EE40 Lec 07 Capacitors and Inductors Prof. Nathan Cheungee40/fa09/lectures/Lec_07.pdf · Stored Energy The energy stored on a capacitor is , w hich h as the dimension of Joules. During

EE40 Lec 07Capacitors and Inductors

Prof. Nathan Cheung

09/17/2009R di H bl Ch t 3Reading: Hambley Chapter 3

Slide 1EE40 Fall 2009 Prof. Cheung

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

difference in potential = Vab

l & it h Q d t=> equal & opposite charge Q on conductors

Q = CVab(stored charge in terms of voltage)

ab

where C is the capacitance of the structure, positive (+) charge is on the conductor at higher potential

Parallel-plate capacitor:• area of the plates = Ap• separation between plates = d• dielectric permittivity of insulator = ε

Aε


=> capacitancedAC ε

=

Capacitor

Symbol: or

C C

+

Electrolytic (polarized)

C

Units: Farads (Coulombs/Volt)

C C

(typical range of values: 1 pF to 1 µF; for “supercapa-

y (p )capacitor

These have high capacitance and cannotsupport voltage drops of the wrong polarity

Current-Voltage relationship:idd

(typical range of values: 1 pF to 1 µF; for supercapacitors” up to a few F!)

+vc

ic

dtdvC

dtdQi c

c ==

–To write this it is important to have use a passive convention, otherwise you need a minus sign.


Note: vc must be a continuous function of time since thecharge stored on each plate cannot change suddenly

Voltage in Terms of Current

)0()()(t

c QdttitQ += ∫)0(1

)0()()(0

t

c

Q

QdttitQ

∫

∫)0()(1)(

0cc C

QdttiC

tv += ∫

)0()(1 t

dtti +∫ )0()(0

cc vdttiC

+= ∫


Th t d it i QV hi h h th

Stored EnergyThe energy stored on a capacitor is QV, which has the dimension of Joules. During charging, you might think the average voltageDuring charging, you might think the average voltage across the capacitor was only half the final value of V for a capacitor.

2CV1)V1(Q =Th t d CV2

)V2

(Q =The stored energy

Example: A 1 pF capacitance charged to 5 Volts Stored energy = ½(5V)2 = 12 5 pJ


Stored energy = ½(5V)2 = 12.5 pJ

A more rigorous derivation

+ic

This derivation holds vc–independent of the circuit!

∫=

=∫ ∫=⋅=FinalFinal vv

dQvdtt dQvdtivw

Finalv

∫=

∫ ∫InitialInitial

cc

vvdQcvdt

t dtcvdt ivw

InitialV

2Cv212Cv

21v

vdv Cv InitialFinal

Final

cc −∫ ==


vInitial

Example: Current, Power & Energy for a Capacitor

–+

v(t) 10 µF

i(t)v (V)

)0()(1)(0

vdiC

tvt

+= ∫ ττ

–

t (µs)

1

dvCi =

(µ )0 2 3 4 51i (µA) vc must be a continuous

function of time; howeverdtt (µs)0 2 3 4 51

function of time; however,ic can be discontinuous.

(µ )0 2 3 4 51Note: In “steady state”(dc operation), timederivatives are zero


derivatives are zeroC is an open circuit

Current, Power & Energy for a Capacitor

vip =p (W)

t (µs)0 2 3 4 51–+

v(t) 10 µF

i(t)

w (J)1t

t (µs)

2

0 21 Cvpdw

t

∫ == τ


0 2 3 4 51t (µs)

Capacitors in Parallel

+i1(t) i2(t)

i(t) v(t)

–

C1 C2

+

21 CCCeq +=i(t)

+

v(t)Ceq

–

dtdvCi eq=


Equivalent capacitance of capacitors in parallel is the sumdt

Capacitors in Series( ) ( )

C

+ v1(t) –

+C

+ v2(t) –

i(t)C1 i(t) v(t)=v1(t)+v2(t)Ceq

C2

–

111 +=


21 CCCeq

+

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?voltage across each individual capacitor?

21 vvv ∆+∆=∆∆Q1=C1∆v1

+∆C1

+v1+∆v1–

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

Q1+∆Q1

-Q1−∆Q1v+∆v

+

+–

Q1 ∆Q1

Q2+∆Q2 2211 vCvC ∆=∆⇒C2 v2(t)+∆v2

–−Q2−∆Q2 vCC

Cv ∆+

=∆21

12


∆Q2=C2∆v221

Note: Capacitors in series have the same incremental charge.

Comment on Capacitive Voltage DividerQ: Can we always claim

vCC

Cv21

12 +=

CC 21 +

AC1

+v1–

Q1

-Q1

Answer:

Nov

+

+–

Q1

Q2

No. Divider formula is valid for total voltage only if

C2 v2–−Q2

g ycapacitors are unchargedwhen v is applied.


Application Example: MEMS Accelerometer

• Capacitive position sensor used to measure acceleration (by measuring force on a(by measuring force on a proof mass)

gg1

g2


FIXED OUTER PLATES

MEMS Accelerometer :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

V

Circuit model

VCCCCV

CCCVV ssso

211 )2(+−

=+

+−=C1

Vs

AAV

CCCC 2121

−

++εεVo

constgg

gggg

AAgg

VV

s

o 12

12

1221 −=

+−

=+

= εεC2

V


gg 21–Vs

Op-Amp Integrator

)0()(1)(0

C

t

INo vdttvRC

tv +−= ∫0

C

R inic– vC +

v

+

–

vo+

+

vn

vin

vp–

n

–


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 thematerials break down and become conductors when the electric field (units: V/cm) is too high.– Real capacitors have maximum voltage ratings

A i i t d ff i t b t t i d


– An engineering trade-off exists between compact size and high voltage rating

The Inductor• An inductor is constructed by coiling a wire around some

type of form.i vL(t)

+iL

vL(t)_

• Current flowing through the coil creates a magnetic field and a magnetic flux that links the coil: LiLWh th t h th ti fl h• When the current changes, the magnetic flux changes

a voltage across the coil is induced:diLtv L=)(


dtLtvL =)(Note: In “steady state” (dc operation), time

derivatives are zero L is a short circuit

Inductor

Symbol:L

Units: Henrys (Volts • second / Ampere)(typical range of values: µH to 10 H)

Current in terms of voltage:i1

(typical range of values: µH to 10 H)

+vL

iL=

t

LL dttvL

di )(1

To write this it is important to useh i fi i –

∫ +=t

tLL tidv

Lti

0

)()(1)( 0ττthe passive configuration.


Note: iL must be a continuous function of timebecause magnetic flux cannot change suddenly

0

Stored EnergyConsider an inductor having an initial current i(t0) = i0

)()()( titvtp ==

)()( dptwt

== ∫ ττ

11

)()(0

pt∫

20

2

21

21)( LiLitw −=


Inductors in Seriesdi+ v1(t) –

+

+ v2(t) – dtdiLv eq=

v(t)L1

v(t)+

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

L2+–

+–

i(t) i(t)

( ) didididi

–

( )dtdiL

dtdiLL

dtdiL

dtdiLv eq=+=+= 2121

21 LLLeq +=


Equivalent inductance of inductors in series is the sum

Inductors in Parallel

L1i(t)

i2i1

L2

+

v(t) Leqi(t)

+

v(t)L1 L2( )

–

eq

–1 t

∫)(1)(1

020121 tidvL

tidvL

iiitt

+++=+= ∫∫ ττ)(1

0

0

tidvL

iteq

+= ∫ τ

[ ])()(110201

21 00

titidvi

LLt

tt

++

+= ∫ τ [ ]

)()()(ith111

)()( 020121 0

iii

titidvLL

it

++

+ ∫ τ


)()()( with 0201021

tititiLLLeq

+=+=⇒

C it I d t

SummaryCapacitor Inductor

ddvCi =

ddiLv =

21 Cvw

dt

= 21 Liw

dt

=

v cannot change instantaneouslyi can change instantaneously

i cannot change instantaneouslyv can change instantaneously

2 2iw

i can change instantaneouslyDo not short-circuit a chargedcapacitor (-> infinite current!)

v can change instantaneouslyDo not open-circuit an inductor with current (-> infinite voltage!)

n 11 ∑n

n cap.’s in series: n ind.’s in series:∑=

=

n

i ieq CC 1

11

∑

∑=

=

ni

ieq LL1

11


n cap.’s in parallel: n ind.’s in parallel:∑=

=i

ieq CC1

∑=

=i ieq LL 1

11

Documents

EE40 Lec 07 Capacitors and Inductors Prof. Nathan Cheungee40/fa09/lectures/Lec_07.pdf · Stored Energy The energy stored on a capacitor is , w hich h as the dimension of Joules. During