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Typically the most expensive (and hard to get) elements in a SMPS are the inductors. Thus I was wondering if it's possible to use inductor-less switching mode power supplies (i.e. charge pumps) for generic use cases, for example a bench-top power supply, fixed high power DC-DC converters (several amperes and some hundred watts power), etc.

All charge pump designs I could find though were for low power applications. What prevents one from designing a high power inductor-less power supply? Are there some inherent physical limitations?

Nick Alexeev
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Ali Alavi
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    I suspect that inductors can store more energy per unit volume/cost than capacitors - try the back-of-an-envelope calculation for what size of capacitors you would need for a hypothetical charge pump. – pjc50 Feb 08 '16 at 15:44
  • There are charge-pump / capacitive switcher ICs but these are usually only for very low power. I also expect that you would need very large capacitors. I also expect inductors to be less lossy. – Bimpelrekkie Feb 08 '16 at 15:48
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    @pjc50 With a 50kHz switching frequency, I could easily come up with a 20A, 24V to 12V step-down converter, with capacitors in the range of 10uf. The simulation also looks promising. Am I doing something wrong? – Ali Alavi Feb 08 '16 at 16:01
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    Have you included some realistic ESR (series resistance) for those capacitors ? Simulators are like paper: you can make anything work/not work on them ;-) – Bimpelrekkie Feb 08 '16 at 16:02
  • @FakeMoustache Yes. And agreed! – Ali Alavi Feb 08 '16 at 16:07
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    A properly built switching power supply needs a PCB and it is the PCB thsat is likely to be the most expensive and hard to get item because you have to design it! – Andy aka Feb 08 '16 at 17:32
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    @Andyaka The PCB is neither the most expensive nor the hardest to get item. – Ali Alavi Feb 08 '16 at 21:58
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    A 10uF capacitor supplying 10A will drop one volt per microsecond. At a 50Khz switching frequency, you'd be looking at 100% ripple. – supercat Feb 10 '16 at 20:16
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    If one wanted to e.g. convert 24V to 9V, one could tolerate a two-volt drop over the course of 20us, so a 100uF cap would work (two volts of ripple, plus other losses, could total under three volts). Efficiency would be about 75%, pretty much independent of ESR or other such issues, but something in the circuit would have to dump 60 watts of heat somewhere. – supercat Feb 10 '16 at 20:41

5 Answers5

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There are two problems with your idea. One practical, and one fundamental.

The practical problem is that per amount of stored energy capacitors are more expensive than inductors, and on top of that the realy high-capacity capacitors (electrolytic) age.

The fundamental problem is that charging a capacitor from a voltage source is fundamentally lossy (you dissipate heat). This might seem counter-intuitive, but is nonetheless true. (There was a question about this some time ago.) Hence a flying-capacitor voltage converter, even an ideal one, is inherently inefficient. (An ideal inductor-based voltage converter is 100% efficient.)

You might think it strange that the world is unfair to capacitors, but that is our human fault: we supply power mostly from voltage sources. For current sources the inverse is true: an ideal current converter from flying capacitors can be 100% efficient, while one from inductors must necessarily be lossy.

Wouter van Ooijen
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    Thanks. I cannot get my head around the fact that capacitors are more expensive than inductors (in a SMPS setting). My experience is that, at least for low quantities, I need to do some calculations, buy specific cores and wires and wind the wire around the core myself. It's very time consuming. While with a capacitor, I just buy a ready made one. On the other hand, I'm an absolute newbie in SMPS domain, so probably there are better ways around. – Ali Alavi Feb 08 '16 at 16:24
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    You surely can buy ready-made inductors! But pay attention to my second point: a capacitor-based voltage converter is inherently lossy. No way around that. – Wouter van Ooijen Feb 08 '16 at 16:36
  • Looks like I said essentially the same thing.. later. oops. – Spehro Pefhany Feb 08 '16 at 17:03
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    @Lazarus there are plenty of inductors available off the shelf. Ti have a tool called webbench that will do the design for you including calculating the efficiency and telling you exactly what parts to buy. The only issue i've found with it is it has a tendency to default to parts that are a PITA to solder. – Peter Green Feb 08 '16 at 17:17
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    Huh, nice (+1's everywhere.) Is this the previous question? http://electronics.stackexchange.com/questions/54992/can-a-charge-pump-be-100-efficient-given-ideal-components. I knew about caps and voltage sources... but never really thought about it! – George Herold Feb 08 '16 at 17:48
  • Yep, thanks George, that is the one I had in mind. – Wouter van Ooijen Feb 08 '16 at 18:06
  • Doesn't the fundamental inefficiency tend towards 0 as the frequency goes to infinity? (Or as the output current goes to 0) – user253751 Feb 08 '16 at 19:13
  • Where have I seen the last paragraph before? – user253751 Feb 08 '16 at 19:13
  • @immibis, an idealised charge pump tends towards perfect efficiency as the freqency tends to infinity or the load tends to zero. Any real converter will of course tend towards zero efficiency as the load tends to zero load because of quiescent current. – Peter Green Feb 08 '16 at 19:21
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    No, an idealised voltage charge pump will NOT converge to perfect efficiency for any non-zero output current. You can't beat nature: charging the capacitor will ALWAYS incurr a fixed lost (and discharging too). Even with all components converging to perfection. – Wouter van Ooijen Feb 08 '16 at 20:15
  • @WoutervanOoijen Isn't the loss per cycle proportional to CdV^2, where dV is the amount of voltage change in one cycle, which is proportional to 1/f? So at a higher frequency you have more cycles, but still less loss: C(1/f)^2 * f = C/f. – user253751 Feb 08 '16 at 22:00
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    If memory serves, charging a capacitor through just a resistor (which will be just the resistance of a wire) is only 50% efficient. – Michael Feb 09 '16 at 01:53
  • @Lazarus Your experience only proves how easily available inductors are - so easy that you can custom-tailor them by hand and avoid compromising efficiency of the design. – Agent_L Feb 09 '16 at 12:22
  • @Agent_L can you point me to any reference for more details please ? – Ali Alavi Feb 09 '16 at 12:24
  • @Lazarus You are my source. You said that you design and produce your own inductors (as many people do). Can't do that with a capacitor, can you? Ergo, parts simpler to produce are cheaper when compared under same circumstances. (comparing custom-made inductors with off-the-shelf caps wasn't fair). – Agent_L Feb 09 '16 at 12:29
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    @Agent_L Oh, I meant a reference for more details on how to custom tailor inductors by hand, not a reference to support your claim :) – Ali Alavi Feb 09 '16 at 12:31
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    This answer compares charge pumps and inductor based DC-DC converters in the view of cost and efficiency. It doesn't answer why charge pumps are used for low current applications. – hkBattousai Feb 18 '16 at 15:22
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    It does: when compared for stored energy, capacitor are much more expensive than inductors, and a capacitor-based converter is less efficient. (only) At sufficient small power levels, either argument is important, and the fact that *small* capacitors are chaper than *small* inductors can be more important. – Wouter van Ooijen Feb 18 '16 at 15:32
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    I don't fully agree that there is some inherent problem in capacitor charge pumps regarding losses. If that was true, this device would not be possible: http://cds.linear.com/docs/en/datasheet/7820fb.pdf – payala Oct 19 '17 at 08:14
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    @payala - I agree. The LTC7820 chip you referenced converts 500 watts plus, at more than 90% efficiency, doubling the voltage or cutting it in half. People won't do what they think they cannot do. Perhaps that is the reason... – MicroservicesOnDDD May 10 '23 at 23:06
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    @MicroservicesOnDDD I have been getting efficiencies around 95% while transferring more than 2000W with a power stage based on the LTC7820 modified with external FETs. It's the closest thing to an ideal 4 quadrant middle rail generator. Compact, and so efficient you don't need heatsinks. – payala Jun 28 '23 at 06:35
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Capacitors would be better if the source and output were constant current. You could charge the capacitor until the voltage rose to a certain level, then discharge the capacitor into the load impedance to maintain a constant output current. You'd use a big inductor as an output filter to maintain the output current constant.

Since our sources are constant voltage and we usually want constant output voltage, using inductors to store energy and capacitors to filter it makes more sense.

Note that all efficient switching supplies have both capacitors and inductors.

Yes, charge pumps (flying capacitor) can take a voltage and move it around, flip it, even multiply by integers and such like, but every time you charge or discharge a capacitor through a resistive switch you lose a portion of the capacitor's energy change in the switch itself - a larger voltage change means more losses. A lower resistance switch just means that the energy lost for a given voltage change is compressed into a smaller slice of time, the total remains constant.

Spehro Pefhany
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    "every time you charge or discharge a capacitor through a resistive switch you lose half the capacitor's energy change in the switch itself." that is true if you fully charge and discharge the capacitor each time. If you only discharge it partially you can do better. – Peter Green Feb 08 '16 at 17:21
  • @PeterGreen "change in energy" not total energy. – Spehro Pefhany Feb 08 '16 at 18:06
  • Lets say a 1 farad capacitor starts at 5V and is charged to 6V through a resistor from a 6V source. Energy in capacitor before = 0.5 * 1 * 5 * 5 = 12.5 . Energy in capacitor after = 0.5 *1 *6*6 = 18 . Energy added to capacitor = 18-12.5 = 5.5. Energy drawn from supply = (6-5) *1 *6 = 6 . Only 0.5 joules of energy are lost to add 5.5 joules of energy to the capacitor. – Peter Green Feb 08 '16 at 18:33
  • When charging a capacitor from zero to full through a resistor you do indeed lose half the energy but the ratio of energy added verses energy lost is not constant. The early stage of the charge is very lossy, the late stage very efficient. – Peter Green Feb 08 '16 at 18:38
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    "You could charge the capacitor until the voltage rose to a certain level, then discharge the capacitor into the load impedance to maintain a constant output current." - [as it turns out, this is just a buck converter with an extra capacitor on the input](http://electronics.stackexchange.com/questions/215985/what-is-this-circuit-called-dual-of-a-boost-converter/). – user253751 Feb 08 '16 at 19:19
  • @immibis [SEPIC](https://upload.wikimedia.org/wikipedia/commons/thumb/c/c8/SEPIC_Schematic.gif/500px-SEPIC_Schematic.gif) topology is along similar lines. – Spehro Pefhany Feb 08 '16 at 22:11
  • @PeterGreen I suppose if the flying capacitor is only slightly discharged the charge pump may be more efficient, but I guess that also means you're drawing a relatively small current from the flying capacitor, or have a very huge capacitor at which point an inductor is smaller and more cost effective. That may be why charge pumps are only used for low-current applications. – Phil Frost Feb 10 '16 at 20:45
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If two capacitors or series strings of capacitors with different voltages are connected together, their charges will average out in a way which reduces the amount of energy stored therein. If they are connected using an inductor, the excess energy will be transferred to that inductor and may subsequently be put to some useful purpose. If the connection is purely resistive, the energy will be 100% converted to heat. Minimizing the resistance will not reduce the energy loss; it will merely reduce the amount of time required for it to occur.

Consequently, in order for a charge pump to be efficient, the capacitors need to be large enough that voltage across them never varies very much. In cases where a charge pump doesn't need to convey much energy, one can use a linear regulator on the output and boost the voltage enough that under worst-case ripple conditions the output voltage will still be high enough to maintain regulation, but efficiency will be limited by the ratio of the load voltage times the boost ratio to the source voltage.

supercat
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There are a few issues with charge pumps.

  1. they can't offer efficiency and voltage regulation at the same time. The only way to regulate the output voltage so that it remains constant during input voltage variations and load variations is to introduce deliberate inefficiency.
  2. Current must pass through two switching elements (diodes or transistors) during both the charging and discharaging parts of the cycle (whereas with a buck or boost converter it only needs to pass through one switching element at a time).
  3. Efficiency is highly dependent on the desired input to output voltage ratio. If you want to make say a 1.5x voltage converter then you would have to either use some complex multi-stage arrangement or make a 2x converter and run it in a deliberately inefficient mode.
Peter Green
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    Re point 3, any integer ratio is possible efficiently without much added complexity. For 1.5x, charge the caps in 2-series connections (so each sees 0.5x the supply voltage), and discharge in 3-series. – Nate S. Dec 04 '18 at 19:09
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Charge pumps are DC circuits, based on capacitors. As capacitors charge, current flow reduces. To increase current, and hence more charge into capacitor, one would have to increase voltage, but capacitors have a voltage limit.

Inductors on the other hand, have very low ohmic resistance so high currents can pass through them even at low voltages. Hence buck/boost converters for power applications are based on both, capacitors & inductors.

Charge pumps are good for information processing though, by manipulating voltages and very efficient since they work with low currents eg. arithmetic and clamper circuits. Not good for power applications though, not bench-top power supply and not fixed high power DC-DC converters.

Zimba
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    Capacitors have zero impedance at DC. Can you explain in what way these are DC circuits? It's my understanding they work in fact by switching the capacitor, so quite many AC frequencies are involved. – Tim Williams May 12 '23 at 16:40
  • AC circuits also have negative voltage, current flow reverses periodically. DC circuits only have positive voltage & current. – Zimba May 16 '23 at 15:55
  • @TimWilliams, capacitor has infinite impedance at DC – Zimba May 17 '23 at 06:50
  • Apologies, conductance is zero, heh. In any case, that doesn't seem very useful. How do you transfer power through an open circuit? – Tim Williams May 17 '23 at 06:53
  • Here's an article showing a charge pump to [invert voltage](https://maker.pro/custom/tutorial/how-to-access-negative-voltage-power-supply). A capacitor makes an open circuit for DC, but fluctuating electric field in capacitor is what allows power transfer – Zimba May 17 '23 at 07:16
  • But it's not really "DC" if there's fluctuations, is it? "DC" implies static, unchanging forever. – Tim Williams May 17 '23 at 07:20
  • DC means it's flowing in one direction, direct. AC voltage inverts/alternates to different direction, so goes negative. DC only has positive voltage – Zimba May 17 '23 at 07:22
  • That is not a well accepted definition of "DC" I'm afraid. – Tim Williams May 17 '23 at 07:22
  • DC does not reverse, it remains DIRECTed in one DIRECTion, with varying amplitudes, even from batteries. You may prove this with oscilloscope; any DC source can be proven to vary. eg. MPPT circuits work by varying voltages to maximize power transfer efficiency from wind turbines or solar panels, which are DC devices. Contrary to your understanding, these are not AC devices. – Zimba Aug 16 '23 at 04:12