Re: Capacitors & conservation of charge

On Sep 11, 1:44 pm, cabraha...@xxxxxxx wrote:
On Sep 1, 12:41 pm, exxos...@xxxxxxxxxxx wrote:





Hi all,

While messing with buck/boost circuits I had some thoughts which don't
hold up in real tests....

A simple example (aside from losses) is that if transfer energy from
say 10V 10uF into 1uF the voltage will increase to preserve the
charge.

Now, in my circuit, I charge 25V 1,000uF capacitor and charge a 10uH
inductor. When the switch turns off, all the energy should be in the
10uH inductance. So if I have a 100pF capacitor, then the voltage
should be like 100,000volts according to my workings out.

Now I ran a computer simulation on this, at best I can only obtain
12KV on the 100pF. So most of the energy is lost in switching losses I
assume.

In realworld tests, I end up with  less voltage than I started out
with, So I am trying to find out why ?

I know charging 22uH inductor at 100khz can be used as a buck/boost
supply, I built a simple 12V to 30V inverter, can switch 10amps
easily. Though I am not running at 100khz, only 100hz. Though the
current pulse rises to something like 500amps over 500uS.

I am not sure I follow all this exactly, Or even if it will work ?
AFAIK, The longer a inductor has current pumped across it the more
charge it obtains over time. So at turn off, all the energy given to a
coil is recovered. It works well, even with my simple buck/boost
circuit.

So I am slightly confused as to why pushing 500A into a coil has no
effect. I can only assume I have a huge loss somewhere, Or I do not
follow the idea correctly ?

Cheers,
Chris

You need to aquaint yourself with the buck-boost topology and theory
of its operation.  If you go to TI's web site and download their buck-
boost app note no. SLVA059, studying this thoroughly, you will
understand why.

In a nutshell, the 1000 uf input capacitor did not transfer ALL of its
energy to the inductor.  The output voltage is related to the input
voltage and the duty cycle as follows:

Vout = Vin * (D/(1-D)) not accounting for losses.

The energy transferred to the inductor is half the inductance times
the difference between the square of the final current and square of
the original current.  This is determined by the voltage across said
inductor multiplied by the on time.  During the power switch off time,
the volt-seconds must equal that of the on time.  The app note covers
this.  The value of the input capacitor does not determine the output
voltage.  The same goes for the output cap value.  The values of both
caps, however, are important regarding ripple. noise, and transient
response.

Did I help?  BR.- Hide quoted text -

- Show quoted text -

i haven't gotten the whole detail, but you seem to start with a bit of
a misunderstanding which i will now endeavor to correct:

you charge a capacitor by loading a voltage into it, then open
circuiting it; but, an inductor being essentially the opposite, you
charge it by loading a current into it and shortcircuiting it; whereas
a charged capacitor holds its energy in the electric field with zero
current, a charged inductor holds its energy in the magnetic field,
with zero voltage. If you short circuit a capacitor it loses its
charge, if you open circuit an inductor it loses its current. In real
life, the series resistance in the inductor windings decays the
charged current a lot more quickly than the leakage resistance in the
capacitor decays the charged voltage.
.

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