Q:

Stepper motor holding current

I am using this 5V Stepper Motor and Driver IC. My motor is working fine but it consumes 10-20mA when there is no pulse. The circuit should consume power only when there is a pulse. I have attached the driver circuit and schematic of the communicator. My question is how to drop the holding current of the motor by that the motor doesn't consume power. Or if you got any idea or any modifications I can do. Please help me with your ideas.

A:

Since you  have spare HC14s you can do it this way. 

simulate this circuit – Schematic created using CircuitLab
Tie the _STEP input to the same place the current HC14 gets it's input from so you take another copy of the Step Signal and do not affect the current one.
The circuit is a simple one-shot that will hold the _ENABLE line low for a period after any STEP pulse arrives. Note, the first step pulse may be ignored by the driver. I have no idea how far apart your step pulses are so I chose values in this circuit to hold the enable low for the half a second ball-park.
However, since I believe you are not just using full steps to move the motor you will lose position each time you stop stepping and the ENABLE goes away.
A better alternative would be to drop the current in the driver. The circuit above can be modified to do that like so.

simulate this circuit
Note the carrier has a spot for an SMT resistor which you could use for R2, or you need to jumper that spot on the carrier. You need to chose a value for R2 to reduce VREF to the minimal level that will still allow the motor to hold position.
Further Explanation...
In case you have not figured this out yet.. This is what you should end up with (Note I changed R2 to a variable resistor so you can adjust the holding current.)

simulate this circuit
On board the carrier they gave you a potentiometer that lets you divide the reference voltage that the driver creates, VREFO, and feeds that smaller voltage back into the current setting reference voltages AVREF and BVREF. That voltage then sets your run time current.
The designer also brought out the wiper on that pot to a pin on the carrier. As such, by attaching another resistor to ground, though a switch, you can change the divider to produce a smaller voltage when desired.
The circuit I have suggested will do exactly that when there have been no step pulses for a while. 
BTW: In case you had not realized, in your current circuit, holding the STEP line high at the source will turn off the driver entirely.

