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Old 26th May 2015, 02:45 PM   #1
oldjack is offline oldjack  United Kingdom
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Default Attenuator input and output impedance

I know this subject has been on many previous threads but I am trying to find out how to calculate the input and output impedance of an attenuator based on a simple type potentiometer, pi type and a pi type with a series resistor at the inlet to the pi network. Its the theory I am after.
These resistor layouts appear on various switched type attenuators.
I think I understand how to work out the input impedance but not completely sure. I am looking at using a switched type attenuator using two 2 pole switches 6 way with pcb mount terminations to give course and fine volume control.
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Old 26th May 2015, 03:30 PM   #2
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okay, show us what you got sofar, and then you can get help working out the rest.
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Old 26th May 2015, 04:03 PM   #3
oldjack is offline oldjack  United Kingdom
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for pi attenuator I add the series resistor with the second resistor to ground and then take parallel sum, ie for r1 (series), r2 first resistor to gnd, r3 2nd resistor to gnd,
input imp = r2 || (r1+r3).
for potentiometer I take input impedance as r1 section of pot where total r = r1 + r2.
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Old 26th May 2015, 04:49 PM   #4
Davey is offline Davey  United States
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I think you should post a diagram so everybody is clear exactly what configuration you're talking about.

A simple voltage-divider scheme is very straightforward. Input resistance is = r1 + r2. Output resistance is = r1 || r2.

Dave.
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Old 27th May 2015, 10:19 AM   #5
oldjack is offline oldjack  United Kingdom
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ok will do some diagrams hopefully today
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Old 27th May 2015, 03:34 PM   #6
oldjack is offline oldjack  United Kingdom
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Have attached attenuator layouts with my best guess of input and output impedance
Attached Images
File Type: jpg attenuator1.JPG (333.4 KB, 70 views)
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Old 27th May 2015, 03:55 PM   #7
sreten is online now sreten  United Kingdom
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Hi,

Your neglecting the load impedance for Rin.
Source impedance can usually be ignored.

For 2 Rout = R3 ||(R2+R1)
For 3 Rout = R3 ||R2||R1
For 4 Rout = R4||(R2+R1||R3)
For 5 Rout = R1

rgds, sreten.
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Old 27th May 2015, 04:23 PM   #8
oldjack is offline oldjack  United Kingdom
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Just to clarify what you are saying; is it that the load impedance will be much greater than Rin meaning Rin can be neglected.
I can understand the theory for Rout from your formula except for 5, why does R2 have no effect.
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Old 27th May 2015, 04:55 PM   #9
sreten is online now sreten  United Kingdom
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Quote:
Originally Posted by oldjack View Post
Just to clarify what you are saying; is it that the load impedance will be much greater than Rin meaning Rin can be neglected.
I can understand the theory for Rout from your formula except for 5, why does R2 have no effect.
Hi,

Rload can be neglected assuming its very high, it usually isn't.
Assuming Rsource is very low, an ideal voltage source is safer.

For 5 Rout = R1 + R2||Rsource and if Rsource is zero ....

rgds, sreten.
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Last edited by sreten; 27th May 2015 at 05:14 PM.
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Old 28th May 2015, 08:25 AM   #10
oldjack is offline oldjack  United Kingdom
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Thanks for the help, very good information
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