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I have battery power (9v) for stepper motor (auto-tracker for telescope mount). I need micro stepping and high torque (0.1 N*m). I decided to use bipolar stepper and A4988 because there are cheap modules from eBay.

But: I need to rotate stepper very slowly (about 1 step per second), so most of time the motor stay without rotation and A4988 produce a holding current, but I need no holding torque (there are no external torque that can rotate my motor).

How to maximize battery life in my case. Maybe just set "enable" pin of A4988 to 1 between rotations? But I have another questions in this case:

1) Can I lost steps if I will set enable pin to 1 between every steps? In data sheet there is note that if I will use sleep pin, then after wake up A4988 go to HOME state, so I will lost steps. But there are no notes about behavior of enable pin.

2) How microcontroller can know when step have executed (before set enable pin to 1)?

Maybe there are another solutions. I need at least 5 hours of 9v alkaline battery, 0.1 N*m, single step each second.

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A4988 data sheet here

(1) The data sheet says that the /ENABLE input does not destroy "state"information. However, as Anindo notes, while there is no external torque, the motor will apply "pull in torque" when depowered. It MAY be possible to use the current limiting setting to reduce holding current to a minimum and to increase it when actually stepping, but this is not a certain solution, and it adds complexity which may make the overall solution unattractive.

On page 10 it says:

Enable Input /ENABLE

This input turns on or off all of the FET outputs.
When set to a logic high, the outputs are disabled.
When set to a logic low, the internal control enables the outputs as required.
The translator inputs STEP, DIR, and MSx, as well as the internal sequencing logic, all remain active, independent of the /enable INPUT STATE.


(2) You may also be able to dynamically alter the effective sense resistors that control maximum current. These MAY latch, on a step by step basis, which would make this method impractical.

Data sheet says:

  • At each step, the current for each full-bridge is set by the value of its external current-sense resistor (RS1 and RS2), a reference voltage (VREF), and the output voltage of its DAC (which in turn is controlled by the output of the translator).
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  • \$\begingroup\$ I know about current limiting, but changing logical level on a chip's pin is more simple solution :) What negatives follows from switching "enable" pin? If no, I have only one problem: I need a way to know a time when step done. \$\endgroup\$ – user54579 Jul 8 '14 at 14:52
  • \$\begingroup\$ "The translator... remain active": so, when I enable chip again, I get currents through coils exactly the same as before disabling chip? And I can continue steps/microsteps as if I didn't disable the chip for a time? \$\endgroup\$ – user54579 Jul 8 '14 at 14:57
  • \$\begingroup\$ Russell, for microstepping, there would be holding current on the coils. If one were to stop this current (by using ENABLE or any other way), the stepper would snap to the closest detent point, would it not? If yes, that isn't suitable for the stated purpose, of telescope control. \$\endgroup\$ – Anindo Ghosh Jul 8 '14 at 16:14
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    \$\begingroup\$ @AnindoGhosh I believe you are correct. The motor will snap to the closest full step position when you remove holding current. It might be better to use a gear reduction drive instead of microstepping. That would allow driving the motor with full steps so it wouldn't move when you remove holding current. It also multiplies the motor torque, potentially allowing the motor to be driven at lower current, and thus extending battery life. \$\endgroup\$ – jwygralak67 Jul 8 '14 at 17:17
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    \$\begingroup\$ @AnindoGhosh - I took the statement "but I need no holding torque" at face value but, as you note,, the motor will apply "pull in torque". It MAY be possible to use the current limiting setting to reduce holding current to a minimum and to increase it when actually stepping, but this is not a certain solution, and it adds complexity which may make the overall solution unattractive. \$\endgroup\$ – Russell McMahon Jul 8 '14 at 23:35

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