Gee Russel, quite brillant on your part for working that out!

----- Original Message -----
From: "Russell McMahon" <apptech@PARADISE.NET.NZ>
To: <PICLIST@MITVMA.MIT.EDU>
Sent: Thursday, October 23, 2003 2:34 PM
Subject: Re: [PIC]: Pic based wattmeter (I forgot the attachment !)


> This circuit is MUCH cleverer than the descriptions so far have given it
> credit for.
> It implements a true four quadrant multipler.
>
> Note that the 25w resistors are not major errors - they are meant to be
> 0.25w and the decimal point has become hard to see. A good example of why
> leading zeroes should be used.
>
> Sadly there doesn't seem to be any trace on the web of the original
article.
> its very sad that the cct Cellar article didn't provide a  proper or even
> superficial description as this is the real genius of the design.
>
> Consider what happens when mains is at full positive half cycle into "hot"
> terminal.
> Top left and bottom right optos will be on driven by same 100k resistor
R7.
> If there is no current flow then both optos are driven equally and the
R5/R6
> midpoint is maintained at half supply. The U5A integrator is not driven
off
> midpoint.
>
> The integrator is "leaky" and is effectively an amplifier with a gain of
> between about 15 and 200 but with its response smoothed by the 4.7uF C5.
>
> Now, if current is flowing the rh end of R8 is more positive than left
hand
> end and there will be less voltage across the bottom left opto than across
> the top left opto. The midpoint of R5/R6 is driven towrds ground and the
> integrator ramaps positively. The amount of midpoint offset is controlled
by
> the product of the drive curret - provided by the input mains voltage, and
> the amount by which this voltage is offset which is caused by the current
> flow. Increasing either voltage or current will linearly increase the
offset
> and thus the integrator drive.
>
> When the voltage reverses the opposite diagonl of optos conduct. Both
> voltage and current sense are swapped in the two optos so a positive
product
> still occurs.
>  If voltage and current are out of phase there will be a period when the
> product is neagtive and the integrator input will be redcued. Overall a
true
> VI summing is provided across the cycle. (What it doesn't (seem to) do is
> provide an RMS sum as the amplifer/integrator does not track the cycle.
I'd
> have to think about this further).
>
> The multiplier gives a true (more or less) VI product across the entire
> waveform.
> I consider this circuit to be utterly brilliant. It may not produce an
> overly accurate result but for very very simple implementation it is very
> hard to beat for cost. Where are my quad optos ... ? :-)
>
>
>         Russell McMahon
>
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