Re: RMS and Average Current

On Tue, 3 Jan 2006 13:21:32 +1100, "Phil Allison" <philallison@xxxxxxxxxx>
wrote:

>
>"John Larkin"
>>
>> The power is the average of the instantaneous series of E*I samples,
>> which can get messy to compute.
>>
>> If it's a clean rectangular pulse, measure pulse current and
>> simultaneous diode voltage drop, and do
>>
>> P = E * I * n
>>
>> where n is the duty cycle.
>
>
>** The power produced by such a rectangular pulse can be measured with an
>average responding meter.
>
> P = E . I
>
>Where E = diode conduction voltage
>
>and I = average amps.

Actually, the power delivered by *any* current waveform (for which an
"average" exists) into a constant voltage drop can be measured by an average
responding meter.

The canonical computation of average power for a time T is:

1/T * Integral from 0 to T of [v(t)*i(t)] dt

If v(t) is constant (call it V), then it comes out of the integrand and we're
left with:

V/T * Integral from 0 to T of [i(t)] dt

The integral is now just the average value of the current waveform.

And, of course, if the current is constant and the voltage is time-varying,
the same formulation applies using the average voltage.

>
>
>
>......... Phil
>
>
>

.

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