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RC circuit

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Electric circuit composed of resistors and capacitors
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Linear analog
electronic filters

Aresistor–capacitor circuit (RC circuit), orRC filter orRC network, is anelectric circuit composed ofresistors andcapacitors. It may be driven by avoltage orcurrent source and these will produce different responses. A first order RC circuit is composed of one resistor and one capacitor and is the simplest type of RC circuit.

RC circuits can be used to filter a signal by blocking certain frequencies and passing others. The two most common RC filters are thehigh-pass filters andlow-pass filters;band-pass filters andband-stop filters usually requireRLC filters, though crude ones can be made with RC filters.

Natural response

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Simplest RC circuit

The simplest RC circuit consists of a resistor withresistanceR and a charged capacitor with capacitanceC connected to one another in a single loop, without an external voltage source. The capacitor will discharge its stored energy through the resistor. IfV(t) is taken to be the voltage of the capacitor's top plate relative to its bottom plate in the figure, then thecapacitor current–voltage relation says the currentI(t)exiting the capacitor's top plate will equalC multiplied by thenegative time derivative ofV(t).Kirchhoff's current law says this current is the same current entering the top side of the resistor, which perOhm's law equalsV(t)/R. This yields alinear differential equation:

(C−dV(t)dt)⏞capacitor current=(V(t)R)⏞resistor current,{\displaystyle \overbrace {{\Biggl (}C{\frac {-\mathrm {d} V(t)}{\mathrm {d} t}}{\Biggr )}} ^{\text{capacitor current}}=\overbrace {{\Biggl (}{\frac {V(t)}{R}}{\Biggr )}} ^{\text{resistor current}},}

which can be rearranged according to the standard form forexponential decay:

dV(t)dt=−1RCV(t).{\displaystyle {\frac {\mathrm {d} V(t)}{\mathrm {d} t}}=-{\frac {1}{RC}}V(t)\,.}

This means that the instantaneous rate of voltage decrease at any time is proportional to the voltage at that time.Solving forV(t) yields an exponential decay curve thatasymptotically approaches 0:

V(t)=V0⋅e−tRC,{\displaystyle V(t)=V_{0}\cdot e^{-{\frac {t}{RC}}}\,,}

whereV0 is the capacitor voltage at timet = 0 ande isEuler's number.

The time required for the voltage to fall to⁠V0/e⁠ is called theRC time constant and is given by:[1]

τ=RC.{\displaystyle \tau =RC\,.}

When using theInternational System of Units,R is inohms andC is infarads, soτ will be inseconds. At any timeN·τ, the capacitor's charge or voltage will be⁠1/eᴺ⁠ of its starting value. So if the capacitor's charge or voltage is said to start at 100%, then 36.8% remains at1·τ, 13.5% remains at2·τ, 5% remains at3·τ, 1.8% remains at4·τ, and less than 0.7% remains at5·τ and later.

Thehalf-life (t½) is the time that it takes for its charge or voltage to be reduced in half:[2]

12=e−t1/2τ⟶t1/2=ln⁡(2)τ≈.693τ.{\displaystyle {\tfrac {1}{2}}{=}e^{-{\tfrac {t_{1/2}}{\tau }}}\longrightarrow t_{1/2}=\ln(2)\,\tau \approx {\text{.693}}\,\tau \,.}

For example, 50% of charge or voltage remains at time1·t½, then 25% remains at time2·t½, then 12.5% remains at time3·t½, and⁠1/2ᴺ⁠ will remain at timeN·t½.

RC discharge calculator

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.000001

1000000

1

1

1

1

1

1

1

.368

36.8

1

0.368

11

1

0.159

1

1

1

1

For instance,1  of resistance with1  of capacitance produces a time constant of approximately1seconds. Thisτ corresponds to acutoff frequency of approximately159millihertz or1radians. If the capacitor has an initial voltageV0 of1, then after1 τ (approximately1seconds or1.443 half-lives), the capacitor's voltage will discharge to approximately368millivolts:

 VC(1τ) ≈ 36.8% of V0 

Complex impedance

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The RC circuit's behavior is well-suited to be analyzed in theLaplace domain, which the rest of this article requires a basic understanding of. The Laplace domain is afrequency domain representation usingcomplex frequencys, which is (in general) acomplex number:

s=σ+jω,{\displaystyle s=\sigma +j\omega \,,}

where

When evaluating circuit equations in the Laplace domain, time-dependent circuit elements of capacitance and inductance can be treated like resistors withcomplex-valued impedance instead ofreal resistance. While the complex impedanceZR of a resistor is simply a real value equal to its resistanceR, thecomplex impedance of a capacitorC is instead:

ZC=1Cs.{\displaystyle Z_{C}={\frac {1}{Cs}}.}

Series circuit

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Series RC circuit

Current

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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 voltageVin{\displaystyle V_{\mathrm {in} }} divided by the sum of the complex impedance of the capacitor and resistor:

I(s)=Vin(s)R+1Cs=Cs1+RCsVin(s).{\displaystyle {\begin{aligned}I(s)&={\frac {V_{\mathrm {in} }(s)}{R+{\frac {1}{Cs}}}}\\&={\frac {Cs}{1+RCs}}V_{\mathrm {in} }(s)\,.\end{aligned}}}

Voltage

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By viewing the circuit as avoltage divider, thevoltage across the capacitor is:

VC(s)=1CsR+1CsVin(s)=11+RCsVin(s){\displaystyle {\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:

VR(s)=RR+1CsVin(s)=RCs1+RCsVin(s).{\displaystyle {\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

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Thetransfer function from the input voltage to the voltage across the capacitor is

HC(s)=VC(s)Vin(s)=11+RCs.{\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

HR(s)=VR(s)Vin(s)=RCs1+RCs.{\displaystyle H_{R}(s)={\frac {V_{R}(s)}{V_{\rm {in}}(s)}}={\frac {RCs}{1+RCs}}\,.}

Poles and zeros

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Both transfer functions have a singlepole located at

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

In addition, the transfer function for the voltage across the resistor has azero located at theorigin.

Frequency-domain considerations

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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 byσ{\displaystyle \sigma } can be ignored in the complex frequency equations=σ+jω{\displaystyle s{=}\sigma {+}j\omega } when only the steady state is of interest. The simple substitution ofs⇒jω{\displaystyle s\Rightarrow j\omega } into the previous transfer functions will thus provide the sinusoidal gain and phase response of the circuit.

Gain

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Amplitude and phase transfer functions for a series RC circuit

The magnitude of the gains across the two components are

GC=|HC(jω)|=|VC(jω)Vin(jω)|=11+(ωRC)2{\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

GR=|HR(jω)|=|VR(jω)Vin(jω)|=ωRC1+(ωRC)2,{\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}}}}\,,}

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

GC→0andGR→1.{\displaystyle 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:

GC→1andGR→0.{\displaystyle G_{C}\to 1\quad {\mbox{and}}\quad G_{R}\to 0\,.}
Operation as either a high-pass or a low-pass filter
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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 alow-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 ahigh-pass filter.

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

GC=GR=12{\displaystyle G_{C}=G_{R}={\frac {1}{\sqrt {2}}}}.

Solving the above equation yields

ωc=1RCorfc=12πRC{\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.

Phase

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The phase angles are

ϕC=∠HC(jω)=tan−1⁡(−ωRC){\displaystyle \phi _{C}=\angle H_{C}(j\omega )=\tan ^{-1}\left(-\omega RC\right)}

and

ϕR=∠HR(jω)=tan−1⁡(1ωRC).{\displaystyle \phi _{R}=\angle H_{R}(j\omega )=\tan ^{-1}\left({\frac {1}{\omega RC}}\right)\,.}

Asω → 0:

ϕC→0andϕR→90∘=π2 radians.{\displaystyle \phi _{C}\to 0\quad {\mbox{and}}\quad \phi _{R}\to 90^{\circ }={\frac {\pi }{2}}{\mbox{ radians}}\,.}

Asω → ∞:

ϕC→−90∘=−π2 radiansandϕR→0.{\displaystyle \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. AtDC (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

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

VC=GCVinejϕCVR=GRVinejϕR.{\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}}}

Impulse response

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The impulse response of a series RC circuit

Theimpulse response for each voltage is theinverse Laplace transform of the corresponding transfer function. It represents the response of the circuit to an input voltage consisting of an impulse orDirac delta function.

The impulse response for the capacitor voltage is

hC(t)=1RCe−tRCu(t)=1τe−tτu(t),{\displaystyle h_{C}(t)={\frac {1}{RC}}e^{-{\frac {t}{RC}}}u(t)={\frac {1}{\tau }}e^{-{\frac {t}{\tau }}}u(t)\,,}

whereu(t) is theHeaviside step function andτ =RC is thetime constant.

Similarly, the impulse response for the resistor voltage is

hR(t)=δ(t)−1RCe−tRCu(t)=δ(t)−1τe−tτu(t),{\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)\,,}

whereδ(t) is theDirac delta function

Time-domain considerations

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This section relies on knowledge of theLaplace transform.

The most straightforward way to derive the time domain behaviour is to use theLaplace transforms of the expressions forVC andVR given above. Assuming astep input (i.e.Vin = 0 beforet = 0 and thenVin =V1 afterwards):

Vin(s)=V1⋅1sVC(s)=V1⋅11+sRC⋅1sVR(s)=V1⋅sRC1+sRC⋅1s.{\displaystyle {\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}}}
Capacitor voltage step-response.
Resistor voltage step-response.

Partial fractions expansions and theinverse Laplace transform yield:

VC(t)=V1⋅(1−e−tRC)VR(t)=V1⋅(e−tRC).{\displaystyle {\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 ischarging; for discharging, the equations are vice versa. These equations can be rewritten in terms of charge and current using the relationshipsC =⁠Q/V⁠ andV =IR (seeOhm's law).

Thus, the voltage across the capacitor tends towardsV1 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 productRC is both the time forVC andVR to reach within⁠1/e⁠ of their final value. In other words,RC is the time it takes for the voltage across the capacitor to rise toV1·(1 −⁠1/e⁠) or for the voltage across the resistor to fall toV1·(⁠1/e⁠). ThisRC time constant is labeled using the lettertau (τ).

The rate of change is afractional1 −⁠1/e⁠ perτ. Thus, in going fromt =Nτ tot = (N + 1)τ, the voltage will have moved about 63.2% of the way from its level att =Nτ toward its final value. So the capacitor will be charged to about 63.2% afterτ, and is often considered fully charged (>99.3%) after about5τ. When the voltage source is replaced with a short circuit, with the capacitor fully charged, the voltage across the capacitor drops exponentially witht fromV towards 0. The capacitor will be discharged to about 36.8% afterτ, and is often considered fully discharged (<0.7%) after about5τ. Note that the current,I, in the circuit behaves as the voltage across the resistor does, viaOhm's Law.

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

Vin−VCR=CdVCdtVR=Vin−VC.{\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 anintegrating factor and the second follows easily; the solutions are exactly the same as those obtained via Laplace transforms.

Integrator

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Consider the output across the capacitor athigh frequency, i.e.

ω≫1RC.{\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 forI{\displaystyle I} given above:

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

but note that the frequency condition described means that

ωC≫1R,{\displaystyle \omega C\gg {\frac {1}{R}}\,,}

so

I≈VinR{\displaystyle I\approx {\frac {V_{\mathrm {in} }}{R}}}

which is justOhm's Law.

Now,

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

so

VC≈1RC∫0tVindt.{\displaystyle V_{C}\approx {\frac {1}{RC}}\int _{0}^{t}V_{\mathrm {in} }\,dt\,.}

Therefore, the voltageacross the capacitor acts approximately like anintegrator of the input voltage for high frequencies.

Differentiator

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Consider the output across the resistor atlow frequency i.e.,

ω≪1RC.{\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 forI again, when

R≪1ωC,{\displaystyle R\ll {\frac {1}{\omega C}}\,,}

so

I≈Vin1jωCVin≈IjωC=VC.{\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,

VR=IR=CdVCdtRVR≈RCdVindt.{\displaystyle {\begin{aligned}V_{R}&=IR=C{\frac {dV_{C}}{dt}}R\\V_{R}&\approx RC{\frac {dV_{in}}{dt}}\,.\end{aligned}}}

Therefore, the voltageacross the resistor acts approximately like adifferentiator of the input voltage for low frequencies.

Integration anddifferentiation can also be achieved by placing resistors and capacitors as appropriate on the input andfeedback loop ofoperational amplifiers (seeoperational amplifier integrator andoperational amplifier differentiator).


Parallel circuit

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Parallel RC circuit

The parallel RC circuit is generally of less interest than the series circuit. This is largely because the output voltageVout is equal to the input voltageVin — as a result, this circuit acts as a filter on a current input instead of a voltage input.

With complex impedances:

IR=VinRIC=jωCVin.{\displaystyle {\begin{aligned}I_{R}&={\frac {V_{\mathrm {in} }}{R}}\\I_{C}&=j\omega CV_{\mathrm {in} }\,.\end{aligned}}}

This shows that the capacitor current is 90° out of phase with the resistor (and source) current. Alternatively, the governing differential equations may be used:

IR=VinRIC=CdVindt.{\displaystyle {\begin{aligned}I_{R}&={\frac {V_{\mathrm {in} }}{R}}\\I_{C}&=C{\frac {dV_{\mathrm {in} }}{dt}}\,.\end{aligned}}}

When fed by a current source, the transfer function of a parallel RC circuit is:

VoutIin=R1+sRC.{\displaystyle {\frac {V_{\mathrm {out} }}{I_{\mathrm {in} }}}={\frac {R}{1+sRC}}\,.}

Synthesis

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It is sometimes required tosynthesise an RC circuit from a givenrational function ins. For synthesis to be possible in passive elements, the function must be apositive-real function. To synthesise as an RC circuit, all the critical frequencies (poles and zeroes) must be on the negative real axis and alternate between poles and zeroes with an equal number of each. Further, the critical frequency nearest the origin must be a pole, assuming the rational function represents an impedance rather than an admittance.

The synthesis can be achieved with a modification of theFoster synthesis orCauer synthesis used to synthesiseLC circuits. In the case of Cauer synthesis, aladder network of resistors and capacitors will result.[3]

See also

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Vrms=kBTC.{\displaystyle V_{\text{rms}}={\sqrt {k_{\text{B}}T \over C}}.}

References

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  1. ^Horowitz & Hill, p. 1.13
  2. ^Hanks, Ann; Luttermoser, Donald."General Physics II Lab (PHYS-2021) Experiment ELEC-5: RC Circuits"(PDF).
  3. ^Bakshi & Bakshi, pp. 3-30–3-37

Bibliography

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  • Bakshi, U.A.; Bakshi, A.V.,Circuit Analysis - II, Technical Publications, 2009ISBN 9788184315974.
  • Horowitz, Paul; Hill, Winfield,The Art of Electronics (3rd edition), Cambridge University Press, 2015ISBN 0521809266.
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