0,2 A. Then the voltage over each chain of 3 LEDs will be 3 x 3.2 = 9.6 volt, so you have to "dissipate" 2.4 volt. Then you need a resistor of 12 ohm.

However a LED driver transistor needs to drive a CURRENT, not a voltage. The BUZ11 single transistor LED-driver is designed for 100 mA peak white. With those white LEDs of 25mA max you may use 4 series strings in parallel. No less. That is about the CURRENT.

Then the voltage: The transistor needs about 5 volt as a minimum voltage for proper operation. If you design series of 3 LEDs then the LEDs alone need 9.6 volt, so in total you will need a supply voltage of 15 volt. This may be higher, not lower.

Then each string should have a resistor. This resistor should have 1 volt at 25 mA, that is 39 ohm. So 4 resistors of 39 ohm, one in each string. The resistors are there to balance the 5 currents. So you have just 12 LEDs in the cluster.

But if you have a higher voltage available, make series of 6 LEDs and then 4 series parallel. Then the total LED voltage will be 19.2 volt, plus the 1 volt of the balancing resistor: 20 volts. Then you need about 25 volt supply voltage.

If you don't want to modify the torch, you will need a new design of the LED-driver. The BUZ11 can handle the 0.6A, no problem. But you need a different resistor from the "source" to ground. I guess 1.0 ohm. There is no real need for a resistor between the LED-array and the drain (top) of the transistor, as there is no balancing of currents possible.

This is needed if you want to have the maximum light output out of the LEDs. If you use the LED-driver unmodified and the torch cluster unmodified, you need no resistor (or leave the 1.5 ohm in series) and you can run from a supply voltage as low as 9 volt (or 12 volt). Klaas Robers
Frankenstein was my uncle.

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### Re: Calculating the LED resistor

Thank you Klaas for the detailed answer, with that I can move on with the torch-LEDs and try out, how it will work best. DrZarkov
I think I've had a cranial implosion.

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Joined: Mon Feb 19, 2007 11:28 pm
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