Wattmeter Location
 

Hi,
I am contemplating a pep reading wattmeter so I can check the output of my HB
amplifier. Consequently, I am putting it right on the output of the amp. The
output of the amp should always see 50 ohms because it will be feeding either a
50 ohm dummy load, or an ATU tuned to 50 ohms. With that in mind, I am simply
using a resistive voltage divider, to get a voltage sample, and squaring it
with an AD633 multiplier. This seems like a simple, cheap way to get watt
info. assuming you will always be working into a near 50 ohm resistive load.
Anyone see any reason why this will not give you a pretty good indication of
your power output? I realize that the load may not always be exactly 50 ohms,
and that there are losses in the ATU.
The reason I have not tried getting a current sample and using the
conventional VI COS Theta with the multiplier is due to the additional
complexity of circuitry. It is also difficult to get accurate current samples
over a wide frequency range. By making everything resistive it somewhat takes
the frequency dependency out of the problem. Thanks.
73 Gary N4AST

The most accurate way of measuring power is to measure volts across a known
resistance.

JGBOYLES wrote:

Hi,
I am contemplating a pep reading wattmeter so I can check the output of my HB
amplifier. Consequently, I am putting it right on the output of the amp. The
output of the amp should always see 50 ohms because it will be feeding either a
50 ohm dummy load, or an ATU tuned to 50 ohms. With that in mind, I am simply
using a resistive voltage divider, to get a voltage sample, and squaring it
with an AD633 multiplier. This seems like a simple, cheap way to get watt
info. assuming you will always be working into a near 50 ohm resistive load.
Anyone see any reason why this will not give you a pretty good indication of
your power output? I realize that the load may not always be exactly 50 ohms,
and that there are losses in the ATU.
The reason I have not tried getting a current sample and using the
conventional VI COS Theta with the multiplier is due to the additional
complexity of circuitry. It is also difficult to get accurate current samples
over a wide frequency range. By making everything resistive it somewhat takes
the frequency dependency out of the problem. Thanks.
73 Gary N4AST

Good luck getting a straight answer here, and I won't try one, because
every technical answer related to power transfer is always wrong by
someone elses measure on this list.

That being said, I'd do it the way you propose, since it's simple and
repeatable, and probably close enough.

tom
K0TAR

On 13 Sep 2004 21:21:01 GMT, (JGBOYLES) wrote:

I am simply
using a resistive voltage divider, to get a voltage sample, and squaring it
with an AD633 multiplier.

Hi Gary,

If you are thinking of E²/R, you should note this is division, not
multiplication. As such you will need an inverter ahead of the first
product and then a second multiplier to complete the squaring. Then
you would have a problem of dynamic range (not much).

I would suggest you look into turning the voltage into a log
representation so you could then have a linear power scale. Adding
logs is much simpler form of multiplication too and logs give much
more dynamic range. You may lose out in resolution, but if you are
driving a common 3½ digit converter for readout, you can recover that.

73's
Richard Clark, KB7QHC

Why complicate matters?

All Gary needs is a diode, a capacitor, a resistor and a micro-ammeter.

Why complicate matters?

All Gary needs is a diode, a capacitor, a resistor and a micro-ammeter.

Hi Reg, Thanks for the reply. That will certainly work, but I would need a
Log scale (or is it square) on the micro-ammeter. I want to be read
0-1500watts pep. Down around the 100 watt level things would get crowded. I
bought some 3-1/2 digit multimeters from a company here in the US for $3 each.
Bought 11 of them. Or I have a 0-1 ma movement that I have re-labeled the
meter face to read watts, but it is a linear scale.
73 Gary N4AST

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. I now have a relative power indicator as suggested by Reg. I can
take care of the squaring and division by putting a Log scale(watts) on a 0-10V
analog meter movement. 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. Checking another point, assume 500 watts, then
the output of the divider will be 5.76 volts, and the output of the multiplier
will be 3.33 volts, which is 33% of 10 volts. The wattmeter will read 33% of
1500 or 500 watts. The scale will be linear.
I have not seen it done this way before, although I am sure it has. For this
method to accurate your load must be near 50 ohms. 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.
73 Gary N4AST

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.