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Week 3b EECS 42, Spring 2005 New topics – energy storage elements Capacitors Inductors

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Page 1: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

New topics – energy storage elementsCapacitorsInductors

Page 2: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

Books on Reserve for EECS 42 in Engineering Library

“The Art of Electronics” by Horowitz and Hill (1st and 2nd

editions) -- A terrific source book on 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

“Electrical Engineering: Principles and Applications” by Hambley (3rd edition) – Backup copy of text for EECS 42

Page 3: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

The EECS 42 Supplementary Reader is now availableat Copy Central, 2483 Hearst Avenue (price: $12.99)

It contains selections from two textbooks thatwe will use when studying semiconductor devices:

Microelectronics: An Integrated Approach(by Roger Howe and Charles Sodini)

Digital Integrated Circuits: A Design Perspective(by Jan Rabaey et al.)

Reader

Page 4: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

The CapacitorTwo conductors (a,b) separated by an insulator:

difference in potential = Vab=> 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

Parallel-plate capacitor:• area of the plates = A (m2)• separation between plates = d (m)• dielectric permittivity of insulator = ε(F/m)

=> capacitance dAC ε

=

(stored charge in terms of voltage)

F(F)

Page 5: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

Symbol:

Units: Farads (Coulombs/Volt)

Current-Voltage relationship:

or

Note: Q (vc) must be a continuous function of time

+vc–

ic

dtdCv

dtdvC

dtdQi c

cc +==

C C

(typical range of values: 1 pF to 1 µF; for “supercapa-citors” up to a few F!)

+

Electrolytic (polarized)capacitor

C

If C (geometry) is unchanging, iC = dvC/dt

Page 6: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

Voltage in Terms of Current; Capacitor Uses

)0()(1)0()(1)(

)0()()(

00

0

c

t

c

t

cc

t

c

vdttiCC

QdttiC

tv

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

Page 7: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

Page 8: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

Schematic Symbol and Water Model for a Capacitor

Page 9: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

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 .221

21 CVQV =

Example: A 1 pF capacitance charged to 5 Volts has ½(5V)2 (1pF) = 12.5 pJ(A 5F supercapacitor charged to 5volts stores 63 J; if it discharged at aconstant rate in 1 ms energy isdischarged at a 63 kW rate!)

Stored EnergyCAPACITORS STORE ELECTRIC ENERGY

Page 10: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

∫=

==∫

=

=∫

=

==⋅=

Final

Initial

c

Final

Initial

Final

Initial

ccc

Vv

VvdQ vdt

tt

tt

dtdQVv

Vvvdt ivw

2CV212CV

21Vv

Vvdv Cvw InitialFinal

Final

Initial

cc −∫=

===

+vc–

ic

A more rigorous derivation

Page 11: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

Example: Current, Power & Energy for a Capacitor

dtdvCi =

–+

v(t) 10 µF

i(t)

t (µs)

v (V)

0 2 3 4 51

t (µs)0 2 3 4 51

1

i (µA) vc and q must be continuousfunctions of time; however,ic can be discontinuous.

)0()(1)(0

vdiC

tvt

+= ∫ ττ

Note: In “steady state”(dc operation), timederivatives are zero

C is an open circuit

Page 12: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

vip =

0 2 3 4 51

w (J)–+

v(t) 10 µF

i(t)

t (µs)0 2 3 4 51

p (W)

t (µs)

2

0 21 Cvpdw

t

∫ == τ

Page 13: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

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 sumdtdvCi eq=

Page 14: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

Capacitors in Series

i(t)C1

+ v1(t) –

i(t)

+

v(t)=v1(t)+v2(t)

–Ceq

C2

+ v2(t) –

21

111CCCeq

+=

Page 15: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

Capacitive Voltage DividerQ: 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+∆vC1

C2

+v2(t)+∆v2–

+v1+∆v1–+

Note that no net charge cancan be introduced to this node.Therefore, −∆Q1+∆Q2=0

Q1+∆Q1

-Q1−∆Q1

Q2+∆Q2

−Q2−∆Q2

∆Q1=C1∆v1

∆Q2=C2∆v2

2211 vCvC ∆=∆⇒

vCC

Cv ∆+

=∆21

12

Note: Capacitors in series have the same incremental charge.

Page 16: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

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

Page 17: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

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

constgg

gggg

gA

gA

gA

gA

VV

VCCCCV

CCCVV

s

o

ssso

12

12

12

21

21

21

21

21

1 )2(

−=

+−

=+

−=

+−

=+

+−=

εε

εεC1

C2

Vs

–Vs

Vo

Circuit model

Page 18: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

• 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

Page 19: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

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)

dtdiLtv L

L =)(

+

_

Note: In “steady state” (dc operation), timederivatives are zero L is a short circuit

Page 20: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

Symbol:

Units: Henrys (Volts • second / Ampere)

Current in terms of voltage:

Note: iL must be a continuous function of time

+vL–

iL

∫ +=

=

t

tLL

LL

tidvL

ti

dttvL

di

0

)()(1)(

)(1

0ττ

L

(typical range of values: µH to 10 H)

Page 21: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

Schematic Symbol and Water Model of an Inductor

Page 22: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

Stored Energy

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

20

2

21

21)(

)()(

)()()(

0

LiLitw

dptw

titvtp

t

t

−=

==

==

∫ ττ

INDUCTORS STORE MAGNETIC ENERGY

Page 23: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

Inductors in Series

21 LLLeq +=

( )dtdiL

dtdiLL

dtdiL

dtdiLv 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

dtdiLv eq=

Page 24: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

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

+=+=⇒

++⎥⎦

⎤⎢⎣

⎡+=

+++=+=

∫∫

τ

ττ

L2

+

v(t)

Leqi(t)

+

v(t)

)(10

0

tidvL

it

teq

+= ∫ τ

Page 25: New topics – energy storage elements Capacitors Inductors

Week 3bEECS 42, Spring 2005

Capacitor

v cannot change instantaneouslyi can change instantaneouslyDo not short-circuit a chargedcapacitor (-> infinite current!)

n cap.’s in series:

n cap.’s in parallel:

Inductor

i cannot change instantaneouslyv can change instantaneouslyDo 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

21 Cvw

dtdvCi

=

=

2

21 Liw

dtdiLv

=

=

=

=

=

=

n

i ieq

n

iieq

LL

LL

1

1

11