Below is a block diagram of a stepper motor controller with an external mosfet h-bridge that drives one of the stepper's coils. Let's say that the coil has 2ohm DC resistance and 3mh inductance, the motor is stationary, and the current through the coil is 2A (stabilized by the stepper controller).

  1. Does the average current from the source Vm depend on Vm's voltage (e.g. doubles when reducing Vm from 24V to 12V)?
  2. Does the power dissipation on the mosfets depend Vm's voltage (e.g. increased X4 when reducing Vm from 24V to 12V due to the fixed Ron of the mosfets )?

enter image description here


EDIT: added A,B,C,D transistor references.

  1. Yes the supply current will be about double on 12V as compared to 24V.

  2. The average current through the pair of active MOSFETs is the coil current, which is fixed. The power dissipation moves from one MOSFET to the other as the supply voltage changes. For example, one might conduct for 10% of the time with 24V and the other conduct 90%, but that would change to 20% and 80% with a 12V supply. Current is only drawn from the supply in the shorter interval in this example, yielding an average coil voltage of 2.4V. There are also losses associated with the switching.

Edit: Imagine just one half-bridge driving one coil. I'll assume the other side of the coil is grounded (in reality it will conduct through the other half bridge). The stepper coil is modeled as a resistor (2.4 ohms in this case) in series with an inductance (assumed to be large enough that the coil current does not change much during the PWM/chopping cycle).


simulate this circuit – Schematic created using CircuitLab

Let's say we set the coil current to 1A in this example, so the voltage across the coil resistance in steady state must be 2.4V. That means we need a 10% duty cycle from a 24V supply. So M1 conducts 1A for 10% of the time, and M2 conducts in reverse 1A for 90% of the time. The total conduction power dissipation of M1 and M2 is thus \$ 1A \cdot 1A \cdot \text {Rds(on)}\$. If you lower V1 then M1 will conduct for more time (in order to maintain 1A) so its dissipation will go up, but M2 will conduct for less time so its dissipation will go down and the total stays constant.

Since there are two half bridges per coil and two coils (8 MOSFETs total) the total power dissipation of all 8 MOSFETs due to conduction is \$4\cdot I_C^2\cdot \text {Rds(on)}\$.

Returning to the total supply current and the half-bridge example above, the 24V supply sees 1A (24W) for 10% of the time, average 100mA or 2.4W. There are two coils in the stepper so that is doubled for the entire motor. If the supply is reduced to 12V the supply sees 1A (12W) for 20% of the time.

The advantage of a higher supply voltage is that during active stepping when the coil current is changing, the higher supply voltage is more quickly able to overcome the motor inductance and thus give you more torque at higher RPMs. For that, you would prefer a motor with a relatively low resistance and thus coil voltage relative to the supply voltage. Typical specs for a small (NEMA 23) motor:

Rated Current: 2.8A

Phase Resistance: 1.1 +/-15% ohm

Phase Inductance: 3.0+/-20% mH

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  • \$\begingroup\$ Thans Spehro. I think I understand #1 but not #2. The Vm current (which is smaller than the coil current) go through the transistors, what additional current goes through the transistors to make the current through them equal to the coil current? I added to the diagram transistor references A,B,C,D, hope it will make it easier to explain. \$\endgroup\$ – user1139880 Aug 11 '19 at 5:50
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    \$\begingroup\$ Thanks Spehro. Excellent explanation! I think I understand it now. I was not aware that M2 is on during the 'off' part of the cycle to maintain the coil current. \$\endgroup\$ – user1139880 Aug 11 '19 at 16:38
  • \$\begingroup\$ @user1139880 It's not always true for H-bridges in general, sometimes a diode (such as a Schottky in parallel with the body diode) is used, but that's usually even more dissipation. In the case of this chip, the body diode only briefly conducts during the "dead time" of the switching transition when both MOSFETs are very briefly off- you can see that referred to in the datasheet). \$\endgroup\$ – Spehro Pefhany Aug 11 '19 at 17:20

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