Controlling a 24V motor

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
Bob Young - Test Engineer - Lapsed Certified LabVIEW Developer
DISTek Integration, Inc. - NI Alliance Member
mailto:[email protected]

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,
BobMessage Edited by Bob Y. on 05-23-2005 11:58 AM
Bob Young - Test Engineer - Lapsed Certified LabVIEW Developer
DISTek Integration, Inc. - NI Alliance Member
mailto:[email protected]

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    Bob Young - Test Engineer - Lapsed Certified LabVIEW Developer
    DISTek Integration, Inc. - NI Alliance Member
    mailto:[email protected]

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    mailto:[email protected]

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