Re: power measure
- From: John Fields <jfields@xxxxxxxxxxxxxxxxxxxxx>
- Date: Mon, 19 Sep 2005 11:56:21 -0500
On Tue, 20 Sep 2005 00:08:55 +1200, Jasen Betts
<jasen-b@xxxxxxxxxxxxxxxxxx> wrote:
>On 2005-09-14, cgrahl@xxxxxxxxxxxxxxxxxxxx <cgrahl@xxxxxxxxx> wrote:
>> Hi!
>>
>> I need to measure the power of some devices. I'm thinking to made a
>> digital ampermeter, since I already know the voltage (220 V), I only
>> need to measure the current to get the power (according to the formula
>> P = I x V), right?
>
>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.
---
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
0.83 amperes from 120V mains it would be dissipating 100 watts.
However, our motor under the same conditions would only be
dissipating 50 watts!
---
>> Does anybody have a schematic or a link to a digital ampermeter?
>> Or have another idea to measure the power of a device?
>> This must be digitally made, since this value will be used in a PIC
>> controller.
>
>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.
---
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.
---
>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.
---
Uh-huh...
Well, then, how about recommending one you _are_ familiar with?
--
John Fields
Professional Circuit Designer
.
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