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### Network # Capacitor charging configurations

M

#### Magneto

Jan 1, 1970
0
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

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

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

Magneto

R

#### Robert Baer

Jan 1, 1970
0
Magneto said:
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

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
** tilt **
I suggest you write the equations instead of guessing; one cannot get
something from nothing.

M

#### Magneto

Jan 1, 1970
0
Robert said:
** tilt **
I suggest you write the equations instead of guessing; one cannot get
something from nothing.

I do plan on writing the equations, but figured this simple case would
be a good conversation piece on this group.

By the way, I'm not trying to get something from nothing. The *constant
voltage source* is always pumping energy into this circuit, just in a
form that builds voltage instead of charge/current.

A good analogy would be to build two towers side-by-side, one rock at a
time, by placing a rock on tower 1, then standing on it so you could
place a rock on tower 2, then standing on that so you could place
another rock on tower 1, etc., etc. As long as you have a constant
source of rocks at your disposal you can incrementally build your
potential in a back-n-forth style.

R

#### Robert Baer

Jan 1, 1970
0
Magneto said:
I do plan on writing the equations, but figured this simple case would
be a good conversation piece on this group.

By the way, I'm not trying to get something from nothing. The *constant
voltage source* is always pumping energy into this circuit, just in a
form that builds voltage instead of charge/current.

A good analogy would be to build two towers side-by-side, one rock at a
time, by placing a rock on tower 1, then standing on it so you could
place a rock on tower 2, then standing on that so you could place
another rock on tower 1, etc., etc. As long as you have a constant
source of rocks at your disposal you can incrementally build your
potential in a back-n-forth style.
Without inductance, there is no "back and forth".

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