[daz450]

Hey fellas, I am currently modding an old cnc mill but as I dont have the specs for the stepper motors, I am having to guess the Voltage and Amps to use however I did find out that they use to run at 27V.

I am running them at 20V and using two 10W 15R in parallel making an equivalent 20W 7.5R on the common wire, oh yeah, the motors are unipolar (6 wires, with both common shunted as in the diagram). The motors are quite big (about 100mm or 3.94inches in diameter and height)

The problem I am having is that the resistance gets baking hot!

I need to know what resistance I should place in between the common wire and the source and more or less what Amps I should be using.

I doubt anyone will be able to help me with such little information but any info is very welcome. By the way, the driver is custom and Im using EMC2

Regards,

Daz

[daz450]

Oh and here is a pic of what is happening to R1

[James Newton]

Is the driver regulating current or no? Need a bit of info about the driver... just the type of driver.

[daz450]

Hi, no the driver has a HEF4028BP that outputs 4 signals to 4 irf540 N-mosfets wich then allow the passage for the 20v supply to the coils.

Regards

[James Newton]

You could just drop your supply voltage. The power resistor is going to do that anyway: It is tossing part of your supply voltage into heat. If you measure the voltage at the motor, it is /not/ 20 volts even now, and with a bigger resistor it will be even less. So why not just reduce the supply voltage? I'm assuming you can't, and that is why you are adding the resistors: To reduce the supply voltage to the motors? Or is it there as a ballast resistor in your linear drive?

Anyway... Ohms law: I = E/R or current(amps) = EMF(volts) / resistance(ohms) can be combined with P = I * I * R or power(watts) = current(amps) squared * resistance(ohms) to find the total wattage that will flow through the resistor.

The trick is that the resistance in the first equation (ohms law) is the TOTAL resistance, including the motor coil /and/ the ballast resistor /and/ the drive transistor /and/ the wires to the power supply, etc... and the resistance in the second equation is just the resistor for which we are finding the power dissipated.

The total resistance is probably very close to just the resistance of the motor coil and the resistor. The transistor is being turned all the way on (I assume) and the wires and power supply probably don't have much resistance at all... if they did, they would also be getting hot. You know the resistor value, and you can use any old ohmmeter to measure the resistance of the motor coil. So you can put in the supply voltage, and the total resistance and find out how much current is flowing through the system. Or, if you have a /really/ good ammeter which can manage large currents, you could directly measure the current flow.

Given the current flowing through the circuit, you can now calculate the power being dissipated by the single resistor using the second formula. And if that power is more than the resistor is rated to manage, you can then see why it is burning up.

But to fix the problem, you must either purchase a resistor that can manage that power, or increase the resistance of the resistor so that less current flows, so that less power is dissipated. If you try really hard, the algebra can be done to reverse the equations so that you can put in the power you want to dissipate and it will find the resistance that gives that result, but I would just try some larger resistance values and you can quickly find the one that gives a good result.

The point is that when you reduce the current flowing through the circuit, you also reduce the power entering the motor coil and reduce the torque of your drive. Doing it this way, you can also calculate how much less power the motor is getting. Just take the current squared times the motor coil resistance.

And the point of all that is: Using a bigger resistor is generally a bad idea. What you really want is a drive circuit that actively regulates the voltage to the motor so that the motor is getting as much current as it can manage when the coil needs to fire up, and then the voltage is backed off so that less current flows and that keeps the motor (and any resistors) from melting when the coil is at full strength. A regulating driver avoids this issue.

[toughtool]

Daz,
That motor looks a lot like mine. They came out of an IBM 5256 printer. My motors are 4.5 volts at 1.5 amps. IBM used a higher voltage and a series resistor, (like you describe) which improved the stepping responses. I tried running them at 5 volts but they didn't work too well. I connect mine to 12 volts using a 5 ohm 25 watt resistor connected in series, to the common wire (Black or White) for each pair of coils (Red/White-Red or Blue/white-Blue). Because only one coil is grounded in each pair of coils as the motor is stepped, there will be 1.5 amps through each 5 ohm series resistor. This means the resistor will dissipate 11.25 watts each so I would recommend a 20 or 25 watt chassis mounted resistor, mounted to a heat sink. They will get hot! Just like a 25 watt light bulb will get hot. The motors I have get hot but not so hot I can not hold my hand on them. This is because there are always two coils "ON", when powered by my full step driver. I have made a circuit using a 555 timer that automatically inserts an additional resistor (unshorts a resistor with a relay ) into the power supply line if the motors are paused longer than a minute. This drops the current to about a third, which is enough to maintain a mechanical detent (so the motor stays at current position). Getting the motor too hot will ruin the magnetic properties of the motor and ruin them. As long as there are steps, the timer is continually reset, keeping the relay closed. If there is a pause of more than a minute (adjustable) the relay opens and the resistor lowers the current, reducing the heating of the motor. Here is a link describing the resistors I used. Digi-Key - KAL25FB5R00-ND (Manufacturer - KAL25FB5R00)

However I did find them cheaper elsewhere.
I have posted the four motor schematic of my driver, which cost me about $50.00 to build, using the IRFZ44, which will handle 45+ amps. They don't even get hot, at all. Hope this helps. Joe


The math: P=E*I, P=7.5(voltage drop R) * 1.5 amps
(one coil)= 11.25 watts.