Can I safely reduce a stepper motor's power consumption by applying a PWM to the driver's enable input?

I have a stepper motor driver breakout board based off the Allegro A4988 chip. I noticed that when the motor is static, it has much better holding torque than when moving. I have a battery operated configuration and a majority of the time the stepper motor is only used to hold a position. The limit setting potentiometer is already on the breakout board so I can't change it with my micro-controller.

Would I able to apply a PWM signal to the enable pin of the motor driver to change the current going into the motor so that I can drive the motor at a higher current level when actively moving and reduce the current for holding position?

I'm mostly interested on the effects a PWM signal will have on the driver circuit's life. Nowhere on the data sheet dose it say you can or can not do something like this.

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I'm not sure what you mean by 'the limit setting potentiometer', but (assuming that you mean the current sense resistors $R_{S1}$ and $R_{S2}$ (which would be an interesting design, those should be matched low-impedance resistors) I read on page 9, in the section Internal PWM Current Control:

The maximum value of current limiting is set by the selection of RSx and the voltage at the VREF pin. The transconductance function is approximated by the maximum value of current limiting, $I_{TripMAX}$ (A), which is set by

$I_{TripMAX} = V_{REF} / ( 8 R_S)$

where $R_S$ is the resistance of the sense resistor (Ω) and V_{REF} is the input voltage on the REF pin (V).

Do you have access to the REF pin (pin 17)? If so, adjusting this with a DAC will produce the same effect as varying the sense resistors.

If not, the Enable input is probably a better choice than cutting the power or using the sleep or shut-down pins, those would incur a 1 ms delay (per the description on p. 10).

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Sorry, I think the sense resistors are permanently set, the potentiometer is probably adjusting the REF pin (don't know though, don't know enough about electronics). I guess the problem here is that I would end up running two overlapping PWM signals. First, the internal one that uses the sense resistor to control the current, second the one that I'm using to shutdown the FETs. –  Faken Mar 26 '12 at 2:29
From what I can gather, the chip is using the sense resistor to detect current and adjust it's PWM so that it averages out to whatever current limit I am setting. However, what happens if I suddenly start interrupting that process by toggling the FETs on and off with enable very quickly? It may not affect the full step modes, but what about the micro stepping modes? I know not to use the shutdown or sleep pins, that messes with the logic circuits and may reset my current step position. –  Faken Mar 26 '12 at 2:32
You'll need to PWM the enable pin slowly. Take a look at the waveforms on p. 11: You want to be slower than $V_{STEP}$'s period but significantly faster than $t_{off}$. Interrupting the system during $t_{fd}$, while the FET is on, will cause an error in your output voltage. I suggest somewhere between $10 t_{off}\ < t_{PWM} < 0.1 t_{STEP}$ as a period for your PWM if that's possible. –  Kevin Vermeer Mar 26 '12 at 14:42

I can see nothing in the datasheet which states that the enable pin cannot be pulsed. It's hard to tell from the internal circuitry if it would cause a problem. The enable pin switches off the external fets, which might mean that the chip is unable to re-circulate current properly, causing voltage spikes. However, I would expect Allegro would have dealt with that situation, since it would happen every time Enable goes high.

If you have a spare driver, then my advice would be to just try it.

Alternatively, you could reduce the Vref voltage, rather than using PWM. The A4988 will then take care of the current reduction for you nicely. This can be achieved by pulling the Vref pin a little lower with a resistor and either a transistor or a the pin of an MCU which can switch between low and high impedance.

An even better solution might be to make it automatic. Use the step signal to rapidly charge a capacitor which pulls the Vref voltage up. An absence of step signals lets the cap slowly discharge, and with it the Vref voltage and the motor current. The faster the steps, the higher the Vref.

Here's a very rough schematic of how it might work. I just guessed the resistor values, and haven't checked that they would work at all. (Exercise left up to the reader).

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Nice answer, I really like the idea of using the pulse to automatically change the current settings. Unfortunately this is on a breakout board and I don't have access to the Vref pin so I can't do anything with it other than adjusting it with a small plastic screwdriver. –  Faken Mar 28 '12 at 22:11
Thanks. It's a shame not to have access to the Vref pin, because it would give you nice smooth current control. Perhaps you could solder a little wire onto the pot? –  Rocketmagnet Mar 28 '12 at 22:15
I'm torn between which answer to accept. Nether answer actually answers my question. Rocket's answer is very well done and has some really cool concepts but kevin's comments gives me at least some insight as to what to do. –  Faken Apr 3 '12 at 22:19
Have you tried it yet? I think that's going to be the only way to get a definitive answer. Since there's nothing in the datasheet about it, nobody can really know for sure without trying it. –  Rocketmagnet Apr 3 '12 at 23:27