EE40 Lec 12 Transfer Function Bode Plots Filters Transfer Function
EE40 Lec 07 Capacitors and Inductors Prof. Nathan Cheungee40/fa09/lectures/Lec_07.pdf · Stored...
Transcript of EE40 Lec 07 Capacitors and Inductors Prof. Nathan Cheungee40/fa09/lectures/Lec_07.pdf · Stored...
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
Slide 2EE40 Fall 2009 Prof. Cheung
=> 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.
Slide 3EE40 Fall 2009 Prof. Cheung
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
+= ∫
Slide 4EE40 Fall 2009 Prof. Cheung
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
Slide 5EE40 Fall 2009 Prof. Cheung
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 −∫ ==
Slide 6EE40 Fall 2009 Prof. Cheung
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
Slide 7EE40 Fall 2009 Prof. Cheung
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
Slide 8EE40 Fall 2009 Prof. Cheung
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=
Slide 9EE40 Fall 2009 Prof. Cheung
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 +=
Slide 10EE40 Fall 2009 Prof. Cheung
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
Slide 11EE40 Fall 2009 Prof. Cheung
∆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.
Slide 12EE40 Fall 2009 Prof. Cheung
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
Slide 13EE40 Fall 2009 Prof. Cheung
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
Slide 14EE40 Fall 2009 Prof. Cheung
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
–
Slide 15EE40 Fall 2009 Prof. Cheung
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
Slide 16EE40 Fall 2009 Prof. Cheung
– 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=)(
Slide 17EE40 Fall 2009 Prof. Cheung
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.
Slide 18EE40 Fall 2009 Prof. Cheung
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 −=
Slide 19EE40 Fall 2009 Prof. Cheung
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 +=
Slide 20EE40 Fall 2009 Prof. Cheung
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
Slide 21EE40 Fall 2009 Prof. Cheung
)()()( 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
Slide 22EE40 Fall 2009 Prof. Cheung
n cap.’s in parallel: n ind.’s in parallel:∑=
=i
ieq CC1
∑=
=i ieq LL 1
11