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RC circuit - Wikiwand

Current

Kirchhoff's current law means that the current in the series circuit is necessarily the same through both elements. Ohm's law says this current is equal to the input voltage $V_{\mathrm {in} }$ divided by the sum of the complex impedance of the capacitor and resistor:

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

Voltage

By viewing the circuit as a voltage divider, the voltage across the capacitor is:

${\begin{aligned}V_{C}(s)&={\frac {\frac {1}{Cs}}{R+{\frac {1}{Cs}}}}V_{\mathrm {in} }(s)\\&={\frac {1}{1+RCs}}V_{\mathrm {in} }(s)\end{aligned}}$

and the voltage across the resistor is:

${\begin{aligned}V_{R}(s)&={\frac {R}{R+{\frac {1}{Cs}}}}V_{\mathrm {in} }(s)\\&={\frac {RCs}{1+RCs}}V_{\mathrm {in} }(s)\,.\end{aligned}}$

Transfer functions

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

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

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

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

Poles and zeros

Both transfer functions have a single pole located at

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

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

Frequency-domain considerations

The sinusoidal steady state is a special case of complex frequency that considers the input to consist only of pure sinusoids. Hence, the exponential decay component represented by $\sigma$ can be ignored in the complex frequency equation $s{=}\sigma {+}j\omega$ when only the steady state is of interest. The simple substitution of $s\Rightarrow j\omega$ into the previous transfer functions will thus provide the sinusoidal gain and phase response of the circuit.

Gain

The magnitude of the gains across the two components are

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

and

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

As the frequency becomes very large (ω → ∞), the capacitor acts like a short circuit, so:

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

As the frequency becomes very small (ω → 0), the capacitor acts like an open circuit, so:

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

Operation as either a high-pass or a low-pass filter

The behavior at these extreme frequencies show that if the output is taken across the capacitor, high frequencies are attenuated and low frequencies are passed, so such a circuit configuration is a low-pass filter. However, if the output is taken across the resistor, then high frequencies are passed and low frequencies are attenuated, so such a configuration is a high-pass filter.

Cutoff frequency

The range of frequencies that the filter passes is called its bandwidth. The frequency 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

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

Solving the above equation yields

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

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

Phase

The phase angles are

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

and

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

As ω → 0:

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

As ω → ∞:

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

While the output signal's phase shift relative to the input depends on frequency, this is generally less interesting than the gain variations. At DC (0 Hz), the capacitor voltage is in phase with the input 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 input signal and the resistor voltage comes to be in-phase with the input signal.

Phasor representation

The gain and phase expressions together may be combined into these phasor expressions representing the output:

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

Impulse response

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

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

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

Similarly, the impulse response for the resistor voltage is

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

where δ(t) is the Dirac delta function

Time-domain considerations

This section relies on knowledge of the Laplace transform.

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. Assuming a step input (i.e. V_in = 0 before t = 0 and then V_in = V₁ afterwards):

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

Partial fractions expansions and the inverse Laplace transform yield:

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

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 V₁ 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.

The product RC is both the time for V_C and V_R to reach within ⁠1/e⁠ of their final value. In other words, RC is the time it takes for the voltage across the capacitor to rise to V₁·(1 − ⁠1/e⁠) or for the voltage across the resistor to fall to V₁·(⁠1/e⁠). This RC time constant is labeled using the letter tau (τ).

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 is often considered 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 is often considered 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:

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

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.

Integrator

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

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

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 $I$ given above:

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

but note that the frequency condition described means that

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

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

which is just Ohm's Law.

Now,

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

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

Therefore, the voltage across the capacitor acts approximately like an integrator of the input voltage for high frequencies.

Differentiator

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

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

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

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

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

Now,

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

Therefore, the voltage across the resistor acts approximately like a differentiator of the input voltage for low frequencies.

Integration and differentiation can also 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).