Author Topic: Attenuator Question  (Read 1154 times)

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Offline Randy222Topic starter

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Attenuator Question
« on: January 04, 2024, 05:25:44 pm »
Rather basic question, but I perhaps don't know the real answer.

Let's say I have a 50ohm -3db attenuator "A", and it has a flat 1.2 VSWR (0.816% reflected)

Does -3db (-50%, or 0.5 factor) add with the fwd loss factor of 0.99184 ?

Eg;
10v(rms) into 50ohm (2watt) as source, insert the "A" device, I should expect to see 2w x 0.5 x 0.99184 = 0.99184w in my ideal 1:1 antenna?

 

Offline pienari

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Re: Attenuator Question
« Reply #1 on: January 04, 2024, 09:14:58 pm »
I think it is correct.
You can use this calculator.
https://www.random-science-tools.com/electronics/dBm-Watts-volts.htm
 

Offline Randy222Topic starter

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Re: Attenuator Question
« Reply #2 on: January 05, 2024, 04:02:06 pm »
I think it is correct.
You can use this calculator.
https://www.random-science-tools.com/electronics/dBm-Watts-volts.htm
Thanks, but that calc is not for real power calculation using attenutaion and VSWR.
 

Online ejeffrey

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Re: Attenuator Question
« Reply #3 on: January 05, 2024, 04:14:40 pm »
Most attenuators are not accurate enough to worry about an additional <1%  loss either way.  If you care about that you probably need a traceable cal, in which case you will have the exact S21 at least for the test fixture used for calibration.

I'd guess that the nominal value of attenuators is generally based strictly on the resistor values, ignoring the additional loss from reflections.

In the real world, the output match of your amplifier is at least as bad so you will have a frequency dependent standing wave with more than 1% ripple anyway.
 

Offline Randy222Topic starter

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Re: Attenuator Question
« Reply #4 on: January 05, 2024, 05:30:33 pm »
Most attenuators are not accurate enough to worry about an additional <1%  loss either way.  If you care about that you probably need a traceable cal, in which case you will have the exact S21 at least for the test fixture used for calibration.

I'd guess that the nominal value of attenuators is generally based strictly on the resistor values, ignoring the additional loss from reflections.

In the real world, the output match of your amplifier is at least as bad so you will have a frequency dependent standing wave with more than 1% ripple anyway.
True, that's all the real math stuff.
But in general, VSWR loss is in addition to attenuation, in terms of rated attenuator.

In other words, if I tell you the DUT (attenuator "A") is -3db with 1.2 VSWR (f=0-6GHz), that -3db is independent of the VSWR factor. I mean it should be, there's no real SWR at 0Hz.

This is why VSWR(f) is important. A spec of "-3db 0-6GHz" is an incomplete spec of the device. I see many time spec sheet that says "-3db VSWR<1.2 0-6GHz". Somewhat better, but we don't know if that means it runs near 1.2 the whole way, or if 1.2 is just a peak in a very narrow band and then runs 1.05 on eitehr side. This is why I like to see VSWR chart before buying.

 

Offline pienari

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Re: Attenuator Question
« Reply #5 on: January 05, 2024, 06:48:23 pm »
You are right but sometimes you have to use what have in hands.
Like my new Diconex 17-0694 1500w 50ohm dummy load.



800w module was much better vswr curve but it got burned .
« Last Edit: January 05, 2024, 06:54:43 pm by pienari »
 


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