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I want to design a circuit which is able to generate 8-10V from a power supply of 5V by chargin parallel caps and switch them to series connection (by using transistors) so they reach the desired voltage. I'm not sure how to do this.

This is what I have thought:

enter image description here

Explanation (please correct me when I'm wrong):

  • The caps. are put in parallel/series by a series of BJTs controlled by a clock signal (2Hz).
  • During the rising edge, Q10, Q12 and Q13 are enabled, and all the other BJTs are off. This puts C7 and C8 in parallel with the 5V power supply, so they charge.
  • Then, during the falling edge, Q11 and Q2 are enabled and the other BJTs disconnected. That leaves C7 and C8 charged at 5V in series (so they should mimic a 10V power supply if I'm not wrong), and in parallel with C9. As far as I understand, at this moment C9 should be charged until it reaches 10V (ignoring losses).
  • When the caps. are in parallel, they are in series with a 220 ohm resistor (R34). That gives a time constant of 0.439ms (220 ohms * 2e-6 farads). Having that the clock is working at 2Hz, they should have more than enough time to charge I think.
  • When the C7 and C8 are in series, and in parallel with C9, there is no resistor so they should charge C9 very fast, right? The discharge will be very quick as well, but for now I only want to see some ~8V moments at C9.
  • RL is a resistor used to simulate a load.

The result I get at C9 after the simulation finishes is that it stays at 4V. If I remove C9 and probe the voltage at C7, it switches between ~4 and ~8V depending on if they are in series or parallel due to the action of the clock.

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    \$\begingroup\$ (1) How, for example, is C8 supposed to pass its charge to C9? Both Q11 and Q13 won't conduct in the direction you want. (Look at their emitter arrows.) (3) If you did succeed in getting the capacitors in series how would R45 disappear so that "they should charge C9 very fast"? (3) What is the 2 Hz signal voltage level? Q2 is wired as an emitter follower so if your 2 Hz is a 5 V high level then the highest possible voltage at its emitter is 4.3 V. \$\endgroup\$
    – Transistor
    Apr 5 '21 at 18:53
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    \$\begingroup\$ Your idea has been invented before: the circuit is called a charge pump. You can look up charge pump designs on the Internet. \$\endgroup\$
    – user253751
    Apr 5 '21 at 19:04
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    \$\begingroup\$ Search the interweb for 'voltage doubler' and you'll see charge pump circuits. You should find a circuit with a square wave frequency coming out on a push-pull driver , two caps and two diodes. You can use a power op-amp, motor driver etc. as your push-pull driver if need more current. A 555 timer will do it but there are more elegant power solutions. Or you use a step-up regulator and do it properly. \$\endgroup\$
    – TonyM
    Apr 5 '21 at 19:07
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Your idea has been invented before, and it works: the circuit is called a charge pump. You can find many charge pump designs on the Internet. They can be simpler than yours (see below).

As for your immediate question, why the voltage on C9 doesn't go higher than 4V, even though the top end of C7 goes all the way up to 8V:

Transistors are not just switches where the switch goes on if the base has voltage - they get way more complicated than that. In particular, note that the transistor detects the base voltage compared to the emitter, not compared to 0V. If the base is 5V and the emitter is 4V, the transistor only detects a base voltage of 1V, not 5V.

And that's exactly what's happening here. Because Q2's base voltage can't be higher than 4.54V (10/11 of 5.0V), Q2's emitter voltage can't go any higher than about 3.9V before Q2 turns off. Transistors are analog devices, so Q2 will find an equilibrium, where the base-emitter voltage is exactly enough to stop the emitter voltage going any higher.

This is used in "emitter follower" or "common-collector" amplifier circuits: the emitter voltage tracks the base voltage, minus about 0.6V. But that's not what you want here.

Note that Q12 doesn't have this problem, because its emitter is always 0V. For good switching, you want all of your transistors to be like Q12. If you want to have a switch connected to +5V, you can also use a PNP transistor, which uses opposite voltages, and will turn on when the base voltage goes below about -0.6V compared to the emitter, which you connect to +5V.

Q11 is used in "reverse active" mode. When it's turned on, current flows from the emitter to the collector, which is backwards. Transistors can operate in this mode, but not as well as they do in the usual "forward active" mode.

Note the whole C8/Q13/Q11/Q12/etc circuit is unnecessary. You can just connect the bottom of C7 to the output of the inverter, unless your inverter is too weak to supply the load current - in this configuration all of the load current comes out of the inverter. Q10 doesn't need active switching and can just be a diode. Q2 also doesn't need active switching and can just be a diode. If you make these changes, you will find yourself with something like the standard charge pump voltage doubler circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

BTW: Different charge pump designs can also be used for voltage tripling, etc, and for creating negative voltages.

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    \$\begingroup\$ Here is a CircuitLab tutorial (with a working simulation) about building a voltage doubler very similar to the schematic you've drawn. \$\endgroup\$
    – compumike
    Apr 5 '21 at 19:44
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    \$\begingroup\$ @compumike That is a different circuit however - that one only works on AC. \$\endgroup\$
    – user253751
    Apr 5 '21 at 19:51
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    \$\begingroup\$ yes that one is slightly different but could be quickly rearranged to match yours. I've just done so for your convenience: Charge Pump Voltage Doubler with a working simulation that matches your schematic. \$\endgroup\$
    – compumike
    Apr 5 '21 at 20:16

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