I have a board with digital output that is based on open collector Darlington driver (providing up to 500mA) and I want to use that output to control WS2812x LED strip. What should the circuit look like in order to provide control signal for LED strip from such output?

  • \$\begingroup\$ Difficult. May we please know which chip it is (ULN2003?) to even start thinking if it is even possible and if it is, then start thinking what the circuit would look like? What is driving the driver - an MCU pin, which MCU and which pin? The pin is important because it determines if MCU hardware can be used to generate the waveform or if it must be done manually in software. \$\endgroup\$
    – Justme
    Jun 22, 2023 at 17:40
  • \$\begingroup\$ If MCU and strip have different power supplies just connect together the grounds. And resistor 10 kOhm from collector to +5V of strips PS and collector to data input of strip. \$\endgroup\$
    – user263983
    Jun 22, 2023 at 20:46
  • \$\begingroup\$ A Darlington circuit may be too slow to generate the required data rate. \$\endgroup\$
    – Jens
    Jun 23, 2023 at 21:12

1 Answer 1


WS2812 is chip, which can be controlled by applying special code, name of protocol is NRZ. Logical level of control signal is Vcc+/- 0.5V, there is Vcc is strip's PS. So open collector is a good way to form control signal. Especially the MCU has different voltage. Just connect collector to data input and pull-up resistor 10kOhm to strip's PS +5V. And connect grounds of power supplies.


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