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'[OT]: BackUp capacitor'
2002\08\05@132917 by =?iso-8859-1?Q?=C5ke_Neehr?=

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Hello!

I need help with the calculation
of the time of charge and discharge
for a BackUp capacitor 0,1 uF 5,5 V

Regards
ÅkeN

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2002\08\05@140128 by Harold M Hallikainen

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On Mon, 5 Aug 2002 19:25:02 +0200 =?iso-8859-1?Q?=C5ke_Neehr?=
<spam_OUTake.neehrTakeThisOuTspamTELIA.COM> writes:
> Hello!
>
> I need help with the calculation
> of the time of charge and discharge
> for a BackUp capacitor 0,1 uF 5,5 V
>

       Well, you can start with I=C dv/dt , which becomes dv/dt = I/C . So, if
you draw 1uA from the 0.1F capacitor, the voltage will drop at 10e-6
volts per second, or 100,000 seconds per volt.

Harold


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2002\08\05@160141 by Herbert Graf

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face
> I need help with the calculation
> of the time of charge and discharge
> for a BackUp capacitor 0,1 uF 5,5 V

       Well, off the top of my head this should be enough for you:

Q=CV (where Q = charge in columbs, C is capacitance in Farads, V is voltage
across the cap)
I=Q/S (where I = current drawn, Q/S is columbs per second)

These calcs are a little rough since there are other things going on (ESR
will have an effect depending on your current draw, I doubt it though unless
you draw relatively large currents). There is also leakage to deal with. But
this should be enough to start. TTYL

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2002\08\05@170313 by Peter L. Peres

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On Mon, 5 Aug 2002, Åke Neehr wrote:

>Hello!
>
>I need help with the calculation
>of the time of charge and discharge
>for a BackUp capacitor 0,1 uF 5,5 V

C = Q/U = I*t/U

F.ex. assuming 0.1F 5.5V and start voltage=5V, end=3V, current=20uA  (=2 *
10^-5A) we have time:

t = C*U/I = 0.1 * (5-3) / ( 2 * 10^-5) = 1 * 10^4 seconds or just short
of a week.

The charge time is determined by the capacitor construction and the power
supply. Usually it takes up to 100mA when empty, and can be considered
charged after 1/2 hour at 5V nominal. This assuming it is a supercap.

Peter

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