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.

Honestly, I would just use a 100 ohm resistor, or higher. The LEDs will be bright enough for normal indicator lamps. If you want to learn more about calculating what resistors you need, I do a lot of examples in this video about halfway through.

## Discussions

Best Answer 3 years ago

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.

Answer 3 years ago

I should have added that knowing the answer is right is not half as important as knowing WHY the answer is right!

https://www.instructables.com/id/How-electricity-an...

Answer 3 years ago

I wanted to know that will this schematic work as in is this wired right? Oh and thx

Answer 3 years ago

yes it will work.. your welcome.

3 years ago

So you've gone for the circuit I recommended after all ?

Answer 3 years ago

Yeah hehe

3 years ago

Or maybe his problem is only with electronics and his evil twin was asking for help? ;)

3 years ago

I don't really need to know but why do you need multiple accounts for yourself on Instructables?

3 years ago

I just started using this schematic program thingy. I thought it was a battery XD

Answer 3 years ago

That's ok any technically trained person can understand your schematic..

3 years ago

Wait a second, how can my schematic be right. Don't parellel circuits use 1 resistor?

3 years ago

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.

3 years ago

Skip to 7:38 in the video, I go over an example with 3 LEDs across a 12V power supply.

3 years ago

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.

Answer 3 years ago

Honestly, I would just use a 100 ohm resistor, or higher. The LEDs will be bright enough for normal indicator lamps. If you want to learn more about calculating what resistors you need, I do a lot of examples in this video about halfway through.