# Power LED from 3.3V Power Pin without Resistor

I tried wiring one white LED to the 3.3V pin on a Pi Zero W with a 330 Ohm resistor but it's not as bright as I would like. If I wire it up without a resister I get a good amount of brightness. Is that a viable solution? Or should I try a smaller resistor?

If it matters, the power supply is a standard 5V 2.5A supply like this one. The LED is this one.

• I'd recommend 10Ohm resistor – Jaromanda X Apr 29 at 23:12

You need to consider that the forward voltage of a white LED is likely in the range of 3.0 V to 3.2 V (according to the linked article). Assuming the best case of 3.0 V and a resistor of 330 Ohms and using Ohm's law R = U / I we find that the current is about 1 mA and thus the brightness of the LED is rather low. Decreasing the resistor will help to some degree but be aware that this is borderline in any case. Say 33 Ohms will get you about 10 mA which might be ok. Best bet is to use the 5 V instead and design the resistor in such a fashion that the current suits the safe operational range of the LED.

Connecting a LED to a voltage source without a current limiting resistor is not a safe way of operating a LED. It is therefore not advisable to do so.

Find a more elaborate description of the calculations here; even though it is aiming at GPIO pins the physics still apply.

GPIOs on the Pi can deliver 8mA maximum by default. They can be configured to deliver between 2mA and 16mA as well.

So likely when you omit the resistor you are getting around 8mA. With the resistor you are limited to about 1mA.

I can't find a datasheet for that LED but 1mA is probably a bit low. Most white LEDs need a bit more than that. But it's also a good idea not to rely on the current limiting built in to the Pi. It's designed to protect the Pi, not regulate current to the LED. A smaller resistor is a better option.

• Pin drive strength (which you seem to refer to) is not a current limitation feature. Setting a pin to 8mA expected current doesn't mean you can pull out more, it only means the pin will be optimized for an 8mA load. – Dmitry Grigoryev Apr 30 at 11:51
• @DmitryGrigoryev yes that's why I don't recommend it. – user Apr 30 at 12:41
• I didn't say you recommended it. I'm saying that your answer is wrong in this regard: 8mA drive strength doesn't mean the current will be anywhere near 8mA. – Dmitry Grigoryev Apr 30 at 12:49
• I see what you mean. Actually it is correct, given a forward voltage of 3.0 from the LED. The GPIOs are rated for 8mA drive at 2.7V output voltage, so when I say "around" I'm basing it off that. – user Apr 30 at 12:56
• I was reading the OP's question different anyways 3V3 is not a GPIO pin, which I hope is what the OP is going for. – Ghanima Apr 30 at 13:04