Electric current Electric current ‐ 2 · Electric current • Electric current = movement of...
Transcript of Electric current Electric current ‐ 2 · Electric current • Electric current = movement of...
Lecture 1Lecture 1
• Charge
• Current, voltage, Power,Current, voltage, Power,
• KCL, KVL
• Voltage and current sources
• ResistorsResistors
• Book: Chapter 1– Sections: 1.2/1.3/1.4/1.5/1.6/1.7
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Electric ChargesElectric Charges
• It all starts with electric charges – Symbol: qy q
– Unit: Coulomb [C]
E g an electron 1 602e 19 C– E.g. an electron = ‐1.602e‐19 C
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Electric currentElectric current
• Electric current = movement of charges
tdq )( Electric current tells us how much
dttdqti )()( =
ect c cu e t te s us o uccharges per unit time flow through an area
– Symbol: i
– Unit: Ampere [A]p [ ]
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Electric current 2Electric current ‐ 2
• Current = movement of charges reference direction!
baab ii −= baab
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Electric current: DC & ACElectric current: DC & AC
DC AC
DC+AC
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
VoltageVoltage
• Voltage = energy per unit charge, needed to move charge from one node to the other nodeg– Symbol = v
Units = Volt [V] = Joule/Coulomb– Units = Volt [V] = Joule/Coulomb
• Reference polarity
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Voltage 2Voltage ‐ 2
• Circuit element can generate (e.g. battery) or dissipate (e.g. resistor) energyp ( g ) gy
Element dissipates energy
Element generates energy
Note: assume vab and iab are both positive values !
energy energy
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Voltage 3Voltage ‐ 3
• Voltage “of a node”– Voltage is always across a circuit elementg y
ReferenceReference node
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PowerPower
• Power =– Units: Watt [W]
ivp ⋅=[ ]
– Symbol: p
Di i t d G t d?• Dissipated or Generated?– With this convention positive power = dissipated, negative power = generated
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EnergyEnergy
∫=2t
dttpE )(∫1t
Is the energy delivered or generated byIs the energy delivered or generated by a circuit element between t1 and t2
Units = Joule [J]
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Kirchoff’s Current Law (KCL)Kirchoff s Current Law (KCL)
• The net current entering a node = 0– Net current entering a node arriving currents g gare positive, leaving currents are negative
0iii 321 =+−
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Kirchoff’s Current Law (KCL)Kirchoff s Current Law (KCL)
• Alternative formulation:– sum of entering currents = sum of leaving currentsg g
∑∑ ∑∑ = leavingenteringii
231 iii =+
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Kirchoff’s Voltage Law (KVL)Kirchoff s Voltage Law (KVL)
• The algebraic sum of the voltages in a loop = 0
0vvvv 4321 =−+− 4321
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Kirchoff’s Voltage Law (KVL)Kirchoff s Voltage Law (KVL)
• The algebraic sum of the voltages in a loop = 0
0vvv + 0vvv 542 =+−
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Kirchoff’s Voltage Law (KVL)Kirchoff s Voltage Law (KVL)
• Alternative formulaton:– Sum of positive arrows = sum of negative arrowsp g
452 vvv =+
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Voltage sourceVoltage source
• Ideal, independent voltage source– Ideal: output voltage is independent of current p g pthrough the source
– Independent: output voltage does not depend onIndependent: output voltage does not depend on other voltages or currents in the network
Voltage is known, but current depends on load network
Direction of Is chosen such that source delivers power, but this is p ,not necessarily so.
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Voltage sourceVoltage source
• Ideal, independent voltage source– Regardless of Is, Vs maintains its valueg ,
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Current sourceCurrent source
• Ideal, independent current source– Ideal: output current is independent of voltage p p gacross the source
– Independent: output current does not depend onIndependent: output current does not depend on other voltages or currents in the network
Current is known, but voltage depends on load network
Direction of Vs chosen such that source delivers power, but this is p ,not necessarily so.
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Current sourceCurrent source
• Ideal, independent current source– Regardless of Vs, Is maintains its valueg ,
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Dependent sourcesDependent sources
• Voltage‐Controlled Voltage Source (VCVS)– = ideal voltage source whose output voltage g p gdepends on another voltage
A l i12V34 vAv ⋅= AV = voltage gain
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VCVS CCVS VCCS CCCSVCVS, CCVS, VCCS, CCCSVCVS VCCS
12V34 vAv ⋅= 12M34 vGi ⋅=
CCVS
AV = voltage gain [‐] GM = trans‐conductance [A/V]
CCVS CCCS
12M34 iRv ⋅= 12i34 iAi ⋅=RM = trans‐resistance [V/A] Ai = current gain [‐](c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Resistors and Ohm’s LawResistors and Ohm s Law
• Resistor– Linear relationship between current through the p gresistor and the voltage across the resistor.
– Resistance of a resistor ; units: Ohm [Ω]Resistance of a resistor ; units: Ohm [Ω]
– Always dissipates energy
Ohm’s Law:
RR iRv ⋅=
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Resistance and ConductanceResistance and Conductance
RR vGi ⋅=
R1G = [S] SiemensRG [ ]
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Resistor graphical representationResistor – graphical representation
RR iRv ⋅= RR
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Resistor and physical parametersResistor and physical parameters
ρ [ ]Ω• Resistivity of a material:
l
ρ [ ]mΩ
Αl
⋅ρ=R Copper: 1.72e‐8 ΩmGold: 2.27e‐8 ΩmΑ
ASilver: 1.63e‐8 ΩmTeflon: 1e19 ΩmC b 3 5 8 ΩA Carbon: 3.5e‐8 Ωm
Exercise: what is theExercise: what is the resistance of a copper wire of 1 meter length and
lof 1 meter length and diameter of 1mm?
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Resistors and powerResistors and power
iivp ⋅=
For resistors, this becomes
iRp 2⋅=
viRp
2
Rp =
Which is always positive if R is positive!(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
HomeworkHomework
• P1.63
• P1.67P1.67
• P1.79
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Lecture 2Lecture 2
• Resistors in series and parallel
• Voltage and current dividerVoltage and current divider
• Thévenin and Norton
• Superposition,
• Book chapter 2Book chapter 2– 2.1/2.2/2.3/2.4/2.5/2.6/2.7
Resistors in SeriesResistors in Series
N21t iiii ==== LKCL:
NN2211
N21t
iRiRiRvvvv+++=
+++=L
LKVL:
( )N21t RRRi +++= L
∑=N
it R
iv ∑
=1iti
Resistors in SeriesResistors in Series
∑N
t RvR ∑=
==1i
it
teq R
iR
Resistors in ParallelResistors in Parallel
N21t vvvv ==== LKVL:
+++
+++=
N21
N21t
vvviiii L
= Nt
11
ivKVL:
⎟⎞
⎜⎛
+++=N
N
2
2
1
1
111
RRRL
∑=
N
1i i
t
R1i
⎟⎟⎠
⎞⎜⎜⎝
⎛+++=
N21t R
1R1
R1v L
Resistors in ParallelResistors in Parallel
1v
∑== N
t
teq 1
1ivR
∑=1i iR
Resistors in ParallelResistors in Parallel
NN
∑∑==
===N
1ii
N
1i ieqeq G
R1
R1GParallel circuits easier to
work with conductances!1i1i ieq
Two Resistors in ParallelTwo Resistors in Parallel
21eq RR
RRR+
=Parallel circuits smallest one dominates!
21 RR +
Resistive voltage dividerResistive voltage divider
inin RR
vi+
=R21 RR +
21
2inout RR
Rvv+
=
2it Riv = 2inout Riv
Resistive current dividerResistive current divider
21RRi21
21inin RR
iv+
=1
inoutRii =
21inout RR +
inout R
vi =2R
Resistive current dividerResistive current divider
iniv =21
in GGv
+=
2inout GG
Gii =
2inout Gvi =21
inout GG +
Node voltage analysisNode‐voltage analysis
• Self‐study
Mesh current analysisMesh‐current analysis
• Self‐study
Thévenin equivalent circuitThévenin equivalent circuitAny two‐port network consisting of …can be replaced by this resistors, current and voltage sources… Thévenin equivalent circuit
RThevR
Thevv
Equivalent Circuit (1)Equivalent Circuit (1)
Thevv
openThev vv =⇒
Equivalent Circuit (2)Equivalent Circuit (2)
ThevR
Thevv
vshort
Thev
Thev iRv
=⇒Thev
openThevThev
vvR ==⇒shortshort
Thev iiR⇒
Thévenin equivalent circuitThévenin equivalent circuit
Norton equivalent circuitNorton equivalent circuitAny two‐port network consisting of …can be replaced by this resistors, current and voltage sources… Norton equivalent circuit
NortiNortR
Nort
Norton equivalent circuitNorton equivalent circuitAny two‐port network consisting of …can be replaced by this resistors, current and voltage sources… Norton equivalent circuit
ExampleExample
R v1
21
2open v
RRRv+
=1
1short R
vi =
ExampleExample
21
21
RRRR+1
1
Rv
21
21
RRRR+
21
21 RR
Rv+ 21 RR +
Another way to determine RAnother way to determine RThev
• Zero all independent sources– Zero voltage source = short circuitg
– Zero current source = open circuit
C l l t th i t b t th t• Calculate the resistance between the two terminals– Connect a voltage source Vtest, calculate Itest, impedance between two nodes is Vtest/Itesttest test
Another way to determine RAnother way to determine RThev
Rinin
21Thevin RRRR //==
Make v1 zero volt
Example 2Example‐2 Maximum power transferMaximum power transfer
ThevRLThev
ThevL RR
vi+
=
Thv LR 2
LThev
iRp =Thevv LR
( )2L
2Thev
LLL
RRRv
iRp
=
=
( )2LThev RR +
Maximum for ThevL RR =
SuperpositionSuperposition
• For linear networks,– The response (voltage and currents) of the p ( g )network for all sources is the sum of the individual responses. p
– Individual response = response of the network for one source only when other sources are zero‐edone source only when other sources are zero ed.
• Zero voltage source = short circuit
• Zero current source = open circuit• Zero current source = open circuit
• This is the superposition principle
Superposition: exampleSuperposition: example
Superposition: exampleSuperposition: example
21
111OUT RR
RVv+
=,21 RR +
21
11
21
211OUT RR
RVRR
RRIv+
++
=
21
2112OUT RR
RRIv+
=,
HomeworkHomework
• Resistive networks:– P2.8
– P2.11
P2 40– P2.40
– P2.87
Lecture 3Lecture 3
• Capacitance
• InductanceInductance
• Mutual Inductance
• Book chapter 3– 3.1/3.2/3.3/3.4/3.5/3.6/3.73.1/3.2/3.3/3.4/3.5/3.6/3.7
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Capacitors and capacitanceCapacitors and capacitance
q+
εq−q−
C CvCq ⋅=ε =permittivity of the dielectric
C = capacitance of the capacitorUnits: Coulomb/Volt = Farad [F]Typical values: fF (10‐15) mF (10‐3)Typical values: fF (10 5)…mF (10 3)
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Capacitors: physicalCapacitors: physicalA
d
A
ε d
AεParallel‐plate capacitor:
dAC ε
=
0r ε⋅ε=ε Relative permittivity values
120 10858 −⋅=ε . F/m
Air: 1Mica: 7Silicon dioxide: 3.9ε = relative permittivity [‐]Water: 78.5rε = relative permittivity [‐]
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Capacitors: exampleCapacitors: example
• Two metal plates of 10cm by 10cm
• Plastic foil in between (Epsr=3) with aPlastic foil in between (Epsr 3) with a thickness of 1mm
Wh i h i ?• What is the capacitance?
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Capacitors: currentCapacitors: current
q+Ci+ q−
Ci+ C
Ci+Ci+
q− q−q
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Capacitors: voltage and currentCapacitors: voltage and current
q+ Ci+q+ Ci+
dtdqi = dvdt
CvCq ⋅= dtdvCi C
C =
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Capacitors: exampleCapacitors: exampleCv
1V
dvCi C
t1us 3us 4us 5us
dtCi C
C =
Draw iC for C = 100nF
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Capacitors: power and energyCapacitors: power and energy
ddtdvvCivtp ⋅⋅=⋅=)(
∫=t
dttptE )()( Energy delivered to the ∫=t0
dttptE )()( capacitor from t0 till t
)()( tvC21dvvCdt
dtdvvCtE 2
vt
⋅⋅=⋅⋅=⋅⋅⋅= ∫∫ 2dt 0t0
∫∫
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Capacitors in seriesCapacitors in series
N21t
N21t
dtdv
dtdv
dtdv
dtdv
vvvv
+++=
+++=
K
K
N21t
Ci
Ci
Ci
Ci
dtdtdtdt
+++= KN21eq
C1
C1
C1
C1
CCCC
+++= KN21eq CCCC
11
∑=
+++= Neq 1
1111
1C∑=
+++1i iN21 CCCC
K
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Capacitors in parallelCapacitors in parallel
N21eq CCCC +++= Kq
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Capacitor: remarksCapacitor: remarks
dvCi C
dtCi C
C =
dv=⇒= C
C 0i0dt
dvfor DC, a capacitor is an open circuit
↑↑⇒ CC i
dtdv for high frequencies, a capacitor is a short circuitdt
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Inductors and inductanceInductors and inductance
• Current and magnetic flux:– Changing current creates a changing magnetic g g g g gfield
– Changing magnetic field induces a voltageChanging magnetic field induces a voltage
– Self‐inductance effect induced voltage across a coil due to its own changing currentcoil due to its own changing current
di L = (self‐)inductance
dtdiLv L
L =L = (self )inductanceUnits: Henry [H]Typical values: pH…mH
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Inductors: practicalInductors: practicalVarious approximate equations depending
N22
pp q p gon the shape:
⎥⎦⎤
⎢⎣⎡ +
πμ=
lD450201l
NrL22
0
.Example: Wheeler long-coil equation (l>D)
⎥⎦⎢⎣ l
μ = permeabilityr = radius (m)N = number of turns ( )N = number of turns (-)D= diameter (m)l = coil length (m)
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Inductors in seriesInductors in series
∑=N
LL ∑=
=1i
ieq LL
Assuming zero magnetic coupling ( l l )(mutual coupling) between the inductors
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Inductors in parallelInductors in parallel
1
∑= Neq 1
1L Assuming zero magnetic coupling ( l l )∑
=1i iL(mutual coupling) between the inductors
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Inductor: remarksInductor: remarks
diL L
dtLv L
L =
di=⇒= L
L 0v0dtdi
for DC, an inductor is a short circuit
↑↑⇒ LL v
dtdi
for high frequencies, an inductor is an open circuitdt
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Mutual inductanceMutual inductance
P t f th ti flPart of the magnetic flux of L1 flows through L2
Part of the magnetic flux f L fl th h Lof L2 flows through L1
didi = mutual inductanceMdtdiM
dtdiLv 2
121
11 += Units: Henry [H]Typical values: pH…mH
M
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Mutual inductance: dot conventionMutual inductance: dot convention
dididtdiM
dtdiLv 2
121
11 +=
dtdiL
dtdiMv 2
21
122 +=dtdt
dtdiM
dtdiLv 2
121
11 −=
diLdiMv
dtdt2
21
122 +−=dtdt 2122
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
HomeworkHomework
• Capacitance and inductance:– P3.13
– P3.53
(c) Prof. Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS
Lecture 4Lecture 4
• Transients– First order RC circuits, ,
– RL circuits,
Second order circuits– Second order circuits
• Book chapter 4– 4.1/4.2/4.3
Discharging a charged capacitor with a current source
SC
C IdvCti −==)(SI SC Idt
Cti )(S
iC V0tv == )( CviV
S
i
ICVt =dttitv
t
CC = ∫ )()( SI
tIV S
0CC ∫ )()(
ttCIV S
i ⋅−=
RC circuits: discharging a capacitor through a resistor
?)( =tvOUT
vdv0ii RC =+
V0t )(0
Rv
dtdvC CC =+
iC V0tv == )(
0vdvRC C =+ tC Ketv α=)(0v
dtRC C =+ C Ketv =)(
RC circuits: discharging a capacitor through a resistor (2)
0vdt
dvRC CC =+
dtt
C Ketv α=)(
V0t )(
C Ketv )(
iC V0tv == )(
KKetv0tVv0t
0iC
==→=
=→=α+
+
)( iVK =KKetv0t C ==→= )(
RC circuits: discharging a capacitor through a resistor (3)
dv 0vdt
dvRC CC =+
tC Ketv α=)(
V0t )(
C )(
iC V0tv == )(
0KeeRCK tt =+α αα
RC1
−=α
RC circuits: discharging a capacitor through a resistor (4)
τ−− ==t
RCt
eVeVtv )( == iiC eVeVtv )(
V)(tvC
iV
tiV370 ⋅.
RC=τt
RC circuits: discharging a capacitor through a resistor (5)
τ−==tiC eVtvti )()( ==R e
RRti )(
RV /)(tiR
RVi /
tRV370 i /. ⋅
RC=τt
Charging a capacitor with a current source
ddt
dvCtiI CCS == )(
tS dtIt ∫)(
0
SC
I
dtC
tv = ∫)()(tvC
iS Vt
CI
+⋅=
0Vi =
t
RC circuits: charging a capacitor through a resistor
?)( =tvOUT
RC
vVdvii =
00t )(
CSC
RvV
dtdvC −
=
00tvOUT == )(SC
C Vvdt
dvRC =+
tt21C eKKtv α+=)(
RC circuits: charging a capacitor through a resistor (2)
SCC Vv
dtdvRC =+
t21C eKKtv α+=)(
dt
St
21t
2 VeKKeKRC =++α ααS212
1α VKdRC
−=α S1 VK =and
RC circuits: charging a capacitor through a resistor (2)
021C eKK0tv α+== )( 21C eKK0tv +)(
0KK 21 =+
S2 VK −=
RC circuits: charging a capacitor through a resistor (2)
⎟⎞⎜⎛= τ−t
e1Vtv )( ⎟⎠⎞⎜
⎝⎛ −= τ
SC e1Vtv )(
SV)(tvout
SV
SV630 ⋅.
tRC=τ
t
RC circuits: charging a capacitor through a resistor (2)
τ−
=−
=tScS
R eRV
RtvVi )(
R RR
)(tiR
RVi /
RV370 i /. ⋅
RC=τt
RL circuitsRL circuits
diSL
L VRidtdiL =+
t21L eKKti α+=)(
tt VRKRKKL ααS
t21
t2 VeRKRKeKL =++α αα
LR
−=αRVK S
1 = RVK S
2 −=L R R
RL circuitsRL circuits
⎞⎛ −tV⎟⎠⎞⎜
⎝⎛ −= τ
tSL e1
RVti )(
τ−
=t
SOUT eVv
L )(tvOUT
R=τ
SV
SV370 ⋅.
RL /=τt
RLC circuitsRLC circuits
• Second‐order differential equations– (Damped) oscillations( p )
HomeworkHomework
• T4.1
Lecture 5Lecture 5
• Steady‐state Sinusoidal Analysis– Sinusoidal currents and voltagesg
– Phasors
Complex impedance– Complex impedance
– Circuit analysis with phasors and complex i dimpedances.
• Book chapter 5p– 5.1/5.2/5.3/5.4/5.6
Sinusoidal voltages and currentsSinusoidal voltages and currents
)cos()cos()( ϕ+π⋅=ϕ+ω⋅= ft2AtAtv)(tv
A2T π=ω
t
A f2π=ω
= period [s]= frequency [Hz]f
T
f1T =
frequency [Hz]= angular frequency [rad/s]= amplitude [v]
ωf
A
Power and RMSPower and RMS
v 2
Rvivp =⋅=
2T ⎤⎡
dtvT1
v11
T
0
2
T 2T ⎥⎥⎦
⎤
⎢⎢⎣
⎡⋅∫
∫∫ Rdt
Rv
T1dtp
T1P
0
00avg
⎥⎦⎢⎣=⋅=⋅= ∫∫
T1 V 2
∫ ⋅=0
2RMS dtv
T1V
RVP
2RMS
avg =
RMS for sinusoidal signalsRMS for sinusoidal signals
A11 TT
( )2
AdttAT1dtv
T1V
T
0
2T
0
2RMS =⋅ϕ+ω=⋅ ∫∫= )cos(
00
A2
R2AP
2
avg =
Complex numbersComplex numbers
ϕ⋅=+= jeAjyxSIm (imaginary part)g y p
y ϕ⋅= Ax )cos(ϕ⋅=ϕ
Ay )sin()(
Re (real part)
x
ϕ
⎟⎞
⎜⎛
+=
yyxA 22
x ⎟⎠⎞
⎜⎝⎛=ϕ
xyarctan
PhasorsPhasors
jyxeAVtAtv j11 +=⋅=⇔ϕ+ω⋅= ϕ)cos()(
is a complex number that represents the sinusoidal voltage
1V)(tv1represents the sinusoidal voltage
with amplitude A and phase ϕ)(tv1
Summation of sinusoidal voltages = vectorial sum of phasorsSummation of sinusoidal voltages = vectorial sum of phasors
Phasors represent sinusoidsPhasors represent sinusoids)(ts
t
y
x
PhasorsPhasors
)cos()( tAtv ϕ+ω j jyxeAV 1 +ϕ
)cos()()cos()(
222
111
tAtvtAtv
ϕ+ω⋅=ϕ+ω⋅=
22j
22
11j
11
jyxeAVjyxeAV
2
1
+=⋅=
+=⋅=ϕ
ϕ
Summation of sinusoidal voltages = vectorial sum of phasors
1V ( ) ( )212121 yyjxxVV +++=+
21 VV +
2V
Complex impedances: CComplex impedances: C
dtdvCi C
C =
)cos()( tAtvC ϕ+ω⋅=)/cos()sin()( 2tCAtCAtiC π−ϕ+ω⋅ω⋅−=ϕ+ω⋅ω⋅−=
( ) ϕπ−ϕπ−ϕ
ϕ
⋅ω⋅⋅=⋅⋅ω⋅−=⋅ω⋅−=
⋅=j2jj2j
C
jC
eCAjeeCAeCAIeAV
//C j
C Z1VLi (b t l ) l ti hiC
C
C ZCjI
=ω
= Linear (but complex) relationship between voltage and current (phasors)
Complex impedances: LComplex impedances: L
dtdiLv L
L =
)cos()( tAtiL ϕ+ω⋅=)/cos()sin()( 2tLAtLAtvL π−ϕ+ω⋅ω⋅−=ϕ+ω⋅ω⋅−=
( ) ϕπ−ϕπ−ϕ
ϕ
⋅ω⋅⋅=⋅⋅ω⋅−=⋅ω⋅−=
⋅=j2jj2j
L
jL
eLAjeeLAeLAVeAI
//L j
L ZLjVLi (b t l ) l ti hiL
L
L ZLjI
=ω= Linear (but complex) relationship between voltage and current (phasors)
Complex impedances: RComplex impedances: R
Rvi R
R =
)cos()( ϕ+ω⋅= tAtvR
)cos(/)( ϕ+ω⋅⋅−= tR1AtiR
ϕ
ϕ⋅=j
jR
R1AIeAV
/ ϕ⋅⋅= jR eR1AI /
RV Li d l l ti hi b tR
R
R ZRIV
== Linear and real relationship between voltage and current (phasors)
Circuit analysis with phasors and complex impedances
• KCL, KVL, series and parallel, Thévenin and Norton…are also valid for complex pimpedances!
Cj
Cj1ZC
−== Cj
Z1YC ω==
CCj ωω ZC
LjZL ω= Lj1YL ω
=jL Ljω
Circuit analysis with phasors and complex impedances
What is the amplitude and phase of the output voltage for C1=10nF, C2=20nF, R1=10kΩ and vIN a sinewave with frequency 1kHz and amplitude 1V?
dvdvC1vdCR INOUT2OUT2
⎟⎞
⎜⎛
dtdtC1
dtCR INOUT
1
22OUT
21 =⎟⎟⎠
⎜⎜⎝
++
Circuit analysis with phasors and complex impedances
• Assume vIN is sinusoidal
• Series and parallel still valid!Series and parallel still valid!
111 Cj
1RZω
+=
22
1
Cj1Z
j
ω=
2j
Circuit analysis with phasors and complex impedances
2INOUT
ZVV =
1
21INOUT
1CV
ZZVV
+
12121
1IN
RCC
CCj1
1CC
CV
+ω+
⋅+
⋅=
21 CC +
( )ϕjAVV O l l b !( )ϕ⋅= jINOUT AeVV Only one complex number!
Circuit analysis with phasors and complex impedances
What is the amplitude and phase of the output voltage for C1=10nF, C2=20nF, R1=10kΩ and vIN a sinewave with frequency 1kHz and amplitude 1V?
( ) 39670jININOUT e30750V11880j28360VV .... −=−⋅= ( ) ININOUT j
)sin( te12Av 3IN π=
).sin(. 39670te1230750Av 3OUT −π=
O t t i i ith lit d f 308 V d i iti lOutput is a sinewave with amplitude of 308mV and initial phase of ‐0.397 rad = ‐22.73 degrees.
Thévenin and NortonThévenin and Norton HomeworkHomework
• P5.44
• P5.46P5.46
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
1
Lecture 6
• Frequency response and bode‐plots– Fourier
– Transfer functions
– Bode plot
• Book chapter 6– 6 1 / 6 2 / 6 3 / 6 4 / 6 5– 6.1 / 6.2 / 6.3 / 6.4 / 6.5
Fourier
• Fourier series: – Periodic signal = sum of sinusoids
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
2
Transfer functions
121
2121
1
21
2
IN
OUT
RCC
CCj1
1CC
CZZ
ZV
V
+ω+
⋅+
=+
=
Example: low‐pass filter
11Cj1
V 1
C
IN
OUT
ffj1
1RCj1
1
Cj1R
CjV
V
+=
ω+=
ω+
ω=RC21fC π
=with
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
3
low‐pass filter: transfer function
1RC21fC π
=
( )1jH )( =ω
( )
C
IN
OUT
ffj1
1jHV
V
+=ω=
( )2Cff1j
/)(
+
⎟⎟⎠
⎞⎜⎜⎝
⎛−=ω∠
Cff0jH arctan)(
low‐pass filter: magnitude
( )2ff11jH
/)(
+=ω 7070
21jH .)( ==ω⇒= Cff( )Cff1 /+ 2
j )(C
Normalized frequency
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
4
low‐pass filter: phase
4jH /)( π−=ω∠⇒= Cff⎟⎟⎠
⎞⎜⎜⎝
⎛−=ω∠
ff0jH arctan)( j )(C⎟⎠
⎜⎝ Cf
j )(
Normalized frequency
Low‐pass filter: magnitude
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
5
Logarithms
( ) x10 x10 =log ( )
( ) x10
110 x10x
10 −=⎟⎠⎞
⎜⎝⎛=− loglog
( ) ( )ABAB ll ( ) ( )ABA 10B
10 loglog =
Bell and deciBell
⎟⎟⎞
⎜⎜⎛
= OUT10
PBell log
⎟⎟⎞
⎜⎜⎛
=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎠
⎜⎝
OUT2
OUT
IN
OUT10
IN10
RV10
PP10dB
P
/log
log
g
⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎠
⎜⎜⎝
=
IN
OUT10
IN2
IN10
VV20
RV10
log
/log
If ROUT = RIN
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
6
deciBell
( ))(log ω= jH20H 10dBdBH)( ωjH( ))(g j10dB
1100.11000.012
0dB+20dB‐20dB+40dB‐40dB
240.5
+6dB+12dB‐6dB+3dB‐3dB
41412 .=707021 ./ =
Low‐pass filter
( )( )( )2
CdB2ff110H
ff11jH /log
/)( +−=⇒
+=ω
( )Cff1 /+
dB3H7070HffdB0H1Hff
dBC
dBC
. −=⇒=⇒==⇒≈⇒<<
( )f20f20HffHff CdBCC log)log(/ −=⇒≈⇒>>
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
7
Bode plot ‐magnitude
Bode plot ‐magnitude
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8
Bode plot – magnitude: asymptotic
( ) 1jHω
=ω dB3H7070HffdB0H1Hff
dBC
dBC
. −=⇒=⇒==⇒≈⇒<<
Cj1ωω
+ ( )f20f20HffHff CdBCC
dBC
log)log(/ −=⇒≈⇒>>
πω
=2
f CC
Bode plot ‐ phase
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
9
Bode plot ‐ phase
Bode plot – phase: asymptotic
4Hff0Hff
C
C
/π−=∠⇒=≈∠⇒<<
⎟⎟⎠
⎞⎜⎜⎝
⎛−=ω∠
Cff0jH arctan)(
2Hff C /π−≈∠⇒>>
πω
=2
f CC
f
⎠⎝ Cf
10fC / Cf10
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
10
Bode plot – asymptotic
( )Cj1
1jHωω+
=ω/
( ) Cj1jH ωω+=ω /( ) KjH =ω
)log(K20
dBH dBH
f f
)(∠ jH )( ω∠ jH
dBH
f
)( ω∠ jH
CfCf
“pole” “zero”
‐20dB/dec
+20dB/dec
f f
)( ω∠ jH )( ω∠ jH
f
)( ω∠ jH
2/π−
2/π+
Bode plot – asymptotic
( )Cj
1jHωω
=ω/
( ) CjjH ωω=ω /( ) KjH =ω
)log(K20
dBH dBH
f f
)(∠ jH )( ω∠ jH
dBH
f
)( ω∠ jH
CfCf
‐20dB/dec+20dB/dec
f f
)( ω∠ jH )( ω∠ jH
f
)( ω∠ jH
2/π−
2/π+
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
11
Using Bode‐plots
( ) ( )21
jHjH20jH20HjHjHjH
)()(log)(log)()()(
ω⋅ω=ω=
ω⋅ω=ω
( ) ( )( ) ( )
dB2dB1
210110
211010dB
HHjH20jH20
jHjH20jH20H
,,
)(log)(log)()(log)(log
+=
ω+ω=
ω⋅ω=ω=
jH )(
dB2dB12
110dB
2
1
HHjHjH20H
jHjHjH
,,)()(log
)()()(
−=⎟⎟⎠
⎞⎜⎜⎝
⎛
ωω
=
ωω
=ω
Using Bode plots: example( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )ω⋅ω⋅ω⋅ω=⋅⋅ωω+⋅=
ωω+⋅ωω+ωω+
=ω
jHjHjHjHj11
j11j1A
j1j1j1AjH
43211C
3C2C
1C
///
///
( ) ( ) ( ) ( ) ( ) ( ) ( )ωω+ωω+
jjjjj1j1
j 43213C2C
1C //
dBH
f
1Cf
2Cf 3Cf
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
12
Using Bode plots: example( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( )
ωω+⋅ωω+ωω+
=ω
11j1j1
j1AjH3C2C
1C
///
( ) ( ) ( ) ( ) ( ) ( ) ( )ω⋅ω⋅ω⋅ω=ωω+
⋅ωω+
⋅ωω+⋅= jHjHjHjHj11
j11j1A 4321
3C2C1C //
/
dBH
)/log()log( 1C2C ff20A20 +
f
1Cf
2Cf 3Cf
Using Bode plots: example( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )ω⋅ω⋅ω⋅ω=⋅⋅ωω+⋅=
ωω+⋅ωω+ωω+
=ω
jHjHjHjHj11
j11j1A
j1j1j1AjH
43211C
3C2C
1C
///
///
( ) ( ) ( ) ( ) ( ) ( ) ( )ωω+ωω+
jjjjj1j1
j 43213C2C
1C //
)( ω∠ jH
1Cf 2Cf 3Cf
90+
f
90−
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
13
Using Bode plots: example( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )ω⋅ω⋅ω⋅ω=⋅⋅ωω+⋅=
ωω+⋅ωω+ωω+
=ω
jHjHjHjHj11
j11j1A
j1j1j1AjH
43211C
3C2C
1C
///
///
( ) ( ) ( ) ( ) ( ) ( ) ( )ωω+ωω+
jjjjj1j1
j 43213C2C
1C //
)( ω∠ jH
1Cf 2Cf 3Cf
90+
f
90−
‐20dB/dec, ‐40dB/dec( ) ( ) ( )
( ) ( )
ωω+⋅ωω+=ω
11j1j1
1jH2C1C //
( ) ( ) ( ) ( )ω⋅ω=ωω+
⋅ωω+
= jHjHj11
j11
212C1C //
dBH
f1Cf 2Cf
‐20dB/dec = ‐6dB/oct‐40dB/dec = ‐12dB/oct
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
14
Using Bode plots (1)
( ) ( ) ( )ω⋅ω=ωω+
⋅=ω jHjHj11AjH 21
1C/
dBH
f
)( ω∠ jH
dBH
f
)( ω∠ jH
1Cf1Cf 2dB1dBdB HHH ,, +=1dBH ,
2dBH ,
f
)( ω∠ jH
f
)( ω∠ jH
2/π− 2/π−
)()()( ω∠+ω∠=ω∠ jHjHjH 21
Using Bode plots (2)( ) ( ) ( ) ( )ω⋅ω=
ωω+⋅ωω+=
ωω+ωω+
=ω jHjHj11j1
j1j1jH 21
2C1C
2C
1C
//
//
dBH
f
)( ω∠ jH
dBH
f
)( ω∠ jH
2Cf2Cf 2dB1dBdB HHH ,, +=
2dBH ,
1Cf1Cf
f
)( ω∠ jH
f
)( ω∠ jH
2/π− 2/π−
)()()( ω∠+ω∠=ω∠ jHjHjH 21
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
15
Using Bode plots (3)( ) ( ) ( ) ( ) ( ) ( ) ( )ω⋅ω=
ωω+⋅
ωω+=
ωω+⋅ωω+=ω jHjH
j11
j11
j1j11jH 21
2C1C2C1C ////
dBH
f
)( ω∠ jH
dBH
f
)( ω∠ jH
2Cf2Cf 2dB1dBdB HHH ,, +=1Cf 1Cf
f
)( ω∠ jH
f
)( ω∠ jH
2/π− 2/π−
)()()( ω∠+ω∠=ω∠ jHjHjH 21
π−
Homework
• P6.25
• P6.52
• P6.65
• T6.3
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
Part 7 1
Lecture 7
• Amplifiers: specifications and external h t i ticharacteristics
• Book chapter 11– 11.1, 11.2, 11.4, 11.5, 11.6, 11.7, 11.10
Ideal voltage amplifier
INVOUT vAv ⋅=
AV = voltage gain (can be negative)
OUTv
v
t
INv
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
Part 7 2
Ideal voltage amplifier
12V34 vAv ⋅=
AV = voltage gain
Realistic amplifier model
INVOUT vGv ⋅=
AV = (internal) voltage gain (can be negative)RIN = input resistance (should be high)ROUT = output resistance (should be low)
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
Part 7 3
Input and output loading effect
LOUT RAv LINOUT RRAvOUTL
LV
IN
OUT
RRA
v +⋅=
RIN = should be large compared to RS
ROUT = should be small compared to RL
OUTL
L
SIN
INV
S
OUT
RRRRA
v +⋅
+⋅=
Voltage gain
ROUTL
LV
IN
OUTV RR
RAv
vG+
⋅==
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
Part 7 4
Current gain
RiL
INV
IN
OUTI R
RGi
iG ⋅==
RIN = should be large compared to RS
ROUT = should be small compared to RL
Power gain
RP ( )L
IN2VIV
IN
OUTP R
RGGGP
PG ⋅=⋅==
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
Part 7 5
Cascading amplifiers
2V1V2OUTL
L2V
1OUT2IN
2IN1V
1IN
2OUTV GG
ZRRA
ZZZ
Av
vG ,,,
,,
, ⋅=+
⋅⋅+
⋅==2OUTL1OUT2IN1IN ZRZZv ,,,, ++
2dBV1dBVdBV GGG ,,,,, +=
Voltage‐amplifier model
AV = (internal) voltage gainRIN = input resistance (HIGH)ROUT = output resistance (LOW)
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
Part 7 6
Current‐amplifier model
Ai = (internal) current gainR = input resistance (LOW)RIN = input resistance (LOW)ROUT = output resistance (HIGH)
Transconductance‐amplifier model
Gm = transconductance [S]RIN = input resistance (HIGH)ROUT = output resistance (HIGH)
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
Part 7 7
Transresistance‐Amplifier model
Rm = transresistance [Ohm]RIN = input resistance (LOW)ROUT = output resistance (LOW)
High or low input impedance ?
Electrodes = high impedance source high input impedance neededhigh input impedance needed
Current measurements = low input impedance needed
OUTL
L
SIN
INV
S
OUT
RRR
RRRA
vv
+⋅
+⋅=
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
Part 7 8
High or low output impedance
Driving a loudspeaker = low output impedance needed ( l d )(voltage drive)
Driving a laser‐diode = high output impedance needed (current drive)
Specific input/output impedance
Transmissionline needs to be terminated by the d flcorrect resistor to avoid reflections
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
Part 7 9
Frequency responseComplex voltage gain = ratio of phasors = transfer function
IN
OUTV V
VjG =ω)(
dBVG ,
High‐frequency
f
low‐frequency or midband region
region
Example
Calculate the transfer function vOUT/vIN and draw the Bode plot LL
LL CRj1
RZω+
=
OUTLC
CRR1
⋅=ω
LOUT
LV
IN
OUTV ZR
ZAV
VjG+
⋅==ω)(
C0v
LOUTL
OUTLLOUT
LV j1
1AC
RRRRj1
1RR
RAωω+
⋅=
+ω+
⋅+
⋅=/
LOUTL
CRR +
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
Part 7 10
DC‐ or AC‐ coupleddBVG ,
High‐frequency region
DC‐coupled amplifier
f
low‐frequency or midband region
region
dBVG ,
Low
f
midband region
High‐frequency region
Low‐frequency region
AC‐coupled amplifier
Example
Calculate the transfer function vOUT/vIN and draw the Bode plot
ZV
( )OUTLC
CLV
LOUT
LV
IN
OUTV
RRCj1CRjA
ZRZA
VVjG
+ω+ω
⋅=
+⋅==ω)(
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
Part 7 11
Narrowband amplifier
Wideband amplifier
dBVG ,
High‐frequency Low‐
frequency
dBVG ,
f
midband region
regionfrequency region
f
Narrowband or Tuned amplifier
Amplifier limits: distortion
OUTvINVOUT vAv ⋅=
t
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
Part 7 12
Amplifier limits: distortion
OUTv
INVOUT vAv ⋅=Maximum output voltage is limited
INv
Amplifier limits: noise
OUTv
INVOUT vAv ⋅=
INv
Noise at the output
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
Part 7 13
Funny circuitsRi
RR
R
ININ i
vZ =
RvAvi INvIN
R−
=V
IN A11RZ−
=
What happens if AV=1, ‐1, ‐10, … ?
Homework
• P11.11
• P11.12
• P11.51
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 1
Lecture 8: OpAmps
• Operational Amplifiers– Amplifiers
– Filters
– Schmitt‐Trigger
• Book chapter 14– 14 1 14 2 14 3 14 4 14 5 14 9– 14.1, 14.2, 14.3, 14.4, 14.5, 14.9
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 2
Ideal OpAmp
vNon‐inverting input
( )nINpINOLOUT vvAv ,, −⋅=
OUTvpINv ,
nINv ,
Non‐inverting input
inverting inputoutput
AOL is very large (105…107)
Ideal OpAmp
OUTvpINv ,
( )nINpINOL vvA ,, −⋅OUTvnINv ,
AOL is very large (105…107)Infinitely high input impedanceZero output impedance
OUTv
( )nINpIN vv ,, −
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 3
Non‐Inverting amplifier
( )nINpINOLOUT vvAv ,, −⋅=OpAmp:
Resistors:
21
1OUTnIN RR
Rvv+
⋅=,
1OUTOLINOLOUT RR
RvAvAv+
−=AOL is very large (105…107)Infinitely high input impedance
21 RR +
1
2
1
21
1OL
21
21
IN
OUT
RR1
RRR
RA
RRRR
vv
+=+
≈+
++
=
Infinitely high input impedanceZero output impedance
Feed‐back
OLAOUTv
pININ vv ,= +
( )A
21
1
RRR+
nINv ,
‐
R( )
⎟⎟⎠
⎞⎜⎜⎝
⎛+
−=
−=
21
1OUTINOL
nINpINOLOUT
RRRvvA
vvAv ,,
1
2
IN
OUTCL R
R1v
vA +==
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 4
Open‐loop gain, closed‐loop gain
OLAOUTv
pININ vv ,= +G
in out+ ε
21
1
RRR+
OL
nINv ,
‐
H
‐
( )
GH1G
inout
outHinGout
+=
⋅−=εε⋅== open‐loop gain
= loop‐gain
= closed‐loop gainGH1
G+
G
GH
Feed‐back
Gin out+
Gout ε⋅=ε
H
‐ ( )
GH1G
inout
outHin
+=
⋅−=ε
H1
inout1GH ≈⇒>>
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 5
Feed‐back
Gin out+ ε
H
‐ ( )
⎟⎠⎞
⎜⎝⎛
+=
⋅−=ε
GH11in
outHin
01GH ≈ε⇒>>
Virtual short‐circuit
nINpIN
nINpIN
vv0vv
,,
,,
≈⇒
≈−=ε
Virtual short circuit
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 6
Voltage follower
0R2 =
Rv v1
2
IN
OUT
RR1
vv
+≈ 1v
vIN
OUT ≈
Ideal voltage amplifier with gain=1Infinitely high input impedanceZero output impedance
Inverting amplifier
Simplified analysis:2Ri
2
OUT
1
IN2R1R
nINpIN
Rv
Rvii
0vv−
=⇒=
=≈ ,,
1Ri
1
2
IN
OUT
RR
vv
−=
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 7
Virtual ground
0vvi 0vv nINpIN =≈ ,,
1Ri
2Ri
1IN1
IN1R RZ
Rvi ≈⇒≈
Inverting amplifier
Exact analysis:
RR
2Ri
nINOLOUT
21
1OUT
21
2INnIN
vAvRR
RvRR
Rvv
,
,
−=+
++
=
RR
1Ri
1
2
1OL
21
2
IN
OUT
RR
RA
RRR
vv
−≈+
+−
=
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 8
Inverting amplifier
21
1OUT
21
2INnIN RR
RvRR
Rvv , ++
+=
21
2
RRR+
in out+
+
ε
nINOLOUT vAv ,−=
OLA−
+
21
1
RRR+
Inverting amplifier
( )+
−−+
==ε21
1OUT
21
2INnIN RR
RvRR
Rvv ,
21
2
RRR+
in out−+ ε
ε==− OLnINOLOUT AvAv ,
OLAout
1−‐
21
1
RRR+
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 9
Summation amplifier
RAi1IN
1R Rvi =
Virtual ground:
2IN2R R
vi = OUTRA R
vi −=
1Ri
RAi1R 2R
Infinitely high input impedance:
RA2R1R iii =+
AR
vvv
2RiA
OUT
2
2IN
1
1IN
Rv
Rv
Rv
−=+
⎟⎟⎠
⎞⎜⎜⎝
⎛+−= 2IN
2
A1IN
1
AOUT v
RRv
RRv
IntegratorCi
vi INR = dt
dvCi OUTC −=
Ri
RiR dtC
INOUT
RC vRC1
dtdvii −=⇒=
0tOUT
t
0INOUT vdtv
RC1v =+−= ∫ ,
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 10
DifferentiatorRi
vi OUTR −= dt
dvCi INC =
Ci
RiR dtC
dtdvRCvii IN
OUTRC −=⇒=
Active filters
1
2
1
2
IN
OUT
ZZ
RR
vv
−⇒−=
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 11
Example: Low‐pass filter
R2=10kR1=1kC2 100 FC2=100nFDraw the Bode‐plot
OpAmp: gain, bandwidth and GBW
dBOLA ,0AjA )(
f
dB0A ,
C
0OL
j1jA
ωω
+=ω)(
C0A2GBW ω⋅⋅π=
BWfC =GBWfA C0 =⋅
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 12
Feed‐back
)( ωjAOL
OUTvINv +
21
1
RRRH+
=
H
‐
21
C
0OL
j1
AjA
ωω
+=ω)(
H1
GH1G
vv
IN
OUT ≈+
= only if GH is large enough!
Closed‐loop Transfer function
)( ωjAOL
OUTvINv +
‐ C
0OL
j1
AjAG
ωω
+=ω= )(
21
1
RRRH+
=H
C
0
1A
G /
( ) C00
0
C0
0
C
0
C
IN
OUT
HA1j11
HA1A
jHA1A
Hj1A1
j1GH1
Gv
v
ω+ω+⋅
+=
ωω++=
ωω++
ωω+=
+=
/
//
/
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 13
Trading gain for bandwidth
)( ωjAOL
OUTvINv +
‐ 1Av
H
( ) C00
0
IN
OUT
HA1j11
HA1A
vv
ω+ω+⋅
+=
/
1A0 ( )HA1f
Low‐frequency gain: bandwidth
( ) H1
HA1A
0
0 ≈+
( )HA1f 0C +
( ) C00C0
0 fAHA1fHA1
AGBW =+⋅+
=GBW is the same!
Trading gain for bandwidth
)( ωjAOL
OUTvINv +
‐ C
0OL
j1
AjA
ωω
+=ω)(
H ( ) C00
0
IN
OUT
HA1j11
HA1A
vv
ω+ω+⋅
+=
/
0Aopen‐loop gain
fBWfC =
GBWfA C0 =⋅
H1
HA1A
0
0 ≈+
( )HA1f 0C +
open loop gain
closed‐loop gain
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 14
Voltage limits
OUTvDDV
pINv ,
nINv ,( )nINpIN vv ,, −
DDV
SSV
Positive feedback
OUTvDDV
( )nINpIN vv ,, −
DD
0v1v
OUT
IN
==
SSV
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 15
Positive feedback
0v
DDOUTIN21
1DDpINDDOUT Vvv
RRRVvVv =⇒>+
=⇒= ,
SSOUTIN21
1SSpINSSOUT Vvv
RRRVvVv =⇒<+
=⇒= ,
0v IN =
Hysteresis / Schmitt‐Trigger
OUTv21
1DDpIN RR
RVv+
=,
INv
DDV21 RR +
SSV
21
1SSpIN RR
RVv+
=,
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 16
Example: temp controlINv
t
OUTv
t
Temp control with hysteresisINv
t
OUTv
t
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8/2 17
Homework
• P14.32
• P14.33
• P14.78
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
1
Lecture 9
• Diodes (Book Chapter 10)– 10.1, 10.2, 10.3, 10.6, 10.7, 10.8
Semiconductor basics
‐‐
‐ ‐N‐type doped Silicium
free charges = electrons (negative)
‐‐
‐
‐
‐
‐‐
‐ ‐‐‐
‐‐
‐
‐
P‐type doped Siliciumfree charges = holes (positive)
+++
+ +
++
+ ++
++
+
+
++ ++
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
2
junction
‐‐
‐ ‐+++
+ +‐
‐
‐
‐ ‐
‐ ‐‐‐
‐‐
‐
++
+
+ +
+ +++
++
+
Electrons and holes will recombine
Diode basics
‐‐
‐ ‐+++
+ Electric ‐
‐
‐
‐ ‐
‐‐‐
‐+
+
+
+ +
+ ++
+
+
Depleted region (no free charges)Electric
l
field
potential
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
3
Diode forward bias
‐‐
‐ ‐+++
+ Electric V+ V‐‐
‐
‐
‐ ‐
‐‐‐
‐+
+
+
+ +
+ ++
+
+
Electric l
fieldV+ V‐
current
potential
Diode reverse bias
‐‐
‐ ‐+++
+ Electric V‐ V+‐
‐
‐
‐ ‐
‐‐‐
‐+
+
+
+ +
+ ++
+
+
Electric l
fieldV‐ V+
current
potential
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
4
Diode symbol
cathodeanode
‐‐‐
‐
‐
‐
‐‐
‐‐‐
‐
‐+++
+
+
+
++
+ ++
+
+
Electric
Electric field
Electric potential
Diode curve
Di
100uA … 1A
Forward bias regionReverse breakdown
Dv0.5V…0.8V
Reverse bias region
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
5
Simplified diode curve
Dv∝SI area
DiTnV
SD eIi ≈ mV25q
kTVT ≈=
n depends on doping profile (1…2)
Dv
≈Di very small (nA)k=1.38e‐23 (Boltzmann Constant)T = temperature in Kelvinq = charge of an electron (1.61e‐19 C)
Solving network with diodes
DRDC
vRivvV+=+=
DD vRi +=
vVi DDC −=
T
DnVv
SD eIi ≈
RiD =
Non‐linear equation!
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
6
Load line analysis: diodeDi
RVi DC
D =
Dv
DCD Vv =
Dv
0iVvR
Vi0v
DDCD
DCDD
=⇒=
=⇒=
RvVi DDC
D−
=
T
DnVv
SD eIi ≈
Iteration: convergenceDi
DvDv
⎟⎟⎠
⎞⎜⎜⎝
⎛≈
S
DTD I
inVv ln 2. calculate vD
Start: chose vDR
vVi DDCD
−= 1. calculate iD
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
7
Iteration: divergenceDi
DvDv
T
DnVv
SD eIi ≈ 1. calculate iD
Start: chose vD
RvVi DDC
D−
= 2. calculate vD
Example
R = 10k Result:VDC = 5VIS = 5*10e‐14 An = 1
Result: vD = 573mViD = 443uA
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8
(very) simplified diode curve
Di
V60vA0i
D
D
.=>
V60vA0i
D
D
.<=
Dv0.6V
Zener diode
Di
DvZV
Typical VZ values: 3.6V, 4.5V, 6.8V, 9.1V, …
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
9
Load line analysis: Zener diodeDi
Dv
DR
DRDC
vRivvV−=−=
DCD Vv −=
DD
DR
vRivRi−−=
RvVi DDC
D−−
= RVi DC
D −=
Rectifier circuits
)(tvIN )(tvR)(IN
t
)(R
~0.6V
Half‐wave rectifier
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
10
Rectifier circuits
Positive input voltage Negative input voltage
Rectifier circuits
)(tvIN )(tvR)(IN
t
)(R
~1.2V
Full‐wave rectifier
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
11
Filtering
t
Filtering
( ) ( )τ−≈−= τ /)( / t1Ve1Vtv 0t
0OUT
tfC2I
RC2TVv L0
OUT ≈≈Δ
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
12
Wave‐shaping circuits
OUTv
INv‐0.6V
Wave‐shaping circuits
OUTv
VZ+0.6V
INv‐0.6V
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
13
Voltage doubler and shifter
• Page 518
Diode: small‐signals
iDi
DI
DvdDDDD vVvVv +=Δ+=
dDDDD iIiIi +=Δ+=DV
DC Small‐signal AC
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
14
Diode: small‐signals
iDi
dDDDD vVvVv +=Δ+=dDDDD iIiIi +=Δ+=
DI
di ⎞⎛Dv
DVD
D
DD v
dvdii Δ⋅⎟⎟
⎠
⎞⎜⎜⎝
⎛=Δ
⎟⎟⎠
⎞⎜⎜⎝
⎛=
D
D
d dvdi
r1
small‐signal equivalent circuit
⎟⎟⎠
⎞⎜⎜⎝
⎛= D
dvdi
r1
T
DnVv
SD eIi = DnVv
SD
nVIe
nVI
dvdi
T
D
==⎟⎠
⎜⎝ Dd dvr TTD nVnVdv
D
Td I
nVr =
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
15
Diode: small‐signals
DiDd
dDvvii Δ
===Δ
ITnVr
2DIdd
dD rr
Dv1DI
1DV
D
Td I
r =
2DV
Larger DC current lower small‐signal resistance larger AC small‐signal current
Example
R1=1kRBIAS 10kRBIAS=10k
Calculate the ratio vout/vin for VBIAS=1V and 5V
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
16
Example: DC calculation
Assume VD=0.6VVBIAS = 1V ID = 40uAVBIAS = 5V ID = 440uA
Example: small‐signal calculation
Assume VD=0.6VVBIAS = 1V ID = 40uA rd1=625Ω AV= 0.384 = ‐8dBVBIAS = 5V ID = 440uA rd2=57Ω AV = 0.054 = ‐25dB
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
17
Homework
• P10.9
• P10.28
• P10.72
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
1
Lecture 10
• MOSFET transistors– Operating principles
– Amplifiers
– Logic gates
• Handbook: Chapter 12– Sections: 12 1 12 2 12 3 12 4 12 5 12 6 12 7– Sections: 12.1, 12.2, 12.3, 12.4, 12.5, 12.6, 12.7
NMOS transistor
MOS = Metal-Oxide-Semiconductor4 terminals:GateDrainSourceBulk
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
2
NMOS Transistor (2)
Cross-sectional view
Top view
NMOS Channel
drain
gate
drainsourceGate-oxide SiO2
no voltage no current can flow from drain to source
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
3
NMOS ChannelGATE: V++
DRAIN: 0VSOURCE: 0V Gate-oxide SiO2
Electrons are attracted by the positive voltage at the gate conductive channel between drain and source!
NMOS CurrentGATE: V++
DRAIN: V++SOURCE: 0V
current
Gate-oxide SiO2
Electrons move from source to drain current flows from drain to source
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
4
NMOS symbol
NMOS: Triode
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
5
NMOS: Triode
DSi
MOS = voltage-controlled resistor
DSv↓GSv
↑GSv
NMOS: Saturation
DSi↑GSv
MOS = voltage-controlled current source
DSv
↓GSv
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
6
NMOS: Triode equation( )[ ]( )[ ]DSTGS
2DSDSTGSDS
vVv2KvvVv2Ki
⋅−≈−⋅−=
( )TGSon
DS
DS
VvK21r
iv
−=≈
DSi↑GSv
( )[ ]DSTGS vVv2K≈ ( )TGSDS
DSv
↓GSv
NMOS: Saturation equation
( )2TGSDS VvKi −=( )2DSDS vKi =
DSi↑GSv
( )TGSDS VvKi =( )DSDS
DSv
↓GSv
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
7
K, W, L, KP
⎟⎠⎞
⎜⎝⎛⋅=
LW
2KPK OXCKP μ= OX
OX tC ε
=⎟⎠
⎜⎝ L2 OX
OXt
Moore’s Law
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
8
Example‐1
R=1kΩ 100WVuA100KP 2/=
V=5V
V=0.8V
R=1kΩ
V50Vum1L
um100W
T .===
What is the current through the transistor?
Load‐line approach
VDDRLDSDD
DS
DS
vVi
i−
=
= NMOS equations
VGSDSi ↑GSv
L
DD
RV
DSi LDS R
i
DSv
↓GSv
DDV
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
9
Load‐line approachDSi ...=GSv
...=GSv
DSv...=GSv
Load‐line approach (2)
Input signal = sinewave with 1V amplitude = 2Vptp
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
10
Load‐line approach (3)
of the 1kOhm resistor
Load‐line approach (4)
Input: 2Vptp
Output: 12Vptp
Gain = -6 = 15.6dB
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
11
Small‐signal equivalent circuit
( )2TGSDS VvKi −= in saturation region
gsGSGS vVv += DC bias voltage + small signal
Small‐signal equivalent circuit
( )2TGSDS VvKi −= in saturation region
gsGSGS vVv += DC bias voltage + small signal voltage
( ) ( )2gsgsmDS
2gsgsTGS
2TGSDS
KvvgI
KvvVVK2VVKi
++≈
+−+−=
gsgsmDS g
dsDSDS iIi += DC bias current + small signal current
gsmds vgi = LINEAR relationship between small-signal gate-source voltage and small-signal drain-source current!
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
12
Small‐signals
dsDSDS iIi +=
gsGSGS vVv +=
DSi
GSv
t
gsGSGS vVv + GSv
t gsmds vgi =
Transconductance( ) ( )
2gsgsmDS
2gsgsTGS
2TGSDS
KvvgI
KvvVVK2VVKi
++≈
+−+−=
gsgsmDS g
( )TGSm VVK2g −=
( )2TGSDS VVKI −=
DSm KI2g = DSm
⎟⎠⎞
⎜⎝⎛⋅=
LW
2KPK
Example: IDS=1mA, W=100um, L=0.5um gm = ?
What is VGS-VT for that transistor?
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
13
Small‐signal equivalent model
In the saturation region and forregion and for
small-signals only
gsmds vgi =
( ) DSTGSm KI2VVK2g =−=
Amplifier calculation
DDV
v
dsDS iI +
LR
outOUTOUT vVv +=
inv
BIASVtwo sources superposition!
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
14
Amplifier calculation: DCDDV
R R
DDV
inv
LR LROUTV
BIASV
DC calculation set all small-signal sources to zero
BIASV
Amplifier calculation: ACDDV
R R
inv
LR LR
inv
outv
BIASV
AC calculation set all DC sources to zero
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
15
Amplifier calculation: AC
outvinv
Amplifier calculation: AC
outv
inLmLgsmout vRgRvgv −=−=
inv
Lmin
outv Rg
vvA −==
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
16
Common‐source amplifier
Common‐source amplifier
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
17
Common‐source amplifier: DC
Common‐source amplifier: AC
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
18
Common‐source amplifier: AC
( )ings
DLgsmout
vvRRvgv
=
−= //2G1G1
2G1G1in RRR
RRvv//
//+
=
ings
( )DLmin
outv RRg
vvA //−==
Common‐source amplifier
Dout RR =2G1Gin RRR //=
Dout
( )DLmv RRgA //−=
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
19
Common‐drain amplifier
Common‐drain amplifier: DC
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
20
Common‐drain amplifier: AC
Common‐drain amplifier: AC
( )SLgsmout RRvgv //=
outings vvv −=
2G1G1
2G1G1in RRR
RRvv//
//+
= ( )( ) 1
RRg1RRg
vv
SLm
SLm
in
out ≈+
=//
//
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
21
Common‐drain amplifier
2G1Gin RRR //=
m
mDout
g1
g1RR
≈
= //
( )( )SLm
SLmv RRg1
RRgA//
//+
=
Common‐drain output impedance
gsmS
testtest
gstest
vgR
vi
vv
−=
−=
mtest
testout g
1ivR ≈=
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
22
PMOS transistors
Logic gates
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 11/2/2011
23
Homework
• P12.22
• P12.40
• P12.27
• P12.53
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 10/27/2011
1
Lecture 11
• Bipolar transistors– Operating principles
– Amplifiers
• Handbook: Chapter 13
NPN structure and symbol
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 10/27/2011
2
NPN: biasing
NPN: Currents
BC ii β=
⎟⎞
⎜⎛ v
100020...=β
⎟⎟⎠
⎞⎜⎜⎝
⎛
= T
BEVv
SB eIi
CCBE iiii ≈+=
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 10/27/2011
3
NPN transistor curves
NPN transistor curves
Linear region: BECE ii β=
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 10/27/2011
4
NPN transistor curves
Saturation region: 3020vv satCECE ....., ==
NPN common‐emitter amplifier
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 10/27/2011
5
Voltage gain
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 10/27/2011
6
distortion
PNP transistor
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 10/27/2011
7
PNP transistor curves
biasing
( ) EB
BEBB R1R
VVI+β+
−=
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 10/27/2011
8
Practical common‐source amplifier
Small‐signal circuit
(c) Patrick Reynaert ‐ K.U.Leuven ESAT‐MICAS 10/27/2011
9
Small‐signal circuit
t
B
BE
BE
VI
vi
r1
=∂∂
=π
Small‐signal analysis