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Author Topic: Calculating SWR with RF directional couplers  (Read 3681 times)
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The people's republic of Massachusetts
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Trying to build a simple SWR / RF Power meter.  I am using a dual directional coupler that supplies a voltage relative to the RF power.  My problem is it is not liner... ( I thought it would be  smiley-red

This is what I get...
100 watts =  2.82  volts
  10 watts = 0.891 volts
    1 watt =  0.281 volts 

My question is how do I use this?
I was originally going to just read an analog value and * by a the value for 1/10 of a watt but that will not work.   smiley-eek-blue
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Montreal
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Power = ( Volts / 0.281 ) ^ 2;
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The people's republic of Massachusetts
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Power = ( Volts / 0.281 ) ^ 2;


What is the ^ ?
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Montreal
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^ is power (mathematical, not electrical ), according to: http://www.cplusplus.com/reference/cmath/pow/
Power = pow(( Volts / 0.281 ), 2);   may works, or
Power = ( Volts / 0.281 ) * ( Volts / 0.281 );
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The people's republic of Massachusetts
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Thanks, that looks promising...  smiley-cool
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Left Coast, CA (USA)
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Do you have a link to the directional coupler(s) circuitry you are using?

If it's a true representation of the voltage of a properly matched 50 ohm system then power is the product of (E X E) / R, where R = 50 ohms.  If simple diodes are being used to rectify and measure the RF amplitude be aware that there will be non-linearity as the lowest power levels until the detected voltage is above the Vf rating of the diode(S). There are 'compensation' circuits that can help with that for diode detection, but one first must look at the total dynamic range one wants to have and what acceptable accuracy tolerances you can live with.

 The better RF SWR meters available these days can give very good performance using log amp detectors, with many popular designs based on the Analog Devices AD8307, good for a claimed  92 DB range and 0-500Mhz bandwidth. With a log amp the output voltage is a straight linear representation of wattage.

 I built a RF milliwatt meter based on one of those chips and the accuracy has been very impressive and tracks +/- 1 db difference between from about -77dbm to +15dbm when driven by a HP signal generator with an accurate selectable decade attenuator output.


Lefty
« Last Edit: November 26, 2012, 10:14:44 pm by retrolefty » Logged

The people's republic of Massachusetts
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I have a Wavenode HF-1
http://www.wavenode.com/minidin.htm

I was looking at using a Bird dual element line section but I thought the HF-1 used the AD8307, but no.

Do you have any info on the meter you made?
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The people's republic of Massachusetts
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Ok this is what I cam e up with after looking into power

Code:
float watts_forward = pow((voltage_forward / 0.281), 2);
while reading 2.13v on the ADC (voltage_forward) it is calculating  57.25 Watts.   As soon as I can get another meter to compare this to it's on hold.

Thanks for the help.
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I have a Wavenode HF-1
http://www.wavenode.com/minidin.htm

I was looking at using a Bird dual element line section but I thought the HF-1 used the AD8307, but no.

Do you have any info on the meter you made?


Well I built it based on a typical circuit application shown in fig 32 in the datasheet.
http://www.analog.com/static/imported-files/data_sheets/AD8307.pdf

It was a very easy chip to work with, still available in a 8 pin DIP package. Trade-off is that is was around a $15 chip when I built it around 7 years ago. It takes a lot of complexity and difficulty of realizing accurate RF measurements out of the builders hands and puts it into a chip, highly recommended.

Lefty
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The people's republic of Massachusetts
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1 μW to 1 kW, 50 Ω Power Meter, wow page 21.

thanks
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