## Reference Designer Calculators

### RC Time Constant Derivation

The circuit shows a resistor of value $R$ connected with a Capacitor of value $C$.

Let a pulse voltage V is applied at time t =0. The current starts flowing through the resistor $R$ and the capacitor starts charging. As a result of this the voltage $v(t)$ on the capacitor $C$ starts rising. The charge $q(t)$ on the capacitor also starts rising. The amount of the charge $q(t)$ at any time $t$ is given by

$q(t) = C.v(t)$

where $C$ is the Capacitance of the capacitor.

Differentiating this equation with respect to time gives

$\frac{dq(t)}{dt} =C.\frac{dv(t)}{dt}$

$\frac{dq(t)}{dt}$ is the rate of the change of charge in the circuit which is nothing but the current flowing. So we can write the equation as

$i(t) =C.\frac{dv(t)}{dt}$

Notice that this current $i(t)$ flows through the resistor

If $v(t)$ is the voltage across the capacitor at time t, then the voltage across the resistor at time t is given by $V-v(t)$.

By Kirchoff's law, we can write

$V-v(t) = i(t). R$

Combining the two equations we get

$\frac{V-v(t)}{R} =C.\frac{dv(t)}{dt}$

Separating the variables.

$\frac{dt}{RC} =\frac{dv(t)}{V-v(t)} $

Solving this equation leads us to

$ \frac{t}{RC}= -ln(V-v(t)) + K$

where K is a constant of integration.

Since, at t =0, v(t) = 0, we get by substitution

$K = ln(V)$

We therefore get

$\frac{t}{RC} = -ln(V-v(t)) + ln(V)$ or

$ ln\frac{V-v(t)}{V} = -\frac{t}{RC}$

Expressing in exponent form, we get

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

which simplifies to

$ V -v(t)= V.e^{-\frac{t}{RC}} $

It is more conveniently expressed as

$ v(t)= ( 1-e^{-\frac{t}{RC}}).V $

when $t = RC$

$v(t)= (1-e^-1).V $

So RC time constant can also be defined at the time at which value of the voltage $v(t)$ reaches to $(1-e^{-1})$ or about 63.2% of final value