Re: Wattmeter

> Terry, Steve, ALL
> Sorry about the miscommunication..  :^C
> 1.  Wind a 50:1 current transformer (CT) on ferrite core.  Pass
>     main power path =insulated= conductor through opening with
>     additional polyethylene tubing insulation.  This will feed
>     a shunt of approximately 12 milliohms.

Why not use one of those Honeywell hall effect current sensors? They
have fairly good bandwidth, a hole through the middle for your wire, and
produce a buffered analog output. Saves winding coils and toroids, etc. 
They were about $20 last time I checked..

Jim Lux                               Jet Propulsion Laboratory
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