RC CIRCUIT

A 'resistor-capacitor circuit' '(RC circuit)', or 'RC filter' or 'RC network', is an electric circuit composed of resistors and capacitors driven by a voltage or current source. The ''1st order RC circuit'', composed of one resistor and one capacitor, is the simplest example of an RC circuit and is the subject of this article.

Contents

Introduction

Complex impedance

Sinusoidal steady state

Series circuit

Transfer functions

Poles and zeros

Gain and phase angle

Current

Impulse response

Frequency domain considerations

Time domain considerations

Integrator

Differentiator

Parallel circuit

See also

External links

Introduction

There are three basic, linear analog circuit components: the resistor (R), capacitor (C) and inductor (L). These may be combined in four important combinations: the RC circuit, the RL circuit, the LC circuit and the RLC circuit with the abbreviations indicating which components are used. These circuits, between them, exhibit a large number of important types of behaviour that are fundamental to much of analog electronics. In particular, they are able to act as passive filters. This article considers the RC circuit, in both series and parallel as shown in the diagrams.
:''This article relies on knowledge of the complex impedance representation of capacitors and on knowledge of the frequency domain representation of signals''.

Complex impedance

The complex impedance ''Z''_''C'' (in ohms) of a capacitor with capacitance ''C'' (in farads) is
:
The angular frequency ''s'' is, in general, a complex number,
:$s\; =\; sigma\; +\; j\; omega$
where

★ ''j'' represents the imaginary unit:
:$j^2\; =\; -1$

★ $sigma$ is the exponential decay constant (in radians per second), and

★ $omega$ is the sinusoidal angular frequency (also in radians per second).
Note: the identity $j^2=-1$ is correctly written as such, and not as $j=sqrt\{-1\}$. The second expression is usually avoided, since the right-hand side has two possible values, one the negative of the other, and it is not desirable for the constant ''j'' to have two values at once.

Sinusoidal steady state

Sinusoidal steady state is a special case in which the input voltage consists of a pure sinusoid (with no exponential decay). As a result,
:$$
sigma = 0

and the evaluation of ''s'' becomes
:$$
s = j omega

Series circuit

By viewing the circuit as a voltage divider, the voltage across the capacitor is:
:$$
V_C(s) = rac{1/Cs}{R + 1/Cs}V_{in}(s) = rac{1}{1 + RCs}V_{in}(s)

and the voltage across the resistor is:
:$$
V_R(s) = rac{R}{R + 1/ Cs}V_{in}(s) = rac{ RCs}{1 + RCs}V_{in}(s)
.

Transfer functions

The transfer function for the capacitor is
:$$
H_C(s) = { V_C(s) over V_{in}(s) } = { 1 over 1 + RCs }
.
Similarly, the transfer function for the resistor is
:$$
H_R(s) = { V_R(s) over V_{in}(s) } = { RCs over 1 + RCs }
.

Poles and zeros

Both transfer functions have a single pole located at
:$$
s = - {1 over RC }
.
In addition, the transfer function for the resistor has a zero located at the origin.

Gain and phase angle

The magnitude of the gains across the two components are:
:$$
G_C = | H_C(j omega) | = left|rac{V_C(j omega)}{V_{in}(j omega)}
ight| = rac{1}{sqrt{1 + left(omega RC
ight)^2}}

and
:$$
G_R = | H_R(j omega) | = left|rac{V_R(j omega)}{V_{in}(j omega)}
ight| = rac{omega RC}{sqrt{1 + left(omega RC
ight)^2}}
,
and the phase angles are:
:$$
phi_C = ngle H_C(j omega) = an^{-1}left(-omega RC
ight)

and
:$$
phi_R = ngle H_R(j omega) = an^{-1}left(rac{1}{omega RC}
ight)
.
These expressions together may be substituted into the usual expression for the phasor representing the output:
:$$
V_C = G_{C}V_{in} e^{jphi_C}

:$$
V_R = G_{R}V_{in} e^{jphi_R}
.

Current

The current in the circuit is the same everywhere since the circuit is in series:
:$$
I(s) = rac{V_{in}(s) }{R+1/ Cs} = { Cs over 1 + RCs } V_{in}(s)

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 delta function.
The impulse response for the capacitor voltage is
:$$
h_C(t) = {1 over RC} e^{-t / RC} u(t) = { 1 over au} e^{-t / au} u(t)

where ''u''(''t'') is the Heaviside step function and
:$$
au = RC
is the time constant.
Similarly, the impulse response for the resistor voltage is
:$$
h_R(t) = delta (t) - {1 over RC} e^{-t / RC} u(t) = delta (t) - { 1 over au} e^{-t / au} u(t)

where ''δ''(''t'') is the Dirac delta function

Frequency domain considerations

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 $omega\; o\; infty$:
:$G\_C\; o\; 0$
:$G\_R\; o\; 1$.
As $omega\; o\; 0$:
:$G\_C\; o\; 1$
:$G\_R\; o\; 0$.
This shows that, if the output is taken across the capacitor, high frequencies are attenuated (rejected) 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 rejected. 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
:.
Solving the above equation yields
:
or
:
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 $omega\; o\; 0$:
:$phi\_C\; o\; 0$
:$phi\_R\; o\; 90^\{circ\}\; =\; pi/2^\{c\}$.
As $omega\; o\; infty$:
:$phi\_C\; o\; -90^\{circ\}\; =\; -pi/2^\{c\}$
:$phi\_R\; o\; 0$
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.

Time domain considerations

:''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 $jomega\; o\; s$. Assuming a step input (i.e. $V\_\{in\}\; =\; 0$ before $t\; =\; 0$ and then $V\_\{in\}\; =\; V$ afterwards):
:$$
V_{in}(s) = Vrac{1}{s}

:$$
V_C(s) = Vrac{1}{1 + sRC}rac{1}{s}

and
:$$
V_R(s) = Vrac{sRC}{1 + sRC}rac{1}{s}
.


Partial fractions expansions and the inverse Laplace transform yield:
:$$
,!V_C(t) = Vleft(1 - e^{-t/RC}
ight)

:$$
,!V_R(t) = Ve^{-t/RC}
.
These equations are for calculating the voltage across the capacitor and resistor respectively while the capacitor is charging; for discharging, the equations are vice-versa. These equations can be rewritten in terms of charge and current using the relationships C=Q/V and V=IR (see Ohm's Law).
Thus, the voltage across the capacitor tends towards ''V'' as time passes, while the voltage across the resistor tends towards 0, as shown in the figures. This is in keeping with the intuitive point that the capacitor will be charging from the supply voltage as time passes, and will eventually be fully charged and form an open circuit.
These equations show that a series RC circuit has a time constant, usually denoted $au\; =\; RC$ being the time it takes the voltage across the component to either rise (across C) or fall (across R) to within $1/e$ of its final value. That is, $au$ is the time it takes $V\_C$ to reach $V(1\; -\; 1/e)$ and $V\_R$ to reach $V(1/e)$.
The rate of change is a ''fractional''
ight) per $au$. Thus, in going from $t=N\; au$ to $t\; =\; (N+1)\; au$, the voltage will have moved about 63.2 % of the way from its level at $t=N\; au$ toward its final value. So C will be charged to about 63.2 % after $au$, and essentially fully charged (99.3 %) after about $5\; au$. When the voltage source is replaced with a short-circuit, with C fully charged, the voltage across C drops exponentially with ''t'' from $V$ towards 0. C will be discharged to about 36.8 % after $au$, and essentially fully discharged (0.7 %) after about $5\; au$. Note that the current, $I$, in the circuit behaves as the voltage across R does, via Ohm's Law.
These results may also be derived by solving the differential equations describing the circuit:
:$$
rac{V_{in} - V_C}{R} = Crac{dV_C}{dt}

and
:$$
,!V_R = V_{in} - V_C
.
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.
:.
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 = rac{V_{in}}{R+1/jomega C}

but note that the frequency condition described means that
:$$
omega C gg rac{1}{R}

so
:$$
I pprox rac{V_{in}}{R}
which is just Ohm's Law.
Now,
:$$
V_C = rac{1}{C}int_{0}^{t}Idt

so
:$$
V_C pprox rac{1}{RC}int_{0}^{t}V_{in}dt
,
which is an integrator ''across the capacitor''.

Differentiator

Consider the output across the resistor at ''low'' frequency i.e.,
:$$
omega ll rac{1}{RC}
.
This means that the capacitor has time to charge up until its voltage is almost equal to the source's voltage. Considering the expression for $I$ again, when
:$$
R ll rac{1}{omega C}
,
so
:$$
I pprox rac{V_{in}}{1/jomega C}

:$$
V_{in} pprox rac{I}{jomega C} pprox V_C

Now,
:$$
V_R = IR = Crac{dV_C}{dt}R

:$$
V_R pprox RCrac{dV_{in}}{dt}

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.

Parallel circuit

The parallel RC circuit is generally of less interest than the series circuit. This is largely because the output voltage $V\_\{out\}$ is equal to the input voltage $V\_\{in\}$ — as a result, this circuit does not act as a filter on the input signal unless fed by a current source.
With complex impedances:
:$$
I_R = rac{V_{in}}{R},

and
:$$
I_C = jomega C V_{in},
.
This shows that the capacitor current is 90° out of phase with the resistor (and source) current. Alternatively, the governing differential equations may be used:
:$$
I_R = rac{V_{in}}{R}

and
:$$
I_C = Crac{dV_{in}}{dt}
.
For a step input (which is effectively a 0 Hz or DC signal), the derivative of the input is an impulse at $t=0$. Thus, the capacitor reaches full charge very quickly and becomes an open circuit — the well-known DC behaviour of a capacitor.

See also

★ Electrical network

★ List of electronics topics

External links

★ RC Filter Calculator