DRV8825 nFAULT pin operation and schematic - Page 1
Forum section: electronics / beginners

author: newtekuser
author_rank: Frequent Contributor
timestamp: April 04, 2024, 12:59:22 am
content: Hi AllI'd like to leverage the nFAULT pin in the DRV8825 motor driver to light up an LED whenever an overcurrent or over-temperature condition occurs. My understanding is this pin needs to be driven high by a pull-up resistor, otherwise it will never light up.Does my schematic look correct to accommodate this?Thanks!

author: DavidAlfa
author_rank: Super Contributor
timestamp: April 04, 2024, 02:59:18 am
content: I think so. Why not make a single thread like "Help with DRV8825 design" and post everything there?I don't think flooding the forum with threads for every little thing is beneficial?

author: newtekuser
author_rank: Frequent Contributor
timestamp: April 04, 2024, 03:41:42 am
content: Quote from: DavidAlfa on April 04, 2024, 02:59:18 amI wasn’t planning on starting other threads on this subject but your suggestion makes sense

author: pcprogrammer
author_rank: Super Contributor
timestamp: April 04, 2024, 06:17:57 am
content: nFault is an open drain output. Only when you need the signal else where, like as an input for a MCU, the pull up is needed. For the LED it does not matter. It will go off when the nFault is not active and on when it is active.The datasheet does not give an absolute maximum current for the nFault pin, but on page 6 it shows test conditions 5mA for low output and 3.3V for high out leakage current. It might be that these are also limits to obey.For a modern LED, current can be fairly low to still be bright though.Another thing to keep in mind is that the chip when in fault due to over current it needs a reset or motor voltage disconnect to return to normal working.

author: newtekuser
author_rank: Frequent Contributor
timestamp: April 05, 2024, 12:58:21 am
content: Quote from: pcprogrammer on April 04, 2024, 06:17:57 amFrom what I've read about open-drain, the output (nFault in this case) is floating when it is off otherwise it's pulled to ground when it's on. We want to pull it high to avoid an undefined state which we want to avoid when connecting to an MCU. Is this correct?

author: DavidAlfa
author_rank: Super Contributor
timestamp: April 05, 2024, 02:43:30 am
content: That's how you handle a open drain output, yes.

author: newtekuser
author_rank: Frequent Contributor
timestamp: April 05, 2024, 03:56:43 am
content: Now I’m a bit confused about the schematic. If nfault is an output pin, shouldn’t I connect it in series with the LED then ground, instead of 5v or 3.3v?I mean, if it’s an output, shouldn’t the IC source the current?

author: DavidAlfa
author_rank: Super Contributor
timestamp: April 05, 2024, 04:03:08 am
content: It's an output, but open drain means it can only connect to ground, in high state it's left floating.This would be a normal push-pull output:Open drain:The purpose of this is compatibility, you don't have to worry about logic voltage (For example, interfacing a 5V output to a 3.3V mcu).As long as you don't exceed Vmax, it'll work with 1.8V, 3.3, 5V by just adding a pullup resistor to the same voltage level.

author: newtekuser
author_rank: Frequent Contributor
timestamp: April 05, 2024, 05:42:14 am
content: Quote from: DavidAlfa on April 05, 2024, 04:03:08 amThank you, your schematic helped! I get it now... the LED is powered from the 5V supply and at the other end, the ground (source) in the transistor completes the circuit which makes the LED turn on.

author: pcprogrammer
author_rank: Super Contributor
timestamp: April 05, 2024, 05:56:07 am
content: Quote from: newtekuser on April 05, 2024, 05:42:14 amCorrect, and that is why you only need a pullup when you connect it to an input that needs defined levels. The LED does not care if that pin floats. It will only turn on when the pin is connect to ground via the transistor.