week 3bee 42 and 100, fall 20051 new topics – energy storage elements capacitors inductors
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Week 3bEE 42 and 100, Fall 2005
1
New topics – energy storage elements Capacitors Inductors
Week 3bEE 42 and 100, Fall 2005
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Books on Reserve for EE 42 and 100 in the Bechtel Engineering Library
“The Art of Electronics” by Horowitz and Hill (2nd edition) -- A terrific source book on practical
electronics“Electrical Engineering Uncovered” by White and Doering (2nd edition) – Freshman intro to aspects of engineering and EE in particular”Newton’s Telecom Dictionary: The authoritative resource for Telecommunications” by Newton (“18th edition – he updates it annually) – A place to find definitions of all terms and acronyms connected with telecommunications. TK5102.N486 Shelved with
dictionaries to right of entry gate.
Week 3bEE 42 and 100, Fall 2005
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The Capacitor
Two conductors (a,b) separated by an insulator:difference in potential = Vab
=> equal & opposite charges Q on conductors
Q = CVab
where C is the capacitance of the structure, positive (+) charge is on the conductor at higher potential
Parallel-plate capacitor:• area of the plates = A (m2)• separation between plates = d (m) • dielectric permittivity of insulator = (F/m)
=> capacitance d
AC
(stored charge in terms of voltage)
F(F)
Vab
+
-
+Q
-Q
Week 3bEE 42 and 100, Fall 2005
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Symbol:
Units: Farads (Coulombs/Volt)
Current-Voltage relationship:
or
Note: Q (vc) must be a continuous function of time
+vc
–
ic
dt
dCv
dt
dvC
dt
dQi c
cc
C C
(typical range of values: 1 pF to 1 F; for “supercapa-citors” up to a few F!)
+
Electrolytic (polarized) capacitor
C
If C (geometry) is unchanging, iC = dvC/dt
Week 3bEE 42 and 100, Fall 2005
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Voltage in Terms of Current; Capacitor Uses
)0()(1)0(
)(1
)(
)0()()(
00
0
c
t
c
t
cc
t
c
vdttiCC
Qdtti
Ctv
QdttitQ
Uses: Capacitors are used to store energy for camera flashbulbs,in filters that separate various frequency signals, andthey appear as undesired “parasitic” elements in circuits wherethey usually degrade circuit performance
Week 3bEE 42 and 100, Fall 2005
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Week 3bEE 42 and 100, Fall 2005
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Schematic Symbol and Water Model for a Capacitor
Week 3bEE 42 and 100, Fall 2005
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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.
Thus, energy is .22
1
2
1CVQV
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!)
Stored EnergyCAPACITORS STORE ELECTRIC ENERGY
Week 3bEE 42 and 100, Fall 2005
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Final
Initial
c
Final
Initial
Final
Initial
ccc
Vv
VvdQ vdt
tt
tt
dt
dQVv
Vvvdt ivw
2CV2
12CV2
1Vv
Vvdv Cvw InitialFinal
Final
Initial
cc
+vc
–
ic
A more rigorous derivation
Week 3bEE 42 and 100, Fall 2005
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Example: Current, Power & Energy for a Capacitor
dt
dvCi
–+
v(t) 10 F
i(t)
t (s)
v (V)
0 2 3 4 51
t (s)0 2 3 4 51
1
i (A) vc and q must be continuousfunctions of time; however,ic can be discontinuous.
)0()(1
)(0
vdiC
tvt
Note: In “steady state”(dc operation), timederivatives are zero C is an open circuit
Week 3bEE 42 and 100, Fall 2005
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vip
0 2 3 4 51
w (J)–+
v(t) 10 F
i(t)
t (s)0 2 3 4 51
p (W)
t (s)
2
0 2
1Cvpdw
t
Week 3bEE 42 and 100, Fall 2005
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Capacitors in Parallel
21 CCCeq
i(t)
+
v(t)
–
C1 C2
i1(t) i2(t)
i(t)
+
v(t)
–
Ceq
Equivalent capacitance of capacitors in parallel is the sumdt
dvCi eq
Week 3bEE 42 and 100, Fall 2005
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Capacitors in Series
i(t)C1
+ v1(t) –
i(t)
+
v(t)=v1(t)+v2(t)
–
Ceq
C2
+ v2(t) –
21
111
CCCeq
Week 3bEE 42 and 100, Fall 2005
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Capacitive Voltage Divider
Q: Suppose the voltage applied across a series combination of capacitors is changed by v. How will this affect the voltage across each individual capacitor?
21 vvv
v+v
C1
C2
+ v2(t)+v2
–
+ v1+v1
–+–
Note that no net charge cancan be introduced to this node.Therefore, Q1+Q2=0
Q1+Q1
-Q1Q1
Q2+Q2
Q2Q2
Q1=C1v1
Q2=C2v2
2211 vCvC
vCC
Cv
21
12
Note: Capacitors in series have the same incremental charge.
Week 3bEE 42 and 100, Fall 2005
15
MEMS Airbag Deployment Accelerometer
Chip about 1 cm2 holding in themiddle an electromechanicalaccelerometer around which areelectronic test and calibrationcircuits (Analog Devices, Inc.) Hundreds of millions have beensold.
Airbag of car that crashed into the back of a stopped Mercedes. Within 0.3 seconds after deceleration the bag is supposed to be empty. Driver was not hurt in any way; chassis distortion meant that this car was written off.
Week 3bEE 42 and 100, Fall 2005
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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
FIXED OUTER PLATES
g1
g2
Week 3bEE 42 and 100, Fall 2005
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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
const
gg
gg
gg
gA
gA
gA
gA
V
V
VCC
CCV
CC
CVV
s
o
ssso
12
12
12
21
21
21
21
21
1 )2(
C1
C2
Vs
–Vs
Vo
Circuit model
Week 3bEE 42 and 100, Fall 2005
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Flexible conducting diaphragm
Sound waves
Cylindrical air-filled cavity
Conducting rigid cup
Condenser microphone Electret microphone
Electret: insulator(e.g., teflon) that wasbombarded with electronsthat remain imbeddedin it to “bias” thecondenser.Widely used in tele-phone handsets;available at RadioShack
G
X1Econst
xVout
Vout ~
x
x Econst
Econst
X1
GG
GVout
Vout ~ x Econst
Application: Condenser Microphone
Week 3bEE 42 and 100, Fall 2005
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• 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
Practical Capacitors
Week 3bEE 42 and 100, Fall 2005
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The Inductor
• An inductor is constructed by coiling a wire around some type of form.
• Current flowing through the coil creates a magnetic field and a magnetic flux that links the coil: LiL
• When the current changes, the magnetic flux changes a voltage across the coil is induced:
iLvL(t)
dt
diLtv L
L )(
+
_
Note: In “steady state” (dc operation), timederivatives are zero L is a short circuit
Week 3bEE 42 and 100, Fall 2005
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Symbol:
Units: Henrys (Volts • second / Ampere)
Current in terms of voltage:
Note: iL must be a continuous function of time
+vL
–
iL
t
t
LL
LL
tidvL
ti
dttvL
di
0
)()(1
)(
)(1
0
L
(typical range of values: H to 10 H)
Week 3bEE 42 and 100, Fall 2005
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Schematic Symbol and Water Model of an Inductor
Week 3bEE 42 and 100, Fall 2005
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Stored Energy
Consider an inductor having an initial current i(t0) = i0
20
2
2
1
2
1)(
)()(
)()()(
0
LiLitw
dptw
titvtp
t
t
INDUCTORS STORE MAGNETIC ENERGY
Week 3bEE 42 and 100, Fall 2005
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Inductors in Series
21 LLLeq
dt
diL
dt
diLL
dt
diL
dt
diLv eq 2121
v(t)
L1
+ v1(t) –
v(t)
+
v(t)=v1(t)+v2(t)
–
Leq
L2
+ v2(t) –
+–
+–
i(t) i(t)
Equivalent inductance of inductors in series is the sum
dt
diLv eq
Week 3bEE 42 and 100, Fall 2005
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L1i(t)
i2i1
Inductors in Parallel
)()()( with 111
)()(11
)(1
)(1
0201021
020121
022
011
21
0
00
tititiLLL
titidvLL
i
tidvL
tidvL
iii
eq
t
t
t
t
t
t
L2
+
v(t)
–
Leqi(t)
+
v(t)
–
)(1
0
0
tidvL
it
teq
Week 3bEE 42 and 100, Fall 2005
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Capacitor
v cannot change instantaneously
i can change instantaneously
Do not short-circuit a chargedcapacitor (-> 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 (-> infinite voltage!)
n ind.’s in series:
n ind.’s in parallel:
Summary
n
iieq
n
i ieq
CC
CC
1
1
11
2
2
1Cvw
dt
dvCi
2
2
1Liw
dt
diLv
n
i ieq
n
iieq
LL
LL
1
1
11
Week 3bEE 42 and 100, Fall 2005
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Transformer – Two Coupled Inductors
N1 turns N2 turns
+
-
v1
+
v2
-
|v2|/|v1| = N2/N1
Week 3bEE 42 and 100, Fall 2005
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Flowing wateror
Steam producedfrom
Fuel oil, Natural gas,
Coal,
Nuclear energy
Turbine Generator
Generating Plant Step-up Transformer(for efficient trans-mission at higher volt-age, lower current)
35,000 volts 130,000 volts
Transmission-line support towersStep-down Trans-former (for local distribution)
21,000 volts
Step-down trans-former mountedon power pole
120-240 volts
Customer
Simplified representation of the transmission and distribution systems that bring electricpower from a generating station to customers. In the generating station, a turbine driven by any of themeans indicated at the upper left is coupled to a generator, turning it to produce three-phase alternatingvoltages.
or
AC Power System
Week 3bEE 42 and 100, Fall 2005
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High-Voltage Direct-Current Power Transmission
http://www.worldbank.org/html/fpd/em/transmission/technology_abb.pdf
Highest voltage +/- 600 kV, in Brazil – brings 50 Hz power from12,600 MW Itaipu hydropower plant to 60 Hz network in Sao Paulo
Week 3bEE 42 and 100, Fall 2005
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Relative advantages of HVDC over HVAC power transmission
• Asynchronous interconnections (e.g., 50 Hz to 60 Hz system)
• Environmental – smaller footprint, can put in underground cables more economically, ...
• Economical -- cheapest solution for long distances, smaller loss on same size of conductor (skin effect), terminal equipment cheaper
• Power flow control (bi-directional on same set of lines)• Added benefits to the transmission (system stability,
power quality, etc.)
Week 3bEE 42 and 100, Fall 2005
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Quantity Variable Unit UnitSymbol
Typical Values
DefiningRelations
ImportantEquations
Symbol
Charge Q coulomb C 1aC to 1C magnitude of 6.242 × 1018 electron charges qe = -1.602x10-19 C
i = dq/dt
Current I ampere A 1A to 1kA 1A = 1C/s
Voltage V volt V 1V to 500kV 1V = 1N-m/C
Summary of Electrical Quantities
01
nodeN
nnI
01
loopN
nnV
Week 3bEE 42 and 100, Fall 2005
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Power P watt W 1W to 100MW
1W = 1J/s P = dU/dt; P=IV
Energy U joule J 1fJ to 1TJ 1J = 1N-m U = QV
Force F newton N 1N = 1kg-m/s2
Time t second s
Resistance R ohm 1 to 10M V = IR; P = V2/R = I2R
R
Capacitance C farad F 1fF to 5F Q = CV; i = C(dv/dt);U =
(1/2)CV2
C
Inductance L henry H 1H to 1H v = L(di/dt); U = (1/2)LI2
L
Summary of Electrical Quantities (concluded)