Rc Circuit

INTRODUCTION

There are three basic, , 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
:

Z_C = rac{1}{Cs}

The Angular Frequency ''s'' is, in general, a Complex Number ,
:

s \ = \ \sigma + j \omega

where

''j'' represents the Imaginary Unit :

j = \sqrt{-1}

$\sigma \$ is the Exponential Decay constant (in Radians Per Second ), and

$\omega \$ is the Sinusoidal angular frequency (also in radians per second).

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

RC circuit]]

By viewing the circuit as a Voltage Divider , we see that 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  }  =  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 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 gains across the two components are:

Solving the above equation yields
:

\omega_{c} = rac{1}{RC}

Rad / S
or
:

f_c = rac{1}{2\pi RC}

Hz
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

The most straightforward way to derive the time domain behaviour is to use the Laplace Transform s of the expressions for

V_C

and

V_R

given above. This effectively transforms

j\omega 	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 Fraction s expansions and the inverse Laplace Transform yield:
:

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

\,\!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 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''

\left(1 - rac{1}{e}
ight)

per

au

. Thus, in going from

t=N	au

t = (N+1)	au

, the votage will have moved about 63% of the way from its level at

t=N	au

toward its final value. So C will be charged to about 63% 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 37% 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 Equation s 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.
:

\omega >> rac{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 = rac{V_{in}}{R+1/j\omega C}

but note that the frequency condition described means that
:

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

so
:

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

which is just Ohm's Law .

Now,
:

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

so
:

V_C \approx 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 << 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 << rac{1}{\omega C}

,
so
:

I \approx rac{V_{in}}{1/j\omega C}

V_{in} \approx rac{I}{j\omega C} \approx V_C

Now,
:

V_R = IR = Crac{dV_C}{dt}R

V_R \approx 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 Amplifier s.

PARALLEL CIRCUIT

RC 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 = j\omega CV_{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

RL Circuit

LC Circuit

RLC Circuit

Electrical Network

List Of Electronics Topics

EXTERNAL LINKS

RC Filter Calculator

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Rc Circuit