Re: Input power measurement

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "by way of Terry Fritz <teslalist-at-qwest-dot-net>" <dhmccauley-at-spacecatlighting-dot-com>


 >  > 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.


This can be accomplished through the use of log and anti-log configured
op-amps and is fairly easy.

The Captain