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.
Last edited by Viann; 16th June 2009 at 09:06 AM.
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".
Last edited by duffy; 16th June 2009 at 11:36 AM.
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".
You can often find a kill-a-watt on eBay. Many sellers ship worldwide. You haven't indicated your location so it's hard to say.
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.
Going to be difficult without a scope. Are you using a microcontroller?
Yes, a PIC16F877A.
I will be interfacing the ACS712 Hall Effect Sensor with it to obtain the Voltage and Current reading.
Last edited by Viann; 16th June 2009 at 04:50 PM.
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've used the ACS712 before, it's great for reading current, but it doesn't read voltage.
You could use an energy measurement IC such as one of these Energy Measurement | Analog to Digital Converters | Analog Devices.
Carl
Curmudgeon Elektroniker
I might be wrong, but isn't the product of (Volts * Amps) True Power?Originally Posted by user_88
Isn't it the other way round?
i.e. the ACS712 receives current and outputs voltage?
Once again, I might be wrong.
I'll have a look up on that.
Thank you.
Last edited by Viann; 17th June 2009 at 02:24 PM.
It OUPUTS voltage but it READS current.
So it doesn't READ voltage.
Yes, but you said you were trying to OBTAIN voltage, which that doesn't do.
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.
Last edited by crutschow; 17th June 2009 at 04:02 PM.
Carl
Curmudgeon Elektroniker
I see, thanks for clearing it up crutschow.
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?
A voltmeter.
Carl
Curmudgeon Elektroniker
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