[Premium-Rx] Measurement of Oscillator Output Power

[George Georgevits]

Hi Ben,

Yes, I was just waiting to see if anyone objected, but clearly not, so here
it is, in all its glory.

For what it is worth, I ended up using a folded dipole to do the
measurements first with the transmitter, and then replacing the transmitter
with a second dipole driven by an HP8657A sig gen (which has a nice
calibrated output) and adjusting the output of the sig gen until I got the
same response in the receiver. Given that I was only after an answer that
was within the right order of magnitude, this worked great and it was easy.
After taking away the cable losses at 433MHz, the answer .. -27dBm.

Now this may be a little high (and from the measured radiation pattern of
the transmitter, I suspect it is), but I was only after an order of
magnitude result, and I got it. When I get a bit more time, I am going to
try Gary's method and try to get a more accurate answer.

de VK2KGG.






> Hi George,
>
> I was really interested in the responses -- as to how you were going to
> measure or derive the output power.  I've ran across that same problem
> several times in my career also.
>
> Ben
>


 Hi George,

I can recommend to use a calibrated signal generator and a resonant dipole.
YOu will find in the classic literature which gain factor is between the
dipole and an isotropic radiator. Then you compare the signal of the low
power transmitter with the signal radiated by the dipole. You rainse the
output power of the sinal generator that both signals have the same field
strength. If you reduce the output power of the signal generator by the gain
(over isotropic) of the dipole, you have the radiated power.

73 de Andreé DD3LY







George,

The best way to take into account horizontal and vertical polarization
is at the receive antenna.  That way you just rotate the receive antenna
90 degrees for each position and take the data twice. Add the readings
together after you compute watts for each...see below.

The other calculation.  I'm going to go one step at a time from theory
to reality.

Let's say you have a transmitter/antenna that emits exactly 1 watt.
Let's say that you have a calibrated field strength receiver/antenna
combination that reads in out directly in watts per square meter.

Now at a constant distance take a large number of readings all about 4
pi solid angle and average them.  If the receive antenna is r meters
from the source, simply multiply the average reading by 4 * pi * r^2.
This will give you the transmit power.  When all is said and done, it
should add up to 1 watt.

But RF measuring equipment is not typically calibrated in watts per
square meter.  It is calibrated in volts per meter.  The conversion is
simply:

Watts/m^2 = (volt/m)^2 / (120 * pi).

But it is usually the case that you have a receiver that is calibrated
in volts (or microvolts).  How do you get the "per meter" term into the
calculation?

Simply put, this is due to the antenna and its size and efficiency.
Some folks do simple things like divide the receiver reading by the
length of the (hopefully isotropic) antenna in meters. I doubt that this
is ever correct or can be made so.

A better approach is to take a calibrated RF source and do the same test
that you would perform on the unknown.  When you get your reading for
the calibrated source, you can compute out a term that some call the
"antenna factor", which is stand in for the "per meter" term in the
readings from the receiver.  This term will allow you to do the
measurement on a known source and get the correct result.  Then you can
do the test on the unknown source and get a good result.

RF professionals use calibrated receivers and antennas with calibrated
antenna factors; as a result, this is all easy for them.

I hope this helps,

Gary WA0SPM

George,

One other thing.

There is a good description of antenna factor and the computations
involved at:
http://www.emctest.com/pdf/Ant11.pdf#search='antenna%20factor'

Unfortunately I found this AFTER I wrote the previous email. I could
have saved a little time if I had looked for this first.  Oh well...

73's,

Gary WA0SPM

George,

Well, I suppose that you could calculate it; however, it is likely that
you will overestimate the RF power output due to AC circuit losses in
the inductors, etc.

We typically solve the problem brute force.  First we decouple the power
leads so that essentially no energy goes down the power wires.  Then we
put the device on a rotary table in an RF anechoic chamber.  The total
RF output is determined by measuring the field strength at many
different angles at a given distance in 4 pi solid angle and then
integrating it.  The result is sort of a trapezoidal approximation in 3
space.  Also, this has to be done twice, once for vertical polarization
and once for horizontal polarization.  Based on initial results, the
measuring distance sometimes has to be increased in order to keep the
standard receive antenna in the far field.

Sorry, I guess I'm not much help here.

73's,

Gary WA0SPM




Hi George:

Off the top of my head  you could start by assuming (not the case but OK for
now) an isotropic radiator.
Then all it's power propagates out as a sphere.  Suppose you measure the
received power when the sphere has a radius of 5 meters.
The surface area of the sphere is A = (Google is my friend here) = 4 * PI *
R * R or in this case 314 sq meters.
If the effective area of the receiving antenna is 1 sq meter and you measure
say 1 milliwatt, then you would measure this same power at any angle (based
on the isotropic assumption) so you have 1 mw/ sq meter * 314 sq meters =
314 mw.

If the antenna is not isotropic then you need to break up the sphere into
sections then add up the powers in each section.
Also need to adjust for the effective area of the receiving antenna.

For more on the capture or effective area of antennas see:
http://k9erg.tripod.com/theory.htm
a dipole has an area that's 1.64 times that of an isotropic radiator.
Isotropic antenna has an area that's 5/16 the wavelength.

Does any of that make sense?

Have Fun,

Brooke

Hi George:

First since I'm interested in precision time keeping, thanks for the
reference to the oscillator web page at NIST.

Regarding your oscillator-transmitter question I would think that you
could measure the signal strength at a distance that's in the far field
(i.e. not too close).  Then knowing the distance and frequency you can
calculate the radiated power.  Depending on what you're trying to
accomplish you may want to do this just for the orientation where the
power is maximum or do it for many orientations and sum all of them.

73,

Brooke Clarke, N6GCE



George,
Solar makes a radiation safety probe that measures mw/cm^2 which could also
be used for field density. fc

George,
Set up a little range with a receive dipole and a fixed distance say 1
meter.
Measure the signal level of your module.
Take a second board and drive the pcb antenna with a signal generator to the
same reference level on the receiver. When you get to the same field
strength you have an idea by looking at the generator output power. Might
want to match the antenna to the generator output???.





Regards,
George Georgevits
Power and Digital Instruments Pty Ltd
+61 2 9411 4442

> -----Original Message-----
> From: benwallace at dslextreme.com [mailto:benwallace at dslextreme.com]
> Sent: Tuesday, 13 June 2006 2:09 AM
> To: George Georgevits
> Subject: RE: [Premium-Rx] Measurement of Oscillator Output Power
>
>