# Last and final LED question? Answered

The picture says it all.

-Prickly Potato

Tags:

## Discussions

Lets try to work it out.

The LEDs are in parallel so effectively the voltages are all the same but the currents will be added.

3.3 v is required for the LED and the battery supplies 4.5 volts, (by the way your symbol is for an AC supply I assumed this was wrong), So 4.5 - 3.3=1.2 volts must be dropped across the resistor.

So you need 20 mA so using Ohms law V=I x R we can work out the resistance that will be needed to pass 20 mA at 1.2 volts.

R=V/I = 1.2/0.020 amps = 60 ohms

in the standard set of resistors there isn't a 60 ohms so we pick the closest 68 ohms. Actually I think 62 ohms is closer and in the set.

So your supply will give you 4.5 volts at a current of 0.02 x 3 = 0.06 amps or 60 mA

Ohms law is about the most useful thing in electronics. there are of course web sites that have calculators to work all this out for you.

Yes it will work fine. I actually just noticed the power supply is AC, that is fine, the diodes will only light up the one half of the AC cycle, and there may be noticeable flickering. You could use a full wave bridge rectifier and a smoothing capacitor.

Again, it should be noted that mAH is the wrong rating, esp. since AC suplies are generally wall adaptors.

LEDs are not rated in mAh (capacity), I think you mean mA, (the current.)

I assume you want the voltage drop across all the LED to be 3.3V, and the current needs to be 20mA. To do this, Just use ohms law, and pretend like the LEDs are small 3.3V batteries connected backwards, so that way we can figure out the voltage across the resistor is, and then we need some sort of resistor, lets pretend that we are not sure, so 68 ohms might be correct, it might be wrong.

Ohms law: V / A = R, or V= I*R

(Vbattery-Vled) / A = R

Plug in the numbers: (4.5V-3.3V) / 0.02 A = R

1.2 / 0.02 = 60

R = 60 ohms to limit the current to 20mA when there is 1.2V across the resistor.