[109jb]

Ok. I have some 6 wire unipolar motors that I want to run using bipolar drives. There are basically 2 ways to do it. Either use center tap and one end wire from each coil (half-coil), or use the end wires and don't use the center tap (full-coil).

I am wondering about a couple things.

First, the current rating. To start with, I will be using a chopper type bipolar driver. The motor ratings are based on using the motor as a unipolar motor, so with the half-coil option it still remains where only 1/2 of each coil is energized at a time, so the current rating will remain at the 2.9A unipolar rating. Using option 2, the resistance will double and the inductance will be 4x. With double the resistance, it seems that from V=I*R that the current rating should be 1/2 of the unipolar rating, or in my case 1.45A. However, in looking at several 8 wire motor specs and the current rating for different wiring schemes, it appears that the current rating for an 8 wire motor in bipolar series connection has a current rating of 0.7 times the unipolar connection rating. So, if I chose the full-coil option what current rating do I use? 1/2 of the unipolar rating or 0.7 times the unipolar rating?

Second, if I use the full-coil option, the inductance will increase by a factor of 4. This obviously means that the motor would require twice the voltage and would be "slower" due to the inductance increase. I seems that it would also have a higher torque at lower speeds though. Is my thinking here correct?

Finally, is one of these methods preferred over the other, and for what reason.

Thanks in advance.

[jalessi]

109jb,

See the attached link:

Support

Jeff...

[109jb]

109jb,

See the attached link:

Support

Jeff...

The link doesn't work, but I read the stepper FAQ on the geckodrive site. In the FAQ they say that if running a 6-wire unipolar motor on half-coil, to run at the full rated current of the motor (unipolar rating). That makes perfect sense to me. They also say that if running a 6-wire motor on full-coil, to run at 1/2 of the rated current. The problem I have with that is that it doesn't match the listed bipolar series connection for 8-wire motors. For example. the Keling KL23H286-20-8B motor has the following specs:

Unipolar: 6V, 2A, 3ohms
Bipolar series 8.4V, 1.4A. 6ohms

Since an 8-wire motor is essentially the same as a 6-wire motor when wired as a unipolar or a bipolar series then according to the geckodrive FAQ, it would follow that the current rating for bipolar series should be 1/2 of the unipolar rating. It isn't though, and this holds true for all of the 8-wire motors I have looked at. The bipolar series rating is always 0.7 * the unipolar rating. I think I have it figured out though.

Take I(u) as the unipolar current rating, R(u) as the unipolar resistance, I(s) as the unknown bipolar series current rating, and R(s) as the bipolar series resistance. If we feed the motor I(u) at the half coil resistance R(u), then we are putting the following amount of power (watts(u)) into the motor:

watts(u)=I(u)*I(u)*R(u) -----> Power=I*I*R

For the bipolar series connection the following is how much power we will feed it:

watts(s)=I(s)*I(s)*R(s)

If we keep power constant, then watts(s)=watts(u) so,

watts(u)=I(s)*I(s)*R(s)

but watts(u)=I(u)*I(u)*R(u) so substituting that we get:

I(u)*I(u)*R(u)=I(s)*I(s)*R(s)

Also, we already know R(s)=2*R(u) so:

I(u)*I(u)*R(u)=I(s)*I(s)*2*R(u)

We want to find I(s), so we re-arrange the equation to solve for I(s):

I(s) = sqrt[{I(u)*I(u)*R(u)}}/{2*R(u)}] = I(u)/[sqrt(2)] = I(u)*(0.7)

I respect the geckodrive folks, but in this case I think they are in error. It is either that or all of the manufacturers of 8-wire stepper motors are wrong. I don't think that is the case.

As an aside, you can get the bipolar series voltage rating by using the formula:

V(s)=I(s)*R(s)

If you perform the above calculations on any 8-wire motor by just starting with the unipolar ratings, you get the same bipolar series ratings as are listed on the motor specs.

Using the Keling motor specs I listed above,

I(s) = I(u)*0.7 = 2*0.7 = 1.4A
R(s) = 2*R(u) = 2*3 = 6 ohms
V(s) = I(s)*R(s) = 1.4*6 = 8.4V

[jalessi]

109jb,


I was going to suggest sending Mariss an email about the broken link and possible discrepancy however the Geckodrive Website is not functioning at this moment.

Maybe Mariss will chime in and clear up your question.

Jeff...

[James Newton]

All that info in a table at: techref.massmind.org/techref/io/stepper/connections.htm