I have noticed on at least a couple of websites and in a few posts here that there is an error in the way the value of the series load resistor is calculated. It seems that people are being advised to take the rated voltage for the motor to be used (V) and divide it by the current per phase (I) to figure the DC resistance of the windings. This is good information. Then, to subtract the motor voltage from the power supply voltage and divide this by the winding current. This would be great if these were the only devices in the current path. However, there is a voltage drop across the switching device, whether using FETs or bipolars transistors. This must be taken into account, especially when using power supplies that are closer to the rated voltage of the motor! It should also be noted that a 5V power supply will not supply 5V across the windings of a motor when put in series with the switching device.
That's the summary. For those who have the time and want proof...
As an example, let's assume I have a 9VDC p/s capable of supplying the needed current, and use steppers rated for 5V and 1.6A/ph.
My series resistor needs to drop 9V(p/s)-5V(winding)=4V.
Using Ohm's law, we can calculate the DC resistance of our windings. R=E/I, so R=5V/1.6A=3.13ohms. We'll use this later.
Going back to Ohm's law, our series resistor needs to be R=E/I, R=4V/1.6A=2.5ohms.
This resistor would need to be able to handle P=IE, P=1.6A*4V=6.4W.
Right? Wrong.
Let's factor in the FET that I will use in my example.
The datasheet says that my FET will have an on drain-source resistance of 1.76ohms(typical).
Since this is in series with the windings, it will have the same 1.6A of current going through it, so it will drop E=IR, or E=1.6A*1.76ohms=2.82V.
If we add that to the 5V that we want our motor to "see" we get a total voltage drop of 7.82V. Now when we subtract that from our 9V power supply we get 9V-7.82V=1.18V.
Back to Ohm's law again, R=E/I, R=1.18V/1.6A=0.74ohms.
This resistor would need to be able to handle P=IE, P=1.6A*1.18V=1.89W.
If we use the original resistor value we came up with, we would have a total series resistance of 1.76ohms(FET)+3.13ohms(winding)+2.5ohms(resistor)= 7.39ohms. With the 9V power supply this results in only, I=E/R, I=9V/7.39ohms=1.2A, or 25% less current than we want! With a bigger difference between the supply and motor voltages this is less of a factor, but should still be taken into consideration.
That's my 2.5 cents.
Feel free to comment.
Dave