Selecting the right driver

[1113562]

You are correct, to meet the holding torque (the force required to force the motor from its current step to the next or cause slip - not good) will require in your example 2.73V across the parallel phase winding which will draw a steady 4.2A. The parallel coil resistance = 0.65 Ohms in this case. However, due to its inductance when switching, the current will not get to 4.2 Amps immediately, there is a time constant due to L and R and your driver source resistance combined. The torque produced is a function of the current not the applied voltage. At time 0 at switching there is no current flow at all so no torque but current quickly builds up over time to meet the 4.2 Amps at time = infinity, it follows an exponential curve. To accelerate the time to get to full current i.e. to achieve maximum torque in a much shorter time (which will allow your motor to go faster) the trick is to apply a much higher voltage to the windings. This will cause a faster ramp rate and the coil will get to full current in a much shorter time but current will carry on going up so the driver needs to measure the current and turn off the voltage when it gets to full current. Then the current will begin to decay but the driver then zapps it with another voltage pulse to get it back up to full current again and so on thus retaining the nominal 4.2 Amps. The PWM type driver does the same except that it follows a current profile as defined by the microstep look-up table and not the motor maximum current for every step. I can send you or link you to a brilliant application note which describes this much better than me here if you would like. It is on the Microchip web site under the stepper motor applications section if you can't wait for my reply.

Cheers, John