0
\$\begingroup\$

I have a USB controlled device (Razer DeathAdder Chroma) that has an RGB LED on it and I would like to use that LED to control a 5050 RGB LED strip about a meter in length. Both the "input" LED and the strip are common anode with the "input" LED's + pin at just over 3V and the strip's at just over 12V.

I planned on following the adafruit RGB LED Strip tutorial with the gate on the MOSFETs being connected to the "input" LED. The problem I run into is which ground do I use? The ground from the 12V supply that's powering the strip or the ground pin on the USB device.

Pictures of the LED and a segement of the LED strip. enter image description here enter image description here enter image description here

Any advice would be greatly appreciated and thank you in advance!

\$\endgroup\$
3
  • \$\begingroup\$ Not enough information about how the device led is wired to give you an answer on how to trigger the mosfet. As to the ground question, you should be tying the two grounds together, so the answer is "both". \$\endgroup\$
    – Passerby
    Jul 30, 2017 at 0:33
  • \$\begingroup\$ Yes, as passerby says ground is intended to be one common value so tie the two together, low side switch your led cable with a logic level FET. Really need actual datasheets/schematics to assist \$\endgroup\$
    – sstobbe
    Jul 30, 2017 at 3:52
  • \$\begingroup\$ MOSFET datasheet: cdn-shop.adafruit.com/datasheets/irlb8721pbf.pdf Crude circuit diagram: photos.google.com/share/… \$\endgroup\$
    – Phillip L
    Jul 30, 2017 at 4:22

1 Answer 1

1
\$\begingroup\$

IN general you need a current sense amplifier or current mirror with gain ... which could be 1 BJT and 1 FET with R's for each. It could be highside sense differential V-controlled output current on the low side output , which needs 2 BJT's.

But specifics depend exactly on your V or I input and V & I output. RGB strings may be 12V or single with 5V using Rs drops from V+ and regulated with PWM on the (-) input so there is no ground except at the driver.

More details needed.

\$\endgroup\$
9

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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