power measure

You are assuming the voltage is 220V, because that's what the supply authority says (I'm assuming this is mains 200VAC) - this may not be the case at any particular time.

It is probably going to be easier and safer to get a digital multimeter with a serial (RS232/485) link to your uP. There are meters which will read power of your load by measuring V and I directly. The more expensive ones even tell you the power factor.

Ken .

Why do you need a digital one? Analog ones work fine, and they are cheap. You need a multimeter. The problem is that you did not say what power consumption range you need. If less than 10 amps worth, you can buy a meter much cheaper than you can make one for.

Don Stauffer wrote this in :

RTFP. He needs it to input to a PIC circuit. Make that with a analogue one. But maybe just opening a cheap DMM, and figuring out a suitable place to get the value is the cheapest way, if, as said you wan't to measure

will be a bit of work to build. But best chance is really to buy a el-cheapo dmm, and figure out where to get a binary value out of it.

Maybe things have changed, but as I recall most (of the cheaper) analogue multimeters don't have an AC current range

David

D> cgrahl@blablabla_nospam.com wrote:

They can do, but they are picky about wave-shape.

Ken

"quietguy" wrote in message news: snipped-for-privacy@REMOVE-TO-REPLYconfidential-counselling.com...

Article on Digital meter with a Serial Port

only if it's 220V DC if it's AC power can flow the opposite direction during some of the cycle meaning that VxA is more than the watts used.

you have to start analogue and convert it wit soome sort of anallogue to digital converter, (I think they make PICs with inbuilt ADCs.)

if you're measuring AC you'll have to measure voltage too and multiply the readings (volts * amps) average that and then take the square root. for this you're going to want a microcintroller that knows how to multiply. (since frequent ADC readings give better accuracy, and software multiply is typically 5-20 times slower than hardware multiply.

not being familiar with the PIC line I can't reccomend a particular device.

that's assuming that the current is a sine wave...

--- If it's AC, the voltage across the load alternates as does the charge flowing through it. Power though, continues to be dissipated in only one direction. For example, an incandescent lamp doesn't light up for one half-cycle and then dim the next.

The reason that VA is sometimes greater than watts is because a load may be reactive and, if it is, then only the resistive portion of the circuit will dissipate power while the circuit will still see VA across it. For example, an incandescent lamp will present a nearly pure resistance to an AC signal since it is not very reactive, while a motor will look like a resistance in series with an inductance.

The result of that will be that the time-varying voltage and current waveforms across and through the lamp will be in phase, while for the motor they won't be.

Consequently, for the lamp, since the phase angle between current and voltage will be zero we can say:

P = VA(cos phi) = V * A * (cos 0°) = V * A * 1 = VA watts

So, for a resistive load, the power dissipation is simply the product of the voltage across it and the current through it.

For the motor, however, (assuming a phase angle of 60° between current and voltage, we have:

P = V * A * (cos 60°) = V * A * 0.5 = 0.5VA watts.

Numerically, that means that if our incandescent lamp was drawing

---

--- No. You need to measure the voltage and current and, if necessary, convert the values read to RMS, then measure the phase angle between the voltage and current, get the cosine of that angle and, finally, multiply the voltage, the current, and the cosine of the phase angle together to get power.

---

--- Uh-huh...

Well, then, how about recommending one you _are_ familiar with?

-- John Fields Professional Circuit Designer

Jasen Betts wrote in news:vmm203-6pn.ln1 @news.compass.net.nz:

actually any wave. E is RMS value...

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I was thinking of an reactive load... I'm surprised you didn't catch that with a reactive load for some period of the cycle the current oposes the voltage, during that period the instantaneous wattage of the device is negative. - energy is flowing from the device into the supply.

how do you compute phase diference when there's no gaurantee that the current looks anything like a sine wave it seems to me that that, it would be harder to compute than calculating the average of a number of voltage.current products.

Atmel ATmega48 - 20 MHZ risc microcontroller with plenty of 10-bit A-D channels.

Bye. Jasen

I would recommend that the OP investigate four quadrant analog multiplier chips, such as the Analog Devices AD633 and AD532, and the OnSemi MC1495, etc. There have been several power meters designed and available on the web using these and similar devices. Start by looking at the design at

Cheers!!!!