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?
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.You will get 24k just from the resistors.
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.
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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errr... somebody...? right/wrong?
It's right. Check my rule of thumb calculations in my previous post. Mind you if there were no overall feedback it would be a bit lower as the input impedance of the transistors would be lower, probably about 50k each.
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