Electronics/RC transient

< Electronics

When the switch is open, the initial voltage across the capacitor is zero. When the switch closes (which we will refer to as time zero) the capacitor charges via the resistor to $V_0$.

When the switch is closed, the circuit must follow the relationship:

$V_0=v_c(t)+i_c(t)R$

$V_0=v_c(t)+\frac{d{v}_c(t)}{dt}RC$

which is derived by analysing the circuit using Kirchoff's Voltage Law.

By letting $\tau =RC$ and rearranging the equation:

$\frac{d{v}_c(t)}{dt}+\frac{1}{\tau }{v}_c(t)=\frac{1}{\tau }{V}_0$

This is a first order linear differential equation with integrating factor:

$e^{\int \frac{1}{\tau }dt}$

$e^{\frac{t}{\tau }}$

Multiplying both sides by the integrating factor:

$\frac{d{v}_c(t)}{dt}{e}^{\frac{t}{\tau }}+\frac{1}{\tau }{v}_c(t)e^{\frac{t}{\tau }}=\frac{1}{\tau }{V}_0{e}^{\frac{t}{\tau }}$

Note that:

$\frac{d}{dt}[e^{t/\tau }{v}_c(t)]=\frac{d{v}_c(t)}{dt}{e}^{\frac{t}{\tau }}+\frac{1}{\tau }{v}_c(t)e^{\frac{t}{\tau }}$

Substituting and integrating both sides:

$e^{\frac{t}{\tau }}{v}_c(t)=V_0{e}^{\frac{t}{\tau }}+K$

where K is the integration constant.

When t=0

$v_c(0)=0$

Therefore:

$K=-V_0$

When t>0 this gives:

$e^{\frac{t}{\tau }}{v}_c(t)=V_0{e}^{\frac{t}{\tau }}-V_0$

$v_c(t)=V_0-V_0{e}^{-\frac{t}{\tau }}$

$v_c(t)=V_0(1-e^{-\frac{t}{\tau }})$

when t<0:

$v_c(t)=0$