Electronics > RF, Microwave, Ham Radio

Does my SWR power meter need to be calibrated?

(1/3) > >>

Hello, I'm looking for feedback to see if I am correct in my findings with regards to a SWR power meter.

I wanted to set the output of my HF amplifier to 5 watts. I checked this with an Oscope and a power meter.

1.) I connected my amp to my Oscope (1 MOhm) via a T-BNC connector.
2.) From the T-BNC connector I connected to the SWR power meter
3.) From the SWR power meter I connected to a dummy load terminated at 50 Ohms.

I increased DC power to my amp until my Oscope read (22.4V | 24dBV) - 5 watts at 50 Ohms
My SWR meter only displayed about 3.5 watts.

This leads me to believe my SWR meter is off. Is my method correct?

Well, 3.5W instead of 5W means a voltage error of maybe 15%.  You said HF so I am assuming short cables.  Most SWR meters (I am guessing) would have a calibration pot so maybe that's what needs adjusting.  If the frequency is much above 10 MHz I would be cautious about what to do about it.  At 30 MHz the dummy load may not be 50 Ohms so it's wise to check it.

I didn't do the math but 22.4 V seems not right.  Is that p-p?  And what kind of amplifier can be running at 5W out?  How much drive are you giving it?  Is the wave sinusoidal?

There's a couple bits of trickery here and I can't read the O-scope display to be sure.  Your SWR meter is supposed to measure only the power delivered to the load, and the SWR measures the power reflected from the load. It appears to be zero reflection or the meter is broken.  But the Oscope measures the total voltage at the point on the line, so if you moved the Tap point along the line you would see a peak and a valley value and the ratio of those two values is the definition of SWR.  But if the SWR meter reflection measure is correct, you don't have any standing wave.  The upper plot on the O-Scope shows the sine wave value and normally you would need to take the peak-to-peak value divided by 2 (to get Peak voltage) and divided by square root of 2 to get RMS value of the voltage.  I can't read the amplitude of the O-scope but if you did take 1/2 peak to peak as the 22.4 volts then the O-scope is reporting a higher power than the SWR meter.  funny thing is you say you see 22.4 V / 24dBV but normally we use 20Log10(V) to get dBV, so 22.4 V peak would compute to 27 dBV.  If the dBV reading is from the FFT, it might be already dBV rms (22.4/sqrt(2)).  You might try to move your T point closer or farther from the antenna (use a shorter or longer length) and see if the voltage changes. If it does, it means you DO have a non-zero reflection and you do have some SWR, and the SWR meter has an issue.  I suspect a bit the SWR meter as zero reflection is a little suspicious. You could try lowering your power to a very safe level and take off the load, to see if the SWR jumps up. If it does not, suspect the SWR meter.

22,4Vp, so 15,8Vrms equals 5W into 50 \$\Omega\$ - so far everything seems to be correct.
my guess would be that the accuracy of the swr/power meter in the lower range (1-10W) is less than in the higher range (10-100W). probably that can be corrected by an internal trimmer, but I don't know the meter; maybe a manual gives more infos about that.

and don't use the FFT function of the scope to get dB-values; they are certainly way off the real world values. This would be a task for a spectrum analyzer or a power meter based on a logarithmic amp.

Thanks for the reply, I'll try moving the T connection down the line and see if it makes a difference.

I used this here calculator, and it appears 5W = 37 dBm = 24 dBV. - https://www.analog.com/en/design-center/interactive-design-tools/dbconvert.html

The SWR has shown reflection in the past, and whenever I use a 50Ohm term, those reflection usually go to zero.


[0] Message Index

[#] Next page

There was an error while thanking
Go to full version