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'[EE]: Capacitor Charge'
2001\02\20@041347 by Roman Black

flavicon
face
Andy, its not hard to do. I've pulled apart a low
cost "personal protection" device, the ones that
zap the attacker with a high voltage.

It was very simple, 2x C cell batteries for high A
low V supply. A simple one-transistor inverter
(self excited) and simple low-turns transformer
with a single rect diode. That charged a 2kv polyprop
cap to a few hundred volts, and very quickly,
about 1/20th of a second.

Then there was a simple sparkgap, so when the cap
volts got high enough it jumped the gap and this
energy dump was fed straight into a low turns
primary of a simple autotransformer. The secondary
of the autotransformer was about 120kV, at about
20 "zaps" per second. This is similar to a tesla
setup.

Interesting that the crude mass-produced thing
just keep charging the first cap, the inverter
was not disconnected at all for the secondary
discharge/sparking.

One transistor, one diode, and a couple of discretes.
And no PIC chips in sight!
-Roman



Andy Faulkner wrote:
{Quote hidden}

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2001\02\20@052443 by Bill Westfield

face picon face
> How can you charge a 450V 160uf Electolytic capacitor
> from a 12V 1A source, quickly as possible to a
> pre-defind charge upto 300V.
>
> Once charged I want to disconnect the charge circuit
> and connect another circuit that will quickly
> discharge the capacitor.
       :
> Also just to make it a bit more complex, I want to
> test that the capacitor has reached it's charge and
> get the result into a micro controller.

What you describe is very similar to the guts of the flash circuitry in many
of today's disposable cameras, ie the Kodak MaX.  There are schematics and
circuit explanations on the web - try starting at:

       http://www.misty.com/people/don/samflash.html

These use a triggered oscillator and transformer.  "Done" is detected via
zener or MOV feeding back and shutting off the oscillator.  It ought to be
easy to detect whether the oscillator is on or off.  You can connect a sidac
across the trigger circuit and a load between the flash tube (which is a
fine high voltage high current switch) and ground and get the sort of
automatic charge dumping you're asking about.  Or see the modified flash
circuit for microprocesor control at the web site for other ideas.

The flash circuit draws a couple amps (limitted by the battery) at 1.5V from
a single alkline AA cell, and charges a 120-160uF cap to about 300 odd volts
in less than 10 seconds.  The piece missing from what you're looking for is
a current limit in the primary side - I'm inclined to believe that something
as simple as a resistor might suffice in an appropriately modified version.
(Current draw isn't constant - it goes down as the cap approaches full
charge.)

BillW

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2001\02\20@114350 by Roman Black

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Hi guys, I'm re-sending this as the dates on my computer
were all stuffed up! :o)

Roman Black wrote:
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2001\02\20@230203 by Charlie Davis

picon face
Andy    
Being stuck with 12 volts and 1 amp greatly increases the time to
charge from 0 to 300 volts.  If we assume 100 % efficiency and a constant current source, the time to charge 160 UFd's to 300 volts
would be 1.2 seconds and the capacitor energy would be 7.2 J.
The energy can be  pushed to12 J by increasing the voltage to 387 V, or increasing the capacitance to 266.67µFd.  Increasing the voltage  to
387 V lowers the available constant current to 12/ [387.3 ] = 31 mA,
which translates to a time of t = CE/I = [.00016 x 387.3] /.031 = 2 SEC.   Increasing the capacity to 266 µFd will give a time of 2.58 SEC.
By the time you figure the increased voltage required to insure current
source compliance and the efficiency of the dc to dc converter you could be
looking at a very significant time period.  With a knowledge of your required
energy and time constraints, the capacitor and the voltage could be
optimized for the most efficient operation.

Charlie, W8AWS

Midland Electronics
  Specialties

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2001\02\21@084621 by Olin Lathrop
face picon face
>>
Being stuck with 12 volts and 1 amp greatly increases the time to
charge from 0 to 300 volts.  If we assume 100 % efficiency and a
constant current source, the time to charge 160 UFd's to 300 volts
would be 1.2 seconds and the capacitor energy would be 7.2 J.
<<

12 volts at 1 amp is 12 watts, or 12 joules/second.  Therefore at 100%
efficiency it would take 7.2 / 12 = 600mS to charge the capacitor.


*****************************************************************
Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, olinspamspam_OUTembedinc.com, http://www.embedinc.com

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2001\02\21@205243 by Charlie Davis

picon face
Olin

Let me clarify my position on the cap charge question.
Neglecting pawer factor and efficiency the power source
can deliver 12 watts.  If 12 V is converted to 300 V, the
original stated voltage, the maximum available current is
limited to 40 mA.  300 V X 40 mA = 12 watts.
The current, I = dQ/ dt.  Q = CE so I = C dE/ dt and
I dt = C dE.  Integrating both sides,  It = CE and t = CE/I
The time t = 160uF X 300 V / .04 A = 1.2 Seconds.
This time was confirmed on Ansoft's program "Serenade"
and Electronic Workbench version 5.1  They show a dV/dt of
250 V/s.  1.2 S X 250 V/S = 300 V.  Of course with lower
duty cycle we may be able to use 80. mA and reduce the
charge time to 0.6 seconds.

Charlie W8AWS

Midland Electronics
    Specialties

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2001\02\21@232416 by Spehro Pefhany

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At 08:52 PM 2/21/01 EST, you wrote:
>Olin
>
>Let me clarify my position on the cap charge question.
>Neglecting pawer factor and efficiency the power source
>can deliver 12 watts.  If 12 V is converted to 300 V, the
>original stated voltage, the maximum available current is
>limited to 40 mA.  300 V X 40 mA = 12 watts.
>The current, I = dQ/ dt.  Q = CE so I = C dE/ dt and
>I dt = C dE.  Integrating both sides,  It = CE and t = CE/I
>The time t = 160uF X 300 V / .04 A = 1.2 Seconds.
>This time was confirmed on Ansoft's program "Serenade"
>and Electronic Workbench version 5.1  They show a dV/dt of
>250 V/s.  1.2 S X 250 V/S = 300 V.  Of course with lower
>duty cycle we may be able to use 80. mA and reduce the
>charge time to 0.6 seconds.

When the capacitor is at zero volts, and is charging at
40mA, the power required is just zero (ignoring
inefficiencies).

That's why your answer is off by exactly 2:1

Best regards,


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2001\02\22@101339 by Olin Lathrop

face picon face
> Let me clarify my position on the cap charge question.
> Neglecting pawer factor and efficiency the power source
> can deliver 12 watts.  If 12 V is converted to 300 V, the
> original stated voltage, the maximum available current is
> limited to 40 mA.  300 V X 40 mA = 12 watts.
> The current, I = dQ/ dt.  Q = CE so I = C dE/ dt and
> I dt = C dE.  Integrating both sides,  It = CE and t = CE/I
> The time t = 160uF X 300 V / .04 A = 1.2 Seconds.

You are off by a factor of two because you assume the current into the cap
is constant over the whole charge time.  There is no reason this needs to be
true, especially since we were talking about the minimum time for a
theoretically perfect converter.  However, even real world converters can,
and most would, have higher current when the cap is at lower voltage.

> This time was confirmed on Ansoft's program "Serenade"

Oh, a computer said it was true.  We can all go home now!

<rant>
What a sorry state of affairs things are in when a computer simulation is
used  to answer a theoretical question instead of applying a little theory.
This is a great example of the dangers of relying too much on simulation.
You get such a nice neat authorotative answer that it distracts you from
realizing you asked the wrong question to begin with.  Real engineering
isn't about cranking thru numbers.  Even a computer can be programmed to do
that.  It is about understanding the problem, then using creativity,
intuition, experience, and intelligence to apply science to solve the
problem.  Computers can be useful tools for number crunching and keeping
track of things, but they can't do real engineering.  Not by a long shot.
</rant>


*****************************************************************
Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, KILLspamolinKILLspamspamembedinc.com, http://www.embedinc.com

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