EECS 105 Fall 2001 Lecture 5
R. T. Howe
Lecture 5
• Last time:
– Bode plots for first-order transfer functions
(low-pass and high-pass filters)
– Rapid sketching techniques
• Today :
– Second order circuits: time domain analysis
Dept. of EECS
University of California at Berkeley
EECS 105 Fall 2001 Lecture 5
R. T. Howe
A Second Order System
+
vL
-
R
vC(t)
vS(t)
+
L
iL
iR
-
iC
C
Where does the inductor come from?
Do step response: vS(t) jumps to VDD at t = 0
Dept. of EECS
University of California at Berkeley
EECS 105 Fall 2001 Lecture 5
R. T. Howe
Step Response of L-R-C Circuit
Initial conditions: vC(t=0) = 0 V; iL (t=0) = 0 A
iL  iC
dvC
 1 t


   vL t dt  C
dt
 L 0
  dvC 

Inductor voltage: vL  VDD    C
 R  vC 
  dt 

Dept. of EECS
University of California at Berkeley
EECS 105 Fall 2001 Lecture 5
R. T. Howe
Second-Order Diff. Eqn. for vC(t)
Substituting for vL:
dvC
  dvC 

 1 t 
 R  vC dt   C
   VDD    C
dt
 L 0 
  dt 

Differentiating both sides:
d 2vC
  dvC 

 1 
 R  vC   C 2
  VDD    C
dt
 L 
  dt 

Dept. of EECS
University of California at Berkeley
EECS 105 Fall 2001 Lecture 5
R. T. Howe
Solving the 2nd Order ODE
2
d vC
dvC
LC 2  RC
 vC  VDD
dt
dt
Steady-state solution: vC,ss = VDD (t  )
Transient solution: vC,tr = ? … guess vC,tr = aest
and substitute:
 
 
LCs 2 aest  RCs aest  aest  0
Dept. of EECS
University of California at Berkeley
EECS 105 Fall 2001 Lecture 5
R. T. Howe
Characteristic Equation
s 2  L / R s  1 /( LC )   0
Use quadratic formula to find the roots:
2
 R
 R  1 
s1, 2        

 2L 
 2 L   LC 
Dept. of EECS
University of California at Berkeley
EECS 105 Fall 2001 Lecture 5
R. T. Howe
Three Cases for s1 and s2
 R
 1 
#1  2 L    LC 
 


… get two negative, real roots
 R
 1 

#2    
 2L 
 LC 
… get a single negative root
 R
 1 

#3    
 2L 
 LC 
… interesting case
Dept. of EECS
University of California at Berkeley
EECS 105 Fall 2001 Lecture 5
R. T. Howe
Underdamped Case
2
s1, 2
 R
 R  1 
 R
 1   R
       



j

 

 
 2L 
 2 L   LC 
 2L 
 LC   2 L 
vC (t )  VDD  a1e
a2e
( R / 2 L )t j (1 / LC ) ( R / 2 L ) 2 t
e
2

( R / 2 L )t  j (1 / LC ) ( R / 2 L ) 2 t
e

d
Form of solution …
Dept. of EECS
University of California at Berkeley
EECS 105 Fall 2001 Lecture 5
R. T. Howe
Qualitative Underdamped Waveform
vC(t) (V)
2.5
2
1.5
1
0.5
0
0
Dept. of EECS
100
200
300
400
500
t
University of California at Berkeley
EECS 105 Fall 2001 Lecture 5
R. T. Howe
Extreme Underdamped Case
Exponential decay time is set by   R /( 2 L)
Small R/L  decay takes a long time and
oscillation has a frequency that’s nearly 1 /( LC )
Number of cycles during “ringdown” is
(1 /  )
2L / R 2 L
N


(1 / 1 /( LC ) )
LC R C
What happens when R = 0 ?
Dept. of EECS
University of California at Berkeley