Connecting Led to push-pull instead of open-drain

Question

I am using the PCA10056 Nordic dev board which integrate a nRF52840.

The following LEDs are wired to some of the MCU GPIOs.

Currently, my MCU firmware configures those GPIOs as Push-pull output.

But I wonder if open-drain mode could be a better solution than push-pull in this configuration ?

In push-pull mode, are they some limitations ? Especially when pushing the GPIO output to HIGH ?

Why do you believe that it's better to configure them as open-drain? — brhans, Jul 19 '22 at 15:11
Seems simpler because I only need to connect to ground to turn on the led, and open the circuit to turn it off. — The_Average_Engineer, Jul 19 '22 at 15:32

score 7 · Accepted Answer · answered Jul 19 '22 at 15:32

As long as VDD_PER is lower than or equal to MCU's logic supply there's no problem (NOTE: The reverse voltage across the LED shouldn't exceed the max ratings).

I know that it is certainly better to configure them as open-drain,

Maybe you should define "better" here. Using Hi-Z or P/P output configuration is no better than the other in your application.

score 5 · Answer 2 · answered Jul 19 '22 at 16:11

In push-pull mode, are they some limitations ?

Not really but Vo max might be an advantage to open drain.

With complementary MOS (CMOS) you can invert logic using LED's by choosing high or low side drive. For cables it also lowers impedance giving more immunity to high impedance stray AC fields preventing the dim issue during OFF.

With Open drain, you have the flexibility of choosing same, higher or lower pull-up V= supply. You can also wire them as "wired NOR". The cost saving is a trivial reason.

All switches are inverting so low side drive is common to use and in special cases, the Pch Nch current limits are not matched to provide an option for high current yet reduce the shoot-thru transition currents. But CMOS tends to be complementary and symmetrical by design.

Connecting Led to push-pull instead of open-drain

2 Answers2