For the purpose of learning electronics I am trying to reverse engineer old Olivetti typewriter. I got to the point where I can understand most things on the board, but one part is still can get around.

Part of the board in question is presented on schematic is this: Schematic

VCC_OUT is used to drive on big solenoid, and small DC motor & small solenoid. From my research I gathered a few things:

  • components T2/R8/D1 are used as voltage regulator.
  • transistor T4 is used as a switch (connecting R12 to GND) based on HIGH/LOW signal on GPIO P74.
  • C10 will will be charged in no time (no mater the capacity) when due to diode D10 and/or T3 being closed.
  • A assume that GPIO P74 (and T4) is used to control VCC_OUT voltage from regulator (at least that what I get when I try to simulate schematic in Circuit JS)

I want to know if my assumptions are correct, and how precisely the circuit works, also what is the reason to include C10 and GND when on pin 2 of T3 and why closing T3 changes the VCC_OUT, when regulator T2 is always on.

  • 1
    \$\begingroup\$ Welcome! Have you tried to simulate it? \$\endgroup\$
    – winny
    Jun 15, 2022 at 13:40

1 Answer 1


The circuit is best understood by redrawing it to make it easier to understand. Q2 is is a simple Emitter Follower buffering the voltage on the Zener. You don't mention the Zener voltages so below I've used a 10V and a 6V8 to demonstrate. Assuming the GPIO is zero/off, the the load voltage would in this case be about 9-9.2V due to Q2 and the 10V zener. C10 will be charged to about 8-8.5V.

If the GPIO goes high/on then Q3 will turn on. The load voltage will then be approximately 24-6V8 or 17.2V. Your values will be different depending on the Zener values in your circuit. C10 simply provides a longer fall time from the boost voltage to the hold voltage, probably done in an attempt to reduce back EMF voltage problems.

Given that you mention the load is a small motor and a solenoid, it appears that Q3 (controlled by the GPIO) is used to boost the voltage to the inductive components, while Q2 provides a holding voltage. You see this done in many old fashioned printers to reduce the power dissipated when not printing. It's done also with stepper motors to reduce power when not stepping. Note here that the Zener in the boost switch is carrying the full load whenQ3 is on. It's likely quite a beefy Zener compared to the other one.

The circuit is not very well designed IMO, and could have been very much simpler with a fraction of the components.

I've added enough values to allow the circuit to be simulated, but they are not equivalent to the components in your circuit.


simulate this circuit – Schematic created using CircuitLab


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