Using 75 ohm attenuators for 50 ohm applications - Page 1
Forum section: electronics / beginners

author: hcripe
author_rank: Newbie
timestamp: December 11, 2017, 08:56:17 pm
content: I have a set of 75 ohm attenuators and was wondering if I can use them in 50 ohm applications like dropping a signal generator's output to test ham radio receiver sensitivity. I will be using them for under 60mhz signals.If I can use them, is there a way to calculate the attenuation difference because of the impedance change.?Thanks,Howard

author: danadak
author_rank: Super Contributor
timestamp: December 12, 2017, 11:43:55 am
content: This might help -Use this calculator by using backing into SWR from laod mismatch. Yo get loss.http://kv5r.com/ham-radio/coax-loss-calculator/http://www.rfcafe.com/references/electrical/coax.htmhttps://www.google.com/search?q=coax+termination+impedance+and+attenuation&tbm=isch&source=iu&ictx=1&fir=MCij8597Ug5frM%253A%252CXmN1MgCdwmXoqM%252C_&usg=__OKW62s4P8xj9cZdmfNfIVldmElE%3D&sa=X&ved=0ahUKEwizrPvlr4TYAhXBQd8KHZSMBnkQ9QEIczAL#imgrc=woTavca5WKeHJM:https://www.belden.com/blog/broadcast/50-ohms-the-forgotten-impedanceRegards, Dana.

author: rfeecs
author_rank: Frequent Contributor
timestamp: December 13, 2017, 01:51:56 am
content: One way to calculate is start with 75 ohm S-parameters for the attenuator, convert to Z-parameters, convert back to 50 ohm S-parameters.The result is very messy algebra. If I did it right:A is 75 ohm attenuation value. You can see the trend.

author: danadak
author_rank: Super Contributor
timestamp: December 13, 2017, 02:37:17 am
content: Some ap notes here might help -http://www.analog.com/en/applications/markets/instrumentation-and-measurement/electronic-test-measurement/rf-power-measurement.htmlRegards, Dana.

author: djnz
author_rank: Regular Contributor
timestamp: December 13, 2017, 03:09:35 am
content: Some pictures, if you are into that sort of thing:https://www.eevblog.com/forum/beginners/oscilloscopefunction-gen-combo-(can-you-mod-a-gen-output-probe)/msg1113821/#msg1113821

author: Kire Pûdsje
author_rank: Regular Contributor
timestamp: December 13, 2017, 05:49:46 am
content: For sensitivity measurements in general lots of attenuation is needed, so one can assume enough isolation is provided to separate the input and output from VSWR interaction.A 50-75 transition will generate a nominal 0.2 dB mismatch loss. In total 0.4 dB total mismatch loss assuming both ends are 50 ohms and the attenuator has a perfect 75 ohm impedance. This is also the value at the end of rfeecs his table. This offset could be corrected for.However it could just be left uncorrected as well. Its effect then could be incorporated in the measurement uncertainty.To be really correct there are some additional steps to be taken, but assuming a system where both the source and load return loss are > 15 dB in 50 ohm and the attenuator is 20 dB in 75 ohm, together with u-shaped distributions of the reflections, it is also boils down to about 0.4 dB at 95% confidence. (no wonder).Assuming all other errors add up to say 1 dB uncertainty, the combined uncertainty will be 1.1 dB. So nothing to worry about.

author: hcripe
author_rank: Newbie
timestamp: December 13, 2017, 05:00:05 pm
content: Thanks for all the replies - this is a great group! This is exactly the information I was looking for.I will use my signal generator / filter / oscilloscope combination to verify the loss due to the impedance mismatch and also to see if the waveform is distorted like the examples djnz posted. If the distortion is too great, I nay just have to get some 50 ohm terminators, but I'm hopeful that I will be able to use these for the few occasions that I need to measure sensitivity.Thanks again,Howard

author: Kire Pûdsje
author_rank: Regular Contributor
timestamp: December 14, 2017, 12:49:08 am
content: I would not worry too much about the 'distortion'. This is not a non-linear effect, so no intermods, etc. The examples of djnz can only be observed for wideband signals. Since a radio receiver (in general) is a narrow-band device, you will not notice anything except the mismatch loss.Also as a rough indication about return losses (for amplifiers etc):10 dB: somewhat acceptable> 15 dB: good> 20 dB: greatPlease note that low-loss devices such as adapters etc need better figuresA 75 ohm load in a 50 ohm system will still give you 14 dB return loss.

author: gibbsmm828
author_rank: Newbie
timestamp: September 09, 2020, 02:46:17 am
content: For a given pi-pad attenuator in a reference impedance, the typical procedure is to calculate the shunt and series resistors required. This question asks for the inverse. One can calculate the required resistors for a given 75 ohm attenuator, then determine the loss that this would present when used in a 50 ohm system. I have done this using an Excel spreadsheet (shown below). I am presenting the results in terms of loss relative to a 75 ohm system. That is, how much additional loss is presented using 75 ohm attenuators in a 50 ohm system on top of the nominal pad value (relative to a 75 ohm system). As you can see, as you increase the pad value, the more it looks like 75 ohms, as expected. The loss asymptotes at just under 2 dB. Long story short, if you are using a 75 ohm pad to knock down a lot of power (in this case >20 dB), then you are only losing about 2 dB. Please note that this is not the same as a minimum loss pad, which targets to present the minimum thru loss while maintaining an input impedance that matches the impedance of the source. For 50/75, that value is roughly 5.7 dB. Cheers.

author: exe
author_rank: Unknown
timestamp: September 09, 2020, 08:07:01 am
content: May be there is a way to modify attenuator? If 75 ohm is determined by a resistor, then resistor can be changed (though this will affect attenuation).