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Voltage Across Motor Exceeds Applied Voltage??
4

Voltage Across Motor Exceeds Applied Voltage??

Voltage Across Motor Exceeds Applied Voltage??

(OP)
Hi,

I have a DC motor (with brushes) that is designed to be operated on 120 VDC. I am taking the 120 VAC line voltage and rectifying it with a full-wave bridge. What I don't understand is that when I measure the DC voltage across the motor terminals (using a Fluke 87 multimeter on the DC volts range), I read about 140 VDC--greater than the AC voltage applied to the bridge rectifier. If I remove the motor and replace it with a resistive load (I'm using a 60W light bulb), I read about 110 VDC. So something in the motor is causing the voltage to read higher than the line voltage.

I have read about "back EMF"... Also, one colleague said that the full-wave bridge causes the DC to be 1.414 times the AC (I understand that in power supplies, but I have no filter capacitor in the circuit)...

Can anyone comment on this? My customer thinks that I am overdriving the motor, but I think not.

Thanks in advance!

Regards,
Mike

RE: Voltage Across Motor Exceeds Applied Voltage??

What size is the motor?
How much load is on the motor? % of full load.
What is the motor rated current and actual current?
What is the rated RPM and the actual RPM?
If the motor is lightly loaded, it may be "picking the peaks" of the sine wave. The back EMF of the motor may be keeping the voltage up between peaks.
The induction of the shunt field makes a great filter. You don't need capacitors to have a filter.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: Voltage Across Motor Exceeds Applied Voltage??

As Bill says. The equivalent diagram for a lightly loaded DC motor actually is a capacitor. So the rectified sine's peaks keep the motor running at higher speed than rated. And the internal counter-EMF keeps the voltage up between peaks. Just like a capacitor does.

If you load the motor, that voltage will go down. I wouldn't say you are overdriving the motor. But make sure that the motor is designed for rectified single-phase AC. Some motors arc heavily when running from simple rectifiers and loaded. And that is not good.

As so many times, if you just had connected and ran the motor, you probably wouldn't have any problem..

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: Voltage Across Motor Exceeds Applied Voltage??

2

Quote:

Also, one colleague said that the full-wave bridge causes the DC to be 1.414 times the AC (I understand that in power supplies, but I have no filter capacitor in the circuit)...Can anyone comment on this?
The peak of an AC wave is 1.414 times the rms, so the peak of 120vac is around 170 volts.  

If you had an unloaded filtered rectifier, the output dc (average) voltage could approach this value (170vdc) as a maximum.

If you have an unloaded unfiltered single-phase full bridge rectifier, the dc (averaged) output would be  (1/[Pi]) * int{154*Sin(theta),  from theta = 0 to Pi} = (1/[Pi])[1 - - 1] = 170*2/Pi= 110.

With the resistor connected you read about 110vac which is about the expected unfiltered value... loading apparently didn't pull it down (and exact input voltage may not be 120).

With motor connected as well described by Bill and Gunnar there is a filtering effect and your voltage is somewhere in between the unfiltered value (110) and the highest expected filtered value (170).

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: Voltage Across Motor Exceeds Applied Voltage??

One correction of an intermediate value in bold below (the final calculated number was correct):

Quote (electricpete):

The peak of an AC wave is 1.414 times the rms, so the peak of 120vac is around 170 volts.  

If you had an unloaded filtered rectifier, the output dc (average) voltage could approach this value (170vdc) as a maximum.

If you have an unloaded unfiltered single-phase full bridge rectifier, the dc (averaged) output would be  (1/[Pi]) * int{170*Sin(theta),  from theta = 0 to Pi} = (1/[Pi])[1 - - 1] = 170*2/Pi= 110.

With motor connected as well described by Bill and Gunnar there is a filtering effect and your voltage is somewhere in between the unfiltered value (110) and the highest expected filtered value (170).

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: Voltage Across Motor Exceeds Applied Voltage??

(OP)
Bill, Gunner and ElectricPete:

Thanks very much for your feedback.  This does answer my questions.

Regards,
Mike

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