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Re: Input power measurement



Original poster: "Crow Leader by way of Terry Fritz <teslalist-at-qwest-dot-net>" <tesla-at-lists.symmetric-dot-net>

 > Original poster: "Paul Nicholson by way of Terry Fritz
<teslalist-at-qwest-dot-net>" <paul-at-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