Measuring DC motor effective voltage

[mawad]

Hello
I am using a DC motor which is drived by H-bridge. The input to the motor coil is 53.1khz PWM signal. Is the effective voltage on the motor coil: the average voltage or the true rms voltage. Is normal digital multimeter can measure the effective voltage on the motor coil correctly?
Thanks in advance

 

[electricpete]

Assuming you want to use voltage for example to predict dc current, I'd go with the average value rather than RMS value. The reason is that at the switching frequency, the motor acts like an inductor. I = 1/L integral (v(t) dt). The integral is an averaging operation. The high frequency stuff tends to get filtered out.



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[electricpete]

Whether your digital meter can read it correctly or not I'm not sure. I'll leave that to someone else.

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(2B)+(2B)' ?

 

[mawad]

Actually, I want to measure the effective voltage on the motor coil to use it to estimate the current electrical power. I am confused is the effective voltage on the motor coil is the rms or the average of the pwm as both are different

 

[electricpete]

Look at it this way, the inductance filters out the vast majority of the ac so you are left with a dc current. (neglect the small ac current *)

If you have a dc current, the associated power is product of that dc current times the dc component of the associated voltage.

the dc component of voltage is the average value of the voltage. That's what you want. rms is not relevant for this purpose.

* If you wanted to refine your calculation to include parasitic losses associated with very small ac currents that might result from the ripple, then you'd need to consider more info but rms voltage would still not be what you'd need.


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(2B)+(2B)' ?

 

[waross]

An old analogue meter with a d'Arsonval movement will respond to the average value of the current.
The ratio between RMS and average for a sine wave is about 1.1 to 1.
The d;Arsonval movement was almost universal in the old multimeters.
Voltage was converted to current with a series resistor.
If you use an old moving meter, d'Arsonval multi meter it should indicate about 110% of average voltage on the AC scale. It should read average on the DC scale.

Bill
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"Why not the best?"
Jimmy Carter

 

[electricpete]

Just to emphasize for op benefit, that factor 1.1 applies to sinusoidal as Bill said, not to pwm.

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(2B)+(2B)' ?

 

[Skogsgurra]

RMS can be anything when you measure a DC armature voltage fed from a PWM H-bridge.

Take stand-still as an example. RMS will usually be very high because it indicates RMS of the square waveform which, if the meter has high enough band-width, is equal to the DC link voltage. The motor has no net DC (or average) voltage in that situation. So it is obvious that you shall not measure RMS at stand-still or low speed. And you shall not measure RMS at other speeds, either.

Same thing applies to a battery charger. If you measure RMS, you will get a much higher current than if you measure AVG. I actually had a job where a (locally) rather well-known manufacturer had a problem with resulting ampere-hours when charging with 10 A_RMS for ten hours, the resulting battery capacity was only around half of the calculated charge. Changed the measurement to AVG and then the calculated and actual charges were in better agreement.

RMS or TRMS was a sales argument for many years. It probably still is, but you have to know what you want to know. For heating effects, RMS is usually OK. But for DC applications, you have to think a little more. And it usually turns out that AVG is the correct choice.

Gunnar Englund

http://www.gke.org

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Half full - Half empty? I don't mind. It's what in it that counts.

 

[mawad]

Actually, I am little bit confused. As I want to measure the electrical power of a DC motor which equal to effective voltage*current. This motor is controlled by PWM motor driver. I was looking online and found some suggest that the effective voltage is the average voltage while the others suggest that it is the RMS voltage. I know there is a formula to convert from average voltage to the rms, but I am not sure which voltage is the effective voltage on the motor in this formula Veff=iR+L(di/dt)+backEMF

 

[electricpete]

If it were an sinusoidal ac circuit, you'd be working with rms. If you have a resistor in a single phase ac cirdcuit, you can compute the power using rms values in identical manner that you could compute power in the same resistor using dc values. That may be the origin of the term "effective" when referring to rms values.

But for your dc circuit (especially pwm), rms voltage has no usefulness.

To see why, look at the power calculation you want to do.

Assume the current is dc (it is well filtered by the inductance).

<p(t)> = <I0 * v(t)> = <I0 * v*t) >
where
< > denotes average value over time
v(t) = (V0 + v1*cos(w1*t) + v2*cos(w2*t) + ….)
where V0 is the dc component = average component

<p(t)> = <I0*(V0 + v1*cos(w1*t) + v2*cos(w2*t) + ….)>
<p(t)> = <I0*V0> + <I0*v1*cos(w1*t)> + <I0*v2*cos(w2*t) + ….)>

what is <I0*v1*cos(w1*t)>? It is 0 (two constants times a sinusoid… the sinusoid has zero average). Same for <I0*v2*cos(w2*t)> etc

<p(t)> = <I0*V0> = I0 * V0
so the relevant portion of v(t) for purposes of calculating power is V0. It is the average value also called the dc component.


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(2B)+(2B)' ?

 

[mawad]

Many thanks electricpete for detailed elaboration
So this means if I used low pass filter connected in parallel with the motor terminals I can measure the effective voltage as the low pass filter will provide the average value. Also, the current I0 is the average value too. Many thanks electricpete

 

[waross]

Actually, I want to measure the effective voltage on the motor coil to use it to estimate the current electrical power.

The current is determined by the difference between the effective applied voltage and the back EMF and the motor resistance.
The back EMF is mostly determined by the RPM.
The RPM is mostly determined by the applied effective voltage and by the load.
This is assuming a permanent magnet field.
A wound field adds some additional factors.

Bill
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"Why not the best?"
Jimmy Carter