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Lets consider for instance this voltage doubler:

enter image description here

It was told that a practical limit of this circuit is the fact that we cannot absorb too much current from it. But I wanted to verify (if it is true) this by analyzing the circuit. Intuitively I may think that, since the maximum power which can be provided by the voltage source V1 is fixed, double voltage will mean half current. But I wanted to have a more precise idea of how much current a load can absorb with respect to that provided by V1.

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    \$\begingroup\$ You have the simulation running...just add a load resistor and see what happens. \$\endgroup\$ – Elliot Alderson Sep 6 at 18:50
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    \$\begingroup\$ What is your question? \$\endgroup\$ – joribama Sep 6 at 18:54
  • \$\begingroup\$ How can we evaluate the maximum current that a generic load R may absorbe from V1 (which can provide a maximum current equal to Imax)? \$\endgroup\$ – Kinka-Byo Sep 6 at 19:01
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The impedance of that circuit will be proportional to the impedance of your capacitors at the frequency of your AC source.

I don't know how exactly it will relate in the circuit you are using. That arrangement isn't one I've fiddled with.

Since you are using LTSpice, you can easily find out how the impedance of the doubler relates to the impedance of the capacitors.

Measure the unloaded voltage of the doubler, then put a resistor from the output to ground and measure the voltage. Keep lowering the resistance until the output voltage is one half of the unloaded voltage. That resistance R is the approximate impedance of the doubler.

Compare that to the impedance of a single capacitor at your AC frequency. R should be some multiple of the capacitor impedance.

Repeat for different capacitor values, and you should find out quickly how the capacitor impedance and the doubler impedance are related.


This site goes into some detail about voltage multiplier impedance and loading. It's about Cockcroft-Walton multipliers, though, which are different from your circuit.

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  • \$\begingroup\$ Thank you for the answer. Why do you say "That resistance R is the approximaye impedance of the doubler."? \$\endgroup\$ – Kinka-Byo Sep 6 at 19:18
  • \$\begingroup\$ If you have a voltage divider with two equal resistors, then the junction of the two resistors is at half the voltage at the input. If you put a resistor on the output of your doubler, you have voltage divider made of the resistor and the impedance of the doubler. When the load resistance is equal to the impedance of the doubler, then the voltage will drop to half the unloaded voltage. \$\endgroup\$ – JRE Sep 6 at 19:26
  • \$\begingroup\$ "Approximate" because it ignores the voltage drop across the diodes. \$\endgroup\$ – JRE Sep 6 at 19:28
  • \$\begingroup\$ Perfect, thank you very much! \$\endgroup\$ – Kinka-Byo Sep 6 at 19:40
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There is no practical limit. The idea that charge pumps are impractical or even inefficient has no basis in reality, at least not today.

It might have been superficially true at one point many many years ago before higher capacity ceramic capacitors existed, and most electrolytic caps were basically garbage. Actually, most electrolytic caps today are still garbage, except for some pricey performance versions.

The current you can draw from a voltage doubler, or to put it more accurately, the output impedance of the voltage doubler is something you can treat like the ESR of a battery for example, or simply a resistor in series with the output. At least for a basic DC approximation. More load current means more voltage dropped due to that impedance. Note however, as this is impedance, that energy is stored rather than dissipated, so this impedance doesn't directly contribute to the losses of the charge pump, but like you say in your question, this can certainly place a practical limit on the power levels reasonably achievable.

However, there is no reason specific to charge pump's operation that is behind this. Inductors have series resistance, and it can often be high for higher value inductors, yet we handily use them to power levels of many kW or more.

The idea that charge pumps are not as capable is not because of the charge pump, but rather how poor capacitor technology used to be. And this wasn't really to the output impedance either - impedance is easily lowered simply by using larger capacitors or higher frequencies. Use 600µF of capacitance at 10kHz and 5A will cause less than 300mV of voltage droop on the output. That's really not bad at all.

The problem was that, as I mentioned earlier, electrolytic capacitors were (are) complete garbage. They're terrible at every part of being a capacitor except for the actual stored charge. Sure, they've got µF for days, but you pay for it in extremely high ESR (unless you use monstrously sized capacitors), and very poor ESL and ripple performance on top of that. Most electrolytic capacitors stop behaving like a capacitor entirely at frequencies above 100kHz (solid polymer capacitors sometimes manage 150kHz..at great additional BOM cost). So it is actually the inductance of electrolytic capacitors, along with the relatively high series resistance and poor cost and volumetric efficiency that, combined, made charge pumps a fairly unattractive option. You would be forced to stay well away from the self-resonant frequency of the capacitor, where the ESL makes them behave like resistors (with reactive impedance of course rather than real resistance) entirely, thus robbing you of any actual voltage doubling. This is why many commercially available charge pump chips use such low switching frequencies. So low they're audible! 5-10kHz is pretty common.

And that lower frequency of course compounds the issue as it increases the reactive impedance due to the capacitance as well... but not nearly as much as one would think.

It is very important to keep in mind that when talking about AC circuits, everything is strictly sinusoidal. The impedance formula for a capacitor is only accurate for a sine wave at a given frequency, and it does not apply to more discontinuous-looking waveforms like MOSFETs switching on, and they switch on hard. Nanosecond rise times are not very sinusoidal at frequencies below 1GHz.

Now, no waveform is truly discontinuous, so these switching waveforms, while appearing squarish, can be deconstructed into a series of continuous, sinusoidal components, starting with the fundamental and then either even or odd harmonics (depending if its a rising or falling edge), spanning the frequency range all the way to the sinusoidal frequencies with periods similar to the rise time of the edge. Like this: enter image description here

As you can see from the bottom graph, only some of the power is contained in the fundamental, the rest is contained in higher frequency harmonics, meaning the capacitor's reactive impedance is not applied evenly to the various components of the switcher waveform. This lets a modest but meaningful increase in the effective power that can be transferred at a given frequency through a given capacitor, simply because the waveform isn't sinusoidal. While the impedance goes up again after 100kHz or so, almost all of the energy is contained in the harmonics lower than that (for 10kHz and typical MOSFET rise times).

And intuitively, this makes sense: if you just slam charge into a capacitor as fast as it will accept it instead of ramping it up in a nice sine wave, then you can charge that capacitor faster and in the context of the charge pump, your voltage drop due to the output impedance is a bit less. Overall, this isn't a game changer or anything, but it is noticeable and I felt was worth mentioning.

Anyway, it isn't the 1980s anymore and we are no longer limited in our selection of capacitors. Ceramic capacitors are no longer only available in nanofarads, they're beginning to match (and supplant) small electrolytics, with values like 100µF or more readily available in ceramic capacitors.

And unlike electrolytic capacitors, ceramic capacitors are actually deserving of that name - capacitor. Their ESR is so low, in the single-digit mΩ typically, and their ESL limited to the parasitic inductance from the height of the SMD end caps (!!!), that these are truly the capacitors we deserve.

So lets make a voltage doubler that switches at over an order of magnitude faster than the nonsense 10kHz devices that predate the DeLorean, 150kHz. This isn't even noteworthy for a ceramic capacitor. And let's build it0 using 6 x 10µF 63V ceramic capacitors hardly unreasonable. And lets double 24V to 48V... at 360W, or 7.5A at 48V output.

This is the performance we get from our so-called 'low power' voltage doubler: enter image description here

98.25% efficient.

So no, charge pumps are not anymore power limited than, say, conventional inductor-based buck or boost converters. Indeed, they actually outperform them in a wide range of power levels and output voltages. Look at this 500W+ voltage doubling/dividing behemoth, the LTC7820. Its data sheet is the source of that efficiency graph.

There is a fundamental problem with them that is likely the true cause of their fairly limited usage in power conversion, and that they cannot regulate their outputs like inductor-based switchers can. They are true doublers/dividers. You get half or double whatever the input voltage is and only that. They are purely ratiometric, and those ratios are fairly limited (like doubling, halving, multiplying). So they can work great as intermediate power conversion, but you'd typically still need a post-regulator (or preregulator, or both). And at that point, you might as well just make a 500W conventional switcher. Depending.

All I am saying is, don't assume charge pumps can't handle some serious wattage - they definitely can.

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  • \$\begingroup\$ Thank you very much for your analysis. There is a last question that I cannot solve (probably the answer is simple but I do not see it): if you consider an AC voltage source which can provide a certain power P, and you connect it to a "standard" diode rectifier (diode in series and then a capacitor in parallel), and in another case you connect it to a voltage doubler, I would say that the conservation of power will establish that the current provided by the second one is the half of that of the previous one. Where is the mistake of this reasoning? \$\endgroup\$ – Kinka-Byo Sep 7 at 23:44

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