# 120 Volt DC Motor Question

Discussion in 'General Electronics Chat' started by technicom, Jun 30, 2010.

1. ### technicom Thread Starter New Member

Jun 29, 2010
5
0
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.

Regards,
Mike

2. ### Dx3 Member

Jun 19, 2010
87
7
Your main error is in the measurement.

The peak voltage of a sine wave is 1.414 times the RMS value, but without a capacitor, you will not develop the peak voltage or apply it to the motor. You are actually applying 120 volts to the motor, but it is in the form of pulses. The meter set on DC can't interpret that correctly.

Forget about the "back EMF". It is not part of the problem.

Jul 7, 2009
1,585
142
I agree with Dx3. Your best bet is to get a digital storage scope and use it to measure the RMS value of the waveform. The two that I have always read the real RMS value. Besides, the scope lets you see what's really going on.

Unfortunately, manufacturers like Fluke and HP have told a little white lie all these years. When they say their DMMs are "true RMS" measuring, it's probably not what you think it means. What they should be saying is that the meter measures the AC-coupled RMS value. This goes back to at least 1964 with the introduction of the HP 3400A RMS voltmeter (a favorite of mine) and potentially even before.

What apparently is the kosher terminology today is that you want an instrument that measures "AC+DC". This is the measurement that all of us would normally call the RMS value of a waveform. This feature can be found on some of the high-end DMMs being sold now. Take a look at the HP 3403C instrument made in the 70's and early 80's -- it was a nice tool that made all three (DC, AC-coupled RMS, and real RMS) measurements. My beloved HP 3456A also makes these measurements, as well as the Agilent instruments descended from it.

So what's a poor body to do? You can still measure the real RMS value of a waveform with your "true RMS" meter like your Fluke 87. Just measure the DC value and the AC value, then add them in quadrature. That means

$RMS = sqrt{DC^2 + AC^2}$

If you don't have a "true RMS" meter, then you simply can't make this type of measurement.

4. ### BillB3857 AAC Fanatic!

Feb 28, 2009
2,450
371
Is it a permanent magnet field, series wound or shunt wound motor? If it is a permanent magnet motor, then counter EMF may come into play since the motor will have a generator effect during the periods of time that the rectified power is dropping off. The result would appear as almost pure DC. Loading the motor should reduce the measured voltage.

5. ### Dx3 Member

Jun 19, 2010
87
7
The "back EMF" is not what you are driving the motor with. I'm afraid discussing back EMF with the customer will only muddy the conversation. However, try putting your meter on AC voltage and see if that reading makes sense. If the AC component is rather small, it would seem that back EMF is filling in the droops between the pulses. If the AC measurement is rather large, you're just reading the pulses from the rectifier. Once you know these things, you will have nothing new to report. 120 VAC rectified to pulses is still 120 V.