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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?

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    \$\begingroup\$ 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. \$\endgroup\$
    – pjc50
    Feb 8, 2016 at 15:44
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    \$\begingroup\$ Have you included some realistic ESR (series resistance) for those capacitors ? Simulators are like paper: you can make anything work/not work on them ;-) \$\endgroup\$ Feb 8, 2016 at 16:02
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    \$\begingroup\$ 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! \$\endgroup\$
    – Andy aka
    Feb 8, 2016 at 17:32
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    \$\begingroup\$ @Andyaka The PCB is neither the most expensive nor the hardest to get item. \$\endgroup\$
    – Ali Alavi
    Feb 8, 2016 at 21:58
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    \$\begingroup\$ A 10uF capacitor supplying 10A will drop one volt per microsecond. At a 50Khz switching frequency, you'd be looking at 100% ripple. \$\endgroup\$
    – supercat
    Feb 10, 2016 at 20:16

7 Answers 7

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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.

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    \$\begingroup\$ 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. \$\endgroup\$
    – Ali Alavi
    Feb 8, 2016 at 16:24
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    \$\begingroup\$ 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. \$\endgroup\$ Feb 8, 2016 at 16:36
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    \$\begingroup\$ Huh, nice (+1's everywhere.) Is this the previous question? electronics.stackexchange.com/questions/54992/…. I knew about caps and voltage sources... but never really thought about it! \$\endgroup\$ Feb 8, 2016 at 17:48
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    \$\begingroup\$ If memory serves, charging a capacitor through just a resistor (which will be just the resistance of a wire) is only 50% efficient. \$\endgroup\$
    – Michael
    Feb 9, 2016 at 1:53
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    \$\begingroup\$ @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... \$\endgroup\$ May 10, 2023 at 23:06
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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.

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    \$\begingroup\$ "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. \$\endgroup\$ Feb 8, 2016 at 17:21
  • \$\begingroup\$ @PeterGreen "change in energy" not total energy. \$\endgroup\$ Feb 8, 2016 at 18:06
  • \$\begingroup\$ 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. \$\endgroup\$ Feb 8, 2016 at 18:33
  • \$\begingroup\$ 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. \$\endgroup\$ Feb 8, 2016 at 18:38
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    \$\begingroup\$ "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. \$\endgroup\$
    – user253751
    Feb 8, 2016 at 19:19
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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.

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As pointed out in the comments of this answer, switched capacitor DC-DC converters, such as one built around an LTC7820, can be extremely efficient (99% efficiency claimed in that datasheet) and can handle relatively high power (500 W claimed in the same datasheet).

The main disincentives to switched capacitor DC-DC converters have been

  • the cost of storing energy in a large peak-energy capacitor, versus storing the same energy in a large peak-energy inductor,

  • the physical bulk of large peak-energy capacitors

  • the limited lifespan of electrolytic capacitors

  • the inability of switched capacitors to efficiently convert DC to DC at ratios other than the ratio of small integers. One cannot smoothly adjust the output voltage of such a switched capacitor DC-DC converter, without a loss in efficiency.

The cost/benefit analysis for capacitors as energy storage devices, however, has shifted somewhat due to the development of super-capacitors.

The bulk of super-capacitors is smaller than previous generation capacitors and their lifespan is longer than electrolytics. Unfortunately they are still quite expensive.

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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.
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    \$\begingroup\$ 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. \$\endgroup\$
    – Nate S.
    Dec 4, 2018 at 19:09
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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.

Instead, see what suitable fixed inductors are available and design the converter around those. That is pretty much how it's done for low volumes unless some extreme circuit optimization is required by the end application.

Inductors for most popular SMPS applications are really jellybean parts, and are plentiful and are extremely cost-optimized. There's big competition in the fixed inductor market and there's quite a pressure to keep the prices down as far as they can go.

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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.

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    \$\begingroup\$ 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. \$\endgroup\$ May 12, 2023 at 16:40
  • \$\begingroup\$ AC circuits also have negative voltage, current flow reverses periodically. DC circuits only have positive voltage & current. \$\endgroup\$
    – Zimba
    May 16, 2023 at 15:55
  • \$\begingroup\$ @TimWilliams, capacitor has infinite impedance at DC \$\endgroup\$
    – Zimba
    May 17, 2023 at 6:50
  • \$\begingroup\$ Apologies, conductance is zero, heh. In any case, that doesn't seem very useful. How do you transfer power through an open circuit? \$\endgroup\$ May 17, 2023 at 6:53
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    \$\begingroup\$ That is not a well accepted definition of "DC" I'm afraid. \$\endgroup\$ May 17, 2023 at 7:22

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