# Capacitor charging problem

Discussion in 'Electronic Design' started by john, Nov 27, 2007.

1. ### johnGuest

Hello,

A constant current source is in series with a capacitor ( 10volts,
0.5uF) and a resistor ( load) of 100kilo ohms.
What happens if

1. A constant current source charges up the capacitor upto its maximum
voltage and keep charging it for a long time. The constant current
source has following specs

a. compliance voltage +/- 15 volts
b. current range 0 to +/- 600uA (ac)
c. biploar
d. dc leakage current of 100nA

2. The capacitor will show high impedance to the current source but
get charged by the DC leakage current constantly. Will the AC flow
through the capacitor anymore?

Regards,
John

2. ### EeyoreGuest

Homework was it ?

Graham

3. ### Helmut SennewaldGuest

Hello John,

The output stage of your current source will run
into saturation beyond the compliance range.
This will result in distortion and less amplitude

Best regards,
Helmut

4. ### Jan PanteltjeGuest

No, it will initially show a short!
That is why it starts at zero volt across the capacitor.
..6mA in 100kOhm makes 60V, you only have 15V, you example makes no sense.

If you omit the resistor for a moment, then the voltage across the capacitor is
The charge Q = C * U = I * t

So U = (T * t) / C
C in Farad, I in ampere, and t in seconds.
The voltage across the capacitor will then rise LINEAR with time, until it reaches
the 15V (theoretically).
However, with the resistor in parallel, the max voltage will be limited by I * Rp, or
the 15V, whatever is lower.
In this case 15V is lower then 60, so 15V
And the charge current in the capacitor is no longer linear, as
as teh voltatge rises, the current through the resistor is increasing,
substract that from the current in the capacitor.

The capacitor will charge like this:
http://en.wikipedia.org/wiki/RC_circuit
see under 'Time domain considerations'.
There is no longer a linear rise in voltage.

5. ### johnGuest

Hello,

The leakage current will charge the capacitor upto its maximum
voltage , because the leakage current does not have any other place to
go it will keep charging the capacitor. Now, my questions are

1. will the current source shut down?

2. for longer times, the charge will break the dielectric and the
leakage current will increase?

John

6. ### Jan PanteltjeGuest

1) provide current diagram
2) what leakage current
3) what ARE you trying to drive, so what IS your load.
4) what frequency are you talkinga bout when you said 'AC' in a previous posting.
5) why do you drive the load, or try to drive the load with a current osurce.

1)
2)
3)
4)
5)

CUL

7. ### EeyoreGuest

That'll depend on the design of the current source but most just continue
delivering the rated current up to theit maximum voltage compliance.

Not as long as the capacitor's voltage rating isn't exceeded.

Graham

8. ### leggGuest

If you're looking for something strange to happen, you have only one
real practical unknown - the leakage current of the capacitor. This
will only be a factor if the leakage current develops a signifigant
voltage across a 100K resistor.

When you're considering practical components, ratings and breakdown,
you'll be dissapointed to know that there's no guarantee that anything
will or will not happen outside paper ratings, unless published
otherwise or practically demonstrated by somebody with the time money
and need to do so.

Leakage in electrolytics can increase with shelf age. There's a
typical method recommended for 'reforming' higher voltage parts
safely.

http://www.dscc.dla.mil/Programs/MilSpec/ListDocs.asp?BasicDoc=MIL-HDBK-1131

RL

9. ### YDGuest

The whole thing is probably unanswerable from a theoretical point of
view. He states that the CS has a compliance of 15 V but the capacitor
has a 10 V rating. So once the cap's charged beyond 10 V it may die
with a bang or a whimper and possibly just take it like a hero. Most
likely just start leaking, had an odd one like that the other day. A
leaky cap making very low frequency noise, it would charge up for a
bit, then drop a few volts, charge up again and so on at an average
period of a couple of seconds, very irregular both in period and
peaks. It was below rated voltage so probably just old age that got
it.

- YD.