I am trying to understand a charge pump circuit here but one part has me scratching my head and this sort of scenario happens a lot with me in electronics. It’s a question of what happened first??

If you look at the image below:
I understand the left image where the capacitor is basically in parallel with Vs and GND so this configuration is charging it up to VS .

In the next drawing, If I understand correctly, the Capacitor is connected in series with VS thus giving you 2VS. Then, because VS goes through the mosfets Drain -> Source thus connecting VS to what was previously the GND node of the capacitor

How did the mosfet turn on to let that through?

It seems like to get 2VS the mosfet has to have BEEN on, So basically a catch22 , which happened first? I don’t understand.

I think an answer might be if you switch S1 to off position first, that way the gate is not grounded and the source is still at ground and then the cap begins to charge the gate and once the mosfet is on then you flip S2 off and it works as described to give you 2VS ???

Any insight would be much appreciated. Thanks.


  • \$\begingroup\$ The MOSFET doesn't need 2Vs to turn on. It needs some voltage between its gate and source. Don't look at the "absolute" voltage on the MOSFET's gate with reference to ground, look at Vgs and you'll see that this is provided by the capacitor no matter what the initial state of the MOSFET is. \$\endgroup\$
    – brhans
    Nov 2, 2018 at 15:37
  • 1
    \$\begingroup\$ imho these are ideal circuits to learn a good simulator and observe what is going on. \$\endgroup\$
    – PlasmaHH
    Nov 2, 2018 at 15:52
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    \$\begingroup\$ In case it isn't apparent, the values shown in figC are the final state values, not the values immediately after opening the switches \$\endgroup\$
    – sstobbe
    Nov 2, 2018 at 15:56
  • \$\begingroup\$ @PlasmaHH yeah I agree I think I would benefit from observing some transients \$\endgroup\$ Nov 2, 2018 at 15:57

1 Answer 1


Similar circuits are used to make a charge pumps (switching with capacitors) to give you more than what your voltage rail can provide. Some DC-DC converters use circuits similar to this to boost the voltage rail for driving N-ch mosfets. With N-ch mosfets the gate voltage needs to be more than the source voltage (Vgs) for the mosfet to be able to turn on. The cool thing about mosfets is the gate needs a voltage (it's a capacitor) so the current is mimimal to get the mosfet to turn on.

In this circuit the switches are driven external to the circuit, so you would need additional circuitry to drive the switches. The switches all switch at the same time.

How did the mosfet turn on to let that through?

The switches turn off and then the capacitor starts to charge the gate (remember the gate voltage would start to charge at Vs and V source of the mosfet would be 0V, so Vgs would still be enough to turn the mosfet on).

As the mosfet turns on, the capacitors negative terminal is always higher than the source of the mosfet, so the mosfet will always be on.

  • \$\begingroup\$ I understand all of that but it doesn’t answer my catch22 question of how do you get 2VS if the mosfet needs to be on in order to get 2VS but the mosfet can’t be on yet. \$\endgroup\$ Nov 2, 2018 at 15:30
  • \$\begingroup\$ Edited, see the answer now \$\endgroup\$
    – Voltage Spike
    Nov 2, 2018 at 15:34

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