Electronic – Capacitor Discharge through Constant Current Source

capacitorconstant-currentdischargetheory

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).

Best Answer

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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