"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

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

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 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
...
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

It's much more likely that the shunt capacitance is in the MFJ.

Calculate the parallel impedance of 11 + j0 (the presumed resistor) and
0 - j600 (the shunt C) and you'll see that you wouldn't be able to see
the shunt C when making the 11 ohm "sanity check".

Even at HF, measurements aren't nearly as simple as they sometimes seem
they should be.

Roy Lewallen, W7EL

Tam/WB2TT wrote:

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

Roy,

You are right (as per usual). I didn't check the MFJ269 open circuit
readings. At 30 MHz, with the N to UHF adapter installed, open circuit Z is
0 & j656. Since my capacitor readings were about 10% less than that, my
guess is that the resistor reactances were actually about 1/10th of that (~1
PF). If I remove the N/UHF adapter, x rises to 1389. So, most of the C is in
the adapter, which was in place. I think the 12 Ohm data is OK. The
capacitance washed out, and it showed a series inductance of 21 nH.

As I said, this is not a precision instrument. Unfortunately, it is less
precision than I thought. I hope that somebody with access to an HP or
similar instrument will feel inspired to measure some resistors at RF.
Meanwhile, I am going to look at some resistor manufacturers web sites.

Tam/WB2TT


"Roy Lewallen" wrote in message
...
It's much more likely that the shunt capacitance is in the MFJ.

Calculate the parallel impedance of 11 + j0 (the presumed resistor) and
0 - j600 (the shunt C) and you'll see that you wouldn't be able to see the
shunt C when making the 11 ohm "sanity check".

Even at HF, measurements aren't nearly as simple as they sometimes seem
they should be.

Roy Lewallen, W7EL

Tam/WB2TT wrote:

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

All radio people suffer from delusions of measuring accuracy.

RF power measurements are the most inaccurate of all.

The accuracy of measurements are a function of the instrument user.

They who attempt to grasp support by stating the manufacturer's type number
of the instruments used are most in need of the self-confidence it falsly
generates.

Either that or the statements are gratuitous adverts.

How cynical can one get at this hour of the day?

All radio people suffer from delusions of measuring accuracy.

RF power measurements are the most inaccurate of all.

The accuracy of measurements are a function of the instrument user.


They who attempt to grasp support by stating the manufacturer's type number
of the instruments used are most in need of the self-confidence it falsly
generates.

Hi Reg. What exactly are you talking about? I had a few minutes in between
Hurricane Ivans wrath to get the Emergency generator cranked up and had a
chance to read this. Lucky you don't have these things in the UK.
73 Gary N4AST