Week 4 Handout
Remaining Schedule:
11/19 - Week 4:
Capacitors, Inductors, DC analysis,
(11/26 - Thanksgiving Holiday – No meeting)
12/3 - Week 5:
1st, 2nd order transient analysis
12/10 - Week 6:
AC Analysis (phasors), Power Balance
(12/17 - Final Exams Week – No meeting)
(12/24 - Christmas Holiday – No meeting)
(12/31 - New Year’s Holiday – No meeting)
1/7 Week 7:
Two-port networks, Transfer functions, s-domain, Filters
1/14Final Meeting: Review Practice Test 2 (S08), Review of all topics, and
any other questions you have
Today’s topic:
Capacitors, Inductors and DC analysis:
The Capacitor
The parallel plate capacitor shown below is the simplest physical form of the capacitor.
When a voltage exists across the capacitor, a uniform electric field exists between the
plates, given by:
E(t) =
vC (t )
d
Where d is the separation between the two plates. This electric field produces a charge
separation, with equal and opposite charges appearing on opposite plates, where the
magnitude of the electric charge on each plate, q(t) is related to the to the electric field
by:
ECE Ph.D. Qualifying Exam Preparation Circuit Fundamentals Review Group
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E(t) =
q(t)
A
Where ε is permittivity of the dielectric between the plates and A is the area of the plates.
These two equations together give:
 A 
q(t)    vC (t )
d 
Where the proportionality constant in the brackets will be defined as the capacitance C
of the capacitor:
C
A
d
Using this relationship in the above equation gives the defining relationship for the
capacitor:
q(t)  C  vC (t )
Differentiating this equation with respect to time gives:
dv (t )
dq(t)
C C
dt
dt
Which gives the I-V characteristics of the capacitor, since
iC (t )  C 
dq(t)
 i C (t ) :
dt
dvC (t )
dt
This relationship also gives the DC response of the capacitor, since the current is zero
when the voltage is constant, i.e:
A capacitor acts like an open circuit (iC = 0) when under DC excitation
Solving for the voltage by integrating both sides gives the capacitor voltage:
vC (t )  vC (0) 
1 t
 iC ( x)dx
C 0
where x is a dummy integration variable.
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The capacitor power is given by:
pC (t )  iC (t )vC (t )
This can also be written in the form:
pC (t )  Cv C (t )
dvC (t ) d  1

2
  Cv C (t )
dt
dt  2

Energy is the time rate of change of power, so the energy stored on the capacitor is:
wC (t ) 
1
2
Cv C (t )
2
The Inductor
The simplest physical form of an inductor is a coil of wire with N loops, as shown below:
When current flows through the wire, lines of flux concentrate along the axis. In a linear
magnetic medium, the flux is proportional to both the current and the number of turns in
the coil. The total flux is then given by:
 (t )  k  N  i L (t )
where k is a constant of proportionality. The magnetic flux links the coils with flux
linkage λ, given by:
 (t )  N (t )  [k  N 2 ]  i L (t )
The quantity in brackets is the inductance L of the coil:
L  k N2
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This gives the defining relationship of the inductor as:
 (t )  L  i L (t )
Differentiating this equation gives:
di (t )
d (t )
L L
dt
dt
Where the time rate of change of the flux linkage is given by Faraday’s law as the voltage
across the inductor, which gives the inductor I-V relationship:
v L (t )  L
di L (t )
dt
This relationship also gives the DC response of the inductor, since the voltage is zero
when the current is constant, i.e:
An inductor acts like an short circuit (vL = 0) when under DC excitation
Solving for the current by integrating both sides gives the inductor current:
i L ( t )  i L ( 0) 
1 t
 v L ( x )dx
L 0
where x is a dummy integration variable.
The inductor power is given by:
p L (t )  i L (t )v L (t )
This can also be written in the form:
 di (t )  d  1

2
p L (t )  [i L (t )] L L    Li L (t )
dt  dt  2


Energy is the time rate of change of power, so the energy stored in the inductor is:
w L (t ) 
1
2
Li L (t )
2
For both the inductor and the capacitor,
If the power p(t) is POSITIVE the element is ABSORBING POWER
If the power p(t) is NEGATIVE the element is DELIVERING POWER
ECE Ph.D. Qualifying Exam Preparation Circuit Fundamentals Review Group
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Dynamic Op-Amp Circuits
The capacitor adds two additional fundamental Op-Amp circuits to our circuit analysis
toolkit, as shown below (Also, see problem 2)
Equivalent Capacitance and Inductance
Capacitors add in parallel, and inductors add in series:
C EQ  C1  C 2    C N
(parallel connection)
1
1
1
1



C EQ C1 C 2
CN
(series connection)
LEQ  L1  L2    LN
(series connection)
1
1
1
1



LEQ L1 L2
LN
(parallel connection)
ECE Ph.D. Qualifying Exam Preparation Circuit Fundamentals Review Group
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Today’s Problems:
ECE Ph.D. Qualifying Exam Preparation Circuit Fundamentals Review Group
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ECE Ph.D. Qualifying Exam Preparation Circuit Fundamentals Review Group
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6.
7.
ECE Ph.D. Qualifying Exam Preparation Circuit Fundamentals Review Group
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