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Re: Input power measurement
Original poster: "Crow Leader by way of Terry Fritz <teslalist@qwest.net>" <tesla@lists.symmetric.net>
> Original poster: "Paul Nicholson by way of Terry Fritz
<teslalist@qwest.net>" <paul@abelian.demon.co.uk>
>
> Hi John,
>
> Yes, to extract the active component of the current means determining
> the portion of the current which is in-phase with the load voltage.
>
> Thus there's got to be some component which can take in both signals,
> and that component has to somehow effect an instant-by-instant
> multiplication of the two waveforms.
sounds like a job for an op-amp.
KEN