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I am designing (simulating in LTSpice and building on a breadboard) a charge pump to power my numitron tube. I need to get ~15-20 V out of 3.3 V. The circuit is as follows:

schematic

simulate this circuit – Schematic created using CircuitLab

All diodes are 1N4148, the V1 source comes from a voltage regulator, while CLK1 from an Atmega OSC1 pin (Atmega powered by the same voltage regulator). The NOT gate is made from a NAND gate on a separate IC.

The circuit gives me about 18-20 V on the output without load, but when loaded, it lowers drastically. I measured the effective output resistance of this charge pump to be about 11 kOhm which is way too much for me. I tried simulating the circuit with smaller/bigger capacitances or frequencies, but it didn't change much.

The question is: how can I lower the output impedance of my charge pump?

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  • \$\begingroup\$ Have you tried to increase the clock speed? \$\endgroup\$
    – diverger
    Nov 2, 2014 at 9:40
  • \$\begingroup\$ In LTSpice only, because I lost my programmer, lol. I increased to clock frequency 10 times and it gave me no visible difference. \$\endgroup\$
    – Mike Spark
    Nov 2, 2014 at 9:43
  • \$\begingroup\$ What's the high and low level of the clock pulse? \$\endgroup\$
    – diverger
    Nov 2, 2014 at 9:55
  • \$\begingroup\$ They are about 0 and 3.3 V. \$\endgroup\$
    – Mike Spark
    Nov 2, 2014 at 9:56
  • \$\begingroup\$ Assuming static drive, how much current is required per filament and how many filaments will be on at once? \$\endgroup\$
    – EM Fields
    Nov 2, 2014 at 20:34

3 Answers 3

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The basic output impedance should be like this

$$ R_{eq} = \frac{N}{f*C} $$

So, increase f or C.

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  • \$\begingroup\$ Hm, I tried both, but only by simulating. I'll try switching capacitors on breadboard. Interesting equation btw, could you possbily give a source, so I could read up more? \$\endgroup\$
    – Mike Spark
    Nov 2, 2014 at 11:16
  • \$\begingroup\$ ftp.dei.polimi.it/outgoing/Massimo.Ghioni/…. And in Sergio Franco's: Design With Operational Amplifiers And Analog Integrated Circuits, 3rd edition, page 187, has similar discussion. \$\endgroup\$
    – diverger
    Nov 2, 2014 at 11:36
  • \$\begingroup\$ I think the impedance of the two drivers is missing from your equation. @MikeSpark: I think you will need beefier drivers. A MOSFET gate driver might be a good choice. \$\endgroup\$ Nov 2, 2014 at 12:08
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    \$\begingroup\$ A driver between the pump and the load would need a power supply of at least the required voltage, which I assume you don't have. (That's the whole point of the charge pump!) Note that you need to drive both 'legs' with a low impedance, not just one. \$\endgroup\$ Nov 2, 2014 at 12:37
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    \$\begingroup\$ @glibg10b 'N' is the stage count, 'C' is the capacitance of each diode-capacitor stage, and 'f' is the frequency of the input signal. \$\endgroup\$
    – diverger
    Jul 15, 2023 at 10:39
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An old thread, but just in case it helps... try using Schottky diodes, so the drop across each is minimized.

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Use Schottky diodes and improve your clock source driving strength if you have tried other ways and they didn't work. If you use Schottky diodes, you need fewer stages to get your desired voltage, which decreases the output impedance.

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