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There's a few resources out there for indcutorless boost circuits, or charge-pump voltage doublers and triplers.

http://www.electroschematics.com/648/555-voltage-doubler/ enter image description here

But this circuit is only able to convert a meager 70mA and the conversion efficiency is not that great. What should be done to increase the amperage into the 2-3A range with efficiencies similar to the charge pump ICs out there

The reason I am looking at these is they are known for their potentially high efficiency, I think I've seen a few that claim voltage conversion efficiency of 99% and power conversion of ~94% such as the TC1240

It would be great to be able to convert DC from a single cell lithium battery to high voltage easily with low loss.

How do I get the max amperage up to 5A on this circuit? Is it possible without an inductor?

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    \$\begingroup\$ Any time you charge or discharge a capacitor, half the energy put into or taken from the capacitor will be dissipated as heat just because of the work done in transferring the charge. However, if you add an inductor, judiciously, the work done will tend toward zero, and that's why almost all switchers can approach 100% efficiency. What's your question, anyway? \$\endgroup\$ – EM Fields Mar 30 '16 at 23:47
  • \$\begingroup\$ I was under the impression that inductors were the main culprit for heat loss, after all theres a lot of resistance in that copper wire. I've seen MLCC's with very low ESR especially with capacitances in the 10s of Farads kemet.com/Lists/TechnicalArticles/Attachments/91/… \$\endgroup\$ – John Evans Mar 31 '16 at 0:12
  • \$\begingroup\$ See my answer, below \$\endgroup\$ – EM Fields Mar 31 '16 at 0:44
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Get an inductor and do this right. Charge pumps are inherently inefficient, and limited to relatively low currents. This is in part due to the low efficiency, and in part due to the large capacitors that would be needed to pump high current. Stepping 12 V to 20 V at 5 A out (100 W) is waaaaaayyyyyy beyond practical charge pump range. Fahgeddaboudit!

The obvious answer is a boost converter, which, of course, will contain a inductor. You have given no justification for avoiding inductors, so this is a viable answer. There are boost converter ICs out there. At this power level, you will need one that manages a external switch.

At 80% efficiency, for example, the average draw from the 12 V supply will be 10.4 A. I would look into multi-phase architectures to split up that current. At this power, you will probably have to design significant parts of the booster yourself. There may not be any suitable off the shelf chips. I'd probably look into a microcontroller driving three PWM outputs 120° out of phase.

At 100 W, you have to wake up and do some real design, not just slap parts together according to a circuit you found in some questionable corner of the internet. This is certainly no place for silly religious convictions, like that inductors are somehow evil.

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Here's a rudimentary boost converter running at about 80% efficiency:

enter image description here

I don't think you can get there with an RC charge pump.

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Any voltage differences translate to losses. That is, using your posted schematic as reference, the voltage of C4 needs to be as close to 12V input as possible. Then the voltage of C5 needs to be as close to 12V input + C4 as possible. Which result in an output of 12V x 2 = 24V.

The circuit as given has voltage drops across D1 and D2, plus Vbe of T1 and T2. You want to replace them with low impedance switches. T1 and T2 are emitter followers, that needs to be redesigned or the drive levels expanded beyond the power rails. D1 and D2 can be replaced with actively driven transistors (probably MOSFETs). Of course, overlapping of the two phases needs to be avoided.

C4 and C5 wants to be big for low ESR and low ripple.

I think one big weakness of switched capacitor power supply is when voltage regulation is necessary. When voltage regulation is introduced, voltage differences are put in on purpose between the input voltage, the voltage of the capacitors and the output voltage, which lead to increased power losses.

Also, if you trace the path of the current from input to output, the current passes through T1, T2, D1, D2, C4 twice, and sources from 12V input twice. So the various series impedances add up, which contributes to power loss and voltage fluctuation due to changing load (when without active voltage regulation).

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