Return loss of an 10dB attenuator? - Page 1
Forum section: electronics / rf-microwave

author: LM21
author_rank: Regular Contributor
timestamp: January 16, 2024, 02:45:38 pm
content: 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.

author: yl3akb
author_rank: Regular Contributor
timestamp: January 16, 2024, 03:37:26 pm
content: And what do you get when measuring 50 Ohm dummy load? Do you get less than 40 dB?

author: xrunner
author_rank: Super Contributor
timestamp: January 16, 2024, 03:43:12 pm
content: Quote from: LM21 on January 16, 2024, 02:45:38 pmOn 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).

author: LM21
author_rank: Regular Contributor
timestamp: January 16, 2024, 04:18:17 pm
content: 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.

author: Roger Need
author_rank: Regular Contributor
timestamp: January 16, 2024, 11:52:26 pm
content: Some comments..Consider what happens with a 50 ohm transmission line that is left open circuit. The return loss of a transmission line will not be exactly twice the attenuation when open circuited. Fringe capacitance at the open end will have an effect on the signal returned. Try this for yourself by measuring the RL with the line open and then short circuited. Measure the attenuation first by using a VNA and measuring S21.The return loss is only 2 x attenuation if the input impedance to the DUT is 50 ohms. In your case a well designed 20 dB attenuator with a 50 ohm load attached will measure close to 50 ohms at the input. Remove the load and the input impedance will not be as close to 50 ohms.If you have an accurate 20 dB attenuator it will only have this value if the load is attached. With no load or a short circuit the signal will return to the input but the RL will not be twice the termiated attenuation.Roger

author: LM21
author_rank: Regular Contributor
timestamp: January 17, 2024, 07:12:44 am
content: By the way. the return loss of 20dB attenuator should be 2*20dB.

author: radiolistener
author_rank: Super Contributor
timestamp: January 17, 2024, 08:01:15 am
content: 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.

author: ejeffrey
author_rank: Super Contributor
timestamp: January 17, 2024, 04:09:27 pm
content: Quote from: radiolistener on January 17, 2024, 08:01:15 amUnless that is what you want to know.

author: virtualparticles
author_rank: Regular Contributor
timestamp: February 07, 2024, 03:21:58 pm
content: The corrected directivity of the Nano is awful. You can't expect to measure much below 15 dB with any accuracy.

author: Bud
author_rank: Super Contributor
timestamp: February 07, 2024, 04:32:32 pm
content: Quote from: ejeffrey on January 17, 2024, 04:09:27 pmWhat would be a practical use of that figure (RL for unloaded att) ?

author: LM21
author_rank: Regular Contributor
timestamp: February 07, 2024, 04:45:13 pm
content: What would be a practical use of that figure (RL for unloaded att) ?For a quick check that everything is working, The VNA and attenuator and theory.

author: ejeffrey
author_rank: Super Contributor
timestamp: February 08, 2024, 02:45:25 am
content: Quote from: Bud on February 07, 2024, 04:32:32 pmTo 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.