I measured a 20dB attenuator with a NanoVNA. The attenuator was connected only to VNA and had nothing on the other end. I think the return loss should have been 2*10dB, but it was only about 24dB. I used a low frequency, less than 5MHz.
And what do you get when measuring 50 Ohm dummy load? Do you get less than 40 dB?
I measured a 20dB attenuator with a NanoVNA. The attenuator was connected only to VNA and had nothing on the other end. I think the return loss should have been 2*10dB, but it was only about 24dB. I used a low frequency, less than 5MHz.
On mine I get a loss of 19.9 dB if I use a 10 dB att. with no load, and 37 dB if I use 20 dB att. with no load (@100 MHz).
(haven't calibrated it for a while though).
I think it was a calibration problem. I took a different setup and after a new calibration, it works. It looks like 20kHz-5MHz with short wires is a bit critical.
By the way. the return loss of 20dB attenuator should be 2*20dB.
it's incorrect to measure VSWR/RL for attenuator with no load on the output. Because when output is not loaded it leads to impedance mismatch and as result VSWR will be high.
You're needs to put 50Ω terminator on attenuator output for proper measurement.
it's incorrect to measure VSWR/RL for attenuator with no load on the output. .
Unless that is what you want to know.
The corrected directivity of the Nano is awful. You can't expect to measure much below 15 dB with any accuracy.
it's incorrect to measure VSWR/RL for attenuator with no load on the output. .
Unless that is what you want to know.
What would be a practical use of that figure (RL for unloaded att) ?
What would be a practical use of that figure (RL for unloaded att) ?
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For a quick check that everything is working, The VNA and attenuator and theory.
it's incorrect to measure VSWR/RL for attenuator with no load on the output. .
Unless that is what you want to know.
What would be a practical use of that figure (RL for unloaded att) ?
To check your understanding of S parameters and VNA measurements? The OP was specifically expecting RL= 2*N for an N dB attenuator. That is the correct "textbook value" for an ideal attenuator that is unloaded. They of course got substantially more reflection because the dominant reflection in a large attenuator is not from the unterminated output but from input mismatch. It's an important concept that responding with "it's an invalid measurement" is counterproductive and wrong.