Controlling a 24V motor

[Bob_Y.]

I have a motion system with a PCI-7344 controlling several actuators through a MID-7654. I need to add one more axis. The actuator that I would like to use is only rated for 24V 0.2A. Is there a way to hook it up to the MID-7654 without overdriving it? I'm thinking of a voltage divider or some other small circuit. Or maybe there is a way to rig it for lower voltage that I am unaware of?

Any advice is appreciated.

Thanks,
Bob Young

[mc-hase]

Hello Bob,
One solution could be that you make a Limit (Tourque Limit) for your DAC Output on the Motion Control Board. With this way, you could prevent your DC Motor from too much Power. (But not from too much Voltage)

[mc-hase]

Hello Bob,
Another simple solution is if you use a serial resistor. 24V 0.2A => 120 Ohm 5 W

[Bob_Y.]

I don't see how series R would help.

In another thread I posted all the specs for my motor. They are:
24Vdc
200mA
1.19mH
54.6ohm

I have been working through the formulas presented here. According to this, if I increase the winding resistance (true, this is not winding resistance, but I would need to consider it as such) I increase the power that will be dissipated. True, most of the power would go through the resistor, but I think it would wind up being more than 5W.

To begin with, hooking up my motor in its raw state would yield:
di = (V/L)*dt = (48/.00119)*.00005 = ~2.0168amps ripple current
P=(I^2)*R_w = (2.0168^2)*54.6 = ~222Watts of power to dissipate.

I am a little sketchy on the current. If it set the limits to the minimum, (.85nom 1.7peak), will it use the ripple current of ~2a or the peak current of 1.7 in this formula? I assume the lower so

P=(I^2)*R_w = (1.7^2)*54.6 = ~157.8Watts of power to dissipate

This is still way too much for this fairly small actuator to handle.

With series R, I think that it would be the same formula for ripple current, but the power formula changes to

P=(I^2)*(R_w+Rseries) = (1.7^2)*(54.6+120) = ~505Watts of power to dissipate, with almost 350Watts being dissipated by the resistor. I either am not doing something correctly, or I would need a bigger resistor than a 5W one. And it won't help my motor.

I think I would need to have parallel resistance, thereby decreasing the winding resistance but I am not sure what happens with the inductance. I assume that it is largely unaffected. Rework the power dissipation formula to find the resistance R=P/(I^2). If I need to get the power down to about 5Watts, the resistance would need to be

R = 5/(1.7^2) = 1.73ohm which would take a parallel resistance of 1.7866ohms assuming that it did not affect the inductance much. Then I think you have problems obtaining a reasonable voltage through the motor V=IR = 1.7*1.73 = 2.941volts.

If I change the inductance without changing the resistance, I think it would change the ripple current without limiting the voltage. So I could increase the inductance (a lot) in order to reduce the ripple current to a value similar to the spec of the motor.

rework the formula di = (V/L)*dt to be L = V*dt/di and put in .2amp for di to get

L=48*.00005/.2 = .012 or 12mH. That would take about 11mH in series to bring the current down to spec. Then the power dissipation would be

P=(I^2)*R_w = .2^2*54.6 = 2.184Watts.

This whole thing has me quite confused however, so I might be wrong entirely. Can someone else run through my logic and numbers to make sure that I have not gone entirely insane . . .

Thanks,
Bob

Message Edited by Bob Y. on 05-23-2005 11:58 AM

[RManion]

Hi Bob,

Unfortunately, it is not possible to use a voltage divider or similar circuitry to 'divide down' a connection to a motor drive. Motors have both a resistive and inductive load which changes based upon the physical load on the motor. As a result the impedance varies as the resistance to spinning changes.

The calculations you have here look to be those needed to calculate the power required by your motor's amplifier. If you are trying to power this motor, you will most likely need a different drive. Maybe it would be cheaper to get a motor that can be used on your current drive?

Hope this help and good luck to you,

Robert