I was trying to understand the charge pump which generates a negative voltage. As far as I understood, the basic operation is as follows:

SquareWaveInput is a 0 to 5V, 10kHz square wave. When the input is high, Q1 is ON and Q2 is OFF. Q1 charges C1 through D1 back to supply. D2 is reverse biased at this time and does not conduct.

When the SquareWaveInput is low, Q1 is OFF and Q2 is ON, supplying a low impedance path to ground. C1 was charged at the previous cycle, and now it will discharge through Q2, ground, C2 and D2. D1 is reverse biased at this time and does not conduct. When the charge at C1 moves through C2, it charges C2 up until both have the same charge, but with the negative voltage. Assume that C1 is in parallel with C2 at this time.

Now, let's look at the schematic below. This circuit is the same as above except for D2 and C2. C1 charges up the same way the above circuit does. But how does C1 get discharged? Will it discharge through D1's reverse leakage current? I was doing a SPICE simulation and it discharged normally, but I don't know how. The current through D1 was very small (about 150nA) when the SquareWaveInput was low. Then I didn't trust SPICE and breadboarded it. It worked! I do not know how. Does it get its way through the oscilloscope probe? The probe was connected to the anode of D1 (Out) and its ground was at circuit ground and when I touched Out node, the oscilloscope showed some weirdness as in the scope-shot below. How did this circuit worked?

Normal operation:

enter image description here

With my hand on Out node:

enter image description here

  • \$\begingroup\$ Why do you think C1 gets discharged? There's no evidence for that in your scope shots. \$\endgroup\$
    – markrages
    Commented May 6, 2012 at 17:58
  • \$\begingroup\$ @markrages, yes, I am not sure whether it gets discharged though, but if not, how will it create that negative voltage? \$\endgroup\$ Commented May 6, 2012 at 18:26
  • \$\begingroup\$ What part of the "working" is a mystery to you? - it sounds like it is doing what it should. When you say "it worked" it depends entirely on what you mean by "worked". You start off with a circuit that provides -ve DC but end up showing a circuit that produces AC and which has it's DC level set by a clamp diode. This circuit is NOT rproviding DC but the first one is. They are different circuits with different results. \$\endgroup\$
    – Russell McMahon
    Commented May 6, 2012 at 19:14
  • \$\begingroup\$ @RussellMcMahon, you are right, I was a bit unclear. I thought the second circuit is the base for the first circuit and the first circuit just takes that AC and produces a DC by storing it in the C2. \$\endgroup\$ Commented May 6, 2012 at 19:18

1 Answer 1


If you look carefully, you will see that C1 doesn't get discharged in the second circuit. After the first positive pulse, it gets charged up, then mostly stays that way. Since the voltage on C1 is then roughly constant, you see the same square wave at Out as you do on the emitters, except that it is offset by the voltage on C1. Note that Out varies from about 0 to negative. This is the same signal you would see on the right side of C1 in the first circuit. The addition of D2 and C2 take the 0 to negative pulses and make a flat negative. You can think of D2 and C2 as being a negative peak detector.


  • \$\begingroup\$ Ah, so after C1 is charged in the positive pulse, what I probe in the Out when there is a negative pulse is effectively C1, because Q1 is OFF, Q2 is ON, D1 is OFF. I end up with just the C1 shorted to the ground on the left side and measuring it on the right side. But since I measure it reversed, I see a negative voltage, right? \$\endgroup\$ Commented May 6, 2012 at 19:25
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    \$\begingroup\$ Btw, I've learnt about this circuit via your designs. Love the way "USBProg2 PIC Programmer" is designed. There is one thing I don't get though, why did you use 3x120 instead of 360, or 2x75 instead of 150? To save the pick-and-place machine's slots or distribute the heat? \$\endgroup\$ Commented May 6, 2012 at 19:28
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    \$\begingroup\$ @abdullah: Yes, I think you've got it now. The places where I used 2 or 3 resistors in series was because the power dissipation was higher than what a single 0805 could handle. I could have used a larger resistor, but I knew the manufacturer routinely stocks 0805 and not larger sizes, so the latter would have to be a separate order. Sometimes you do things for logistic and manufacturing reasons, not purely for electronic ones. \$\endgroup\$ Commented May 6, 2012 at 20:36

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