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September 14th 04, 05:32 PM

On 14 Sep 2004 13:17:36 GMT, (JGBOYLES) wrote:

If you are thinking of EÂ²/R, you should note this is division, not
multiplication.

Thanks Richard. What I am thinking is: assuming a perfect 50 ohm load, scale
the voltage divider rectifier combo to give 10.0 VDC at full scale watts, 1500
in my case.

Hi Gary,

1500W into 50 Ohms yields 274V which you then divide by 27.4 to obtain
10V.

But, what I intend to do is run the voltage divider
signal to a multiplier whose output will be V**2/10. If V is 10. volts for
1500 watts then the multipier output will be 10.0 volts. let 10.0 volts equal
1500 watts on the meter scale.

As you already have 10V by divider action, what do you need to
multiply? The chip offers two inputs one of which I presume is a
constant 10V? Something is missing here because you are still
dividing and if this through a divider at the output of the
multiplier, then you have jacked the multiplier's output to the rail
BEFORE the division (loss of dynamic range).

Checking another point, assume 500 watts, then
the output of the divider will be 5.76 volts,

500W into 50 Ohms yields 158V which you then divide by 27.4 to obtain
5.77V. OK

and the output of the multiplier
will be 3.33 volts, which is 33% of 10 volts.

More missing discussion. Like how does 10 Â· 5.77 / 10 = 3.33? The
10's drop out just like before (a useless operation).

The wattmeter will read 33% of
1500 or 500 watts. The scale will be linear.

I find it difficult to observe how a non-linear (power) response
becomes linearized without an inverse (log) operation.

I have not seen it done this way before, although I am sure it has.

Not by your description.

For this
method to accurate your load must be near 50 ohms.

That value is immaterial to anything but ONE value of power
representation. The nonlinearity of the response would force you to
have a table of Logs nearby to translate any other reading.

The AD633 is a cheap,
easy-to-use multiplier chip. Using a +/- 15 VDC supply, it looks like it will
have the dynamic range I need.

No, it won't as I've already shown. There is a MUCH better solution
and far cheaper (FREE):
http://www.analog.com/UploadedFiles/...06767AN304.pdf

To obtain:
http://www.analog.com/productSelecti...les/index.html

73's
Richard Clark, KB7QHC

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