RC circuit

Sinusoidal steady state is a special case in which the input voltage consists of a pure sinusoid (with no exponential decay). As a result, ${\displaystyle \sigma =0}$ and the impedance becomes

${\displaystyle Z_{C}=-{\frac {j}{\omega C}}\,.}$

The transfer function from the input voltage to the voltage across the capacitor is

${\displaystyle H_{C}(s)={\frac {V_{C}(s)}{V_{\mathrm {in} }(s)}}={\frac {1}{1+RCs}}\,.}$

Similarly, the transfer function from the input to the voltage across the resistor is

${\displaystyle H_{R}(s)={\frac {V_{R}(s)}{V_{\rm {in}}(s)}}={\frac {RCs}{1+RCs}}\,.}$

Both transfer functions have a single pole located at

${\displaystyle s=-{\frac {1}{RC}}\,.}$

In addition, the transfer function for the voltage across the resistor has a zero located at the origin.

The magnitude of the gains across the two components are

${\displaystyle G_{C}={\big |}H_{C}(j\omega ){\big |}=\left|{\frac {V_{C}(j\omega )}{V_{\mathrm {in} }(j\omega )}}\right|={\frac {1}{\sqrt {1+\left(\omega RC\right)^{2}}}}}$

and

${\displaystyle G_{R}={\big |}H_{R}(j\omega ){\big |}=\left|{\frac {V_{R}(j\omega )}{V_{\mathrm {in} }(j\omega )}}\right|={\frac {\omega RC}{\sqrt {1+\left(\omega RC\right)^{2}}}}\,,}$

and the phase angles are

${\displaystyle \phi _{C}=\angle H_{C}(j\omega )=\tan ^{-1}\left(-\omega RC\right)}$

and

${\displaystyle \phi _{R}=\angle H_{R}(j\omega )=\tan ^{-1}\left({\frac {1}{\omega RC}}\right)\,.}$

These expressions together may be substituted into the usual expression for the phasor representing the output:

${\displaystyle {\begin{aligned}V_{C}&=G_{C}V_{\mathrm {in} }e^{j\phi _{C}}\\V_{R}&=G_{R}V_{\mathrm {in} }e^{j\phi _{R}}\,.\end{aligned}}}$

The current in the circuit is the same everywhere since the circuit is in series:

${\displaystyle I(s)={\frac {V_{\mathrm {in} }(s)}{R+{\frac {1}{Cs}}}}={\frac {Cs}{1+RCs}}V_{\mathrm {in} }(s)\,.}$

The impulse response for each voltage is the inverse Laplace transform of the corresponding transfer function. It represents the response of the circuit to an input voltage consisting of an impulse or Dirac delta function.

The impulse response for the capacitor voltage is

${\displaystyle h_{C}(t)={\frac {1}{RC}}e^{-{\frac {t}{RC}}}u(t)={\frac {1}{\tau }}e^{-{\frac {t}{\tau }}}u(t)\,,}$

where u(t) is the Heaviside step function and τ = RC is the time constant.

Similarly, the impulse response for the resistor voltage is

${\displaystyle h_{R}(t)=\delta (t)-{\frac {1}{RC}}e^{-{\frac {t}{RC}}}u(t)=\delta (t)-{\frac {1}{\tau }}e^{-{\frac {t}{\tau }}}u(t)\,,}$

where δ(t) is the Dirac delta function

These are frequency domain expressions. Analysis of them will show which frequencies the circuits (or filters) pass and reject. This analysis rests on a consideration of what happens to these gains as the frequency becomes very large and very small.

As ω → ∞:

${\displaystyle G_{C}\to 0\quad {\mbox{and}}\quad G_{R}\to 1\,.}$

As ω → 0:

${\displaystyle G_{C}\to 1\quad {\mbox{and}}\quad G_{R}\to 0\,.}$

This shows that, if the output is taken across the capacitor, high frequencies are attenuated (shorted to ground) and low frequencies are passed. Thus, the circuit behaves as a low-pass filter. If, though, the output is taken across the resistor, high frequencies are passed and low frequencies are attenuated (since the capacitor blocks the signal as its frequency approaches 0). In this configuration, the circuit behaves as a high-pass filter.

The range of frequencies that the filter passes is called its bandwidth. The point at which the filter attenuates the signal to half its unfiltered power is termed its cutoff frequency. This requires that the gain of the circuit be reduced to

${\displaystyle G_{C}=G_{R}={\frac {1}{\sqrt {2}}}}$.

Solving the above equation yields

${\displaystyle \omega _{\mathrm {c} }={\frac {1}{RC}}\quad {\mbox{or}}\quad f_{\mathrm {c} }={\frac {1}{2\pi RC}}}$

which is the frequency that the filter will attenuate to half its original power.

Clearly, the phases also depend on frequency, although this effect is less interesting generally than the gain variations.

As ω → 0:

${\displaystyle \phi _{C}\to 0\quad {\mbox{and}}\quad \phi _{R}\to 90^{\circ }={\frac {\pi }{2}}{\mbox{ radians}}\,.}$

As ω → ∞:

${\displaystyle \phi _{C}\to -90^{\circ }=-{\frac {\pi }{2}}{\mbox{ radians}}\quad {\mbox{and}}\quad \phi _{R}\to 0\,.}$

So at DC (0 Hz), the capacitor voltage is in phase with the signal voltage while the resistor voltage leads it by 90°. As frequency increases, the capacitor voltage comes to have a 90° lag relative to the signal and the resistor voltage comes to be in-phase with the signal.

This section relies on knowledge of e, the natural logarithmic constant.

The most straightforward way to derive the time domain behaviour is to use the Laplace transforms of the expressions for V_C and V_R given above. This effectively transforms jω → s. Assuming a step input (i.e. V_in = 0 before t = 0 and then V_in = V afterwards):

${\displaystyle {\begin{aligned}V_{\mathrm {in} }(s)&=V\cdot {\frac {1}{s}}\\V_{C}(s)&=V\cdot {\frac {1}{1+sRC}}\cdot {\frac {1}{s}}\\V_{R}(s)&=V\cdot {\frac {sRC}{1+sRC}}\cdot {\frac {1}{s}}\,.\end{aligned}}}$

Capacitor voltage step-response.

Resistor voltage step-response.

Partial fractions expansions and the inverse Laplace transform yield:

${\displaystyle {\begin{aligned}V_{C}(t)&=V\left(1-e^{-{\frac {t}{RC}}}\right)\\V_{R}(t)&=Ve^{-{\frac {t}{RC}}}\,.\end{aligned}}}$

These equations are for calculating the voltage across the capacitor and resistor respectively while the capacitor is charging; for discharging, the equations are vice versa. These equations can be rewritten in terms of charge and current using the relationships C = Q/V and V = IR (see Ohm's law).

Thus, the voltage across the capacitor tends towards V as time passes, while the voltage across the resistor tends towards 0, as shown in the figures. This is in keeping with the intuitive point that the capacitor will be charging from the supply voltage as time passes, and will eventually be fully charged.

These equations show that a series RC circuit has a time constant, usually denoted τ = RC being the time it takes the voltage across the component to either rise (across the capacitor) or fall (across the resistor) to within 1/e of its final value. That is, τ is the time it takes V_C to reach V(1 − 1/e) and V_R to reach V(1/e).

The rate of change is a fractional 1 − 1/e per τ. Thus, in going from t = Nτ to t = (N + 1)τ, the voltage will have moved about 63.2% of the way from its level at t = Nτ toward its final value. So the capacitor will be charged to about 63.2% after τ, and essentially fully charged (99.3%) after about 5τ. When the voltage source is replaced with a short circuit, with the capacitor fully charged, the voltage across the capacitor drops exponentially with t from V towards 0. The capacitor will be discharged to about 36.8% after τ, and essentially fully discharged (0.7%) after about 5τ. Note that the current, I, in the circuit behaves as the voltage across the resistor does, via Ohm's Law.

These results may also be derived by solving the differential equations describing the circuit:

${\displaystyle {\begin{aligned}{\frac {V_{\mathrm {in} }-V_{C}}{R}}&=C{\frac {dV_{C}}{dt}}\\V_{R}&=V_{\mathrm {in} }-V_{C}\,.\end{aligned}}}$

The first equation is solved by using an integrating factor and the second follows easily; the solutions are exactly the same as those obtained via Laplace transforms.

Consider the output across the capacitor at high frequency, i.e.

${\displaystyle \omega \gg {\frac {1}{RC}}\,.}$

This means that the capacitor has insufficient time to charge up and so its voltage is very small. Thus the input voltage approximately equals the voltage across the resistor. To see this, consider the expression for ${\displaystyle I}$ given above:

${\displaystyle I={\frac {V_{\mathrm {in} }}{R+{\frac {1}{j\omega C}}}}\,,}$

but note that the frequency condition described means that

${\displaystyle \omega C\gg {\frac {1}{R}}\,,}$

so

${\displaystyle I\approx {\frac {V_{\mathrm {in} }}{R}}}$

which is just Ohm's Law.

Now,

${\displaystyle V_{C}={\frac {1}{C}}\int _{0}^{t}I\,dt\,,}$

so

${\displaystyle V_{C}\approx {\frac {1}{RC}}\int _{0}^{t}V_{\mathrm {in} }\,dt\,,}$

which is an integrator across the capacitor.

Consider the output across the resistor at low frequency i.e.,

${\displaystyle \omega \ll {\frac {1}{RC}}\,.}$

This means that the capacitor has time to charge up until its voltage is almost equal to the source's voltage. Considering the expression for I again, when

${\displaystyle R\ll {\frac {1}{\omega C}}\,,}$

so

${\displaystyle {\begin{aligned}I&\approx {\frac {V_{\mathrm {in} }}{\frac {1}{j\omega C}}}\\V_{\mathrm {in} }&\approx {\frac {I}{j\omega C}}=V_{C}\,.\end{aligned}}}$

Now,

${\displaystyle {\begin{aligned}V_{R}&=IR=C{\frac {dV_{C}}{dt}}R\\V_{R}&\approx RC{\frac {dV_{in}}{dt}}\,,\end{aligned}}}$

which is a differentiator across the resistor.

More accurate integration and differentiation can be achieved by placing resistors and capacitors as appropriate on the input and feedback loop of operational amplifiers (see operational amplifier integrator and operational amplifier differentiator).