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if I have a 3.6 volt battery should I use a 180 ohm resistor to get 20 mA to a 3.6 volt LED? I got that answer from 3.6 = (180)(0.02)

I'm asking this because online LED calculators disagree with me and say to use no resistor. also the battery might output 4.5 volts when extremeley fully charged, so I want the LED to have protection of not blowing out.

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I'm afraid the online calculator is right. Your mistake was to use value of 3.6V, this is the voltage dropped across the LED. To calculate the resistor you need to subtract the LED forward voltage drop from the battery voltage and divide by the required current (20mA). In this case it would be (3.6-3.6)/0.02 = 0 Ohms

The voltage(4.5V) of the fully charged battery is likely to be the no load voltage, it will drop when you apply a load to it. However, you need to ensure that the voltage of the battery (under load) doesn't exceed the maximum forward voltage drop of the LED (Vmax for an LED would be, typically, about half a volt more than Vforward (4.1V in this case)).

To recap: if the maximum battery voltage(under load) exceeds the maximum forward voltage for the LED then you would need to add resistor.

Cheers,

Pat. Pending

Thanx, that helps a lot. So I should use 4.5 - 3.6 = 0.9 0.9=(45)(0.020)

45 ohms?

Yes the calculation is fine, but are you sure the load voltage is actually 4.5V. It might be worth connecting the battery up and measuring the voltage under load. Cheers, Pat. Pending

I'm no electronic whiz, but that seem alright by me. And yes, the resistor would provide some kind of protection to the LED.