Capacitor

Archiver

Many people I know have been confusing me latey. If i connect two
10F 2.5V Capacitors in series, i know the total capacitance will
drop, but will i be able to double the total voltage and send 5V
through the two of them?

-Chris

Archiver

Chris,

Your question as stated has no meaning, unfortunately. I don't
think this works. But, why believe me - try it and see what
happens.

DLC

Chris wrote:
>
> Many people I know have been confusing me latey. If i connect two
> 10F 2.5V Capacitors in series, i know the total capacitance will
> drop, but will i be able to double the total voltage and send 5V
> through the two of them?
>
> -Chris

--

---

Dennis Clark http://www.verinet.com/~dlc
dlc@v...

---

Archiver

> Many people I know have been confusing me latey. If i connect two
> 10F 2.5V Capacitors in series, i know the total capacitance will
> drop, but will i be able to double the total voltage and send 5V
> through the two of them?
>
> -Chris
>
Para agregar un poco mas de complejidad voy a contestar en castellano.

Si conectas dos capacitores iguales de 10F 2.5V en serie, obtenes un
capacitor equivalente (mirando el conjunto completo), de 5F 5V. ¿Simple no?

Now in pseudoEnglish:

If you connect two equal capacitors of 10F 2.5V in series, you obtain
(looking the whole set), a 5F 5V capacitor.

If you need the equations:

Cseries=C1.C2/(C1+C2) if C1=C2 then Cseries=C/2

Vseries1=V.C2/(C1+C2) if C1=C2 then Vseries1=V/2 (and Vseries2=V/2 too, of
course)

i.e if you apply 5V to the array of caps, each one woks with 2.5V


I hope this helps.

Arístides Alvarez
Curso de Robótica
Mar del Plata
Argentina
aristides@c...

Archiver

I don't think you can do that, actually. I am not absolutely positive
though.. I know the area between the capacitors will be a dead area, I am
uncertain if it will deaden your whole circuit or not. But hey, go buy two
cheap,low voltage caps, put them in series, and check it out.. see what
voltage they fry at.

Archiver

> I don't think you can do that, actually. I am not absolutely positive
> though.. I know the area between the capacitors will be a dead area, I am
> uncertain if it will deaden your whole circuit or not. But hey, go buy two
> cheap,low voltage caps, put them in series, and check it out.. see what
> voltage they fry at.
>
In the theory, with two equal capacitors, you can.

In the implementation of this connection you have to consider certain items:

It doesn't exist two capacitors with the same parameters.
(although both said 10 F, they haven't 10F)
The dispersion generated by tolerance (% of error) is too big on its.
Any variation of the C value modify the voltage drop of both capacitors.
By V1= V.C2/(C1+C2), and V2=V.C1/(C1+C2).

In applications of audio power sources, where this connection is used
frequently,
the easier solution is two-equal-resistor voltage divider, in parallel with
capacitors,
with their middle point interconnected, to ensure V/2 on each capacitor.
This solution is viable only in a circuit that is connected to an
alternating current outlet,
because if you interrupt the outlet connection, the capacitors are
discharged by the
resistor voltage divider.

You could modify this circuit replacing resistors with Zener diodes (with
limiting resistor!),
in parallel with the caps. This ensures the V/2 voltage drop on each cap
when charging,
and auto disconnect when discharging.

Another soution is like Ray McArthur said:

"In practice, they should be charged individually, because mismatch in
capacitance will result in one of the caps reaching full rated voltage first
if charged in series. Charging further would overvoltage the lower
capacitance unit. So charge them individually '(or parallel)', then series
them to run the
load. No loss of stored charged occurs when they are used in series."

But you have to make the parallel to series change .

Another item is the caps discharge graph.
You charge them with 5V, but when you use this charge,
this voltage drops exponentially.
Especially on high value caps that can't deliver too much current.

By the way, what's the origin of this question?
Why you want to use caps to store energy, Chris?

Hopping this helps,


Ar

Archiver

> By the way, what's the origin of this question?
> Why you want to use caps to store energy, Chris?
>
> Hopping this helps,
>
>
> Ar