0
\$\begingroup\$

I'm using the LED driver PCA9633 (8-pin.) After reboot, with default settings, the output of the LED pins are high, turning the LEDs on at reboot.

  1. Is there a good reason why I should want this?
  2. Is there a similar chip with default output state low?

Relevant default values:

  • SLEEP = 1 (low power mode, oscillator off)
  • INVRT = 0 (output logic state not inverted, value to use when no external driver used)
  • OUTDRV = 1 (LED outputs are configured with a totem pole structure)
  • LEDOUT = 00000000 (LED driver is off)

The 8-pin version doesn't have an OE pin.

The datasheet states:

If LEDOUT = 00, INVRT = 0, and OUTDRV = 1, LED state = VDD.

I'm using 12V LEDs, connected as shown at LED3 of the typical application:

typical application pca9633

I don't want to turn off the LEDs immediately after startup, the peak current at startup might cause problems. I'd like to avoid using additional hardware inverting the output signal.

\$\endgroup\$
1
  • \$\begingroup\$ Could you flip the LED's around and low side switch them instead? Then, the default state would be off. \$\endgroup\$
    – Klas-Kenny
    Commented Sep 12, 2023 at 11:25

2 Answers 2

2
\$\begingroup\$

You would normally connect the LEDs with a current-limiting resistor from the output to Vdd, so output = Vdd is 'off'. From the datasheet:

enter image description here

\$\endgroup\$
3
  • \$\begingroup\$ This will not work for me, as I'm using 12V LEDs connected as shown at LED3. I will add this information to my question, thank you. \$\endgroup\$
    – Jay
    Commented Sep 12, 2023 at 12:01
  • \$\begingroup\$ In that case, you're scuppered since the reset condition of INVRT is 0 and no OE pin. You would need an extra inverter or a high-side driver per pin or some other additional circuitry. If you use a ULN2803 you can get 4 drivers and 4 inverters in one package (one additional quad resistor network required for pullups). \$\endgroup\$ Commented Sep 12, 2023 at 12:08
  • \$\begingroup\$ Thank you, I will consider replacing the chip. \$\endgroup\$
    – Jay
    Commented Sep 14, 2023 at 7:59
1
\$\begingroup\$

Replacing the chip with a PCA9632 solved this problem. It's almost the same chip as the PCA9633, but the default value of OUTDRV = 0. The LED output is by default High-Z.

\$\endgroup\$

Your Answer

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

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