0
\$\begingroup\$

I have a strip of RGB LEDs. If I attach a multimeter between the common pins and another pin like the red power, I measure -2.4V . I think this is what is described as common anode.

I have an L293D motor driver relay chip, but its 4 outputs must be positive voltage, so I would have to connect the LED's common pin to the driver output, and all the LED power pins to the ground. This prevents me from controlling red, green, and blue independently.

I would like to find a relay/driver chip that supports multiple independent outputs and supports negative voltage. I also want to be able to use pulse width modulation (PWM) to control brightness. I am pretty sure it would be a bad idea to wire this directly to the pins of a microcontroller since it seems to draw 200 mA when I measured the red wire.

Something like the DS3680 supports negative voltage, but the schematic on the datasheet seems a bit complicated, and I do not have enough training to guess why.

If I search in Relays>Solid State Relays at digikey, the Voltage-Input criteria only lists positive numbers. How would I search for a relay that supports negative voltages, and allows the supply voltage (-2.4V) to be smaller than the logic voltage (3.3V)?

\$\endgroup\$
2
  • \$\begingroup\$ Most LED strips include resistors. But does yours include resistors? Or does it have to be current driven? 2.4V is not a standard voltage for LED strips. \$\endgroup\$
    – bobflux
    Commented Jul 27, 2022 at 19:01
  • \$\begingroup\$ I checked again and it looks like 3.5V. Maybe I had the remote control on the existing controller set to something other than pure red. Considering how non-linear this mess is, it could be something else that I don't understand. \$\endgroup\$
    – Mutant Bob
    Commented Jul 27, 2022 at 19:19

1 Answer 1

2
\$\begingroup\$

You don't need support for negative voltages. If you attach the positive lead to common anode, and negative lead to LED pin, you'll read +2.4V.

You do like any other LED driver, you give a positive common, and pull current via each LED down to ground.

You can do that with three of the L293D channels and leave the fourth unused.

\$\endgroup\$
1
  • \$\begingroup\$ It did not occur to me to use the driver chip's independent outputs as configurable current sinks. \$\endgroup\$
    – Mutant Bob
    Commented Jul 27, 2022 at 19:11

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.