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I'm trying to understand how a voltage doubler using a switched capacitor circuit affects the charge being distributed from the input to the output.

schematic

simulate this circuit – Schematic created using CircuitLab

The circuit operates so that during first phase \$V_{IN}\$ is charging the capacitor \$C_{1}\$ and during second phase the \$C_{1}\$ is connected in series and between \$V_{IN}\$ and \$V_{OUT}\$ and the voltage is doubled. However, theoretically, the maximum output current gets halved.

Now, if current is really halved, does this mean that the charge initially transferred to \$C_{1}\$ halves too (since this relation applies \$Q = I*t\$), or does it stay that same and the time needed to distribute the charge from \$C_{1}\$ to load doubles at the halved current?

P.S.: This might seem a bit weird question but I'm really eager to know the actual answer to this. Also, this has something to do with designing a circuit for charge transfer between battery cells. That is why I'm interested what happens to the amount of charge transferred over to load.

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    \$\begingroup\$ @Jens Are the high current switching transients the only problem when compared to inductor counterpart for cell balancing? Some research has been put into reducing such transients: ieeexplore.ieee.org/document/4443901 \$\endgroup\$
    – lucenzo97
    Commented Jun 12, 2022 at 8:58
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    \$\begingroup\$ @Keno I think I may cover how to approach an analysis in this prior response on EESE. +1 on the question! \$\endgroup\$
    – jonk
    Commented Jun 12, 2022 at 16:28
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    \$\begingroup\$ @Keno Unless you use several MHz switching frequency you need electrolytic capacitors of high quality and even these degrade sooner or later. Good conducting switches are a problem here as well. The signal preparation for 4 FET can be a hassle for multiple cell chains. If you just want to compensate different cell leakage currents, an unregulated voltage double charge pump chip will work. \$\endgroup\$
    – Jens
    Commented Jun 12, 2022 at 16:44
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    \$\begingroup\$ @jonk I will surely go through each section of your answer, as I assume (from previous experiences) that you provided with quality content :) \$\endgroup\$
    – lucenzo97
    Commented Jun 13, 2022 at 10:04
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    \$\begingroup\$ @Keno Sounds to me as though Jens has specific, expert knowledge on the topic and would be able to write up an excellent answer. I could write, but it would feel like copying what I wrote before. If you find something that is still missing when reading that answer, feel free to mention it. Or if you need an LTspice simulation that demonstrates what I wrote delivers quantitative prediction, using ideal switches rather than ideal diodes, that also can be easily shown. \$\endgroup\$
    – jonk
    Commented Jun 13, 2022 at 16:55

1 Answer 1

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When C1 is switched across V_IN, it consumes current (charge) as it (re)charges.

When C1 switches in series with V_IN (thus doubling it), it also consumes current from V_IN as it (dis)charges and replenishes the load.

Note that current is consumed from V_IN in both states, but (new) charge is delivered to the output only in the 2nd state -- this is the origin of the current halving -- consider that roughly equal currents are consumed in each stage from the input, but only on the 2nd state is current delivered to C2.

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