1
\$\begingroup\$

I have a project in mind using an RGB LED (here: http://amzn.com/B007RO9X82).

I would like it to be powered by USB. Since the LED is rated at 3.0-3.6V, I understand I'd need a resistor. What resistor would I need and would it throw much/any heat in this situation? The feed wiring and resistor would be enclosed in a hole drilled in a wood plaque...don't want to create a fire hazard.

USB Power specs from WikiPedia: [5.00±0.25 V (pre-3.0); 5.00+0.25-0.55 V (USB 3.0) Max. current 0.5-0.9 A (general)] what resistor would I need? I assume if I sized the resistor for standard USB and it were to be plugged into a charge port (5A), that would cause trouble, correct?

\$\endgroup\$
  • \$\begingroup\$ Thanks guys! Please forgive my ignorance, but would connecting the LED to a 5A USB Charge port effect what resistor would need to be used, or is it voltage and not amperage that dictates the resistor? Also, I'm thinking of buying a multimeter to help me see this stuff with my own eyes rather than just on paper. Which of these do you think is better? amzn.com/B007FZFTZO or amzn.com/B003YHCFTU \$\endgroup\$ – Davidnwa Nov 29 '14 at 12:54
2
\$\begingroup\$

$$LED\;current = \frac{input\;voltage - LED\;forward\;voltage} {LED\;resistor}$$

$$I_{LED} = \frac{V_{USB} - V_{f_{LED}}} {R_{LED}}$$

Let us assume a current of 20mA for the LED.

$$0.02A = \frac{5V - 3.0V} {R_{LED}}$$ $$R_{LED} = 100\Omega$$

$$0.02A = \frac{5V - 3.6V} {R_{LED}}$$ $$R_{LED} = 70\Omega$$

I would use a 100 Ω LED.

At 20mA per LED, you could have \$ \dfrac{0.5A}{0.02A} = 25\$ LEDs in parallel.

The kit is supplied with 200 Ω resistors. Using them you could have 50 LEDs in parallel.

\$\endgroup\$
1
\$\begingroup\$

Those LEDs come with 200 ohms resistors with the pack. Assuming the USB's 5V as Vin with 3.3V drop (mid value, spec states 3-3.6V) on the LED, the current passing through the 200 ohm resistor and the LED would be 8.5mA which may work just well. IF not, because it is too dim, try reducing the resistance to 100 or 80 ohms as geometrikal suggests, which would increase the current to 17 or 21mA respectively.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.