Transcript Lecture 13

Lecture 13
RC/RL Circuits, Time
Dependent Op Amp Circuits
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Circuits
The steps involved in solving
simple circuits containing dc
sources, resistances, and one
energy-storage element
(inductance or capacitance) are:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
1. Apply Kirchhoff’s current and voltage
laws to write the circuit equation.
2. If the equation contains integrals,
differentiate each term in the equation
to produce a pure differential equation.
3. Assume a solution of the form K1 +
K2est.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
4. Substitute the solution into the
differential equation to determine the
values of K1 and s . (Alternatively, we
can determine K1 by solving the circuit
5. Use the initial conditions to
determine the value of K2.
6. Write the final solution.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysis
Find i(t) and the voltage v(t)
i(t)= 0 for t < 0 since the switch is open prior to t = 0
Apply KVL around the loop:
 VS  i(t ) R  v(t )  0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysis
 VS  i (t ) R  v(t )  0
di(t )
i(t ) R  L
 VS
dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysis
i (t ) R  L
di (t )
 VS
dt
di(t )
i (t ) R  L
 VS
dt
Try i(t )  K1  K 2e st
Try i(t )  K1  K 2e st
RK 1  ( RK 2  sLK 2 )e  VS
st
VS 100V
RK 1  VS  K1 

 2A
R 50
L
RK 2  sLK 2  0  s 
R
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysis
i(t )  2  K 2e

 tR / L
i(0 )  0  2  K 2e  2  K 2  K 2  2
0
i(t )  2  2e
 tR / L
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysis
L
Define  
R
i(t )  2  2e
 t /
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysis
v(t )  VS  i (t ) R  100  50i (t )  100  50(2  2e t / )  100e t /
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RC and RL Circuits with General
Sources
First order differential
equation with constant
coefficients
di (t )
L
 Ri (t )  vt (t )
dt
vt (t )
L di (t )
 i (t ) 
R dt
R
dx(t )
Forcing

 x(t )  f (t ) function
dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RC and RL Circuits with General
Sources
The general solution consists
of two parts.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
The particular solution (also called the
forced response) is any expression that
satisfies the equation.
dx(t )

 x(t )  f (t )
dt
In order to have a solution that satisfies the
initial conditions, we must add the
complementary solution to the particular
solution.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
The homogeneous equation is obtained by
setting the forcing function to zero.
dx(t )

 x(t )  0
dt
The complementary solution (also called
the natural response) is obtained by
solving the homogeneous equation.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Step-by-Step Solution
Circuits containing a resistance, a source,
and an inductance (or a capacitance)
1. Write the circuit equation and reduce it to a
first-order differential equation.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
2. Find a particular solution. The details of this
step depend on the form of the forcing function.
3. Obtain the complete solution by adding the
particular solution to the complementary solution
xc=Ke-t/ which contains the arbitrary constant K.
4. Use initial conditions to find the value of K.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit
with a Sinusoidal Source
q (t )
Ri (t ) 
 2 sin( 200t )
C
t
1
Ri (t )   i (t )  vc (0) 2 sin( 200t )
C0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit
with a Sinusoidal Source
Take the derivative:
di (t ) 1
R
 i (t )  400 cos( 200t )
dt
C
di (t )
RC
 i (t )  400C cos( 200t )
dt
Try a particular solution :
i p (t )  A cos( 200t )  B sin( 200t )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit
with a Sinusoidal Source
di(t )
 i (t )  400 x10 6 cos(200 t )
dt
 A sin(200 t )  B cos(200 t )  A cos(200 t )  B sin(200 t )  400 x10 6 cos(200 t )
equating the coefficients for sin terms :
 A B  0  A  B
5 x10 3
equating the coefficients for cos terms :
B  A  400 x10 6
Solving :
A  200 x10 6  200 A
B  200 x10 6  200 A
i p (t )  200 cos(200 t )  200 sin(200 t )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit
with a Sinusoidal Source
The complementary solution is given by:
ic (t )  Ket /
  RC  (5k)(1F )  5ms
The complete solution is given by the sum of the
particular solution and the complementary solution:
i(t )  200 cos( 200t )  200 sin( 200t )  Ket / A
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit
with a Sinusoidal Source
Initial conditions:
2sin(0+) = 0
vC(0+) = 1V
vR(0+) + vC(0+) = 0  vR(0+) = -1V
i(0+) = vR/R = -1V/5000 = -200A
= 200cos(0)+200sin(0)+Ke0 = 200 + K  K= -400A
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit
with a Sinusoidal Source
i(t )  200 cos(200t )  200 sin( 200t )  400e
t /
A
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit
with an Exponential Source
What happens if we replace the source with 10e-t and the
capacitor is initially charged to vc(0)=5?
q (t )
Ri (t ) 
 10e t
C
t
1
Ri (t )   i (t )  vc (0) 10e t
C0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit
with an Exponential Source
Take the derivative:
di (t ) 1
t
R
 i (t )  10e
dt
C
di (t )
t
RC
 i (t )  10Ce
dt
Try a particular solution :
i p (t )  Ae t
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit
with an Exponential Source
 RCAe t  Ae t  10Ce t
A  RCA  10C
A(1  RC )  10C
10C
(10 )( 2 x10 6 )
A

 20 A
RC  1
2 1
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit
with an Exponential Source
The complementary solution is given by:
ic (t )  Ke t / 
  RC  (1M)( 2F )  2s
The complete solution is given by the sum of the
particular solution and the complementary solution:
t
i(t )  20e  Ke
t / 2
A
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit
with an Exponential Source
Initial conditions:
10e0+ = 10
vC(0+) = 5V
vR(0+) + vC(0+) = 10  vR(0+) = 5
i(0+) = vR/R = 5V/1M = 5A
= 20e0 + Ke0 = 20 + K  K= -15A
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit
with an Exponential Source
t
i(t )  20e  15e
t / 2
A
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Integrators and Differentiators
Integrators produce output voltages that are
proportional to the running time integral of
the input voltages. In a running time integral,
the upper limit of integration is t .
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
vin
iin 
R
t
t
q 1
1
vc    iin dt 
vin dt

C C0
RC 0
1
 v o  v c  0  v o  v c 
vin dt

RC 0
t
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
1
vo t   
RC
t

vin t dt
0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
If R = 10 k, C = 0.1F  RC = 0.1 ms
t
vo t   1000 vin t dt
0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
t
t
vo t   1000 vin t dt  1000 5dt  5000 t


0
0
t
1ms

 1000  5dt   5dt 
0

1ms




for 0  t  1ms
for 1ms  t  3ms
 10  5000 t
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Differentiator Circuit
dvin 0  vo
dvin
dq
iin 
C

 vo   RC
dt
dt
R
dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Differentiator Circuit
dvin
vo t    RC
dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.