One easy way to do it is to just measure the secondary of a transformer. This will be fixed-proportional to the input voltage, so if it's 110Vac the output will be, say, 3Vac, and at 120Vac it would be 3.27Vac.

In the past I've used the same step-down transformer that drives the processor power supply to do this... but that's not as accurate as a separate transformer with a fixed resistive load.

 

Be careful using the transformer method: small transformers give a significantly higher voltage off load, so you need to calibrate your measurements to account for this.

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Couldn't you just give it a small fixed load to stabilize it? Like a 10k resistor, or likewise?

 

10k is far too small.

A transformer's secondary voltage is normally specified at full load.

There are always more turns on the secondary than the ratio of specified secondary to primary voltage, to make up for the copper losses. A 120V to 120V transformer will not be 10:1, it'll be more like 10:1.2.

To calculate the reall turns ration, connect the primary to a known AC voltage and measure the secondary voltage and divide the primary by the secondary readings to get the real turns ratio.

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Understand that, but in this case it's just used as a voltage reference, not power source. I wouldn't think it would need to be loaded down much at all just to make the voltage stable. It doesn't have to be at the correct secondary voltage, just needs to be stable and linear to the primary.

As in, it wouldn't matter if a 12V secondary had an output of 16V due to no load, as long as it's still linear with what the voltage is on the primary.

 

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.

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Ok, that's what I thought. Just wanted to clarify.

 

I do have a circuit that can measure 120 Vac for your uP using a differential Op Amp. No isolation transformer required.
Larry
lbeaty3@tampabay.rr.com

 

You going to block it with a cap? It will pick up all the high-frequency noise on the line.

 

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

 

Sorry for the post

 

Larrybeaty Can you please post the circuit

 

Here is a power factor measurement circuit i found on the internet:

EDN Access — 04.28.94 3½-digit DVM IC measures power factor

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

Attached Files

ACin Differential.doc (37.5 KB, 6 views)

 

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.