I have a question about how best to completely power off the A4988 carrier board.
On the board which the carrier will be integrated with, there’s a 5v buck regulator based on the LM2675, which will be switched on and off using a pushbutton controller.
VMot on the A4988 carrier board is connected to the main board’s Vin before the LM2675.
So, I can cut the power to the A4988’s VDD easily enough by switching off the 5v regulator – but will the driver elements of the A4988 continue to draw current if the logic is off?
I’ve got a pull-up resistor on the A4988’s reset pin connected to VDD. With the 5v regulator switched off, would this be enough to put the chip into reset mode and disable it….or do I need to manually sever the VMot line with a mosfet or similar to ensure there is no current draw?
I can provide a schematic if that would help.
Thanks in advance for any advice!
I tested it here with one of the A4988 carriers I had on my desk, and was not able to notice a draw of more than 1uA under those conditions. However, we are not entirely sure what is going on inside that IC with just motor power and a stepper motor connected. All of the information we have available for it can be found inside its datasheet. So, while I do not expect there to be any significant amount of current drawn from the motor supply to the A4988 when logic power is disconnected, I also cannot be entirely sure I am accounting for everything.
Thanks for taking the time to test that out. The datasheet does seem to get a bit vague when it comes to explaining exactly how much power the chip uses in sleep/ disable/ reset state. Phrases like “disables much of the internal circuitry” are suggestive but slightly non-commital
I’ve got a board on the way which I’ll do some tests with too. 1uA sounds fine if I can replicate it - would I be safe to disconnect the motor in this state though - I presume yes as the coils wouldn’t be energised?
When motor power is connected and logic supply disconnected, it should be okay to disconnect the stepper motor.