Help! What would be the input inpedance of this input stage? - diyAudio
 Help! What would be the input inpedance of this input stage?
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diyAudio Member

Join Date: Mar 2009
Location: Near the Ganges
Help! What would be the input inpedance of this input stage?

Picture attached. Symmetrical emitter feedback.

What is the input impedance of this stage at the arrow-marked point?

Attached Images
 Input.GIF (4.4 KB, 115 views)

 10th September 2011, 06:38 PM #2 diyAudio Member   Join Date: Nov 2005 About 15k? Can't you look at the current of V3?
 10th September 2011, 06:40 PM #3 diyAudio Member     Join Date: Mar 2009 Location: Near the Ganges Just tested the current. its 40uA.
diyAudio Member

Join Date: Mar 2009
Location: Near the Ganges
Quote:
 Originally Posted by shaan Just tested the current. its 40uA.
Opps made a mistake in the measurements. The correct value is 19.5uA @1V peak.

So, it is ~15K?

 10th September 2011, 06:53 PM #5 diyAudio Member   Join Date: Nov 2005 40uA is 25k. 19.5u is about 50k at 1V. 15 was just a quick (dumb) guess. I got 25k when just loading the emitters at 10k, using 3904/06.
 10th September 2011, 06:56 PM #6 diyAudio Member     Join Date: Mar 2009 Location: Near the Ganges Hmm... So it's equal to or higher than say, 47k? Safe here?
 10th September 2011, 06:58 PM #7 diyAudio Member   Join Date: Nov 2005 If you get 19.5uA with your assumedly more complete circuit, yeah.
 10th September 2011, 07:12 PM #8 diyAudio Member     Join Date: Mar 2009 Location: Near the Ganges I see. THANKS A TON FOR THE HELP Andrew!
 10th September 2011, 07:48 PM #9 diyAudio Member   Join Date: May 2007 You will get 24k just from the resistors.
diyAudio Member

Join Date: Jun 2005
Location: Kent, UK
Quote:
 Originally Posted by DF96 You will get 24k just from the resistors.
Don't think so. Don't forget there's the same voltage at each outer end of the 12k resistors. Ohms law and all that.

Roughly, I make it the following resistive (mostly) impedances in parallel..........

12k//12k in series with 47k=> 53k
100k//100k assuming low impedance to earth supplies => 50k
i/p Z of 2 transistors in // - not sure here, may be very high due to feedback so I'll ignore it. No, call it 1M.

So, 53k//50k//1M = approx 25k (to an AC signal) (40uA @ 1v)

Last edited by sbrads; 10th September 2011 at 08:26 PM.

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