Stepper motor enable line

Question

I am using DRV8834 IC driver to run the 5v stepper motor. I've managed to run the 5v stepper motor using DRV8834 IC by using a breadboard. But I've found we need 3 pins (Step, Direction and enable) to enable and step the device. I came to know that by making the enable pin to high makes the driver disable where there is no power consumption in the whole system.

But the requirement is only to use 2 pins (step and direction). If I don't control the enable pin via microcontroller then the driver and motor (whole system) consume 50mA. I have to find a solution where the driver and motor shouldn't consume any power until enable line receives the step output pulse from the microcontroller.Also, the step output will always stop in the low state. Is there any idea you can help me with to use any MOSFET/any other idea/pulse stretcher to control the power consumption without using the enable pin? Please let me know if you guys got an idea. Your help is much appreciated.

---

Update to clarify @Transistor's answer.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. SLEEP control.

How it works:

• When STEP goes low D1 discharges C1. Q2 turns off and R2 pulls /SLEEP high enabling the output.
• When STEP stops high C1 charges through R1, Q2 turns on and the stepper goes into sleep mode.
• C1 can only be discharged to about 0.5 V with D1 and that would tend to turn on Q2. Adding D2 prevents this happening (hopefully).

Choose C1 so that R1 x C1 is the time delay you require.

Answer 2

Expanding on Transistor's answer, this circuit monitors both the DIR and the STEP lines for rising edges and will turn off the sleep signal for a period set by the values or C2 and R3.

R1,C1, R2 act as a high pass filter to turn on Q1 on rising edges of the XOR gate for a short period. This charges C2 down to the ground rail. R3 discharges C2 back up to Vcc. M1 will turn on when the voltage decays back up to Vgs.

Being edge triggered the micro has to actually toggle something to keep it alive. When holding, toggle the DIR line both directions before the RC timer times out. When running the step signal "should" keep it awake naturally.

^{simulate this circuit – Schematic created using CircuitLab}

The XOR could be replaced with discrete parts, but take care not to distort the DIR and STEP signals.

Answer 3

Here is a different solution based on the assumption that you can make the micro outputs high impedance.

^{simulate this circuit – Schematic created using CircuitLab}

Note each signal line is biased to half rail when the output from the micro is high impedance. These biased signals are then fed to two window comparators in a quad LM339. The biased signals are then compared with 1/3 and 2/3 rail voltages and the outputs orred together with diodes and the pull down. If any of them are outside the gray logic range the sleep signal is deactivated.

Since DIR and STEP change states at different times, there should be no time that both are transitioning through the grey logic area at the same time, so this configuration should work.

This means you can put the driver to sleep just by changing the configuration of the outputs to high impedance.

You could do a similar thing with four opto-couplers if you have the drive for the LEDs.

^{simulate this circuit}