Here is a simple dickson inverter:

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The simulation shows this:

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Is it a problem with the simulator? I don't see why the capactior was charged to the voltage of the square wave signal? Why is the push-pull pair not working?

UPDATE: I have replaced the push-pull pair with a single NPN transistor, as suggested.

enter image description here It works as expected, but the current is limited.

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Is there a way to do this without a series resistor and single supply? (E.g. eliminate power disciplation on the resistor).

UPDATE: finally! Frequency increased to 100K, voltage source drives a voltage amplifier, and then the push-pull pair.

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Simulation result:

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Frequency had to be increased because the load resistor was changed to 100 ohms. Now it can deliver about 150mA. Great!


5 Answers 5


V1 in your push pull circuit should be ranging from 0V to +15 volt (15 Vp-p) to properly drive the push-pull pair. These are configured as emitter followers and you are probably only driving them at the moment with a 5 Vp-p square wave hence you only see about -3 volts coming from the charge pump.

With just Q1 used, you have wired it as a common emitter and hence the collector voltage will drive nearly from 0 volts to +15 volts.

  • \$\begingroup\$ Changed my vote because of the detailed explanation, the effort taken to give practical considerations. :-) \$\endgroup\$
    – nagylzs
    Jan 3, 2017 at 6:22
  • \$\begingroup\$ @nagylzs yes I can see that but did the newly accepted answer address the question "Is it a problem with the simulator?" OR "Why is the push-pull pair not working?" OR "I don't see why the capactior was charged to the voltage of the square wave signal?" \$\endgroup\$
    – Andy aka
    Jan 3, 2017 at 8:30
  • \$\begingroup\$ Well you are right. So I changed my mind again. Tony's explanation is still very useful, because it gives practical help to the design. (But you are right, it is not strictly the answer to the question.) \$\endgroup\$
    – nagylzs
    Jan 3, 2017 at 11:40

There are many factors that affect efficiency and Load Regulation in the Dickson Voltage Doubler, inverting or not.

Impedance ratios at fundamental frequency of the the square wave and highest harmonic are the primary factors.

This includes impedance of;

  1. supply, (battery ESR)
  2. diodes in conducting (ESR) and reverse mode (Cjcn)
  3. Zout of the Emitter Follower which is Rb/hFE. or RdsOn of CMOS
  4. switching frequency of driver
  5. clock driver Zout ( 25~50 typ. for [email protected] to 74HC at 5V)

    • The criteria of concern here was % Load Regulation.
      • i.e. excessive % voltage drop with load R.
      • 10% max Load error is consider max.while 2% is good.

So how to choose the optimal parameters.

What not to do

  • Drive the Q's too hard results in I²Rce losses
  • Drive the base with too low a Rb and excessive current spikes & EMI.
  • Use Caps with excessive ESR and too large ESR*C value compared to 1/f .
    • Caps should be ESR*C <<1us
    • load R * C should be 10* 1/f for low ripple.
    • thus ESR or Zout of Dickson Doubler should be < 10% of R load

So what is the Zout of the Doubler?

  • first if you know the load R or current then R you want a Zout = <10% or load R

    • many factors but a simple approximation is the impedance of the two output shunt Caps \$Zc= 2* { \frac{1}{2\pi fC}} \$
    • so if Zc <=10% of Load R then
    • \$Zc =\frac{1}{\pi*f*C} \ <=10R_L\$
    • the best Dickson IC CMOS chips can supply 150mA typ. using 0.1uF
    • 1st make Rb for bipolars 3% to 30% of Load R affects Zout and load regulation and efficiency.

      • I suggest Rb=10% of load using hFe of 100 min.
    • then \$f = \frac{10\pi \ R_L }{C}\$

    • e.g. 10V with no load expect 1V drop to -9V

    • thus 10% drop with max load current drop Vout to -8.1V out
    • Load = 600 Ω, f = 100 kHz , C = 1 uF ESR <70 Ohms or ...
    • Load = 600 Ω, f = 1.0 Mhz , C = 0.1uF ESR <70 Ohms

Due to all the losses dont' expect more than 10% efficiency and compare with commercial IC's. These are intended for low current because of charge transfer losses.

Java Simulator


First, that's not a "inverter". It's a charge pump, which in this case is making a negative voltage.

The first circuit worked exactly as expected. The double emitter follower gives the signal more drive, but looses about 1.4 V of amplitude due to the B-E drops of the two transistors. The two Schottky diodes will eat up another 700 mV or so between them. If the voltage source is putting out a 0-5 V square wave, then 3 V output is about expected.

The basic idea of increasing the current capability with the double emitter follower is a good idea. However, if you want a larger negative output voltage, you have to give it a higher amplitude square wave as input.

Your second circuit creates a higher drive level because Q1 is used to amplify the signal voltage. However, the resulting output drive will be weaker due to the Q1 output having a higher impedance.

For the best result, combine the two. Keep the emitter followers of the first circuit, but drive them with the output of Q1 of the second circuit. Tie the two emitter follower bases together directly and to the output of the Q1 amplifier. That should give you -12 V or so, but with better drive.

For even better drive, use a higher frequency square wave. 1 kHz is very slow for a charge pump. I often run charges pumps from a spare clock output of a microcontroller that is 1 MHz or more.

  • 1
    \$\begingroup\$ it is an "inverting" charge pump or "a +ve clamp Dickson Voltage Doubler" i.e. Vcc minus 2*Vcc=-Vcc (negl. diode drops of 1V) \$\endgroup\$ Jan 2, 2017 at 23:25

Your push-pull buffer config is correct but V1 and V2 should have equal magnitudes.


As far as I can see, there seems to be a problem with the position of your PNP and NPN Transistors. Invert their positions, i.e. put the NPN one between the common point and ground; and the PNP between the common point and +15V point. Your circuit causes the transistors not to switch on completely. It may be the reason why the voltage at the output is not near the input square wave value. Inverting them will enable them to be completely on/off and increase the magnitude of the output voltage.

Also you will get less voltage than desired because of the diode forward voltage drop, although this difference will be around 0.5 V.

Please tell us of your results after the suggested change.

Update: After some comments and some simulating on my side, I would like to inform viewers that my solution is wrong and that the original PNP and NPN configuration is the correct one. Inverting them as suggested by me earlier does not make the circuit perform as it should.

  • \$\begingroup\$ After switching npn and pnp on high/low side, volprobe1 was constant zero volt. \$\endgroup\$
    – nagylzs
    Jan 2, 2017 at 19:42
  • 2
    \$\begingroup\$ Your suggestion won't work. Push pull config in OP's question is correct. It provides very low output impedance and nearly unity voltage gain and that's why it's placed there. But V1 and V2 should have equal magnitudes to have it work correct. \$\endgroup\$ Jan 2, 2017 at 19:45
  • \$\begingroup\$ I am trying to but am not getting anything on the net regarding Dickson Inverter circuits. I ll try to simulate the same and update my answer. Also I would like to know how would the low output impedance and unity gain help in the circuit \$\endgroup\$ Jan 2, 2017 at 19:57
  • 1
    \$\begingroup\$ To have higher output current. In a common emitter driver, output current is limited due to high output impedance (i.e. the resistor connected between Vcc and collector). \$\endgroup\$ Jan 2, 2017 at 20:11

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