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26th May 2015, 02:45 PM  #1 
diyAudio Member
Join Date: Aug 2011
Location: West Midlands, UK

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. 
26th May 2015, 03:30 PM  #2 
diyAudio Member
Join Date: Feb 2011

okay, show us what you got sofar, and then you can get help working out the rest.
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26th May 2015, 04:03 PM  #3 
diyAudio Member
Join Date: Aug 2011
Location: West Midlands, UK

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. 
26th May 2015, 04:49 PM  #4 
diyAudio Member
Join Date: Dec 2001
Location: Bremerton, WA.

I think you should post a diagram so everybody is clear exactly what configuration you're talking about.
A simple voltagedivider scheme is very straightforward. Input resistance is = r1 + r2. Output resistance is = r1  r2. Dave. 
27th May 2015, 10:19 AM  #5 
diyAudio Member
Join Date: Aug 2011
Location: West Midlands, UK

ok will do some diagrams hopefully today

27th May 2015, 03:34 PM  #6 
diyAudio Member
Join Date: Aug 2011
Location: West Midlands, UK

Have attached attenuator layouts with my best guess of input and output impedance

27th May 2015, 03:55 PM  #7 
R.I.P.
Join Date: Nov 2003
Location: Brighton UK

Hi,
Your neglecting the load impedance for Rin. Source impedance can usually be ignored. For 2 Rout = R3 (R2+R1) For 3 Rout = R3 R2R1 For 4 Rout = R4(R2+R1R3) For 5 Rout = R1 rgds, sreten. 
27th May 2015, 04:23 PM  #8 
diyAudio Member
Join Date: Aug 2011
Location: West Midlands, UK

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. 
27th May 2015, 04:55 PM  #9  
R.I.P.
Join Date: Nov 2003
Location: Brighton UK

Quote:
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 + R2Rsource and if Rsource is zero .... rgds, sreten. Last edited by sreten; 27th May 2015 at 05:14 PM. 

28th May 2015, 08:25 AM  #10 
diyAudio Member
Join Date: Aug 2011
Location: West Midlands, UK

Thanks for the help, very good information

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