## General Motion Control

# Bipolar chopper drive circuit III: Controlling current

**Bob Parente**| April 25, 2012

We left off with the last blog explaining that the “voltage waveform at Sense Resistor” in our schematic shows what the voltage levels would look like as the motor is micro stepped. There are discrete changes in the signal, a staircase up and down, as the DAC converts the digitized signals from the “current level control.”

We also assumed that our one ohm, one millihenry, one amp motor had its phase A current following a sine wave and its phase B current following a cosine wave. And both phase A & B were not allowed to exceed one amp, which of course is the motor’s rating. Or is it?

Let’s backup for a minute and take a look at the power that the motor dissipates.

A simple calculation using I^{2}R=Power (in watts) for both phases tells us that the motor can dissipate two watts total.

**1 ^{2} amp * 1 ohm = 1 watt.**

One watt in phase A and one watt in phase B equals two watts total for the motor.

So let’s keep the maximum current at one amp for now and repeat the calculation using the sine/cosine currents in the appropriate phases.

The sine of zero degrees is zero and the cosine of zero degrees is 1.

Since phase A follows a sine wave, then we have zero power being dissipated in phase A and using I^{2}R to calculate the power in phase B we have 1^{2}*1 = 1 watt

One watt? That’s half the power the motor is capable of dissipating.

What’s up with that?

So let’s be creative here and force the current in the B phase to dissipate two watts.

I^{2} * 1 ohm = 2 watts. Solve for I:

**I = (2/1) ^{1/2} = 1.414 amps.**

Does the value of 1.414 ring any bells with you?

OK, OK, besides being the square root of two.

It’s the number you use to convert the RMS value of a sine wave to its peak value.

The one amp in both windings represents the DC current, or the RMS value, if we use a sine wave. Thus, we just have to set the winding current to its peak value instead of using the RMS value.

So now the power at Sine (0) and cosine (0) is two watts.

Let’s check another point and see if it still holds true.

Let’s chose the sine (90) and the cosine (90) for our second test.

Sine (90) = 1 and the cosine (90) = 0 and we end up with 2 watts of power in the A phase and zero watts in B. Just the opposite of what we had before.

Now here comes an important point: no matter what current angle you try, the total power dissipated by the motor is always two watts.

Go ahead and try some oddball angles. The general equation is:

**1.414 ^{2}(sine (Ø))*R + 1.414^{2}(cosine (Ø))*R = 2 watts**

Where the value of R is one ohm.

Pretty slick, huh? Some major stepper motor drive manufacturers didn’t realize this when they first introduced micro stepping drives and their drives ended up producing 70.7% of the output torque (can you guess where that number came from?) when compared to the DC full step two-phase on rating or the micro stepping drives that allowed you to set the current to the peak value instead of the RMS value.

In the next blog posting we’ll take a look at the torque each micro step produces and see where this might get you in trouble if you don’t fully understand how it works.

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