RC circuit

A resistorcapacitor circuit (RC circuit), or RC l-ter or RC network, is an electric circuit composed ofresistors and capacitors driven by a voltage or currentsource. A rst order RC circuit is composed of one re-sistor and one capacitor and is the simplest type of RCcircuit.RC circuits can be used to lter a signal by blocking cer-tain frequencies and passing others. The two most com-mon RC lters are the high-pass lters and low-pass l-ters; band-pass lters and band-stop lters usually requireRLC lters, though crude ones can be made with RC l-ters.

1 IntroductionThere are three basic, linear passive lumped analog cir-cuit components: the resistor (R), the capacitor (C), andthe inductor (L). These may be combined in the RC cir-cuit, the RL circuit, the LC circuit, and the RLC cir-cuit, with the acronyms indicating which components areused. These circuits, among them, exhibit a large numberof important types of behaviour that are fundamental tomuch of analog electronics. In particular, they are ableto act as passive lters. This article considers the RC cir-cuit, in both series and parallel forms, as shown in thediagrams below.

This article relies on knowledge of the compleximpedance representation of capacitors and onknowledge of the frequency domain representa-tion of signals.

2 Natural responseThe simplest RC circuit is a capacitor and a resistor inseries. When a circuit consists of only a charged capac-itor and a resistor, the capacitor will discharge its storedenergy through the resistor. The voltage across the ca-pacitor, which is time dependent, can be found by us-ing Kirchhos current law, where the current chargingthe capacitor must equal the current through the resistor.This results in the linear dierential equation

CdV

dt+

V

R= 0

---->

C R

I

RC circuit

Solving this equation for V yields the formula forexponential decay:

V (t) = V0e tRC ;

where V0 is the capacitor voltage at time t = 0.The time required for the voltage to fall to V0e is calledthe RC time constant and is given by

= RC :

3 Complex impedanceThe complex impedance, ZC (in ohms) of a capacitorwith capacitance C (in farads) is

ZC =1

sC

1

2 4 SERIES CIRCUIT

The complex frequency s is, in general, a complex num-ber,

s = + j!

where

j represents the imaginary unit:

j2 = 1

is the exponential decay constant (in radians persecond), and

! is the sinusoidal angular frequency (also in radiansper second).

3.1 Sinusoidal steady stateSinusoidal steady state is a special case in which the inputvoltage consists of a pure sinusoid (with no exponentialdecay). As a result,

= 0

and the evaluation of s becomes

s = j!

4 Series circuit

Vin VcCR

IVR

Series RC circuit

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

VC(s) =1/Cs

R+ 1/CsVin(s) =

1

1 +RCsVin(s)

and the voltage across the resistor is:

VR(s) =R

R+ 1/CsVin(s) =

RCs

1 +RCsVin(s)

4.1 Transfer functionsThe transfer function from the input voltage to the voltageacross the capacitor is

HC(s) =VC(s)

Vin(s)=

1

1 +RCs

Similarly, the transfer function from the input to the volt-age across the resistor is

HR(s) =VR(s)

Vin(s)=

RCs

1 +RCs

4.1.1 Poles and zeros

Both transfer functions have a single pole located at

s = 1RC

In addition, the transfer function for the resistor has a zerolocated at the origin.

4.2 Gain and phaseThe magnitude of the gains across the two componentsare:

GC = jHC(j!)j = VC(j!)Vin(j!)

= 1q1 + (!RC)

2

and

GR = jHR(j!)j = VR(j!)Vin(j!)

= !RCq1 + (!RC)

2

and the phase angles are:

C = \HC(j!) = tan1 (!RC)and

R = \HR(j!) = tan1

1

!RC

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

VC = GCVinejC

VR = GRVinejR

4.6 Time-domain considerations 3

4.3 CurrentThe current in the circuit is the same everywhere sincethe circuit is in series:

I(s) =Vin(s)

R+ 1Cs=

Cs

1 +RCsVin(s)

4.4 Impulse responseThe impulse response for each voltage is the inverseLaplace transform of the corresponding transfer function.It represents the response of the circuit to an input voltageconsisting of an impulse or Dirac delta function.The impulse response for the capacitor voltage is

hC(t) =1

RCet/RCu(t) =

1

et/u(t)

where u(t) is the Heaviside step function and

= RC

is the time constant.Similarly, the impulse response for the resistor voltage is

hR(t) = (t) 1RC

et/RCu(t) = (t) 1et/u(t)

where (t) is the Dirac delta function

4.5 Frequency-domain considerationsThese are frequency domain expressions. Analysis ofthem will show which frequencies the circuits (or lters)pass and reject. This analysis rests on a consideration ofwhat happens to these gains as the frequency becomesvery large and very small.As ! !1 :

GC ! 0GR ! 1As ! ! 0 :

GC ! 1GR ! 0This 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 asa low-pass lter. If, though, the output is taken across theresistor, high frequencies are passed and low frequenciesare attenuated (since the capacitor blocks the signal as itsfrequency approaches 0). In this conguration, the circuitbehaves as a high-pass lter.The range of frequencies that the lter passes is calledits bandwidth. The point at which the lter attenuatesthe signal to half its unltered power is termed its cutofrequency. This requires that the gain of the circuit bereduced to

GC = GR =1p2

Solving the above equation yields

!c =1

RC

or

fc =1

2RC

which is the frequency that the lter will attenuate to halfits original power.Clearly, the phases also depend on frequency, althoughthis eect is less interesting generally than the gain vari-ations.As ! ! 0 :

C ! 0

R ! 90 = /2c

As ! !1 :

C ! 90 = /2c

R ! 0So at DC (0 Hz), the capacitor voltage is in phase with thesignal voltage while the resistor voltage leads it by 90. Asfrequency increases, the capacitor voltage comes to have a90 lag relative to the signal and the resistor voltage comesto be in-phase with the signal.

4.6 Time-domain considerations

This section relies on knowledge of e, thenatural logarithmic constant.

4 4 SERIES CIRCUIT

The most straightforward way to derive the time domainbehaviour is to use the Laplace transforms of the expres-sions for VC and VR given above. This eectively trans-forms j! ! s . Assuming a step input (i.e. Vin = 0before t = 0 and then Vin = V afterwards):

Vin(s) = V1

s

VC(s) = V1

1 + sRC

1

s

and

VR(s) = VsRC

1 + sRC

1

s

Partial fractions expansions and the inverse Laplace

0 2 3 4 5Time

63.2%

86.5%95.0% 98.2%99.3%

cV

Capacitor voltage step-response.

0 2 3 4 5Time

1.8% 0.7%5%

13.5%

36.8%

rV

100%

Resistor voltage step-response.

transform yield:

VC(t) = V1 et/RC

VR(t) = V e

t/RC

These equations are for calculating the voltage across thecapacitor and resistor respectively while the capacitor is

charging; for discharging, the equations are vice versa.These equations can be rewritten in terms of charge andcurrent using the relationships C=Q/V and V=IR (seeOhms law).Thus, the voltage across the capacitor tends towards Vas time passes, while the voltage across the resistor tendstowards 0, as shown in the gures. This is in keeping withthe intuitive point that the capacitor will be charging fromthe supply voltage as time passes, and will eventually befully charged.These equations show that a series RC circuit has a timeconstant, usually denoted = RC being the time it takesthe voltage across the component to either rise (across C)or fall (across R) to within 1/e of its nal value. That is, is the time it takes VC to reach V (1 1/e) and VR toreach V (1/e) .The rate of change is a fractional

1 1e

per . Thus,

in going from t = N to t = (N +1) , the voltage willhave moved about 63.2% of the way from its level at t =N toward its nal value. So C will be charged to about63.2% after , and essentially fully charged (99.3%) afterabout 5 . When the voltage source is replaced with ashort-circuit, with C fully charged, the voltage across Cdrops exponentially with t from V towards 0. C will bedischarged to about 36.8% after , and essentially fullydischarged (0.7%) after about 5 . Note that the current,I , in the circuit behaves as the voltage across R does, viaOhms Law.These results may also be derived by solving thedierential equations describing the circuit:

Vin VCR

= CdVCdt

and

VR = Vin VCThe rst equation is solved by using an integrating factorand the second follows easily; the solutions are exactly thesame as those obtained via Laplace transforms.

4.6.1 Integrator

Consider the output across the capacitor at high frequencyi.e.

! 1RC

This means that the capacitor has insucient time tocharge up and so its voltage is very small. Thus the in-put voltage approximately equals the voltage across theresistor. To see this, consider the expression for I givenabove:

5I =Vin

R+ 1/j!C

but note that the frequency condition described meansthat

!C 1R

so

I VinR which is just Ohms Law.

Now,

VC =1

C

Z t0

Idt

so

VC 1RC

Z t0

Vindt

which is an integrator across the capacitor.

4.6.2 Dierentiator

Consider the output across the resistor at low frequencyi.e.,

! 1RC

This means that the capacitor has time to charge up untilits voltage is almost equal to the sources voltage. Con-sidering the expression for I again, when

R 1!C

so

I Vin1/j!C

Vin Ij!C

= VC

Now,

VR = IR = CdVCdt

R

VR RC dVindt

which is a dierentiator across the resistor.More accurate integration and dierentiation can beachieved by placing resistors and capacitors as appropri-ate on the input and feedback loop of operational ampli-ers (see operational amplier integrator and operationalamplier dierentiator).

5 Parallel circuit

Vin VoutC

IR IC

R

Parallel RC circuit

The parallel RC circuit is generally of less interest thanthe series circuit. This is largely because the output volt-age Vout is equal to the input voltage Vin as a result,this circuit does not act as a lter on the input signal unlessfed by a current source.With complex impedances:

IR =VinR

and

IC = j!CVin

This shows that the capacitor current is 90 out of phasewith the resistor (and source) current. Alternatively, thegoverning dierential equations may be used:

IR =VinR

and

IC = CdVindt

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

VoutIin

=R

1 + sRC

6 6 SEE ALSO

6 See also RL circuit LC circuit RLC circuit Electrical network List of electronics topics Step response RC Circuit and continuous-repayment mortgage

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IntroductionNatural response Complex impedance Sinusoidal steady state

Series circuitTransfer functionsPoles and zeros

Gain and phaseCurrentImpulse responseFrequency-domain considerationsTime-domain considerationsIntegratorDifferentiator

Parallel circuitSee also Text and image sources, contributors, and licensesTextImagesContent license