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A recent question about cyclically charging a capacitor reminded me of something I read once. As I remember, it demonstrated that it's impossible to construct a charge pump that is 100% efficient with ideal components, but it is possible to build a 100% efficient boost converter with an inductor if components are ideal.

Does this resonate (no pun intended) with anyone else? Any way to demonstrate or refute the truth of this?

To be clear: we are assuming we have ideal components. I realize no real circuit will be 100% efficient with real components. Diodes may have zero voltage drop. Transistors may be ideal switches that take no energy to change state. Wires may have zero resistance.

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Phil Frost
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4 Answers4

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It is all about dualism. With ideal components, you can make an ideal SMPS type voltage converter (= using an inductor to do the work). You can't make an ideal voltage converter using switched (flying) capacitors. That is not the universe being unfair to capacitors: you can make an ideal current converter using switched capacitors, which is not possible using inductors.

The problem with capacitors and a voltage source is like this: take a voltage source \$V\$ with a certain source impedance (= series resistor) \$R\$. Connect a capacitor \$C\$ to it and load it for an infinite time (any finite time will do too). The loading current will be $$ I = \frac{V}{R}\exp\left(\frac{-t}{RC}\right)$$ and so the power dissipated over the resistor will be $$ P = I^2 R = \frac{V^2}{R}\exp\left(\frac{-2t}{RC}\right)$$ Consequently, the total energy cost shall be $$ E = \int_0^\infty P \,dt = \frac{V^2}{R} \int_0^\infty \exp\left(\frac{-2t}{RC}\right) \,dt = \frac{CV^2}{2} $$ which, as you can see, is both always positive and completely independent of \$R\$. Therefore, there is always a cost, and even with an ideal capacitor! Intuitively, this is because a smaller resistor causes a higher initial loading current, and hence a higher RI2 loss.

Management summary:

You can't connect an ideal voltage source to a capacitor, because that would result in an infinite current which is impossible in itself and would cause an infinite magnetic field which would destroy the universe (just kidding, remember this is the management summary). But you can approach this ideal as closely as you like, and the result will still be the same: a fixed amount of energy is lost while charging the capacitor. Hence: sorry boss, no ideal flying capacitor voltage converter.

ManRow
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Wouter van Ooijen
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  • Excellent answer. Now that I understand the problem is unavoidable energy losses in charging capacitors, I was able to find [the article I remember](http://www.smpstech.com/charge.htm). – Phil Frost Jan 14 '13 at 16:41
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    Actually, you can't get infinite current. Any circuit of nonzero area has nonzero inductance, and this will limit the current even if there's no resistance. But electromagnetic energy will radiate away from the circuit, so you still can't get 100% efficiency (but this applies to inductor-based switching converters, too). – Dave Tweed Jan 14 '13 at 17:16
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    I guess Phil would cheat out of that problem by requiring zero-sized components and conductors :) – Wouter van Ooijen Jan 14 '13 at 19:40
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    Also consider the common exam question of connecting a discharged capacitor to an equal value charged one, comparing the total stored energy before connection and after equalization time. – Chris Stratton Jan 14 '13 at 20:12
  • @ChrisStratton: Yes, the "trick" there is if you assume ideal capacitors (no resistance), they never equalize; the charge just keeps oscillating back and forth forever. If there is any resistance, that's where the missing energy went. – Dave Tweed Jan 15 '13 at 05:20
  • Nowhere in the question was destroying the universe disallowed. – gbarry Jan 15 '13 at 06:47
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    @DaveTweed, no, they will not oscillate without an inductance, as otherwise there is no "inertia" to continue current beyond the point where voltage differential first reaches zero. – Chris Stratton Jan 15 '13 at 13:16
  • @ChrisStratton: See my earlier comment. "Any circuit with nonzero area has nonzero inductance." I have yet to see a capacitor that has zero area. – Dave Tweed Jan 15 '13 at 13:24
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    @gbarry: Destroyng the universe would require that the field does work on the matter of the universe. The energy available to do this is no more than what the capacitor held in the first place. – Dave Tweed Jan 15 '13 at 13:28
  • You're right, and moreover, the efficiency wouldn't be 100% if some energy were wasted in its destruction. – gbarry Jan 15 '13 at 15:54
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    @DaveTweed - an ideal capacitor is non-inductive. That you cannot build an ideal capacitor is an entirely different topic, irrelevant to how an ideal capacitor would theoretically behave in a circuit. Two connected ideal capacitors **will** equalize - the equations governing their ideal behavior require it. – Chris Stratton Jan 15 '13 at 18:16
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An inductor-less charge pump cannot be 100% efficient when powering a constant-voltage load from a constant voltage source. An inductor-less charge pump made with ideal components may be 100% efficient if the source current and voltage waveforms have the proper relationship with the load current and voltage waveforms. It is possible for either the source or the load voltage to be constant DC, but not both (except in the trivial case where both voltages are the same and the charge pump doesn't have to do anything).

Note: a charge-pump which contained an internal current source could be 100% efficient at converting input power from a constant-voltage source to an external constant-voltage load, with any energy that was drawn from the internal current source in one cycle being replaced on the next. On the other hand, such a current source would simply be taking the place of an inductor.

supercat
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    Can you elaborate on what the "proper relationship" would be? – Phil Frost Jan 14 '13 at 18:00
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    There are an infinite number of possible relationships, and I'm not sure there's any particularly nice way to characterize them. On the other hand, I can offer an example though: suppose one has two capacitors in series, one of which is charged to five volts and the other of which is discharged. Across the two capacitors is a 5K resistor (which will draw 1mA). If one connects a 2mA source to the cap which is discharged, it will charge from 0 to 5 volts at the same rate as the other cap discharges. If one then switches the 2mA source to the other cap, one may effectively repeat the process. – supercat Jan 14 '13 at 18:14
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    The 2mA source will see the voltage across it rise from 0 volts to 5 volts, then drop essentially instantly to zero, then rise to five, etc. During the time when the input voltage is below 2.5 volts, the circuit will be taking less energy from the source than is going to the load; the difference between input and output energy to that point will match the change in the total energy of the two caps. While the input voltage exceeds 2.5 volts, the input energy will exceed the output energy, with the differential energy replenishing the caps. – supercat Jan 14 '13 at 18:18
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    If what one has is not a constant current source, but at an AC voltage source whose voltage waveform matches the waveform that would have been yielded by a constant current source, the circuit behavior will be the same as it would have been with a constant-current source. Note that while this example for simplicity a voltage which went from 0 to 5 volts, it could have used a voltage that swung from -5 to +5; if one added a polarity switch, one could then accommodate a triangle wave rather than a sawtooth. – supercat Jan 14 '13 at 18:22
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For a boost converter you can design one with idealised components and all the equations still make sense, voltages and currents remain finite. From these voltages and currents you get an efficiency of 100%.

A charge pump with zero stray resistance simply cannot be analysed in this way Trying to do so results in absurd answers. What happens when you connect a perfect capacitor to a perfect voltage source through a perfect switch? Trying to calculate the current results in a divison by zero. The same problem applies to connecting two perfect capacitors.

Lets say we have a capacitor charged to a given voltage and connect it to a voltage source of a higher voltage via a resistor. Lets assume for now that we let it charge fully (ignoring for a moment that doing so would take infinite time). We find that changing the value of the resistor does not change the efficiency, the total energy drawn from the voltage source remains the same. The efficiency is however dependent on the ratio between the starting voltage of the capacitor and the voltage of the voltage source. A smaller voltage difference leads to a higher efficiency tending towards 100% as the voltage difference tends to zero.

In our charge pump there is not infinite charging/discharging time so resistance does affect efficiency but as resistance tends to zero efficiency (for a finite voltage difference) tends towards a finite number less than 100%.

The charge transferred on each switching cycle is related to the change in voltage on the capacitor by the capacitance. To transfer a finite average current to the load we either need to transfer a finite charge per cycle or we need to have an infinite number of cycles.

So making your 100% efficient charge pump would require either an infinitely big capacitor or an infinitely high switching frequency.

Peter Green
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Well it really depends upon how far we go with "ideal components". If diodes had a forward voltage drop of 0 volts, BJTs had a base threshold of 0 volts a saturation of 0volts and infinite current gain, and FETs have a gate threshold of 0volts and a Rds of 0 ohms then it may very well be likely be possible to realize a 100% efficient change pump.

Even in the case of the boost converter it will not be 100% efficient unless the switch FET and flyback diode are ideal in the sense that I described above. Likewise the inductor has to have a DCR that is equal to 0.

Michael Karas
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    We are going _all the way_ with ideal components. FETs that are ideal switches and take no energy to change state and diodes with no voltage drop are both fair. – Phil Frost Jan 14 '13 at 16:16
  • @PhilFrost - OK Then. I cannot think why a charge pump could not be 100% efficient then...as long as all the wires are zero ohms resistance too. :-) – Michael Karas Jan 14 '13 at 16:23
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    The only way to transfer energy between two capacitors or series-connected groups of capacitors is for there to exist a potential difference between the points where they are connected. Any such scenario may be modeled as connecting two capacitors C1 and C2, charged to voltages V1 and V2. The energy before the connection will be (C1·V1·V1 + C2·V2·V2)/2. The voltage after the connection will be (C1·V1+C2·V2)/(C1+C2), and the energy after will be (C1·V1+C2·V2)·(C1·V1+C2·V2)/2(C1+C2). The only time the two energies are equal is if V1=V2, meaning nothing happened. – supercat Jan 14 '13 at 17:05
  • There's a way for an inductorless charge pump to be 100% efficient, but only if certain other conditions are met regarding the inputs and outputs. – supercat Jan 14 '13 at 17:45