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## PowerPedia:RC circuit

Lasted edited by Andrew Munsey, updated on June 14, 2016 at 10:10 pm.

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A resistor-capacitor circuit (RC circuit), or RC filter or RC network, is one of the simplest There was an error working with the wiki: Code[1] There was an error working with the wiki: Code[2] or in There was an error working with the wiki: Code[3], driven by a There was an error working with the wiki: Code[22].

#### Introduction

There are three basic, There was an error working with the wiki: Code[4] circuit components: the There was an error working with the wiki: Code[5]. This article considers the RC circuit, in both There was an error working with the wiki: Code[6] and There was an error working with the wiki: Code[7] as shown in the diagrams.

:This article relies on knowledge of the complex There was an error working with the wiki: Code[8] and on knowledge of the There was an error working with the wiki: Code[23] representation of signals.

#### Complex impedance

The There was an error working with the wiki: Code[9]) of a capacitor with capacitance C (in There was an error working with the wiki: Code[24]) is

:Z_C = \frac{1}{sC}

The There was an error working with the wiki: Code[25] s is, in general, a There was an error working with the wiki: Code[26],

:s \ = \ \sigma + j \omega

where

j represents the There was an error working with the wiki: Code[27]:

: j = \sqrt{-1}

\sigma \ is the There was an error working with the wiki: Code[28] constant (in There was an error working with the wiki: Code[29]), and

\omega \ is the There was an error working with the wiki: Code[30] angular frequency (also in radians per second).

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 There was an error working with the wiki: Code[31], we see that the Voltage across the capacitor is:

:

V_C(s) = \frac{1/Cs}{R + 1/Cs}V_{in}(s) = \frac{1}{1 + RCs}V_{in}(s)

and the voltage across the resistor is:

:

V_R(s) = \frac{R}{R + 1/ Cs}V_{in}(s) = \frac{ RCs}{1 + RCs}V_{in}(s)

.

##### Transfer functions

The There was an error working with the wiki: Code[32] for the capacitor is

:

H_C(s) = { V_C(s) \over V_{in}(s) } = { 1 \over 1 + RCs } = G_C e^{j \phi_C}

.

Similarly, the transfer function for the resistor is

:

H_R(s) = { V_R(s) \over V_{in}(s) } = { RCs \over 1 + RCs } = G_R e^{j \phi_R}

.

##### Poles and zeros

Both transfer functions have a single There was an error working with the wiki: Code[10] located at

:

s = - {1 \over RC }

.

In addition, the transfer function for the resistor has a There was an error working with the wiki: Code[11] located at the There was an error working with the wiki: Code[12].

##### Gain and phase angle

The gains across the two components are:

:

G_C = | H_C(s) | = \left|\frac{V_C(s)}{V_{in}(s)}\right| = \frac{1}{\sqrt{1 + \left(\omega RC\right)^2}}

and

:

G_R = | H_R(s) | = \left|\frac{V_R(s)}{V_{in}(s)}\right| = \frac{\omega RC}{\sqrt{1 + \left(\omega RC\right)^2}}

,

and the phase angles are:

:

\phi_C = angle H_C(s) = \tan^{-1}\left(-\omega RC\right)

and

:

\phi_R = angle H_R(s) = \tan^{-1}\left(\frac{1}{\omega RC}\right)

.

These expressions together may be substituted into the usual expression for the There was an error working with the wiki: Code[13] representing the output:

:

V_C \ = \ G_{C}V_{in} e^{j\phi_C}

:

V_R \ = \ G_{R}V_{in} e^{j\phi_R}

.

##### Current

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

:

I(s) = \frac{V_{in}(s) }{R+1/ Cs} = { Cs \over 1 + RCs } V_{in}(s)

##### Impulse response

The There was an error working with the wiki: Code[33] for each voltage is the inverse There was an error working with the wiki: Code[34] of the corresponding transfer function. It represents the response of the circuit to an input voltage consisting of an impulse or There was an error working with the wiki: Code[35].

The impulse response for the capacitor voltage is

:

h_C(t) = {1 \over RC} e^{-t / RC} u(t) = { 1 \over \tau} e^{-t / \tau} u(t)

where u(t) is the There was an error working with the wiki: Code[36] and

:

\tau \ = \ RC

is the There was an error working with the wiki: Code[37].

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 \tau} e^{-t / \tau} u(t)

where ?(t) is the There was an error working with the wiki: Code[38]

##### Frequency domain considerations

These are There was an error working with the wiki: Code[39] 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 \to \infty:

:G_C \to 0

:G_R \to 1.

As \omega \to 0:

:G_C \to 1

:G_R \to 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 There was an error working with the wiki: Code[40]. 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 There was an error working with the wiki: Code[41].

The range of frequencies that the filter passes is called its There was an error working with the wiki: Code[42]. The point at which the filter attenuates the signal to half its unfiltered power is termed its There was an error working with the wiki: Code[43]. 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_{c} = \frac{1}{RC}There was an error working with the wiki: Code[14]/There was an error working with the wiki: Code[15]

or

:f_c = \frac{1}{2\pi RC}There was an error working with the wiki: Code[16]

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 \to 0:

:\phi_C \to 0

:\phi_R \to 90^{\circ} = \pi/2^{c}.

As \omega \to \infty:

:\phi_C \to -90^{\circ} = -\pi/2^{c}

:\phi_R \to 0

So at Direct current (0 There was an error working with the wiki: Code[17]), 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 There was an error working with the wiki: Code[18].

The most straightforward way to derive the time domain behaviour is to use the There was an error working with the wiki: Code[19] (i.e. V_{in} = 0 before t = 0 and then V_{in} = V afterwards):

:

V_{in}(s) = V\frac{1}{s}

:

V_C(s) = V\frac{1}{1 + sRC}\frac{1}{s}

and

:

V_R(s) = V\frac{sRC}{1 + sRC}\frac{1}{s}

.

There was an error working with the wiki: Code[44]s expansions and the inverse There was an error working with the wiki: Code[45] yield:

:

\,\!V_C(t) = V\left(1 - e^{-t/RC}\right)

:

\,\!V_R(t) = Ve^{-t/RC}

.

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 There was an error working with the wiki: Code[46].

These equations show that a series RC circuit has a There was an error working with the wiki: Code[20], usually denoted \tau = 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, \tau 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 \left(1 - \frac{1}{e}\right) per \tau. Thus, in going from t=N\tau to t = (N+1)\tau, the votage will have moved about 63.2 % of the way from its level at t=N\tau toward its final value. So C will be charged to about 63.2 % after \tau, and essentially fully charged (99.3 %) after about 5\tau. 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 \tau, and essentially fully discharged (0.7 %) after about 5\tau. Note that the current, I, in the circuit behaves as the voltage across R does, via There was an error working with the wiki: Code[21].

These results may also be derived by solving the There was an error working with the wiki: Code[47]s describing the circuit:

:

\frac{V_{in} - V_C}{R} = C\frac{dV_C}{dt}

and

:

\,\!V_R = V_{in} - V_C

.

The first equation is solved by using an There was an error working with the wiki: Code[48] 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 >> \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_{in}}{R+1/j\omega C}

but note that the frequency condition described means that

:

\omega C >> \frac{1}{R}

so

:

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

which is just There was an error working with the wiki: Code[49].

Now,

:

V_C = \frac{1}{C}\int_{0}^{t}Idt

so

:

V_C \approx \frac{1}{RC}\int_{0}^{t}V_{in}dt

,

which is an There was an error working with the wiki: Code[50] across the capacitor.

##### Differentiator

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

:

\omega