Help me in this SWR meter!!!

[Willen]

www.angelfire.com/electronic2/qrp/swrpwr.html
I want to use it in my 5 watt FM Tx at 100MHz (Can't I?). Here are two output- Forward and Reflect, connected in releted meter display. I don't have such meter display so I should have to connect it into multimeter. Can anybody guide me to check watt and and SWR (SWR between 1:1 to 1:6) in multimeter? It's not necessary to be accurate, I just want best match.
Or do you have other simpler ideas? Tell me something.
Thanks

 

[davenn]

That will give you a good indication
You cant get too much simpler than that
It wont give you specific numbers unless you calibrate it
Yes you could connect it out to your multimeter in mA range but you would be better off keeping it all self contained (shielded) in a metal box with its own meter

cheers
Dave

 

[Willen]

That will give you a good indication
You cant get too much simpler than that
It wont give you specific numbers unless you calibrate it
Yes you could connect it out to your multimeter in mA range but you would be better off keeping it all self contained (shielded) in a metal box with its own meter

cheers
Dave

Testing with non-shielded probe of multimeter will affect badly in reading?
And it will be difficult to know that How much ...mA representing 1 or 5 watt and How ...mA representing SWR between 1:1 to 1:6. How ..mA is the lowest SWR like 1:1? It is my main problem!! Do you have any idea?
(I think the mA in multimeter won't be universally equal because there are potentiometer. So it will be difficult to find that ''...mA'' is the indication of the lowest SWR.)

 

[ronsimpson]

I did some thinking about your meter.
Your transformers are 1:12 turns.
There are two 50 ohm resistors. One for forward and one for reverse.
This meter does not read 'power' but current or voltage and you have to x^2 to get power.
The voltage across the 50 resistor is: 1V for 2.88 watts, 2V=11.52W, and 0.5V=0.72W. (notice the squared function) (12V on the coax, 2.88 watts, 50 ohm antenna, 1V on the output.)

Now to calibrate the meter which is not easy. When the output (across the 50 resistor) is 1 volt average the meter will see 1.414 volts because it is a peak meter not average. Take the meter, 1N34A, 0.01uf and the 25k pot (set to mid range) and apply 1.414 volts DC to the circuit. Use the pot to adjust the meter to read anywhere you want maybe 50%. (a well used battery will output about 1.4V) Mark this place on the meter 2.88watts. Now apply 2.818 V and later 0.707V DC and mark 11.5W and .72W.

These numbers are strange. You can work the math backwards. What DC voltage comes form 5W?, 4W? 3W? 2W? 1W?

Note the meter will not read linear because there is a 'squared' function and the diode makes the meter non-linear.

Please, some one, check my math.

 

[davenn]

Testing with non-shielded probe of multimeter will affect badly in reading?
And it will be difficult to know that How much ...mA representing 1 or 5 watt and How ...mA representing SWR between 1:1 to 1:6. How ..mA is the lowest SWR like 1:1? It is my main problem!! Do you have any idea?
(I think the mA in multimeter won't be universally equal because there are potentiometer. So it will be difficult to find that ''...mA'' is the indication of the lowest SWR.)

well there's the probability that the long probe leads will act as nice antennas and pic up RF and screw up the readings, specially when dealing with small currents like this

Dave

 

[ronsimpson]

One problem about long leads is that they could easily be 1/4 or 1/2 wavelength and add strange resonant effects.

 

[Willen]

I did some thinking about your meter.
Your transformers are 1:12 turns.
There are two 50 ohm resistors. One for forward and one for reverse.
This meter does not read 'power' but current or voltage and you have to x^2 to get power.
The voltage across the 50 resistor is: 1V for 2.88 watts, 2V=11.52W, and 0.5V=0.72W. (notice the squared function) (12V on the coax, 2.88 watts, 50 ohm antenna, 1V on the output.)

Now to calibrate the meter which is not easy. When the output (across the 50 resistor) is 1 volt average the meter will see 1.414 volts because it is a peak meter not average. Take the meter, 1N34A, 0.01uf and the 25k pot (set to mid range) and apply 1.414 volts DC to the circuit. Use the pot to adjust the meter to read anywhere you want maybe 50%. (a well used battery will output about 1.4V) Mark this place on the meter 2.88watts. Now apply 2.818 V and later 0.707V DC and mark 11.5W and .72W.

These numbers are strange. You can work the math backwards. What DC voltage comes form 5W?, 4W? 3W? 2W? 1W?

Note the meter will not read linear because there is a 'squared' function and the diode makes the meter non-linear.

Please, some one, check my math.

I am little confuse in ''squired function'' because you didn't give detailed example of this math. How you solve 1V^2=2.88watt!?

And you didn't say about SWR. It is my main issue. How may I know lowest SWR seeing multimeter?

 

[ronsimpson]

I am little confuse in ''squired function'' because you didn't give detailed example of this math. How you solve 1V^2=2.88watt!?

And you didn't say about SWR. It is my main issue. How may I know lowest SWR seeing multimeter?

If you measure voltage and want watts there is a squared think happening. Example if the voltage goes from 1V to 2V the wattage goes from 1 to 4. W=V^2/R

There is more than one way to measure how well your transmitter---coax---antenna are connected. You can measure SWR or measure reflected power. The meter you are building measures forward and reflected power. You want reflected power to be small.

SWR= (I can't remember the formula but it comes from Forward and Reflected power. )

You want good SWR or small reflected power. Same thing.

 

[Willen]

If you measure voltage and want watts there is a squared think happening. Example if the voltage goes from 1V to 2V the wattage goes from 1 to 4. W=V^2/R

There is more than one way to measure how well your transmitter---coax---antenna are connected. You can measure SWR or measure reflected power. The meter you are building measures forward and reflected power. You want reflected power to be small.

SWR= (I can't remember the formula but it comes from Forward and Reflected power. )

You want good SWR or small reflected power. Same thing.

Theoretically, if I matched SWR 100% perfectly, can I get around zero mA in multimeter as a reflected power?

Other practical problem- can I use 1n4148 diode in this SWR?

 

[ronsimpson]

Theoretically, if I matched SWR 100% perfectly, can I get around zero mA in multimeter as a reflected power?

Yes

Other practical problem- can I use 1n4148 diode in this SWR?

They are not the same diodes. I think the 1N4148 will work fine. The calibration will be different. I think the 1N4148 will not work as well at low power levels. (below 1 watt)

 

[Mikebits]

The 1N34 diode is germanium with lower voltage drop, better suited for your application. You might try a schottky diode.

 

[Willen]

Why can't I leave SWR meter keep connected anytime in my FM Tx?
But look this schematic, how I can connect this SWR detector anytime? www.electro-tech-online.com/attachments/ Both has same feature. This is a schematic to auto-control voltage of final amplifier according to peflection (SWR) level.

 

[Mikebits]

You can leave your SWR meter connected all the time. It should be transparent to your transmit path, why do you think otherwise?

A properly made SWR meter just samples the transmit path via mutual coupling.

 

[Willen]

It should be transparent to your transmit path.

The SWR of my original post and the SWR controlled power circuit of previous post (#12) both has transparent path, isn't it? Wow! Then i will make it in same PCB inside my Tx! Can't I?

And I am thinking that why shematic of previous post (#12) detects only Reflected, but not forwarded??