Announcement:

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Lecture 7

• Announcement
• Prof. Ming Wu will substitute for me on 2/10/05.
• Prof. Chang will be out of town on Wed.-Thurs.
• No office hour on Wed.
• Prof. Changs Friday office hour 3-4 will remain
  the same.
• OUTLINE
• Potential plots for resistive circuits
• Water models for voltage source, resistors
• The capacitor
• The inductor
• Reading
• Chapter 3, Chap 4.1

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Potential Plots for a Single Resistor and Two
Resistors in Series (Potential is Plotted
Vertically)

Arrows represent voltage drops
3
Potential Plot for Two Resistors in Parallel
Arrows represent voltage drops
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The Capacitor

• Two conductors (a,b) separated by an insulator
• difference in potential Vab
• gt equal opposite charge Q on conductors
• Q CVab
• where C is the capacitance of the structure,
• positive () charge is on the conductor at higher
  potential

(stored charge in terms of voltage)

• Parallel-plate capacitor
• area of the plates A (m2)
• separation between plates d (m)
• dielectric permittivity of insulator ? (F/m)
• gt capacitance

F
(F)
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Capacitor
Symbol Units Farads (Coulombs/Volt) Current-V
oltage relationship
C
or
C
Electrolytic (polarized) capacitor
(typical range of values 1 pF to 1 mF for
supercapa- citors up to a few F!)
ic
vc
If C (geometry) is unchanging, iC C dvC/dt
Note Q (vc) must be a continuous function of
time
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Voltage in Terms of Current
Uses Capacitors are used to store energy for
camera flashbulbs, in filters that separate
various frequency signals, and they appear as
undesired parasitic elements in circuits
where they usually degrade circuit performance
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Stored Energy
CAPACITORS STORE ELECTRIC ENERGY

• You might think the energy stored on a capacitor
  is QV CV2, which has the dimension of Joules.
  But during charging, the average voltage across
  the capacitor was only half the final value of V
  for a linear capacitor.

Example A 1 pF capacitance charged to 5 Volts
has ½(5V)2 (1pF) 12.5 pJ (A
5F supercapacitor charged to 5
volts stores 63 J if it discharged at a
constant rate in 1 ms energy is
discharged at a 63 kW rate!)
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A more rigorous derivation
ic
vc
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Example Current, Power Energy for a Capacitor
i(t)
v (V)

v(t)
10 mF
1
t (ms)
0
2
3
4
5
1
vc and q must be continuous functions of time
however, ic can be discontinuous.
i (mA)
t (ms)
0
2
3
4
5
1
Note In steady state (dc operation),
time derivatives are zero ? C is an open circuit
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p (W)
i(t)

v(t)
10 mF
t (ms)
0
2
3
4
5
1
w (J)
t (ms)
0
2
3
4
5
1
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Capacitors in Series
v1(t)
v2(t)
v(t)v1(t)v2(t)
C1
C2
i(t)
i(t)
Ceq

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Capacitive Voltage Divider

• Q Suppose the voltage applied across a series
  combination of capacitors is changed by Dv. How
  will this affect the voltage across each
  individual capacitor?

DQ1C1Dv1
Note that no net charge can can be introduced to
this node. Therefore, -DQ1DQ20
Q1DQ1
v1Dv1
C1
-Q1-DQ1
vDv

v2(t)Dv2
Q2DQ2
C2
-Q2-DQ2
DQ2C2Dv2
Note Capacitors in series have the same
incremental charge.
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Application Example MEMS Accelerometerto
deploy the airbag in a vehicle collision

• Capacitive MEMS position sensor used to measure
  acceleration (by measuring force on a proof mass)
  MEMS micro-
• electro-mechanical systems

g1
g2
FIXED OUTER PLATES
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Sensing the Differential Capacitance

• Begin with capacitances electrically discharged
• Fixed electrodes are then charged to Vs and Vs
• Movable electrode (proof mass) is then charged to
  Vo

Circuit model
Vs
C1
Vo
C2
Vs
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Practical Capacitors

• A capacitor can be constructed by interleaving
  the plates with two dielectric layers and rolling
  them up, to achieve a compact size.
• To achieve a small volume, a very thin dielectric
  with a high dielectric constant is desirable.
  However, dielectric materials break down and
  become conductors when the electric field (units
  V/cm) is too high.
• Real capacitors have maximum voltage ratings
• An engineering trade-off exists between compact
  size and high voltage rating

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Inductor
Symbol Units Henrys (Volts second /
Ampere) Current in terms of voltage
L
(typical range of values mH to 10 H)
iL
vL
Note iL must be a continuous function of time
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Stored Energy
INDUCTORS STORE MAGNETIC ENERGY

• Consider an inductor having an initial current
  i(t0) i0

)
(
)
(
)
(
t
i
t
v
t
p
t
ò


t
t
)
(
)
(
d
p
t
w
t
0
1
1
2
-

2
)
(
Li
Li
t
w
0
2
2
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Summary

• Capacitor
• v cannot change instantaneously
• i can change instantaneously
• Do not short-circuit a charged
• capacitor (-gt infinite current!)
• n cap.s in series
• n cap.s in parallel

• Inductor
• i cannot change instantaneously
• v can change instantaneously
• Do not open-circuit an inductor with current (-gt
  infinite voltage!)
• n ind.s in series
• n ind.s in parallel

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Summary 2

• Steady-state ? nothing is time varying.
• In steady state, an inductor behaves like a short
  circuit
• In steady state, a capacitor behaves like an open
  circuit

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First-Order Circuits

• A circuit that contains only sources, resistors
  and an inductor is called an RL circuit.
• A circuit that contains only sources, resistors
  and a capacitor is called an RC circuit.
• RL and RC circuits are called first-order
  circuits because their voltages and currents are
  described by first-order differential equations.

R
R
i
i

vs

vs
L
C
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Response

• The natural response of an RL or RC circuit is
  its behavior (i.e., current and voltage) when
  stored energy in the inductor or capacitor is
  released to the resistive part of the network
  (containing no independent sources).
• The step response of an RL or RC circuit is its
  behavior when a voltage or current source step is
  applied to the circuit, or immediately after a
  switch state is changed.

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Natural Response of an RL Circuit

• Consider the following circuit, for which the
  switch is closed for t lt 0, and then opened at t
  0
• Notation
• 0 is used to denote the time just prior to
  switching
• 0 is used to denote the time immediately after
  switching
• tlt0 the entire system is at steady-state and the
  inductor is ? like short circuit
• The current flowing in the inductor at t 0 is
  Io and V across is 0.

t 0
i
v
L
Ro
R
Io
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Solving for the Current (t ? 0)

• For t gt 0, the circuit reduces to
• Applying KVL to the LR circuit
• v(t)i(t)R
• At t0, iI0,
• At arbitrary tgt0, ii(t) and
• Solution

i
v
L
Ro
R
Io
I0e-(R/L)t