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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 '16 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\$ – Bimpelrekkie Feb 8 '16 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 '16 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 '16 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 '16 at 20:16
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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 '16 at 16:24
  • \$\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\$ – Wouter van Ooijen Feb 8 '16 at 16:36
  • \$\begingroup\$ Looks like I said essentially the same thing.. later. oops. \$\endgroup\$ – Spehro Pefhany Feb 8 '16 at 17:03
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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\$ – George Herold Feb 8 '16 at 17:48
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    \$\begingroup\$ @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 :) \$\endgroup\$ – Ali Alavi Feb 9 '16 at 12:31
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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\$ – Peter Green Feb 8 '16 at 17:21
  • \$\begingroup\$ @PeterGreen "change in energy" not total energy. \$\endgroup\$ – Spehro Pefhany Feb 8 '16 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\$ – Peter Green Feb 8 '16 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\$ – Peter Green Feb 8 '16 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 '16 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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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 - Reinstate Monica Dec 4 '18 at 19:09

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