Stepper motor drive circuit design

Hi all,

I am trying to work out a circuit to drive the windings of a stepper motor. I have done plenty of digging around and know the basics that I can use wither a FET or a Darlington pair to do the power switching. The problem that I am having is working out the power supply and matching it to the motor I have.

The motor I have is a 4V 1A bipolar stepper and I want to drive it with a +5V PC power supply. (Since I have it.) What I am trying to figure out is how to drop the voltage to 4V reliably for the motor. Will the drop across the junction of a Darlington pair be enough to do this? Or is that a bad design? Will a power resisitor in series with the winding and using a FET be a better solution?

Also, if there are any other considerations I need to make I'll take any information.

FYI, I plan on switching the transistors with either +5V or +3.3V logic levels.

Thanks in advance!

The drop voltage in a FET transistor is around 0.7Vdc. If you use a Darlington pair the drop will be around 1.4Vdc, so I think this will be perfect for what you try to do.

Stepper motors ratings should be thought of mostly on terms of current. What voltage it takes to achieve that current is almost irrelevant. The rating of 4 V on your motor means that it can take a steady state of 4 V in stopped mode without burning up. But there is nothing wrong with briefly applying 30 V to overcome the inductance and get the current up to 1 A in a hurry. In fact that is exactly how high-performance stepper drivers work. They have a higher supply voltage but they regulate the current by chopping with the FET or other switch through active current feedback. You can get your stepper to work by applying a fixed 4 V, but the start-up torque will be quite poor. You will have to ramp up the stepper pulse frequency gradually so that the motor can follow it.

If you are set on using a 5 V supply then you could just add a 1 Ohm resistor in series with each winding. This will allow the full 5 V to be applied at the beginning of each step. Then as the current rises according to the L/R time constant it will stabilize at 1 A when the motor comes to a stop and then the windings will see only 4 V.

Now if you can manage to use a 12 V supply, then add a 8 Ohm resistor in series with each winding. Then each winding will see the full 12 V at the start of each step, but it will still stabilize at 4 V and 1 A when the motor stops.

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pedroromanvr said:

The drop voltage in a FET transistor is around 0.7Vdc. If you use a Darlington pair the drop will be around 1.4Vdc, so I think this will be perfect for what you try to do.

Click to expand...

You are right about the Darlington, but the drop across a power MOSFET is essentially I*R, and some of those FETs have on-resistance values in the low milliOhms. There is no diode drop involved.

So I'm now wondering if this would be a good part to use for the Darlington pair: **broken link removed**

I was wanting to avoid using a power resistor because of the expense, but if it is necessary I will do it.

I can go with a 12V supply because of using an old PC power supply.

So I was able to trace out the circuit on the board that drives the steppers I "harvested." It was quite a challenge since the board drives 8 steppers and the circuit is quite interesting.Below is the circuit for one stepper:



THe really odd thing is that the 4 MOSFETS that go to ground are common to ALL 8 steppers. So the board is designed to only operate one at a time. Also the MOSFET goint to power is on the common tap of each winding.

My biggest myesery at this point is the diodes. I'm not able to tell what kind they are. The only markings on the diodes are ED6A and in my internet searches I was unable to turn up anything. Any clues on what these diodes are? FYI they are in an SMD package and fairly large if that helps.

Thanks!

I was able to desolder one of the diodes and then power it up in a simple circuit with a resistor and determined that it is a standard type diode.

It seems to have very high current handling capacity, because I powered it up with +5V and a 5ohm power resistor in series and it didn't even get warm.