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

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Picture attached. Symmetrical emitter feedback.

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

Thanks in advance.
 

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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)
 
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Hmmm... seems interesting. So I ran some plots at the 1K base resistors of each input transistors.

If I'm not wrong then the base impedance is more than 1.5M for both transistors (~0.6uA/V) and the total input impedance is ~25Kohm as the input current is ~40uA/V.

First image - Total input current.
Second image - Current in base resistors.
 

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  • imped3.GIF
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