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### Topic: Calculating SWR with RF directional couplers (Read 9547 times)previous topic - next topic

#### 2632

##### Nov 27, 2012, 02:05 am
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

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.

#### Magician

#1
##### Nov 27, 2012, 02:45 am
Power = ( Volts / 0.281 ) ^ 2;

#### 2632

#2
##### Nov 27, 2012, 02:52 am

Power = ( Volts / 0.281 ) ^ 2;

What is the ^ ?

#### Magician

#3
##### Nov 27, 2012, 03:45 am
^ 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 );

#### 2632

#4
##### Nov 27, 2012, 04:09 am
Thanks, that looks promising...

#### retrolefty

#5
##### Nov 27, 2012, 04:12 amLast Edit: Nov 27, 2012, 04:14 am by retrolefty Reason: 1
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

#### 2632

#6
##### Nov 27, 2012, 04:48 am
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?

#### 2632

#7
##### Nov 27, 2012, 05:11 am
Ok this is what I cam e up with after looking into power

Code: [Select]
`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.

#### retrolefty

#8
##### Nov 27, 2012, 05:18 am

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.

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

#### 2632

#9
##### Nov 27, 2012, 07:02 am
1 ?W to 1 kW, 50 ? Power Meter, wow page 21.

thanks

#### zryder

#10
##### Oct 15, 2014, 09:23 pm

1 ?W to 1 kW, 50 ? Power Meter, wow page 21.

thanks

Sorry to resurrect an old thread, but how would you suggest making an arduino based SWR meter using one of these AD8307 chips? 2 chips from ports on a directional coupler?

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