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Capacitor charging configurations

I was thinking about the classic voltage doubler using caps today, and this got me thinking about the basics of capacitor charge/discharge math, and eventually I thought up a simple design but wasn't quite sure what the effects would be:

First (basic) case: Take two caps of equal capacitance, I'll pick 1 farad for simplicity, and one (C1) is charged at 1 volt while the other (C2) is at 0 volts. When connected together C1 will charge C2, and the steady-state voltage of each will be 0.5 volts.

+----------+ + | | + C1 === 1V 0V === C2 1F | | 1F V V

Now the simple math that gets us there is this: Initial: Q(C1) = C(C1) * V(C1) = 1 Farads*Volts = 1 Coulomb Q(C2) = 0

Steady state: Q(C1+C2) = Q(C1) + Q(C2) = 1 Farads*Volts = 1 Coulomb C(C1+C2) = C(C1) + C(C2) = 2 Farads V(C1+C2) = V(C1) = V(C2) = Q(C1+C2) / C(C1+C2) = 0.5 Volts

and in lamens terms we can describe it by saying the charge in C1, which produces 1 volt there, will distribute itself evenly among the total capacitance and result in half of the charge left in C1 and half in C2, thus producing 0.5 volts in each capacitor.

Second case: Building on the first case, add a constant voltage source of 1 volt on the left side, boosting the voltage from C1:

E |+ +----||----+ | | | | 1V | + | | + C1 === 1V 0V === C2

1F | | 1F V V

And, of course, the catch is that there is an unobtrusive controller & sensor (not shown obviously), that will break the circuit the moment C1 loses all its charge (if that's what's going to happen), so as to avoid inducing a reverse voltage on it, or when steady state is achieved (if that's what's going to happen).

At that point, when C1 loses its charge or steady-state is achieved and the circuit is stopped, what will the charge and voltage be on C2?

Why? The reason I ask is that, if the voltage source simply boosts the final charge in C2 like I think it might, then a rather simple & dynamic charge pump can be made to go up to any voltage with only a 1V input (say it's from a cheap solar panel), 2 caps, a few FETs, and a tiny micro. It would work like this:

After C2 is charged as far as it can go, the resting voltage should be above one volt while C1 will be less than one volt (likely 0v). At that point the micro will switch off a couple of FETs and switch on a couple of other FETs, reversing the polarity of the battery. Example:

E +| +-----||-----+ | | | | 1V | + | | + C1 === 0V 1.5V === C2

1F | | 1F V V

And of course the cycle continues. It would be akin to a pendulum, with the constant voltage source just barely offsetting the voltage at the right time (because of the micro) to push the current in the desired direction, each cycle building up the overall voltage, albeit tenths of a volt at a time with this example.

Once the target voltage has been reached, the caps could be put together in parallel or series, without the voltage source (maybe it would be switched over to pumping another pendulum circuit), to feed into a holding capacitor or battery.

The beauty of this, if it works, is that if the voltage source increases for a short while (say the cloudy day turns into a sunny day, hitting the solar panel harder) the pendulum simply swings higher faster, nothing has to be recompensated or clamped, the frequency of the switching is completely dependent on when the micro senses a steady-state condition.

Does this make sense? If there is a flaw in this design please speak up!

Magneto

Reply to
Magneto
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And each capacitor has had a total voltage change of 0.5 volt.

Just keep in mind that any charge that flows through C2 must also flow through C1, so the total change of voltage across C2 must equal the total voltage change across C1.

Neglecting oscillations, the circuit will reach equilibrium when C2 has 1 volt across it and C1 has zero volts across it. At that point, each capacitor has had a 1 volt total change.

Yes, exactly zero.

Except that C2 cannot experience more voltage change from the common current than C1 did.

Sorry, with perfect components, the 1 volt just sloshes back and forth between the capacitors, and energy is consumed from the 1 volt source with each slosh. You might as well hook a resistor across the 1 volt supply to waste the power.

Done.

Reply to
John Popelish

Almost forgot about this post. Yes, I see your point. I drew up several different arrangements and did the calculations, you're right, the system can never gain more than a volt because it's closed loop. In fact, all the voltage source does is either speed up each cycle or allow it to reach full swing. My flaw was that I forgot that in a simple circuit using just a voltage source and a capacitor the same thing happens, a charge is still forced from one plate and moved to another, up to the potential of the source.

Thanks

Reply to
Magneto

Magneto wrote: (snip)

I have been wondering what happened to you. I didn't want to change the subject till you understood the principles involved in your question.

But now I would like you to think about the case where you charge one capacitor with a 1 volt source, and then you disconnect the source and reconnect it in series with the capacitor and use that series combination to charge the second capacitor. And repeat that cycle a few times.

Reply to
John Popelish

How is that different from what I was suggesting?

Because the two caps are of equal capacitance, all the charge of the full cap, that produces 1 volt in that cap, is moved completely over to the other cap, producing 1 volt in that cap as well. Current stops at that point because the source doesn't have enough force to keep it going.

If the 2nd cap had twice the capacitance, or if the source was 2 volts in the 2nd phase, then the first cap would discharge and then recharge in reverse polarity.

Reply to
Magneto

Agreed. but the second cycle, the first cap again charges to 1 volt but the second one charges to 1.5 as the first one discharges to 0.5. The third cycle, the first cap discharges to .75, and the second one charges to 1.75, etc. After many cycles, if the load current is very small, the second cap approaches 2 volts. Now, if, after several of these cycles, the second cap that is approaching 2 volts were used the same way the original voltage source was, you could, after many cycles, have a third cap (doing the job the first one does) approach 2 volts, also, and a fourth one approach 4 volts. This sort of thing is the basis of charge pump chips like the 766:

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You missed that the negative side of the voltage source is reconnected to either the top or bottom of the first cap, but the positive end is connected either to the top of the first cap, or the top of the second one.

Reply to
John Popelish

Oh you mean that in each cycle you keep the "storage" capacitor out of the circuit while you re-charge the other one. That's actually the effeciency I was looking for, yes.

I knew about the 3-cap charge pump, where two caps alternate between charging in parallel and discharging in series, each time dumping a little more charge into the storage capacitor, in fact that's what started my brainstorm. It was bothering me that the voltage source was sitting there disconnected while the two stacked caps drained out into the storage. So my first thought was maybe find a design that kept all parts in the circuit all the time (minus switching time) while still raising the voltage at each swing, but that's seems to be impossible...like a guy on a sailboat trying to get somewhere by blowing on his own sail.

The only way to avoid wasting the source's time (leaving it disconnected for a while) without creating a backup pump in parallel is what you're describing: *disconnect* the storage cap while you re-use the source as the boosting cap.

Yeah, I was glancing at some of TI's stuff too, but only found a select few that take the wide range of input voltage like this one. This is actually really nice part. Do you know of any that will take a wide input like 1.5v-15v and output a fixed voltage of ~13v rather than a fixed multiplier? Those seem to be non-existent or at least rare.

Reply to
Magneto

Charge pumps based on capacitors and switches are inherently integer multipliers. You need to think in terms of switching regulators if you want efficient regulation.

Have you seen this switching regulator tutorial?

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Reply to
John Popelish

No I hadn't seen that one yet...but very helpful, thanks!

It sounds like there is no easy way, relatively, to do what I was looking for with an inductor-less system. I figured I'd end up back at the inductor-based system but thought I'd put my thinking cap on (no pun intended) and see if I could design something else.

Thanks

Reply to
Magneto

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