Capacitor charging configurations

[Magneto]

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

 

[John Popelish]

Magneto wrote:

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.

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

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).

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.

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?

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.

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).

Yes, exactly zero.

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

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

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.

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.

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

Done.

 

[Magneto]

John Popelish wrote:

Magneto wrote:

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.


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

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).


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.


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?


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.

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).


Yes, exactly zero.

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


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

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.


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.

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

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


Done.

 

[John Popelish]

Magneto wrote:
(snip)

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

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.

 

[Magneto]

John Popelish wrote:

Magneto wrote:
(snip)

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


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.

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.

 

[John Popelish]

Magneto wrote:

John Popelish wrote:

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.


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.

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:
http://rocky.digikey.com/WebLib/Microchip/Web%20Data/TC7662B.pdf

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.

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.

 

[Magneto]

John Popelish wrote:

Magneto wrote:

John Popelish wrote:


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.



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.


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.
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.

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:
http://rocky.digikey.com/WebLib/Microchip/Web%20Data/TC7662B.pdf

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.

 

[John Popelish]

Magneto wrote:

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.

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?
http://www.national.com/appinfo/power/files/f5.pdf

 

[Magneto]

John Popelish wrote:

Magneto wrote:

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.


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?
http://www.national.com/appinfo/power/files/f5.pdf

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