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How would I measure Power Factor?

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The transformer doesn't have to have any load to be stable, it just needs to have a constant load. The problem is that unless you're going to fully load the transformer, you'll need to determine the secondary voltage for a given primary voltage empirically.
 
Ok, that's what I thought. Just wanted to clarify.
 
You going to block it with a cap? It will pick up all the high-frequency noise on the line.
 
AC voltage measurement

If you look at an oscilloscope presentation of the AC line, you see a pretty clear sine wave with little noise on it.
Yes, it uses large caps to pass the 60 HZ faithfully (0.1 uF). I have used it in several microprocessor circuits and works as expected.
If you have ever tried to transformer couple AC voltage, you will find the voltage waveform bears little resemblance to the incoming waveform.
I used Xfmr step down for a product before i came up with this IC approach. Product went out, sold a bunch and worked, but I always had this nagging that the waveform was not what i wanted.
The circuit is yours for the asking, free is good!
Larry
 
Here is a power factor measurement circuit i found on the internet:

**broken link removed**

Looks simple enough. You can eliminate the isolation transformer with a differential input op amp circuit. I have done this successfully on several projects. The op amp circuit is isolated from the power line with 1 meg Ohm resistors.

Keep up the good work!

Larry
 
I am trying to attach the circuit to this reply. The information on this site does not give me confidence that I have accomplished this.

the file is from my computer and labeled "ACin Differential" . If not done right i will keep trying to attach the file.

I see it attached correctly. Note R24 and R25 are for centering the output voltage at 1/2 reference voltage. If you want zero crossing at zero volts, change R25 to 10K and remove R24.

You can see the high resistance isolation. Two resistors are used to increase the voltage capability of the surface mount resistors.

This circuit eliminates the transformer and the transformer distortion. The input leads do not care which is at neutral and hot of the AC line. Only your output phase is dependent on the lead placement.

Larry
 

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You need to measure both the voltage waveform and current waveform (since you are making a wattmeter you would be doing this anyway) and compare the difference in phase.
That only works in cases where both the voltage and current waveforms are purely sinusoidal, which is rarely the case for the current. Measuring power factor accurately is complex, the cheap meters use a mathmatical approximation which is reasonably accurate as long as the peak-to-average value of the current isn't more than about 3X.
 
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