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

[shaan]

Picture attached. Symmetrical emitter feedback.

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

Thanks in advance.

[Andrew Eckhardt]

About 15k? Can't you look at the current of V3?

[shaan]

Just tested the current. its 40uA.

[shaan]

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?

[Andrew Eckhardt]

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.

[shaan]

Hmm... So it's equal to or higher than say, 47k? Safe here?

[Andrew Eckhardt]

If you get 19.5uA with your assumedly more complete circuit, yeah.

[shaan]

I see. THANKS A TON FOR THE HELP Andrew!

[DF96]

You will get 24k just from the resistors.

[sbrads]

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)