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Finding the power usage of my project

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goofeedad

New Member
I have built a small device that runs off of a simple power cord that I salvaged from a hair dryer. I am very interested in how much power it uses but I don't have access to expensive equipment. I do have a digital voltmeter, that checks ohms, amps, etc...

Any way of doing this with some degree of accuracy?
 

birdman0_o

Active Member
It should essentially be your line voltage (240 or 120 ) depending on where you are from multiplied by the current you read by placing the voltmeter in series with the circuit, as easy as that!
 

bountyhunter

Well-Known Member
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bountyhunter

Well-Known Member
The power factor of normal household mains is cos(90) = 1 No?
NO. The "typical" power factor of an average household with all it's various appliances was about 0.85 back when I was in school.

However, that's the shift that all the appliance loads in the house put back on the source. Your specific device will have it's own power factor based on what it is and that will determine how to calculate the power. Devices which have a bridge rectifier and transformer can have bizarre current waveforms which don't read accurately on digital RMS voltmeters anyway.
 
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goofeedad

New Member
Thanks, I was happy with the first reply. I don't have the equipment to find the rms values. Can I get a rough value by useing the first reply?

I'm pretty sure it will give me a value I can use.

Again, Thanks.
 

bountyhunter

Well-Known Member
Thanks, I was happy with the first reply. .
You can use it as long as you are OK with the wrong value.

Can I get a rough value by useing the first reply?

I'm pretty sure it will give me a value I can use.
.
If you are happy with the value, go ahead and use it. But it may or may not bear any resemblance to the real number. It depends on the power factor and that depends on the circuit configuration. You should also realize any meters you have will not be able to read any waveform other than a sine wave. If the current is not sinusoidal in shape, the reading will be gibberish.

For example: if your power comes into your project through a bridge rectifier and capacitor, the current waveforms will be Haversine pulses which are not sinusoidal and will not read accurately on any meter that didn't cost thousands of dollars.


You could learn a lot of useful info researching how AC power is measured and why it has to be measured that way. If you just plug in a meter which can't measure what you want to know and blindly take a completely false reading, you are not really learning.

Some fool at national Semi wrote a document with some good info on the subject. Check page 52:

http://www.national.com/appinfo/power/files/f5.pdf
 
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crutschow

Well-Known Member
Most Helpful Member
You might be interested in a "Kill A Watt" power meter. It plugs into an AC outlet and you plug any device into it and displays RMS voltage, RMS current, power factor, V-A, power, and total power consumed (KWH). It can be purchased for about $20-25. I have one and it seems to work well.
 

bountyhunter

Well-Known Member
You might be interested in a "Kill A Watt" power meter. It plugs into an AC outlet and you plug any device into it and displays RMS voltage, RMS current, power factor, V-A, power, and total power consumed (KWH). It can be purchased for about $20-25. I have one and it seems to work well.
Amazing if it works. We used to have to rent power analyzers to get power data (they cost about $5k each so Skinflint Joe would never buy one).

Does look tempting....

http://www.industrialproducts.com/P3-Kill-A-Watt-Electricity-Monitor-p/5020-402.htm
 
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jrz126

Active Member
I just bought one a couple months ago. they're pretty handy to have, I dont think they are as accurate as a power analyzer.
 

crutschow

Well-Known Member
Most Helpful Member
I just bought one a couple months ago. they're pretty handy to have, I dont think they are as accurate as a power analyzer.
It's not surprising that they may not be as accurate as a commercial device for 20 times the price. It claims 0.2% accuracy and even if it was only 2% that's still more than good enough for measuring household appliance power consumption.
 

goofeedad

New Member
Thanks Crutschow, sounds like that will do the trick. Just need to find one.

Man! All I need right now is a guesstimation. I don't have nor can I get my hands on a $5K device. I was happy with a reminder of the PVI formula. I'll probably use that for now. I just want an approximate value.

Thanks All.
 

The Electrician

Active Member
It's not surprising that they may not be as accurate as a commercial device for 20 times the price. It claims 0.2% accuracy and even if it was only 2% that's still more than good enough for measuring household appliance power consumption.
I have one of these, too, and the main deficiency that I have encountered is that it can't measure small power with much resolution. For example, if you want to how much power a wall wart draws when unloaded, you won't get an accurate reading; the Kill-a-watt only has one digit with no digits after the decimal point at those low power levels.
 

bountyhunter

Well-Known Member
It's not surprising that they may not be as accurate as a commercial device for 20 times the price. It claims 0.2% accuracy and even if it was only 2% that's still more than good enough for measuring household appliance power consumption.
But they don't say under what conditions they get that accuracy. I would be willing to bet it's only usable for sinusoidal type waveforms. The hardest part of measuring power is to be able to get the true RMS current because some current waveforms are bizarre in shape, not sinusoidal at all. The better meters (we're talking several hundred dollars) have "true RMS" circuitry to try to get an RMS value of waveforms of different shapes. Their problem is called "crest factor" limitations which means if the peak of the wave is more than maybe 3X the average value, they are not accurate. The super good meters (thousands of dollars) use a calibrated thermal device to measure RMS current based on the amount of heat it produces in a resistive element.

A $25 meter is not going to have much in the way of "true RMS" hardware which means it probably works with sine waves, which is OK for most appliances.
 

bountyhunter

Well-Known Member
Thanks Crutschow, sounds like that will do the trick. Just need to find one.

Man! All I need right now is a guesstimation. I don't have nor can I get my hands on a $5K device. I was happy with a reminder of the PVI formula. I'll probably use that for now. I just want an approximate value.

Thanks All.
If you post the schematic and specs, I could have a guess at the power factor. If it uses a single phase bridge rectifier to create a DC supply, the PF is close to 0.6 ballpark. Without knowing what the thing is, hard to say.
 
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goofeedad

New Member
If you post the schematic and specs, I could have a guess at the power factor. If it uses a single phase bridge rectifier to create a DC supply, the PF is close to 0.6 ballpark. Without knowing what the thing is, hard to say.
Well, it's a mist machine that drives 4 piezos and their driver circuits as-well-as 3 - 4" pc fans, a microprocessor, a 24VAC solenoid, internally it has 2 - 36VDC transformers to drive the piezo drivers and a 24VDC transformer for the solenoid, a 5VDC cell phone power supply for a relay it has. I don't have a schematic, I assembled it (obviously) from off the shelf parts. The 4 piezo driver circuit boards all have a bridge rectifier.

Sorry you asked now I bet. I appreciate everyones help, I really do. At this stage I'm looking for something close would be nice but I don't really expect miracles.

Thanks.
 
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