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Measuring Power Factor with an oscilloscope?

F

Frank White

Jan 1, 1970
0
I would think it is a simple matter of a current probe and a voltage
probe, one on each channel and measuring the offset to calculate
the phase angle.

Is this a reasonable method given that your probes are properly
calibrated?


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A

Andrew Gabriel

Jan 1, 1970
0
I would think it is a simple matter of a current probe and a voltage
probe, one on each channel and measuring the offset to calculate
the phase angle.

Is this a reasonable method given that your probes are properly
calibrated?

Yes. You'll probably need to take a picture of a steady waveform
so you can go off and measure it accurately. Measuring the offset
is only useful if both are sine waves. For many loads nowadays, the
current waveform can look very strange, and on many mains supplies
the voltage waveform is noticably flattened at the peaks, in which
case you will need to integrate by hand the product of the voltage
and current waveforms over a cycle to work it out.

OTOH, true power meters are the tool designed for the job, and
you can just read off the answer.
 
D

daestrom

Jan 1, 1970
0
Frank White said:
I would think it is a simple matter of a current probe and a voltage
probe, one on each channel and measuring the offset to calculate
the phase angle.

Is this a reasonable method given that your probes are properly
calibrated?

If the two are sine waves and not some chopped up waveform caused by
non-linear devices.

If your scope has some nice feature for measuring period/time-delay, you
could measure the period of the voltage waveform to find frequency (if it's
not known), then measure the time delay between voltage and current peaks.
Figure out the phase displacement and you're practically home free.

daestrom
 
| Frank White wrote:
|> I would think it is a simple matter of a current probe and a voltage
|> probe, one on each channel and measuring the offset to calculate
|> the phase angle.
|>
|> Is this a reasonable method given that your probes are properly
|> calibrated?
|
| Put the "voltage" on the vertical and the "current" on the horizontal.
| You'll get a 1:1 Lissajou figure. If the scope's axis gains are equal,
| two sine waves in phase will show a line at 45 degrees. At 90 degreees
| they will form a circle. Phases in between will form an ellipse, from
| which you can calculate the phase angle.
|
| As has been noted, this will only work correctly with sine waves.

But you can still see the variations between voltage and current with
this method. It just won't be directly easy to figure out specifically
what is happening is your voltage is not a sine wave. But with a voltage
sine wave and harmonic currents, it should be rather obvious.
 
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