microelectronic circuits

RF Microelectronic Design Lab.

Microelectronic Circuits 5th edition Sedra/Smith

담당교수 : 김병성

전화: [email protected]

TA: 안영규

전화:031-290-7225


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)


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


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=


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)


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)


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


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

= + + +


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= + + +


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


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

→

→


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→ ≡ ×

•

→

→


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− −

••

≤ ≤


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

=


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

−

− +

− + =

•

= − > ≥

= =


1.4.9 Symbol convention

Total instantaneous quantityIncremental signal quantitiesDirect current quantitiesmagnitude of sinusoidal signal

C

c

C

c

ii

II

→→→→


(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


(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


(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 ( )


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


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

=+ +


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


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


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


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


(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


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


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


RC

L R L R

C

R

C RR

L

High pass STC

vo vo

vo

io io

iovI vI vI

iIiI iI


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


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


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


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


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


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


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


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


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


Homework

Problems– 17,18,44,46,48,51,54,56,58,63,65,69,77,79

microelectronic circuits

Documents

methods direct measurement

low pass stc

open circuit voltage

circuit source impedance

short circuit current

voltage source

signal source

current source