microelectronic circuits
TRANSCRIPT
RF Microelectronic Design Lab.
Microelectronic Circuits 5th edition Sedra/Smith
담당교수 : 김병성
TA: 안영규
전화:031-290-7225
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IntroductionMicroelectronics– the integrated-circuit (IC) technology implementing passive (R,L,C) and active
device (transistor) in the same chip area– Millions of components – Area on the order of 100mm2
Scope of Electronic Circuits I– Design and analysis of Op Amp application circuits based on the terminal
characteristics→ Op amp is composed of about 20 transistors. We will return to the internal circuitry of
Op amp in Electronic Circuits II– Physical operation and terminal characteristics of diode and its application circuits– Physical operation and terminal characteristics of MOSFET and BJT– Design and analysis of single stage amplifier using MOSFET and BJT– Frequency response of single stage amplifier– Single stage IC amplifier– Differential and multistage amplifiers(* if time is enough)– CMOS inverter and digital logic circuits (topics concerning digital circuit will be
treated in the last lecture)
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Chapter 1. Introduction to Electronics
1.1 Signals1.2 Frequency Spectrum of Signals 1.4 Amplifiers1.5 Circuit Models for Amplfiers1.6 Frequency Response of Amplifiers1.8 Spice
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1.1 Signals
Electric signals– Voltage or current
All signal source can be represented by – A current source with a shunt source resistance : Norton equivalence– A voltage source with a series source resistance : Thevenin equivalence– Two representations are equivalent and their parameters are related by
– Norton equivalence is preferred when Rs is high and Thevenin when Rs is low
Figure 1.1 Two alternative representations of a signal source: (a) the Thévenin form, and (b) the Norton form.
( ) ( )s s sv t R i t=
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Thevenin’s theorem
Vt is a open circuit voltage of A circuitSource impedance can be determined by two methods– Direct measurement of impedance with source reduced to zero– Zt = (open circuit voltage)/(short circuit current)
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Norton’s Theorem
In is a short circuit current of A circuitSource impedance can be determined by two methods– Direct measurement of impedance with source reduced to zero– Zt = (open circuit voltage)/(short circuit current)
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R1 R3
CR4R2
R3
CR4
C
vo
vo
vo
vI
v RR RI ( )2
1 2+
v RR R
RR R R RI ( )(
( / / ))2
1 2
4
4 3 1 2+ + +
( / / )R R1 2
R R R R4 3 1 2/ / ( / / )+
Ex. Ex. TheveninThevenin equivalenceequivalence
RF Microelectronic Design Lab.
1.2 Frequency Spectrum of Signals
Why frequency spectrum?– Practical circuit normally operates in the finite bandwidth. – To avoid signal distortion, the bandwidth of a circuit enough for the signal
bandwidth.
How can we get a frequency spectrum of a time domain waveform?– Periodic signal : Fourier series– Arbitrary signal(nonperiodic) : Fourier transform– Audio band 20Hz – 20kHz, Video band 0MHz – 4.5MHz
Example
Figure 1.4 A symmetrical square-wave signal of amplitude V. 0 0 04 1 1( ) (sin sin 3 sin5 ....)
3 5Vv t t t tω ω ωπ
= + + +
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1.4.1 Amplification
Linear amplification
Nonlinear distortion– If the output is not linear to the input– It can be approximated by power series of the input signal
– Harmonic distortion → For the sinusoidal input x(t)=sinωt, nth order power series term generate nω frequency
components at the output
Class of amplifiers– Small signal amplifier : (ex) preamplifier in home stereo– Power amplifier : (ex) amplifier driving speakers in home stereo
( ) ( )( ) input signal (current or voltage)( ) output signal (current or voltage) amplifier gain, constant independent of the amplitude of ( )
o
y t Ax tx tv tA x t
=→→
→
2 32 3( ) ( ) ( ) ( ) ....y t Ax t A x t A x t= + + +
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1.4.2 Amplifier Circuit Symbol
Amplifier– Two port network – Symbol
→ (a) input port has two distinct terminals from the two output terminals→ (b) when a common terminal exists between the input and output ports of
the amplifier
– Circuit ground→Common terminal used as a reference point between the input and
output ports
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1.4.3-5 Gain
Voltage gain , Current gain , Power gain O O O Ov i p v i
I I I I
v i v iA A A A Av i v i
= = = =
Gain in decibel
log , log , log
is used because gain may be a negative number
Negative gain means a 180 phase difference between input and output sign
10
al
20 20
s
v i pA dB A dB A dB
→
→
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1.4.6 The amplifier Power Supplies
dc 1 1 2 2
dc I L dissipated
dc Power deliveled to the amplifier from the supply P = V I + V IPower balance
P + P (input power) =P (power delivered to the load) + P (dissipated power in the amplifer)
Effi
• →•
•
rms rms rms rms
ciency 100
Solve example 1.1ˆ ˆV or I represent the peak values.
ˆ ˆV IV = , I = ,P=V I2 2
L
dc
PP
η→ ≡ ×
•
→
→
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1.4.7 Amplifier Saturation
An amplifier transfer characteristic that is linear except for output saturation.The range of input signal swing for linear amplification
Iv v
L LvA A− −
••
≤ ≤
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1.4.8 Nonlinear Transfer Characteristics and Biasing
Nonlinear region
Nonlinear region
•Biasing the circuit to operate at a point near the middle of the transfer characteristics for a linear amplification•Q point is an abbreviation for the quiescent point : Operating point or bias point •Av (voltage gain) is a slope or derivative at Q point
Q
Ov
I at
dvAdv
=
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Ex 1.2
4011
,
0
I O
Conditions
10 10 , 0 , 0.3( ) and corresponding ?
0.3 , 10
( )Input bias voltage V for V =5V and the voltage gain?
I
I
vO I O
I
O v V
v e v V v Vi L L v
L V L v V
ii
−
− +
− + =
•
= − > ≥
= =
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1.4.9 Symbol convention
Total instantaneous quantityIncremental signal quantitiesDirect current quantitiesmagnitude of sinusoidal signal
C
c
C
c
ii
II
→→→→
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(cf) Voltgae Source
Ideal source : 공급전류 에 관계없이 일정전압을 유지 (무한 전력원) Rs=0Practical source: 공급전류가 증가함에 따라 출력전압이 감소 (공급 전력에 제한) , Rs>0 Rs가 작을수록 더 좋음
Rs
RL
Vs
iL
Vs'
ideal practical
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(cf) Current source
Ideal source :부하 전압(VL)에 관계없이 일정전압을 유지(무한 전력원), Rs = infinitePractical source:부하 전압(VL)이 증가하면 공급전류 감소(공급전력에 제한), Rs = finiteRs가 클수록 더 좋음
RLRS
IsVL
Ideal is'
practical
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(cf) Voltage meter and Current meter
Voltage meter– Ideal voltmeter 내부저항이무한대 Ri =infinite– Practical voltmeter : 내부저항이유한함 Ri = finite– Ri가클수록더좋음
– Oscilloscope 약 1MΩ
Current meter– Ideal current meter: 내부저항이 Ri=0– Practical current meter : 작은내부저항을갖고있음 Ri>0 – Ri가작을수록더좋음
V
R ideali = ∞ ( )
A
R ideali = 0 ( )
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1.5 Circuit Models for Amplifiers
출력 : dependent voltage source or current source입력 : voltmeter or current meterAssume unilateral amplifiers– No reverse transmission of the output voltage or current to the input
Voltage amplifier– Characterized by three parameters
→Avo : open circuit voltage gain, unloaded voltage gain→Ri : input resistance→Ro: output resistance
– Resulting gain is affected by the source and load resistances
Ri
Ro
voltmeter voltage source
VoltageAmplifier
ivvo iA v
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1.5.1 Voltage Amplifiers
Overall voltage gain is
Even though Avo is large, the overall gain will be low if Ro is high and RL is low.– Need a buffer amplifier which has a unity voltage gain but a high input
resistance and a low output resistanceEx 1.3 : Solve it. There are four types of amplifiers. They are equivalent to each other. The choice of equivalent circuit is for convenience.
o
o i Lv
s i s L o
v R RAv R R R R
=+ +
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Table 1.1 The four amplifier types
Ri
Ro
Ri
Ro
Ri Ro
Ri Ro
A vvvo
o
i io
==0
A iiiso
i vo
==0
G ivmo
i vo
==0
R vimo
i i
==0
RR
i
o
= ∞= 0
RR
i
o
== ∞
0
RR
i
o
= ∞= ∞
RR
i
o
==
00
Voltage amplifier
Current Amplifier
Transconductanceamplifier
TransresistanceAmplifier
(V/V)
(A/A)
(A/V)
(V/A)
Gain parameterIdeal
characteristic
Short-circuit current gain
Open-circuit voltage gain
Short-circuit Transconductance
Open-circuit Transresistance
vi A vvo i
ii
io
A iis i
G vm i
R im i
io
io
vo
vo
vo
vo
Circuit models
ii
io
iv
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1.5.4 Relationships Between the Four Amplifier Models
Conversion of amplifier gains
Determination of the input and output resistances
0
(open circuit voltage)o is o o m i o m i mv is m o
i i i i i i i i
v A R R G v R R i RA A G Rv i R R v i R R
= = = = = = =
ii
vivoRi Ro
Vx
x x
0 0
Input resistanceForce v and measure i and take the ratio
Output resistance open circuit voltage orshort circuit current
i i
xi
x
o
x
x v or i
vRi
R
vi
= =
•→
=
•
=
=
iX
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Ex 1.4 Common Emitter Equivalent Circuit of Bipolar Junction Transistor
voltage gain in (b)
( || )
in (b)
Solve excercise 1.20
om L o
s s
b m be m b m
v r g R rv r R
i g v g r i g r
π
π
π π
ββ β
•
= −+
•= = → =
•
As a transconductance amplifier
As a current amplifier
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1.6 Frequency Response of Amplifiers
Tm dB
|T(jω)|dB
Tm−3 dB
ωL ωH ω(rad/s)
ωL ωH
ω(rad/s)
−90o
−180o
중간 주파수 영역
φ
Linear Amp
tVv Ii ωsin= )sin( φω += tVv Oo
bandwidth
Transfer function |T(jω)|=VO/VI , ∠T(jω) = φ(위상차)3dB bandwidth = ωH - ωLEvaluating frequency response using complex frequency s– -Capacitor → ZC =1/(sC) , Inductor → ZL=sL– -T(jω)=Vo (s) /VI (s)|s=j ω이상적 위상관계– 모든주파수성분이같은시간지연으로증폭단을통과해야신호왜곡이없다.
즉, sin(ω(t-td))이되어야한다.– 이상적위상관계는 φ =-ωtd
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(cf) Bode Plot
Bode Plot이란? :주파수 응답의 크기와 위상을 점근선을통하여 개략적으로 표현한 도표
단일 pole을 갖는 전달함수의 Bode Plot 예
0 dB
Gain
−20 dB
0.1ωP ωP 10ωP
0o
Phase Shift
−90o
0.1ωP ωP 10ωP
−45o
ω
ω
p
p
j)A(j
ωωω
ω+
=
-6dB/octave or -20dB/decade
A j
A jA j
dB
dB decade
p
p
p
pp
( )( )
( )
,
log ( )( )
log
/
ωω
ω ω
ω ωωω
ω ω
ω ω ω ω
ωω
ωω
=+
≅ <
≅ >
RS|T|
= =
= = −
−
2 2
1 2
2
1
1
2
1
10
20 20 20
20
-45°/decω ω ω
ω ω ωωω
< ≅ →
> ≅ − → −
RS|T|
0 1 1 0
10 90
0. ( )
( )
p
pp
A j
A j j
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1.6.4 Single Time Constant Circuits
STC circuit : 한 개의 저항과 한 개의 reactance 소자(L 또는C)로 등가 가능한 회로(appendix F 참조)Time constant(시정수) τ– RL circuit τ=L/R– RC circuit τ=RC
Reduction to STC circuit and rapid evaluation of τ– 전원의영향을제거(전압원은 short, 전류원은 open) 하고
– 한개의 C와여러개의저항으로구성된경우: C에서보이는등가저항Req를 계산 τ=ReqC
– 한개의 R과여러개의 C로구성된경우: R에서보이는등가캐패시턴스 Ceq를구함 τ=RCeq
– RL회로의경우에도마찬가지임
– 여러개의 R과여러개의 C(또는 L)로구성된회로 : 회로의변형이필요함
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Classification of STC circuit Classification of STC circuit --Low pass, High passLow pass, High pass
Test at Replace circuit is LP if Circuit HP if
ω = 0C by o.cL by s.c
Output is finite Output is zero
ω = ∞ C by s.c.L by o.c
Output is zero Output is finite
Table 1
R R3
C
L R
RC
RC
Low pass STC
vo
vo
vo
io
io ioiI iI iI
vI vI vI
LR
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RC
L R L R
C
R
C RR
L
High pass STC
vo vo
vo
io io
iovI vI vI
iIiI iI
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τ의 계산 예
τ =R4||(R3+(R1||R2))C
τ =R(C1+C2)τ =(C1+C2)(R1||R2)
R1 R3
R2 R4 C Vo
V0
C1
C2 R
R1
R2
C1
C2V0
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Frequency response of STC circuit
Low passC
R
Vs vo
T ssRC
Ks
RC RC K
T j j
T j
T j T j
T j T jdB
o
o
o
o
o
( )
, ,
( )
( )
( ) , ( )
log ( ) log ( )
=+
=+
= = = =
=+
=
== ∞ =
−
= −
=
11 1
1 1 1
1
1
12
0 1 0
20 20 03
0
ω
τ ωτ
ω ωω
ω
ω
ω ω
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Frequency response of STC circuit
High pass C
Vs R vo
T s ss RC
Kss
RC RC K
T j j
T j
T j T j
T jT j
dB
o
o
o
o
o
( )
, ,
( )
( )
( ) , ( )
log ( )( )
=+
=+
= = = =
=−
=
== ∞ =
∞= −
1
1 1 1
1
1
12
0 0 1
20 3
ω
τ ωτ
ω ωω
ω
ω
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Example 1.5
Low pass STC
Transfer function T s Ks
o
( ) =+1 ω
vv
R sCR R sC
RR R
RR
sC R RR R
RR
C R RR R
K RR
RR
o
s
ii
S ii
L
L o
S
i
i i s
s i
o
L
i i S
S i S
i
o
L
=+ +
=+ + + +
⇒ =+
=+ +
/ /
/ /
( )
,
1
1
1
1
1
1
1
1
1
1
1
1
µ
µ
τ µ
①
②
τ
ω τ
µ
ωω
=
=
= =+ +
=+
( / / )
( )
( ) ,
R R C
K T j RR R
RR R
T s Ks K
S i i
o
i
s i
L
o L
o
o
1
0
1 1
Low pass STC 라는것을알고있음
대입
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Step response of STC
Step functionFormally solve first order differential equation with an initialconditionGeneral solution form of simple STC(one R, one C(or L), no reduced STC)
/0
0
( ) ( ) , : the final value of STC response: the initial value of STC response at t=0+
ty t Y Y Y eYY
τ−∞ ∞ +
∞
+
= − −
S
X(t)
t0
0/
0/
LP STC : , 0
( ) (1 )HP STC : 0,
( )
t
t
Y S Y
y t S eY Y S
y t Se
τ
τ
∞ +−
∞ +−
• = =
→ = −• = =
→ =
St
y(t)Initial slope = S/τ
S
y(t)
Initial slope = -S/τ
RF Microelectronic Design Lab.
Example appendix D4VI 10V step input
circuit (a) and (b) are equivalentcircuit (c) is Thevenin equivalent circuit of (b)Vo of (c) is superposition of response of (d) and (e)VO=VO1+VO2(d) low pass STC, (c ) high pass STC
+-
R1
R2
C1
C2vO
vI vI+-
R1
R2
C1
C2vO
+-
vI
+
-
+-
+-
+
-
R1||R2 C1+ C2
VI(C1/(C1+ C2))VI(R2/(R1+ R2))
x
x'
vo
+-
+
-
R1||R2
C1+ C2
VI(R2/(R1+ R2))
vo1
+-
+-
R1||R2
C1+ C2
VI(C1/(C1+ C2))
vo2
(a) (b)
(c)
(d)
(e)
10V
/ /1 2
/
2 110(1 ), 101 2 1 2
( 1 2)( 1 || 2)2 1 210 10 ( )
1 2 1 2 1 2Compensated Attenuator
2 : attenuated1 2
1 2Set perfect step response 1 2 1 2
used in the oscillo
t to o
to
R Cv e v eR R C C
C C R RR C Rv e
R R C C R R
RR R
C RC C R R
τ τ
τ
τ
− −
−
= − =+ +
= +
= + −+ + +
•
→+
→ = →+ +
→ scope probe
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Pulse response of low pass STC
Pulse : sum of two step functions
P
X(t)
t0
P
X(t)
t0
P
X(t)
t0
T
τ<<T
τ≈T
τ>>T
Low pass STC response
P
P
P
T
tr
tf tr : rising time 0.1P to 0.9Ptf: falling time 0.9P to 0.1Ptr = tf≈2.2τ
RF Microelectronic Design Lab.
Pulse response of high pass STC response
τ<<T
τ≈T
τ>>T
P
P
P
T
P
-Almost linear, slope P/τ, ∆P= (P/τ)T-출력펄스파형의왜곡도sag[%]= ∆P/P×100= T/τ ×100
∆P
∆P
RF Microelectronic Design Lab.
1.6.5 Classification of Amplifiers Based on Frequency Response
DC coupleda capacitively coupled amplifier
a tuned or bandpass amplifier
Gain reduction at low frequencies due to theimpedance of the coupling capacitor increases
Gain reduction at high frequencies due to the internal parasitic capacitances of devices
RF Microelectronic Design Lab.
Homework
Problems– 17,18,44,46,48,51,54,56,58,63,65,69,77,79