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How would I measure Power Factor?

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Viann

New Member
I'm working on a Digital Wattmeter project, where the wattmeter is plugged to the mains, and an appliance is plugged to the wattmeter.

My question is, is there a way to measure Power Factor accurately?

EDIT:
I have heard of a Power Factor IC, which conveniently measures the power factor when connected. However, I could not seem to come across it.
 
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duffy

Well-Known Member
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. The power factor is the proportional difference between the two.

By the way - you can buy a wattmeter that works like that and measures power factor for $20. The "Kill-a-watt".
 
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Viann

New Member
Since I cannot use an osilloscope, what would you recommend I use though, to measure the difference in phase?

And sadly, where I live, there is no "kill-a-watt".
 

Viann

New Member
There we go.

Anyway, the purpose is building a digital wattmeter, sort of one like the "kill-a-watt" meter. It would be fantastic if the schematic of said meter is readily available, that would really help with information.

If not, I would still like to know how to accurately measure Power Factor without the help of an oscilloscope.
 

duffy

Well-Known Member
Going to be difficult without a scope. Are you using a microcontroller?
 

Viann

New Member
Yes, a PIC16F877A.

I will be interfacing the ACS712 Hall Effect Sensor with it to obtain the Voltage and Current reading.
 
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user_88

Member
Besides the phase method, you can measure the apparent power going to the load, which is the product of volts * amps ... and then measure the watts to the load, and take the ratio of these two quantities. .... If you are capable of taking these measurements.
 

duffy

Well-Known Member
I've used the ACS712 before, it's great for reading current, but it doesn't read voltage.
 

Viann

New Member
Besides the phase method, you can measure the apparent power going to the load, which is the product of volts * amps ... and then measure the watts to the load, and take the ratio of these two quantities. .... If you are capable of taking these measurements.
I might be wrong, but isn't the product of (Volts * Amps) True Power?


I've used the ACS712 before, it's great for reading current, but it doesn't read voltage.
Isn't it the other way round?
i.e. the ACS712 receives current and outputs voltage?



Once again, I might be wrong.


You could use an energy measurement IC such as one of these Energy Measurement | Analog to Digital Converters | Analog Devices.
I'll have a look up on that.
Thank you.
 
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duffy

Well-Known Member
It OUPUTS voltage but it READS current.

I will be interfacing the ACS712 Hall Effect Sensor with it to obtain the Voltage and Current reading.
So it doesn't READ voltage.

Viann said:
Isn't it the other way round?
i.e. the ACS712 receives current and outputs voltage?
Yes, but you said you were trying to OBTAIN voltage, which that doesn't do.
 

crutschow

Well-Known Member
Most Helpful Member
I might be wrong, but isn't the product of (Volts * Amps) True Power?
For DC. For AC the product of volts times amps is apparent power. The product of volts times amps times the cosine of the phase angle between them is true power. For example a capacitor will draw AC current when placed across an AC source but it will draw no real power from the source since the current and voltage are 90 degrees out of phase.
 
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Viann

New Member
For DC. For AC the product of volts times amps is apparent power. The product of volts times amps times the cosine of the phase angle between them is true power. For example a capacitor will draw AC current when placed across an AC source but it will draw no real power from the source since the current and voltage are 90 degrees out of phase.
I see, thanks for clearing it up crutschow.

Yes, but you said you were trying to OBTAIN voltage, which that doesn't do.
Ah, I'm terribly sorry, I misunderstood how the ACS712 functions.

With that note, is there a way to measure voltage from the mains connecting to an appliance?
 

crutschow

Well-Known Member
Most Helpful Member

duffy

Well-Known Member
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.
 

Hero999

Banned
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.
 

smanches

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

Hero999

Banned
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.
 

smanches

New Member
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.
 
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