M

#### Magneto

- Jan 1, 1970

- 0

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