EECS 105 Fall 2003, Lecture 5
Lecture 5:
2nd Order Circuits in the Time Domain
Physics of Conduction
Prof. Niknejad
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Lecture Outline

Second order circuits:
–

Time Domain Response
Physics of Conduction
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Series LCR Step Response



Consider the transient response of the following
circuit when we apply a step at input
Without inductor, the cap charges with RC time
constant (EECS 40)
Where does the inductor come from?
–
–
Intentional inductor placed in series
Every physical loop has inductance! (parasitic)
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
LCR Step Response: L Small


We know the steady-state response is a constant
voltage of Vdd across capacitor (inductor is short,
cap is open)
For the case of zero inductance, we know solution
is of the following form:
v0 (t )
Vdd
v0 (t )  Vdd (1  e t / )
t

Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
LCR Circuit ODE

Apply KVL to derive governing dynamic equations:
vs (t )  vC (t )  vR (t )  vL (t )

Inductor and capacitor currents/voltages take the
dvC
di
form:
ii C
v L
C
L
dt
dt
d 2vC
d  dvC 
vL  L  C
  LC 2
dt  dt 
dt
dvC
vR  iR  RC
dt

We have the following 2nd order ODE:
dvC
d 2vC
vs (t )  vC (t )  RC
 LC 2
dt
dt
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Initial Conditions

For the solution of a second order circuit, we need
to specify to initial conditions (IC):
v0 (0)  vC (0)  0 V
i(0)  iL (0)  0 V


For t > 0, the source voltage is Vdd. We can now
solve for the following non-homogeneous equation
subject to above IC:
dvC
d 2vC
Vdd  vC (t )  RC
 LC 2
dt
dt
d
Steady state:
0
dt
Vdd  vC ()
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Guess Solution!

Let’s subtract out the steady-state solution:
vC (t )  Vdd  v(t )
d 2v
Vdd  Vdd  v(t )  RC
 LC 2
dt
dt
dv
d 2v
0  v(t )  RC
 LC 2
dt
dt
dv

Guess solution is of the following form: v(t )  Ae st
  

1  RCs  LCs 
0  Aest  RC sAe st  LC s 2 Aest
0  Aest
2
0  1  RCs  LCs 2
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Again We’re Back to Algebra

Our guess is valid if we can find values of “s” that
satisfy this equation:
1
Q
0  1  RCs  LCs



2
1  ( s )2  ( s )  0
2
2
1

0
The solutions are: s     2  1
This is the same equation we solved last lecture!
There we found three interesting cases:
 1

 1
 1

Department of EECS
Underdamped
Critically Damped
Overdamped
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
General Case

Solutions are real or complex conj depending on if
ζ > 1 or ζ < 1
 s1
1
2
s  (    1)  

 s2
vC (t )  Vdd  A exp s1t   B exp s2t 
vC (0)  Vdd  A  B  0
dvC (t )
i(0)  C
 0  As1 exp s1t   Bs 2 exp s2t  t 0  0
dt t 0
As1  Bs 2  0
A  B  Vdd
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Final Solution (General Case)

Solve for A and B: A   s1 A  Vdd
s2
s1
s1
 Vdd (1  )  Vdd
Vdd
 Vdd
s2
s2
A
B  Vdd  A 

s1
s1
s1
1
1
1
s2
s2
s2
s1
Vdd
 Vdd
s2
vC (t )  Vdd 
exp s1t  
exp s2t 
s1
s1
1
1
s2
s2




1  s1t s1 s2t  

 e  e 
vC (t )  Vdd 1 


s1 
s
2

 1 s

2


Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Overdamped Case

ζ > 1: Time constants are real and negative
 s1
s  (    1)    0

 s2
1
2
 1
Vdd  1
 2
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Critically Damped

ζ > 1: Time constants are real and equal
1
1
s  (   2  1)  


lim vC (t )  Vdd 1  e t /  tet / 
 1
 1
Vdd  1
 1
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Underdamped

Now the s values are complex conjugate
s1  a  jb
s2  a  jb
vC (t )  Vdd  A exp (a  jb)t   B exp (a  jb)t 
vC (t )  Vdd  e at  A exp  jbt   B exp  jbt 
s1
Vdd
 Vdd
Vdd
s2
*
A 


B
*
s1 s2  1 1  s1
1 *
s2
s2 s1


vC (t )  Vdd  e at A exp  jbt   A* exp  jbt 
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Underdamped (cont)

So we have:


vC (t )  Vdd  e at A exp  jbt   A* exp  jbt 
vC (t )  Vdd  e at 2 ReA exp  jbt 
vC (t )  Vdd  eat 2 A cost   
Vdd
A
s1
1
s2
Department of EECS
Vdd
 
s1
1
s2
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Underdamped Peaking

For ζ < 1, the step response overshoots:
 1
Vdd  1
  .5
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Extremely Underdamped
 1
Vdd  1
  .01
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Ohm’s Law

One of the first things we learn as EECS majors is:
V  I R

Is this trivial? Maybe what’s really going on is the
following:
V  f ( I )  f (0)  f ' (0) I  f ' ' (0) I 2 / 2  ...  f ' (0) I


In the above Taylor exansion, if the voltage is zero
for zero current, then this is generally valid
The range of validity (radius of convergence) is the
important question. It turns out to be VERY large!
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Ohm’s Law Revisited

In Physics we learned:
J  E

Is this also trivial? Well, it’s the same as Ohm’s
law, so the questions are related. For a rectangular
solid:
I
V
L
J  
V
I  RI
A
L
A


Isn’t it strange that current (velocity) is
proportional to Force?
Where does conductivity come from?
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Conductivity of a Gas


Electrical conduction is due to the motion of
positive and negative charges
For water with pH=7, the concentration of
hydrogen H+ ions (and OH-) is:
10 7 mole/L  10-10 mole/cm 3  10-10  6.02 10 23 cm 3  6 1013 cm 3


Typically, the concentration of charged carriers is
much smaller than the concentration of neutral
molecules
The motion of the charged carriers (electrons, ions,
molecules) gives rise to electrical conduction
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Collisions in Gas




At a temperate T, each charged carrier will move in
a random direction and velocity until it encounters
a neutral molecule or another charged carrier
Since the concentration of charged carriers is much
less than molecules, it will most likely encounter a
molecule
For a gas, the molecules are widely separated (~ 10
molecular diameters)
After colliding with the molecule, there is some
energy exchange and the charge carrier will come
out with a new velocity and new direction
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Memory Loss in Collisions

Schematically
Neutral
Molecule

Key Point: The initial velocity and direction is lost
(randomized) after a few collisions
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Application of Field


When we apply an electric field, during each “free
flight”, the carriers will gain a momentum of Eqt
Therefore, after t seconds, the momentum is given
by:
Mu  Eqt

If we take the average momentum of all particles at
any given time, we have:
Time from Last
1
Mu 
N
Number of
Carriers
Department of EECS
 Mu
j
 Eqt j 
Collision
j
Initial
Momentum
Before Collision
Momentum
Gained from
Field
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Random Things Sum to Zero!

When we sum over all the random velocities of the
particles, we are averaging over a large number of
random variables with zero mean, the average is
zero
1
Mu 
N
 Mu
1
Mu 
N
 Eqt j 
j
 Eqt
j
 Eq
Average
Time
Between
Collisions
j
 Eq
J  Nqu  Nq
 M
Department of EECS
j

2 
E  E
  Nq
M

University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Negative and Positive Carriers

Since current is contributed by positive and
negative charge carriers:






N
e


N
e



E
J  J  J  e



M
 M

 
 

N

N
 
2

  e   
 
M 
 M
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Conduction in Metals




High conductivity of metals is due to large
concentration of free electrons
These electrons are not attached to the solid but are
free to move about the solid
In metal sodium, each atom contributes a free
electron: N  2.5 1022 atoms/cm 3
From the measured value of conductivity (easy to
do), we can back calculate the mean free time:
 m 1.9 1017 9 1028 
14
 2

3

10
sec
22
 20
2.5 10 23 10 
Ne
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
A Deep Puzzle


This value of mean free time is surprisingly long
The mean velocity for an electron at room temperature is
about:

mv2 3
 kT v  3 107 cm / sec
2
2
At this speed, the electron travels v  3 10 7 cm

The molecular spacing between adjacent ions is only
3.8  10 8 cm

Why is it that the electron is on average zooming by 10
positively charged ions?
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Wave Nature of Electron




The free carrier can penetrate right through positively
charged host atoms!
Quantum mechanics explains this! (Take Physics 117A/B)
For a periodic arrangement of potential functions, the
electron does not scatter. The influence of the crystal is that
it will travel freely with an effective mass.
So why does it scatter at all?
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Scattering in Metals

At temperature T, the atoms are in random motion
and thus upset periodicity

Even at extremely low temperatures, the presence
of an impurity upsets periodicity
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Summary of Conduction
 
 

N

N
 
2

  e   
 
M 
 M





Conductivity determined by:
Density of free charge carriers (both positive and negative)
Charge of carrier (usually just e)
Effective mass of carrier
Mean relaxation time (time for memory loss … usually the
time between collisions)
–
This is determined by several mechanisms, e.g.:


Department of EECS
Scattering by impurities
Scattering due to vibrations in crystal
University of California, Berkeley
EECS 105 Fall 2003, Lecture 5
Prof. A. Niknejad
Reference

Edward Purcell, Electricity and Magnetism,
Berkeley Physics Course Volume 2 (2nd Edition),
pages 133-142.
Department of EECS
University of California, Berkeley