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

If you are thinking of E²/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 · 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

1500W into 50 Ohms yields 274V which you then divide by 27.4 to obtain
10V.
As you already have 10V by divider action, what do you need to
multiply?

Hi Richard, I want 10.00 VDC to indicate 1500 watts on my 0-10v meter with a
linear scale. If I run the 10 volts thru the multiplier (squaring) circuit I
get (10)**2/10=10 volts. So far so good, and you don't need a multiplier.
500W into 50 Ohms yields 158V which you then divide by 27.4 to obtain
5.77V. OK

If I run 5.77v to my 10 volt full scale meter it will read 5.77/10*1500=865.5
watts which is not 500 watts. However, if I run 5.77 thru my squaring circuit
I get (5.77)**2/10=3.33 volts. Now my meter reads 3.33/10*1500=500 watts which
is correct.
If you plot the voltage across a 50 ohm resistor vs power V**2/50=power, you
get a squared relationship. You can use volts to represent watts only if you
scale the meter face correctly. I have seen wattmeters that did this. I
believe the Drake model did. 0-100 watts took up the first half of the meter,
and 100-1000 watts took up the other half.
If you linearize the voltage by squaring and scaling you get a nice readout
with 100 watts being on the left hand side of the meter, and 1500 on the right
hand side, and linear in between.
The AD633 probably uses log amps in realization of multiplication, I haven't
looked at the internals. I have done multiplication with Log amps. I have to
disagree on the AD538 being a better solution. One could use the 538, but the
633 is much simplier to use. Also I already had some AD633s
73 Gary N4AST

"JGBOYLES" wrote in message
...
1500W into 50 Ohms yields 274V which you then divide by 27.4 to obtain
10V.
As you already have 10V by divider action, what do you need to
multiply?


It all works for me in Excel. Squaring the voltage eliminates the need for
any log stuff.

E = Root (P*R)

Scaled V is V / 273 but you'll have to go further to stay in the dynamic
range of the multiplier. V^2 should be around 10V or whatever the mult can
output.

W V scaled V V^2 V^2/10 Output ratio
1500 273.9 10 100 10 1
500 158.1 5.77 33.33 3.33 0.33
100 70.71 2.58 6.67 0.67 0.07



--
Steve N, K,9;d, c. i My email has no u's.

E = Root (P*R)

Scaled V is V / 273 but you'll have to go further to stay in the dynamic
range of the multiplier. V^2 should be around 10V or whatever the mult can
output.

Thanks for checking my calcs. Steve. I had to do what your spreadsheet did by
hand. I should note that since I have to convert the voltage to the multiplier
to DC, at 1500 watts we are working with 273*SQRT 2 or 386 volts. I size the
divider so that 386 this gives 10.0 volts to the multiplier. With a dual
polarity 15VDC supply, the multiplier has enough dynamic range.
73 Gary N4AST

Oh, yea! I forgot the bit about Peak...silly me.
73, Steve K9DCI

"JGBOYLES" wrote in message
...
E = Root (P*R)

Scaled V is V / 273 but you'll have to go further to stay in the
dynamic
range of the multiplier. V^2 should be around 10V or whatever the mult
can
output.

Thanks for checking my calcs. Steve. I had to do what your spreadsheet
did by
hand. I should note that since I have to convert the voltage to the
multiplier
to DC, at 1500 watts we are working with 273*SQRT 2 or 386 volts. I size
the
divider so that 386 this gives 10.0 volts to the multiplier. With a
dual
polarity 15VDC supply, the multiplier has enough dynamic range.
73 Gary N4AST

Hi Gary,

I like the topic of discussion. I've been working with the AD633's for
several years now. I'm wondering why you vered away from the V*I idea.
That would make your calculation good for any load. Instantaneous V
times instantaneous I will give the answer whatever the relative phase
happens to be. Maybe I'm missing something?

Hi Jim, this didn't show up on my ng reader for some reason, hope you don't
mind me replying on the ng.
The reason I didn't the V*I route, which as you say, is the obvious way to
go: I did not feel I could get accurate V and I samples over a wide range of
RF frequencies, with my limited knowledge and resources. I thought a simple
resistive voltage divider would make it real simple, and somewhat frequency
independent. I am now finding through the responses I have gotten, that
resistors will be frequency dependent as well. Darn.
I recently tried to use 633s in an application at 60hz to detect when a very
large generator went from producing power (generator) to absorbing power
(motor). I found that the potential and current transformers I was using
introduced an unacceptable phase shift from primary to secondary. I tried
several different types but finally took another approach. This is probably
another reason that steered me away from the V*I solution.
If you have a simple way to get accurate V and I samples for say 3-30MHZ I am
all ears, and forever in your debt.
73 Gary N4AST

"JGBOYLES" wrote in message
...
1500W into 50 Ohms yields 274V which you then divide by 27.4 to obtain
10V.
As you already have 10V by divider action, what do you need to
multiply?

Hi Richard, I want 10.00 VDC to indicate 1500 watts on my 0-10v meter
with a
linear scale. If I run the 10 volts thru the multiplier (squaring)
circuit I
get (10)**2/10=10 volts. So far so good, and you don't need a multiplier.
500W into 50 Ohms yields 158V which you then divide by 27.4 to obtain
5.77V. OK

If I run 5.77v to my 10 volt full scale meter it will read
5.77/10*1500=865.5
watts which is not 500 watts. However, if I run 5.77 thru my squaring
circuit
I get (5.77)**2/10=3.33 volts. Now my meter reads 3.33/10*1500=500 watts
which
is correct.
If you plot the voltage across a 50 ohm resistor vs power V**2/50=power,
you
get a squared relationship. You can use volts to represent watts only if
you
scale the meter face correctly. I have seen wattmeters that did this. I
believe the Drake model did. 0-100 watts took up the first half of the
meter,
and 100-1000 watts took up the other half.
If you linearize the voltage by squaring and scaling you get a nice
readout
with 100 watts being on the left hand side of the meter, and 1500 on the
right
hand side, and linear in between.
The AD633 probably uses log amps in realization of multiplication, I
haven't
looked at the internals. I have done multiplication with Log amps. I
have to
disagree on the AD538 being a better solution. One could use the 538, but
the
633 is much simplier to use. Also I already had some AD633s
73 Gary N4AST

Gary,
Comment I have is in your resistive divider. To keep the power dissipations
in the resistors low, you will have to use large resistor values, but at RF
these can be off 50 % or more. So, you will want to build what looks like a
'scope probe, with each of the resistors shunted with a capacitor, where the
two RC products are the same. For starters, try something like 2 - 5 PF for
the smaller capacitor. Do not use ceramic capacitors, except for NPO. I
would build the thing with at least 2 scales, say 0 - 150 and 0 - 1500 W.

Tam/WB2TT

Comment I have is in your resistive divider. To keep the power dissipations
in the resistors low, you will have to use large resistor values, but at RF
these can be off 50 % or more.

Well, I hadn't thought of that, so I'm glad I asked the question. I had
planned on using a 200k 1/2 watt carbon and about 5k 1/2 watt for the divider,
keep the leads short, and locate it a close as possible to what I think is 50
ohms in my system.
Do you think it would be worth the time and effort to do an rf frequency
sweep of the divider to see how accurate it is? I realize this method is not
going to be dead accurate, but I was hoping for better than 50%.
73 Gary N4AST

"JGBOYLES" wrote in message
...
Comment I have is in your resistive divider. To keep the power
dissipations
in the resistors low, you will have to use large resistor values, but at
RF
these can be off 50 % or more.

Well, I hadn't thought of that, so I'm glad I asked the question. I had
planned on using a 200k 1/2 watt carbon and about 5k 1/2 watt for the
divider,
keep the leads short, and locate it a close as possible to what I think is
50
ohms in my system.
Do you think it would be worth the time and effort to do an rf frequency
sweep of the divider to see how accurate it is? I realize this method is
not
going to be dead accurate, but I was hoping for better than 50%.
73 Gary N4AST

Hi Gary,

I just measured some random 1/2W carbon resistors with an MFJ at 30 MHz.
This is not a precision instrument, but shows a trend.

Nominal Measured
5.6 K 0 - j586
220K 0 -j 600
1.8K 99 - j539 (convert this to parallel form)

As a sanity check

11 Ohms 12 + j4 (some lead inductance here)

What this is tending to show is that the resistors are showing a shunt
capacitance whose reactance is about 600 Ohms at 30 MHz. That is about 9 PF,
which seems high. I was expecting more like 1 PF. I want to redo this at a
higher frequency, might be out of range for the MFJ.

I notice my Kenwood power meter uses a capacitive divider for the voltage
sample. A friend of mine built a meter along the lines of what you want to
do. I will ask him what he did.

Tam

Hi Gary,

I just measured some random 1/2W carbon resistors with an MFJ at 30 MHz.
This is not a precision instrument, but shows a trend.

Hi Tam, I can't thank you enough for your efforts. I will be waiting for
your results. Unfortunately, I am waiting for Hurricance Ivan also, so I won't
be in the shop for a few days.
73 Gary N4AST