I took apart a PC power supply to extract the capacitors (I shorted
them with a grounded wire to ensure they were discharged - they were).
I removed them from the circuit board easily enough using my soldering
iron.

I would like to charge them up for some experiments I'm doing. They
are labeled with "200v 470(micro)F" on one side of the can, and
"2202AZ" on the other. They say "NEG." next to a patterned circle; I'm
assuming this means the post with the same pattern on it is the
negative terminal.

I tried hooking one up to a 6v lantern battery for a good 15 minutes,
but when I hooked a test light up to the capacitor nothing happened;
it does not appear that I charged it at all.

Exactly what sort of power source do I need to use to successfully
charge this capacitor? Does it have to be high voltage? Will an AC
source charge it? If I want to use a lantern or car battery, do I need
to step up the voltage? Etc.

Thanks!

jjsonp...

I was an Electronics Technician in the Navy.

First of all, the capacitor you're using is a 200 volt capacitor,
and is designed to be charged by a higher voltage, hence the label
""200v 470(micro)F".

Secondly, the "capacity" of that capacitor is very small, meaning
it won't store that voltage for very long, and will discharge very
quickly, so even if it was charged to full capacity (200V), it 
would likely discharge so quickly that it would barely cause the
light to flicker. Another way to look at this is that, although
a capacitor this size could be charged to 200V, the amount of 
current it can provide will be very tiny.

A larger capacitor, one capable of providing adequate current
for enough time to see the light lit for awhile, will resemble
the battery it is acting like, and be larger in size. It will
be about the size of a C cell battery, or larger.

So you need to find a larger capacitor that is rated at 6V,
with a larger capacitance, more along the lines of 1F.
47uf (microfarads) is a mere 47 millionths of 1 Farad (F).

Another improvement to the experiment would be to use a 6V
LED bulb instead of a regular flashlight bulb. These draw
less current and will stay lit longer.

A 1F 6V capacitor should light an LED bulb for at least 15
whole minutes!

by using an LED bulb, you may be able to use capacitors with
a lower capacitance, say as small as 15000uf (6V, still).
These won't be as large, and won't keep the bulb lit as long
(maybe 15 seconds).

You can't charge a capacitor with AC voltage.

As you guessed, just hook the negative (-) terminal of the
battery to the negative terminal of the capacitor, and the
same with the + terminals. The larger the capacitor, the 
longer it will take to reach a full charge (think of how
long the whine takes when a camera's flash is recharging).
A 1F capacitor might take a few minutes, while a 15000uf
should take only a few seconds. You don't have to worry
about overcharging the capacitor - once it is charged,
it will act as a second battery and hold the charge that
is left in the battery.


If anything's unclear, or you need further explanation,
just let me know.

Please do not rate this answer until you are satisfied that  
the answer cannot be improved upon by way of a dialog  
established through the "Request for Clarification" process. 
 
A user's guide on this topic is on skermit-ga's site, here: 
http://www.christopherwu.net/google_answers/answer_guide.html#how_clarify 
 
sublime1-ga

---

Request for Answer Clarification by jjsonp-ga on 23 Sep 2005 06:06 PDT

Thanks for the clarification sublime. 

However, my question remains: how do I charge up the capacitor I have
in my possession?

I understand that I may not be able to see my test-light glow as the
discharge will happen too rapidly. But I still want my original
question answered - how do I successfully charge up my 200V
470micro-Farad capacitor?

---

Clarification of Answer by sublime1-ga on 23 Sep 2005 12:41 PDT

jjsonp...

I implied the answer in what I told you - to charge a capacitor
with the 6V lantern battery you have, you need a capacitor rated
at 6V.

Likewise, to charge a 200V capacitor, you need a source for 200VDC. 
In the simplest terms, there is no solution to charge the capacitor
with what you've described as having on hand.


You could build a power supply, using a schematic such as the one
on this page in the HeadWize Projects Library:
http://www.headwize.com/projects/showfile.php?file=busbridge_prj.htm

If you wanted to use lantern batteries, you would need to build,
or buy, a step-up converter designed to take 6V and convert it to
200V, and you would probably drain the lantern battery before the
capacitor was charged, so a car battery would be a better input
device. Such a circuit would be hard to find, due to the unusual
output voltage of 200VDC. A more common one would be 12VDC to 120VAC
to convert car battery voltage for use with AC appliances. It would
be much simpler to build a 200V power supply. If you want to give
it a try, Dallas Semiconductor has a Power Supply Cookbook page
which provides free schematics for power supplies for various
purposes:

"Choose the specifications that most closely meet your needs.
 Then view a circuit design, complete with schematic diagram
 and bill of materials."
http://dalsemi.com/cookbook/powersupply/index.cfm

There's a 15V to -180V rated at 0.005A, which probably isn't
enough current for your purposes, in the CB76 PDF:
http://dalsemi.com/cookbook/powersupply/pdfs/CB76.pdf

And there's a 5V to -24V and -100V at 0.12A and 0.075A,
repectively, in the CB79 PDF:
http://dalsemi.com/cookbook/powersupply/pdfs/CB79.pdf

That wouldn't give you the full voltage you need, but
you might be able to modify the circuit in a way that
would.


Another device you could build, which is sometimes used to charge
capacitors in teaching environments is the Wimshurst Electrostatic
Machine, as described on this page by Dr. Antonio Carlos M. de 
Queiroz:
http://www.coe.ufrj.br/~acmq/wimshurst.html


All in all, the simplest and least expensive solution is to obtain
a larger capacity capacitor, such as a 1 farad capacitor, rated at
6V, and use your lantern battery to charge it, which is why I tried
to steer you in that direction.

If anything's unclear, or you need further explanation, let me know.

sublime1-ga

Researcher clarified my answer fully. Thanks!

As the answerer said, a 470uF capacitor won't hold much energy at 6V. 
The total energy in a capacitor is 0.5 x C x V^2, where C is the
capacitance in Farads, V is the voltage in Volts, and the energy is in
Joules.  If you connect a resistive load, it will draw current
according to V=I x R, where V is the capacitor voltage, I is current,
and R is resistance.  The power in this case is P = V x I, and the
energy is power times time, or P x T.  Of course the voltage will
start decreasing as soon as you start drawing current out of the
capacitor, so these are instantaneous values, but you can get the
idea.

So, if you charge the 470 uF cap to 6V, you have 8.5 mJ, not very much
energy.  Even if you charge it to 200V, you only get 9.4 Joules. 
Compare this with a typical AAA battery, which stores about 4300
Joules.  In general, capacitors store much less energy than batteries
for the same volume.

Right. But I am not concerned with how many Joules of energy it can
hold - I just want to know how to charge it as fully as possible.

What I get from your comment is that if I want to charge it fully I'll
need a 200V power source. So, my original question remains: how do I
charge this capacitor?

Do I somehow step up the voltage from a car battery (and how)? Do I
use the original PC power supply? Do I get something that plugs into
110 AC and converts?

I appreciate your comment.

To fully charge a capacitor, you need a DC voltage source to bring the
voltage up to the maximum the capacitor can take, in this case 200V. 
You can't do this directly with a car battery.  In the PC supply
circuit, it's probably part of the input rectification section, and
would be charged to the peak of the AC line voltage (normally sqrt(2)
x 120, or 167V).

If you did get it charged to 200V, it still wouldn't hold much energy
to do anything useful, although it would be enough energy and voltage
to shock someone .

I would caution against playing around with voltages like that - you
can very easily shock or electrocute yourself, especially if you're
poking around in a live power supply.  Even after the supply is off
it's possible for the capacitors to hold their charge for quite a
while.  Even if the energy in the capacitor is small compared to a
battery, it can still be lethal at high voltages.

If I was you I would forget about charging your 200V capacitor.  If
you do manage to fully charge it, you'll likely destroy your lantern
bulb(by the way it will work just fine at 6V, that 200V rating just
describes maximum voltage it is rated for and running it at max
voltage will actually shorten its lifespan).  For some real fun go to
Digikey and order some Panasonic gold capacitors.  They are generally
rated for 3.3-6V and hold >1F.  Let them charge for a few hours with a
battery and then connect them to your bulb.  I think they only cost a
few bucks.