Attenuator input and output impedance
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 26th May 2015, 03:45 PM #1 oldjack   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, 04:30 PM #2 Arty diyAudio Member   Join Date: Feb 2011 okay, show us what you got sofar, and then you can get help working out the rest. __________________ V*A=VA V*A*PowerFactor = WATT
 26th May 2015, 05:03 PM #3 oldjack   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, 05:49 PM #4 Davey   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 voltage-divider scheme is very straightforward. Input resistance is = r1 + r2. Output resistance is = r1 || r2. Dave.
 27th May 2015, 11:19 AM #5 oldjack   diyAudio Member   Join Date: Aug 2011 Location: West Midlands, UK ok will do some diagrams hopefully today
oldjack
diyAudio Member

Join Date: Aug 2011
Location: West Midlands, UK
Have attached attenuator layouts with my best guess of input and output impedance
Attached Images
 attenuator1.JPG (333.4 KB, 92 views)

 27th May 2015, 04:55 PM #7 sreten   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 ||R2||R1 For 4 Rout = R4||(R2+R1||R3) For 5 Rout = R1 rgds, sreten.
 27th May 2015, 05:23 PM #8 oldjack   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.
sreten
R.I.P.

Join Date: Nov 2003
Location: Brighton UK
Quote:
 Originally Posted by oldjack 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.

Last edited by sreten; 27th May 2015 at 06:14 PM.

 28th May 2015, 09:25 AM #10 oldjack   diyAudio Member   Join Date: Aug 2011 Location: West Midlands, UK Thanks for the help, very good information

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