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I was just thnking of how to model the voltage decay from a fully charged capacitor through a constant current source (CCS). A good approximation to this would be to model the constant current source as a resistor sized by the initial voltage divided by the current of the CCS, giving the formula:

$$ V(t) = V(0) * e ^{\frac{-t}{RC}} $$

... but is there a closed form analytical formula for the CCS case?

   +------------+ V(0)
   |            |
   | C          |
 --+--          /\
 --+--          CCS (I)
   |            \/
   |            |
   +------------+
   |
  -+-
  GND

Some ASCII circuit art for good measure...

Obviously I'm only interested in the model up to the point where the current that the capacitor is able to supply is still above the demand of the current source, and that the voltage is greater than GND (i.e. the realizable time).

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In general voltage on the capacitor with respect to the current is governed by the equation:

\$v(t)= \frac{q(t)}{C} = \frac{1}{C}\int_{t_0}^t i(\tau) \mathrm{d}\tau+v(t_0)\$,

By the definition for CCS:

\$ i(\tau) = I \$,

from this we can derive that:

\$v(t)= \frac{1}{C}(I t - It_0) + v(t_0)\$

now assuming \$t_0 = 0\$ this simplifies to:

\$v(t)= \frac{1}{C}I t + v(0)\$.

What this means is simple! The voltage across capacitor will change linearly with time. The "rate" of change (or "slope") depends on the current magnitude and the capacitance:

  • The bigger the capacitance the slower voltage changes.
  • The bigger the current the faster voltage changes.
  • The sign of the change (voltage rising or falling) depends on the sign or direction of the current. Obviously if current is flowing into capacitor voltagwe will rise if flowing out of capacitor voltage will fall.
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  • \$\begingroup\$ very nicely done - this link I had forgotten is that charge is the integral of current over time! \$\endgroup\$ – vicatcu Nov 30 '11 at 20:19
  • \$\begingroup\$ Hi This is very nice explanation. it will applicable where current is a kind of DC source! What about if current source becomes pulsed one? For example if i want to discharge the 1uF cap with 1A/10uS or 1A/1uS or 1A/0.5uS then what is voltage equation across that capacitor? surely voltage decaying rate will be different for all the cases but I would like to know the relationship between this dI/dT and dV/dT. can anybody suggest some equation for this? \$\endgroup\$ – user17229 Dec 25 '12 at 6:45
  • \$\begingroup\$ Isn't the internal serial resistance affecting this ? \$\endgroup\$ – Damien Oct 16 '18 at 8:25
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You can derive it from the charge equation for a capacitor:

Q=C*V

Rearranging it you have

V=Q/C

Since some of the values will be changing over time we have to express this equation in terms of time:

V(t) = Q(t) / C(t)

C(t) is a constant - capacitance never changes, so the equation can be simplified:

V(t) = Q(t) / C

Here's the fun part: Current is charge per unit time:

I(t) = Q(t)/t

Or, rearranged:

Q(t)= I(t)*t

So we've expressed the charge function in terms of a current function. Replacing the Q(t) with the new value gives us:

V(t) = (I(t)*t )/ C

But since this is the constant current source, I(t) is just a number. We'll call it M for magnitude of the current source:

V(t) = (M*t)/C

So you can see the relationship is linear in the constant current situation. Of course, this assumes initial conditions of no charge. The integral approach suggested in another question handles the initial condition, but this is another way of understanding the solution in a more physical sense rather than a strict mathematical one.

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  • \$\begingroup\$ good answer, very similar to other one - i picked the other one because of use of LaTex and integral forms, but I did upvote :). \$\endgroup\$ – vicatcu Nov 30 '11 at 20:27

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