Capacitor as a power supply

[M. Hamed]

This is more of a thought experiment triggered by handling of a huge cap. Would it be possible to charge a big cap and discharge it through a load resistor without dropping voltage much over a significant portion of the discharge time?

One way I could think of is to use a Zener followed by an emitter follower to act like a current source. However the Zener needs a load resistor and non-zero current which means energy (charge) will be wasted there.

A linear regulator would also have the same power wasting problem. Switching regulator? DC-DC converter?

 

[whit3rd]

On Sunday, February 1, 2015 at 2:59:49 AM UTC-8, M. Hamed wrote:

...Would it be possible to charge a big cap and discharge it through a load resistor without dropping voltage much over a significant portion of the discharge time?

If the energy source is a charged capacitor, value C
and if the applied load is accurately characterized as a resistor, value R

Then the system as described 'doesn't drop voltage much' only during discharge times
which are a small fraction of R*C.

When one wants constant voltage until the energy source is nearly empty,
batteries are preferred.

 

[Tim Wescott]

On Sun, 01 Feb 2015 02:59:45 -0800, M. Hamed wrote:

This is more of a thought experiment triggered by handling of a huge
cap. Would it be possible to charge a big cap and discharge it through a
load resistor without dropping voltage much over a significant portion
of the discharge time?

One way I could think of is to use a Zener followed by an emitter
follower to act like a current source. However the Zener needs a load
resistor and non-zero current which means energy (charge) will be wasted
there.

A linear regulator would also have the same power wasting problem.
Switching regulator? DC-DC converter?

I'm not sure what line you draw between a switching regulator and a DC-DC
converter.

Yes, you could use a big cap followed by a switching regulator. But you
might find it a worthwhile exercise to take a cap as big as your hand and
calculate the amount of energy you can store in it (using its rated
voltage, and W = C * v^2/2). Then take a NiMH cell or similar the size
of a finger, and do the same calculation (using rated capacity and
nominal voltage).

I'm not sure where you'll hit energy parity, but when you do the cap will
be much larger than the battery.

--
www.wescottdesign.com

 

[John Larkin]

On Sun, 1 Feb 2015 02:59:45 -0800 (PST), "M. Hamed"
<mhdpublic@gmail.com> wrote:

This is more of a thought experiment triggered by handling of a huge cap. Would it be possible to charge a big cap and discharge it through a load resistor without dropping voltage much over a significant portion of the discharge time?

One way I could think of is to use a Zener followed by an emitter follower to act like a current source. However the Zener needs a load resistor and non-zero current which means energy (charge) will be wasted there.

A linear regulator would also have the same power wasting problem. Switching regulator? DC-DC converter?

A boost/buck switching regulator could apply the constant desired
voltage to the resistor, as the capacitor discharges. The C voltage
would droop slowly at first, and then faster and faster as the cap
discharges, because the load resistor would be running at constant
power.

Once the cap gets down to, say, 20% of its initial voltage, there's
not much energy left to extract. So a buck switcher could be good
enough. Or a boost, depending on the voltages.


--

John Larkin Highland Technology, Inc
picosecond timing laser drivers and controllers

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[M. Hamed]

On Sunday, February 1, 2015 at 8:55:51 AM UTC-7, Tim Wescott wrote:

I'm not sure what line you draw between a switching regulator and a DC-DC
converter.

to me a switching regulator is more sophisticated. the other day I came upon a 2 transistor inverter circuit. Put a rectifier afterward it will become a DC-DC converter. I wouldn't dare call it a switching regulator.

Yes, you could use a big cap followed by a switching regulator. But you
might find it a worthwhile exercise to take a cap as big as your hand and
calculate the amount of energy you can store in it (using its rated
voltage, and W = C * v^2/2). Then take a NiMH cell or similar the size
of a finger, and do the same calculation (using rated capacity and
nominal voltage).

I'm not sure where you'll hit energy parity, but when you do the cap will
be much larger than the battery.

I understand but it's still more fun to see steady voltage coming out of a capacitor

 

[M. Hamed]

On Sunday, February 1, 2015 at 2:29:09 PM UTC-7, John Larkin wrote:

Once the cap gets down to, say, 20% of its initial voltage, there's
not much energy left to extract. So a buck switcher could be good
enough. Or a boost, depending on the voltages.

Shouldn't this be dependent on how much voltage is required to operate the switching regulator rather than how much percentage of the capacitor voltage?

 

[Tim Wescott]

On Sat, 07 Feb 2015 12:52:47 -0800, M. Hamed wrote:

On Sunday, February 1, 2015 at 2:29:09 PM UTC-7, John Larkin wrote:
Once the cap gets down to, say, 20% of its initial voltage, there's not
much energy left to extract. So a buck switcher could be good enough.
Or a boost, depending on the voltages.


Shouldn't this be dependent on how much voltage is required to operate
the switching regulator rather than how much percentage of the capacitor
voltage?

John was speaking of basic physics. When a cap's voltage is down to 20%
of it's initial voltage, the amount of energy stored is down to 4% of the
initial energy.

This is REALLY BASIC STUFF for anyone willing take the trouble to
actually do some REALLY SIMPLE MATH instead of just letting their fingers
dribble all over a keyboard.

--
www.wescottdesign.com

 

[John Larkin]

On Sat, 7 Feb 2015 12:52:47 -0800 (PST), "M. Hamed"
<mhdpublic@gmail.com> wrote:

On Sunday, February 1, 2015 at 2:29:09 PM UTC-7, John Larkin wrote:
Once the cap gets down to, say, 20% of its initial voltage, there's
not much energy left to extract. So a buck switcher could be good
enough. Or a boost, depending on the voltages.


Shouldn't this be dependent on how much voltage is required to operate the switching regulator rather than how much percentage of the capacitor voltage?

There are all sorts of possibilities. It would help to have a
specified requirement.


--

John Larkin Highland Technology, Inc
picosecond timing laser drivers and controllers

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

 

[M. Hamed]

On Saturday, February 7, 2015 at 4:12:21 PM UTC-7, Tim Wescott wrote:
> This is REALLY BASIC STUFF

I'm glad I posted to the right group!!

 

[M. Hamed]

On Saturday, February 7, 2015 at 4:12:21 PM UTC-7, Tim Wescott wrote:

This is REALLY BASIC STUFF for anyone willing take the trouble to
actually do some REALLY SIMPLE MATH instead of just letting their fingers
dribble all over a keyboard.

I know the math. I just don't usually think of caps in terms of energy. It just doesn't come to me automatically.