At 09:59 AM 1/7/2004 +0000, you wrote:


>Vc(t) = Vo e^(-t/RC )  (1)
>
>Where:
>Vc(t) is the voltage across the capacitor at time t
>Vo is the voltage applied across the resistor/capacitor.
>R resistance in Ohms
>C capacitance in Farads.

That is the discharge (towards zero) of a parallel RC with an initial
capacitor voltage of Vo.

In general, if the initial capacitor voltage is Vo, and the
final capacitor voltage is Vx:

Vc(t) = Vx + (Vo - Vx) * e^(-t/tao)  , where tao = R*C  (2)

In the case Vx = 0, it gives you (1) . In the case Vo =0,
you get:

Vc(t) = Vx (1 - e^(t/tao))   (3)

For example, a 0.1uF capacitor charged to 5V is connected to a 3.3V supply
through 1M ohm. How long does it take to get to 4V?

Vc(Tx) = 4 = 3.3 + 1.7 * exp(-Tx/tao)

ln(0.7/1.7) ~= -.8873 = -Tx/tao => Tx ~= 88.7ms

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads