Electrical – what is the formula for charging a capacitor with constant current

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.

From the equation here we have

\$ I_{ref} = I_c(1 + \frac{2}{\beta})\$,so

\$I_c = \frac{I_ref}{1 + \frac{2}{\beta}}\$

where \$I_{ref} \approx \frac{12 - 0.6}{100k} = 0.114 mA\$.

From \$I = C\frac{dv}{dt} \$ we get \$ V(t) = \frac{I}{C}t\$.

Assuming the transistors in the current mirror have a high \$\beta\$ then \$ I_c \approx I_{ref}\$. The 555 timer upper comparator will fire when \$V_c = \frac{2}{3}V_{cc} = 8v.\$ Plugging these numbers in to the equation above along with the capacitor value and solving for t we get:

\$ 8 = \frac{1.14*10^{-5}}{470*10^{-6}}t \implies t \approx 330\$ seconds.

Even assuming that the capacitor is "ideal", high transistor beta, and that the 555 timer inputs draw no current, the actual charging time will probably be greater than this, due to the Early effect of the right hand transistor in the current mirror.