Simulation of Simple

RC Circuits

Article 2

ARTICLE 1

Simulation of
Simple RC Circuit

CONTENTS
1.1 Introduction to Simple RC Circuit
1.2 Mathematical Modeling of RC circuit
1.3 Simulation with MATLAB
1.4 Simulation with SIMULINK
Appendix

1.1 Introduction to Simple RC circuit: We consider a circuit containing a dc voltage

source (battery) and a resister. The circuit contains only a dc voltage source V and
resister R is called a simple R circuit. In the simple R circuit the current and the voltage
are steady in all components.

Section 1.1

Introduction to simulation

For example when you turn on a flashlight current stare to flow almost immediately
through the bulb, batteries, and connecting metal parts. The current continues to flow
steadily until you turn of the switch. When you insert the capacitor in to the simple R
circuit, this picture changes. A capacitor introduces time dependence. The circuit that
contains a dc voltage source V, resister R and a capacitor C is called a simple RC circuit.
In simple RC circuit quantities change more gradually because of the capacitor. The Vc
builds up as charge accumulates on the plates of the capacitor. And the polarity of the
voltage Vc is such that it opposes the applied voltage V. In the simple RC circuit, we
change the capacitor charge by moving charge on or off the capacitor plate. Because the
charge and voltage are proportional in a capacitor it is not possible to change the voltage
across the capacitor with out changing the charge. Therefore, we conclude that the
voltage across the capacitor cannot change instantaneously, and this makes a simple RC
circuit a time dependent circuit.. As more time pass, the voltage Vc increase, so the
current through the capacitor ic decrease until such time as C if fully charged (Vc=V).
The main objectives of this article are:
To

learn (by simulation) how the voltage and the current changes with

time when the dc voltage source are switched into them

To

observe the effect of changes in circuit time constants on the

behavior of transient voltages

To

verify (by simulation) the equations the describe voltage and current

changes in simple RC circuits

1.2 System Analysis: Consider the electrical circuits of fig 1.1. If the out-put terminal
are connected to a high- impedance load, then the current through the resistor and the
current through capacitor are essentially equal.

s
Ein

Eout

Figure 1.1 Simple RC circuit

Article1

Simulation of Simple RC Circuit

Let i represent the current that pass through the resistor and the capacitor. For the
resistor, the current (i) is equal to the voltage difference (Ein - Eout) divided by the
resistance (R).
i

E in E out
R

(2.0)

For the capacitor, the current i is equal to the capacitor C times the instantaneous rate of
change of the capacitor voltage.
iC

dE out
dt

(2.1)

By equating the right-hand side of equation (2.0) and (2.1) we obtain that
RC

dE out
E out Ein
dt

(2.2)

We can easily see that equation (2.2) is a first order ordinary differential equation of type:
dx
x
dt

(2.3)

The general solution of the differential equation of (2.3) is given by*:

x e t e t dt Ke t

(2.4)

Where K is a constant and depend on initial condition of the system.

The quantity RC in equation (2.2) is called the time constant of the circuit and has the
units of time, in seconds, when R is in ohms and C is in farads. The conventional symbol
for a time constant is Greek letter Tau (). Let RC= , and compare equation (2.2) with
equation (2.3) we obtain the following relat
x Eout

Ein

(2.5)

We plug equation (2.5 ) in to equation (2.4) we obtain that

The proof is given in appendix A, and for more detail explanation the reader my consult any differential
equation book ,

Eout e

E out

Ein
e

E out

dt Ke

E out

(2.6)
Section 1.2

System Analysis

After some simplification we obtain

t

E out Ein (1 e )
(2.7)
From equation (2.1) and (2.7) it follow that current through the capacitor C is given by
t

Ein
e
R

(2.8)

Now we exam the significant of the time constant of the circuit. To do so we make the
table shown below with different values of .
t

iout in e
R
Ein 1
E
1
e 36.6 in
R
R
Ein 2
Ein
2
e 13.5
R
R
Ein 3
E
3
e 5 in
R
R
Ein 5
5
e 0
R

The table 1.1 shows the values of E

E out Ein (1 e )
Eout Ein (1 e 1 ) 63.2 Ein
Eout Ein (1 e 2 ) 86.5Ein

E out Ein (1 e ) 95Ein

Eout Ein

Table 1.1,
out ( the capacitor voltage ) with different values of

the time constant . From the table we observe the significant of the time constant . In
fact, one time constant after the application of the source voltage Eout ( the capacitor
voltage has risen to 63.2% of its steady- state value, and the current i has decay to 36.8%
of its initial value. In similar way, we can observe that at t = 2, Eout = 86.5% of it steadystate value and i=13.5% of its initial value. At t = 5 the capacitor is essentially fully

charged (Eou = Ein) and i 0, so we say that steady-state conditions have been reached
after an elapse of time equal to 5 time constants.
If we replace the voltage source Ein with a short-circuit, after the capacitor has been
allowed to charge fully, i.e., after the steady state condition have been reached when this
Article1

Simulation of Simple RC Circuit

happen, the charge on the capacitor flows from one plate through the resistance R to other
plate , and thus discharge. Note that the discharge current is opposite direction from
which that it followed when the capacitor was charging. When the capacitor discharges
the voltage Eout decay toward zero volts and the current decays toward zero amperes.
The equation that describe decay of voltage and current are respectively:
t

E out Ein e

(2.9)

Ein
e
R

(2.10)

1.3 System Simulation: In preceding paragraph we discussed the significant of the time
constant . Now by using the following MATLAB code we simulate the step response of

thevalue
step
response
the%Script
system with (1.1)
three different
of time
constant of E out with

different time constant

ein=1; % input value
t=0:0.1:10;% vector of time points
% for each value of the time constant tau, build,
%solve and plot the step response
for tau= 0.5:5:15.5
eout=ein*(1-exp(-t/tau));
subplot(2,1,1), plot(t,ein),
subplot(2,1,2),plot(t,eout),
xlabel ('time[sec]'), ylabel(Amplitude [Volt]')
title(' the step response with different value
of time constant')
grid
% identify the three-step response
text(1.5,0.8,'tau=0.5');
text(2.5,0.3,'tau=5.5');
text(5.5,0.5,'tau=10.5');
text(8.5,0.3,'tau=10.5');
6
% ishold on
if ishold~=1,hold on, end % put hold on after
the first curve

Section 1.3

System Simulation

Figure 1.2 The rise of the capacitor voltages when =0.5, 5.5,10.5,15.5
The following MATLAB code (script 1.2) computes the step response of the output
current (the capacitor current) with different values of time constant .

% Script (1.2) the step response of Iout with

different time constant
ein=1; % input value
R=10000;
t=0:0.1:10;% vector of time points
% for each value of the time constant tau,
build, %solve and plot the step response
for tau= 0.5:5:15.5 iout=(ein./R)*exp(-t/tau);
subplot(2,1,1), plot(t,in),
ylabel('Amplitude [Volt]')
title('Step input')
grid

Articler1

Simulation of Simple RC Circuit

% continue script (1.2)...

subplot(2,1,2),
plot(t,iout)
ylabel('Amplitude [Amp]')
title('Step response of the current with
different value of tau')
grid
% ishold on
if ishold~=1,hold on, end % put hold on
after the first curve
end
hold off % turn hold of after the last curve

%Script
(1.3)
Charge
and
discharge
step
%responses of the circuit. This program plots
%the step responses of the circuit when the
%capacitor charges and discharges.
ein=1; % input value (step)
t=0:0.1:10;% vector of time points
R=10000; % the value of the restore
% for each value of the time constant tau, plot
the step response
for tau= 0.5:5:20;
eout_charge=ein*(1-exp(-t/tau));% the
voltage across the capacitor.
subplot(2,2,1)
plot(t,ein);
ylabel('Amplitude [Volt]')
Figure 1.3 The decay of the capacitor current when =0.5,5.5,10.5,15.5
title('step input')
subplot(2,2,2)
plot(t,eout_charge)
Section 1.3
System Simulation
ylabel('Amplitude[Volt]')
title('The step responses')
if ishold~=1,hold on,end
eout_dis=ein*exp(-t/tau);
subplot(2,2,3)
plot(t,ein),
xlabel('time
[sec]'),ylabel('Amplitude[Volt]')
title('step input')
subplot(2,2,4)
plot(t,eout_dis),
xlabel('time
[sec]'),ylabel('Amplitude[Volt]')
title('The step responses')
if ishold~=1, hold on
end,
grid
end
9
hold off % turn hold of after the last curve

10

Article 1

Simulation of Simple RC circuit

Figure 1.4 The rise and decay of the capacitor voltage

Figure 1.4 shows the rise of the capacitor voltage during charge and the decay of the
capacitor voltage during discharge with different values of the time constants. More
precisely we pick the upper left corner to plot the rise of the capacitor voltage during
charge, and we pick the lower left corner to plot the decay of voltage on the capacitor
during discharge. We use the red, brown, blue and green colors to indicate the outputs
when = 15.5, 10.5, 5.5 and 0.5, respectively. By comparing the above graphs we can
easily observe that the circuit has fast rise (or decay) response when the time constant is
small, and it has slow rise (or decay) response when the time constant is big. So one can
adjusts the speed of the output response of a RC circuit by choosing the value of R and C.
Similarly the following MATLAB code, Script 1.3, evaluate the step response of the
capacitor current when the capacitor is charging and discharging .

11

Section 1.3

System Simulation

%Script (1.4) Charge discharge step responses of

%the
circuit.
This
program
plot
the
step
responses %of the capacitor current when the
capacitor %charge and discharge.
ein=1; % input value (step)
t=0:0.1:10;% vector of time points
R=10000;
% for each value of the time constant tau plot
the step response
for tau= 0.5:5:20;
iout_charge=(ein/R)*exp(-t/tau);% the voltage
% accross the capacitor.
subplot(2,2,1)
plot(t,ein);
ylabel('Amplitude[Volt]')
title('step input')
subplot(2,2,2)
plot(t,iout_charge)
ylabel('Amplitude[Amp]')
title('The step responses')
if ishold~=1,hold on,end
iout_dis=-1*(iout_charge);
subplot(2,2,3)
plot(t,ein),
xlabel('time [sec]'),ylabel('Amplitude[Volt]')
title('step input')
subplot(2,2,4)
plot(t,iout_dis),
xlabel('time [sec]'),ylabel('Amplitude[Amp]')
title('The step responses')
if ishold~=1, hold on
end,
grid
end
hold off % turn hold of after the last curve

12

Article1

Simulation of Simple RC circuit

Figure 1.5 The decay of the capacitor current during charge and discharge
We pick the upper left corner to plot the decay of the capacitor current during charge, and
we pick the lower left corner to plot the decay of the capacitor current during discharge.
We use the red, brown, blue and green colors to indicate the outputs when = 0.5, 5.5,
10.5, and 15.5 respectively. By comparing the above graphs we can easily observe that
the circuit has fast decay response when the time constant is small, and slow decay
response when the time constant is big.

13

Section 1.3

System Simulation

Now we apply the MATLAB Control Tool Box to analyze the circuit. To do so we need
to build the simulator block by using SIMULINK.To simulate a dynamic system with
SIMULINK we need to calculate the transfer function of the system in exam. The
frequency-domain algebraic equation is obtained by applying the Laplace transformation
to each term in equation (2.2).

deout
eout ein )
dt

(2.11)

Where L is the Laplace operator. Solving for the ratio Eout/Ein.

sE out E out Ein

(2.12)

Note: it is understood that e is a function of t and E is a function of s even thought the t

and the (s) are not used. We will often drop the (t) and the (s) in writing long expressions.
The transfer function of the electric circuit is given by
E out
1

E out 1 s

(2.13)

Thanks to MATLAB object- oriented programming capabilities, we can easily obtained

the three step responses of the electrical system by applying the following MATLAB
code The code tf (num, den) creates transfer function with numerator num and
denominator den.

% script (1.5) this program plot the step

response of the RC circuit with three different
time constant
sys1=tf(1,[1,0.2])% 1/1+5s
sys2=tf(1,[1,0.1])%1/1+10s
sys3=tf(1,[1,0.667])%1/1+15s
sys=[sys1, sys2, sys3];
step (sys) %the step response
14

Articler2
Circuit

Simulation of Simple RC

Step response for different value of time constant

For: tau=5

For: tau=15

For tau = 10)

1
0.9
0.8
0.7

A
M
P
L
I
T
U
D
E

0.6
0.5
0.4
0.3
0.2
0.1
0

30

60

900

30

60

900

30

60

90

Time
(sec.)

Now we use the simulation block diagrams to observe the step response in scope. The
three step responses obtained by using three different time constant are shown below.
Figures 1.1d. 1.2d, and 1.3d show the step response of the system when the time constant
is equal to 5, 10 and 15 respectively.
1

The block diagram for the step response for different value of time constant
0.2s+1
Output Point 1
Transf er Fcn1

1
Step Input Point

0.1s+1
Output Point 2
Transf er Fcn2

Scope1

1
0.0667s+1
Transf er Fcn3

Output Point 3

15

Section 1.3

Step response for different values of time constant

1
To:
Output
Point3

A
m
p
L
i
t
u
d
e

System Simulation

0.5

0
1
To:
Output
0.5
Point2
0
1
To:
Output
Point1 0.5

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Time (sec.)
We can also obtain the three step responses of the system by connecting the block
diagram as shown below.
In the block diagram shown above we used the transfer function method to simulate our
system. We can obtain the same result by using equation (1.2d). For commodity we
rewrite it below
sE out E out Ein

16

Article 1

Simulation of Simple RC Circuit

The block diagram for the step response for different value of time constant tau
1

0.2

s
Integrator
Gain

Gain1

Output Point1

0.2
1

0.1
Step Input Point

s
Integrator1

Gain3

Output Point2

Scope

Gain2
0.1
1

0.667

s
Integrator2

Gain4

Output Point3

Gain5
0.667

Step Response with different value of time constant tau

1
To:
Output
Point2

To:
Output
Point3

To:
Output
Point1

0.5

A
m 0
p 1
L
i 0.5
t
u 0
d 1
e
0.5

We already
visualize the effect of change in the
0
0

10

15

20

25

30

35

Time (sec.)

17

Section 1.3

Section 1.3

System Simulation

System Simulation

We already visualize the effect of change in the time constant of the transfer function on
the system. Now We create the script file called meshpolt.m to demonstrate the effect of
change in the time constant value on the step response of the system in 3-dimennsion.

% script (1.6) Meshplot.m

% This Matlab codes creates a mesh plot showing the
effect of increasing the time constant
n=[1:1:15];
y=zeros(2000,1); %
for i=1:14
num=1/n(i);
den=[1 num];
t=[0:.01:19.99]';
y(:,i)=step(num,den,t);
i=i+1;
end
mesh(fliplr(y),[-120 30])
xlabel( time), ylabel( inputs),zlabel(amplitiud)
title ( Meash plots showing Step Responce for 14 time constants)

18

With more sampling points the mesh plots look like shown below
time
inputs

19