How much hold current is consumed when there is no load?

Question

Given a 24V 1A motor with no actual load connected (just a piece of orange tape so we can see the shaft moving). How much current will the motor use if I configure the driver board for 2A maximum hold current?

This is the scenario:

We have a breadboard that is configured for 2 motors at 24V 2A. However, we only have 1 motor and axis configured for now. So until we get the second motor, I asked for just any motor to be connected to the second channel so that I can see that the software is working. The project EE found a 1A motor and connected that. The EE then warned me to reduce the maximum hold current to 1A. I was told that the driver board would "push" 2A of current to the motor.

This is where I get confused:

1. I don't understand how the motor could draw 2A when there is no load, even if there was no current limit at all.

2. I thought that the "maximum hold current" was the current limit. Not that the board would push 2A to the motor. At 24V, supposing that the motor was drawing say 200mA just sitting there idle, wouldn't the driver board have to supply 24V x 2000mA/200mA = 240V to get 2A of current into the motor?

Answer 1

A stationary stepper motor supplied by DC is just a resistive load, so the maximum current which will be drawn is the driver's maximum output voltage divided by the resistance of the relevant coils.

When you're operating the motor at speed, changing coil current very quickly, you need a lot of voltage to impose the rapid changes of current. For this reason, stepper motor drivers tend to have lots of voltage available, much more than you need to maintain a suitable holding current.

Because of this, the holding current is something which the driver has to control - it's not a matter of the motor 'drawing' it, so much as the driver imposing it.

It's going to be the case that a 1A motor has a higher resistance than a 2A motor, so if you had a constant voltage supply you'd see lower holding current when the 1A motor was connected without changing anything. However, the mere fact that you've been told 'set the holding current lower' implies that actually you have a more sophisticated driver, which will have lots of excess voltage and will need to be told what a safe holding current is.

Answer 2

It depends on the resistance of the motor, the supply voltage, and the driver limits, which one will limit first.

A stepper driver usually has lots of extra voltage in hand for slewing the current quickly against the inductance of the motor coils, most of the voltage ends up getting lost in the driver circuit.

This means that if you have a motor that is rated for 1A operation, and your driver is configured for a 2A hold current, there could well be enough extra voltage available at the driver to push 2A through the motor. This would generate 4x the rated heat in the motor, and would probably overheat it.

You ought to be able to find out what the motor coil resistances are, if not from a specification then by measuring with a meter. Then calculate what voltages would be needed at the terminals to make 1A or 2A flow. What voltage is available at the driver?

Do what your EE says and reduce the driver max current to the motor's rated current. I've melted a motor during a lab, it can be quite embarrassing.

Answer 3

You're right. The EE is wrong.

When both voltage and current are specified, they are in fact limits, not forced values. Whichever it hits first is where it stops at each moment in time. V = IR still holds.

Answer 4

A driver has a chopper and countinously feeds the motor with set current. You say 1A is needed. When motor is not spinning, there is no back EMF and the voltage applied is V = R_winding * I_set. When you drive the motor at higher speeds, the driver has to compensate for BEMF that is induced from rotating motor, but the current is still the same.
As for holding current, this is a feature of your driver, that detects no pulses - standstil and it reduces the current to a lower value.