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This picture is part of a circuit designed to measure capacitance. The control circuit would apply +12V or -12V on "DTR pin 20" and the two resistors give a minimum and maximum voltage reference which is used to compare with the charging/discharging capacitor's voltage.

The book says T1 & T2 are used to ensure a stable voltage drop on R2, which in turn ensures a constant current flow through the base-emitter junctions of T3 & T4. It specifically asks for BSV90 transistors which I've found out that are for high speed applications.

I can't figure out how current is supposed to flow through those transistor pairs, nor how T1 & T2 are supposed to stabilize the voltage across R2. I also don't know what the point of T3 & T4 are considering 1A is supposed to be a sort of voltage follower to ensure the +12V or -12V going towards the capacitor.

enter image description here

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The transistors have their base and collector shorted together, so act as diodes. In the case of a transistor, for the few mA that would come thought R1 at 12 V, the forward VBE will be about 0.7 V. The 'diode' of a transistor has a rather low breakdown voltage, often around 5 V. The pair of diodes will therefore conduct to limit the voltage across themselves to about 6 V, in either direction.

R1/T1,2 therefore sends about a 6 V squarewave to R2, regardless of whether DTR is a 12 V signal, or a few volts more or less.

1A is not a follower, but an inverting amplifier. With T3,4 in its feedback loop, its output voltage will be about +/- 6 V.

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  • \$\begingroup\$ Thank you for the explanation, I got a little confused with the voltages that happen in there. \$\endgroup\$
    – Ilisuan Iannis Patriciu
    Commented Nov 4, 2022 at 20:27
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    \$\begingroup\$ @IlisuanIannisPatriciu IIRC, the temperature coefficient of reverse breakdown voltage for the EB junction is similar in magnitude, and opposite to, the tempco of forward voltage, so the pair has near zero tempco, and the output voltage will be stable with temperature. I'm not confident enough to include that in the answer though, it's dimly remembered (or not) from many decades ago, and I've not found a reference yet on the net. \$\endgroup\$
    – Neil_UK
    Commented Nov 4, 2022 at 20:35

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