Electronic – Can an electrolytic capacitor discharge through its top

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.

With small capacitors up to 1 mF, there is little to worry about. I suppose it's a good idea to make sure they are discharged before plugging them in where the voltage that could be on the cap could damage something, but this is something not generally worried about until you get to some real energies or high voltages.

For small electrolytic caps like what you are working with, just short them against something metal, like a bare component lead, metal chassis, or handy screw driver.

Don't waste brain cycles thinking about this for anything small enough to be a ceramic cap you can plug into a breadboard. By the time you plug it in, your fingers will have discharged it. Even if not, do the math. 1 µF at 10 V is only 50 µJ. Yes microJoules. Big deal.