I am struggling to understand how the jumpers at the bottom of this diagram change the voltage of the circuit in this diagram. The input J1 is a 9V wall wart.
By my understating the MOSFET only amplifies current, not voltage, so how is the output changing with the jumpers?
Edit: I understand now that the transistors are working as an amplifier to give a negative voltage potential to the FET. I am now struggling to grasp how the current changing from the FET changes the output voltage.
Best Answer
This is a feedback circuit:
The voltage divider formed by R3 and R7 (with R6 and R5 potentially in parallel) samples the output voltage and provides the scaled voltage to the base of Q2.
The differential pair formed by Q1 and Q2 is an amplifier that compares the sampled voltage at the base of Q2 with a reference voltage on the base of Q1 (1.2V).
If there is any difference between the two, the gain of the circuit will drive the gate of the FET in such a way as to reduce the difference such that it's close to zero. So the base of Q2 = base of Q1 = 1.2V.
Therefore the output voltage is approximately 1.2V + R3*1.2V/R7 (or replace the value of R7 with R7 in parallel with whichever resistors are jumpered in to get the other voltage options).
This is essentially a discrete linear regulator.