Led circuit

>>934027
Each branch will get 12v as they are in parallel.
Voltage drop across the resistors = 12 - (3*3.4) = 1.8v
Resistance = Voltage/Current = 1.8/0.02 = 90 ohms

Of course, since you're using a battery the supply voltage will drop slightly but it shouldn't be much of an issue.

>>934027
You just calculate the resistor for one of the series arms, then put as many as you like of these arms in parallel.

Exactly like in your circuit diagram.

You don't need any extra resistors, just the ones that're already in the serieses.

>>934030
>>934031
Now i understand. Thanks a lot!

The other thing you should be aware of its the power rating of the resistor, resistor drops 2.4v@20mA so the power out dissipates is 48mW. Normal resistors are 1/4W so this is fine but if you used higher power leds (100mA possibly?) then you would need a fatter resistor.

>>934027
That's a nice exercise, but in real life just use BCR402W or similar.

>>934086
But your parts count just went up. That's a sense resistor plus the IC.

File: led.png (26KB, 516x322px) Image search: [Google]

26KB, 516x322px

>>934069
36mW is the power dissipated in each resistor since 20mA through 90 ohm is 1.8v. I^2*R

Still your point is valid as like you say heat generated in a dropper resistor can become serious business once you move onto higher power LED's/higher supply voltages.

>>934737
Nope, for 20mA no external components are needed. You just leave one pin floating, check the datasheet for details.

>>935696
i was working with vf = 3.2 which is worst case for range op specified
so don't try and school me you cheeky wee cunt!