chapter 1 ac analysis skee 2253

12/15/2009

1

SEE 2253: SEE 2253: ELECTRONIC CIRCUITSELECTRONIC CIRCUITS

LECTURER:

1

LECTURER:

CAMALLIL BIN OMARP05-415

[email protected]: 07- 5535241

SEE 2253 ELECTRONIC CIRCUITS

GENERAL KNOWLEDGE : GENERAL KNOWLEDGE : What is ?ElectronicsElectronic components or devices, Electronic systemsElectronic systems

2

12/15/2009

2


Electronics :Electronics :i) the branch of physics that deals with the

emission and effects of electrons ; and the use of electronic devicesuse of electronic devices

ii) Science of the motion of charges in a gas, vacuum or semiconductor.

3


Electronic components or devices:Electronic components or devices:

An electronic building block packaged in a discrete formwith two or more connecting leads or metallic pads.

Components are connected together to create anelectronic circuit with a particular function (for examplean amplifier, radio receiver, or oscillator). Componentsmay be packaged singly (resistor, capacitor, transistor,diode etc.) or in more or less complex groups asintegrated circuits (operational amplifier resistor array

4

integrated circuits (operational amplifier, resistor array,logic gate etc).

Active components are sometimes called devices.

12/15/2009

3


Electronic systems :Electronic systems :Composed of subsystems or electronic circuits, which may include amplifiers, signal sources, power supplies etcetc…Laptop, DVD players, iPOD, PDA, mobile phones

5


What we are learning?

Electronic Circuits:

BJTsFETs Amplifiers

AC analysis to determine gains, Impedances

and frequency response

6

12/15/2009

4

SEE 2253: ELECTRONIC CIRCUITS

Chapter 0:Chapter 0:DC Biasing DC Biasing -- BJT (revision)BJT (revision)

7

SEE 2253 DC BIASING

0.1 Biasing

Biasing refers to the DC voltages applied to a transistor in order to turn it on so that ita transistor in order to turn it on so that it can amplify the AC signal.

8

12/15/2009

5

SEE 2253 DC BIASING

0.2 Operating Point

The DC input establishes an operating or quiescent point called the Q-point.p p

9

SEE 2253 DC BIASING

0.3 Biasing and the Three States of Operation

Active or Linear Region OperationActive or Linear Region OperationBase–Emitter junction is forward biasedBase–Collector junction is reverse biased

Cutoff Region OperationCutoff Region OperationBase–Emitter junction is reverse biasedBase–Collector junction is reverse biased

Saturation Region OperationSaturation Region Operation

10

Saturation Region OperationSaturation Region OperationBase–Emitter junction is forward biasedBase–Collector junction is forward biased

12/15/2009

6

SEE 2253 DC BIASING

0.4 DC Biasing Circuits

Fixed-bias circuitEmitter stabilized bias circuitEmitter-stabilized bias circuitVoltage-divider bias circuitCollector-feedback bias circuit

11

SEE 2253 DC BIASING

0.5 Fixed Bias Circuit

12

12/15/2009

7

SEE 2253 DC BIASING

Base-Emitter Loop

F Ki hh ff’ ltFrom Kirchhoff’s voltage law:

+VCC – IBRB – VBE = 0

Solving for the base t

13

current:

B

BECCB R

VVI −=

SEE 2253 DC BIASING

Collector-Emitter Loop

The collector current isThe collector current is given by:

IC = βIB

From Kirchhoff’s voltage law:

14

VCE = VCC – ICRC

12/15/2009

8

SEE 2253 DC BIASING

0.6 Emitter-Stabilized Bias CircuitAdding a resistor (RE) to the emitter circuit stabilizes the bias circuit. Stability refers to a bias circuit in which the currents and voltages will remain fairly constant for a wide range of temperatures and transistor Beta (β) values.

15

SEE 2253 DC BIASING

Base-Emitter Loop

F Ki hh ff’ ltFrom Kirchhoff’s voltage law:+VCC – IBRB – VBE – IERE = 0

Since IE = (β + 1)IB:

+VCC – IBRB – VBE – (β+1)IBRE = 0

16

Solving for IB:

EB

BECCB R)1(R

VVI+β+−

=

12/15/2009

9

SEE 2253 DC BIASING


From Kirchhoff’s voltageFrom Kirchhoff s voltage law:+ IERE + VCE + ICRC – VCC = 0

Since IE ≅ IC:Also:V = I R

17

VE = IERE

VC = VCE + VE = VCC – ICRC

VB = VCC – IBRB = VBE + VE

SEE 2253 DC BIASING

0.7 Voltage-Divider Bias CircuitThis is a very stable bias circuit.The currents and voltages are almost independent of variations in β.β

18

12/15/2009

10

SEE 2253 DC BIASING

Approximate Analysis

Where I << I and I and I ≅ I :Where IB << I1 and I2 and I1 ≅ I2 :

Where βRE ≥ 10R2:VE = VB – VBE

From Kirchhoff’s voltage law:

CC21

2B V

RRRV ×+

=

19

From Kirchhoff s voltage law:VCE = VCC – ICRC – IEREIE ≅ ICVCE = VCC – IC(RC + RE)

Exact Analysis

Using Thevenin Theorem:

SEE 2253 DC BIASING

Using Thevenin Theorem:

From Kirchhoff’s voltage law:VCE = VCC – IC(RC + RE)VTH – IBRTH – VBE – IERE = 0

CC21

2TH V

RRRV ×+

= 21TH R//RR, =

20

VTH IBRTH VBE IERE 0

Solving for IB:

ETH

BETHB R)1(R

VVI+β+

−=

12/15/2009

11

SEE 2253 DC BIASING

0.8 Collector-Feedback Bias CircuitAnother way to improve the stability of a bias circuit is to add a feedback path from collector to base. In this bias circuit the Q-point is only slightly p y g ydependent on the transistor beta, β.

21

SEE 2253 DC BIASING

Base-Emitter Loop

F Ki hh ff’ lt lFrom Kirchhoff’s voltage law:VCC – IC′RC – IBRB – VBE – IERE = 0IC′ = IB + IC = IE

Since IC′ ≅ IE and IC = βIB:VCC – (β+1)IBRC – IBRB – VBE

–(β+1)IBRE = 0

22

(β ) B E

Solving for IB:

)RR)(1(RVVI

ECB

BECCB ++β+

−=

12/15/2009

12

SEE 2253 DC BIASING


From Kirchhoff’s voltageFrom Kirchhoff s voltage law:+ IERE + VCE + IC′RC – VCC = 0

Since IC′ ≅ IE and IC = βIB:+ IC(RC + RE) + VCE – VCC =0

23

Solving for VCE:VCE = VCC – IC(RC + RE)

SEE 2253 DC BIASING

0.9 PNP TransistorsThe analysis for pnp transistor biasing circuits is the same as that for npn transistor circuits.The only difference is that the currents are flowing y gin the opposite direction.

24

12/15/2009

13

ExerciseFor the circuits shown below, determine the value of IB, IC and VCE. (a) 35.35µA, 4.24mA, 20.92V

SEE 2253 DC BIASING

( ) µ , ,(b) 10.49 µA, 1.05mA, -13.96V

RC

+20V

2.7k

=120

8.2k

(a) (b)

25

-20V

RE 1.8k2.2k

=120(a) (b)


chapter 1:chapter 1:Introduction to AmplifiersIntroduction to Amplifiers

26

12/15/2009

14

1.0 Basic Voltage Amplifier ModelThe ideal voltage amplifier may be described as having the following attributes:

Infinite input impedance

SEE 2253 AMPLIFIER MODEL

Zero output impedanceInfinite voltage gainInfinite bandwidth (a voltage gain that is not affected by the signal frequency)A zero output offset voltage (

27

(zero volts output with zero volts applied at the input)No drift in these parameters when a change in temperature occurs

1.1 Unloaded Voltage GainVoltage amplifier with no input or output loading.

vout = Av(oc)vin

vin = vs


+ ++

Zo

Rs = 0

28

-Av(oc)vin

--

vin

Zi+

-

vout = Av(oc)vinvs

12/15/2009

15

1.2 Input loading EffectsThe amplifier’s input voltage vin will be less than vsbecause of the voltage division between Rs and Zi.


siS

iin v

ZRZv+

=

+

-Av(oc)vin

++

vin

Zi

Zo

+

-vout = Av(oc)vin

Rs

vs

si

)oc(

in)oc(o

vZR

ZA

vAv

v

vut

=

=

29

iS

i)oc(

s

o

ZRZA

vvA v

utvs +

==

--s

iS)oc( ZRv +

1.3 Output loading EffectsOutput voltage produced by controlled source will be divided between RL and Zo.


+

-Av(oc)vin

+

vin

Zi

Zo

+

-

Rs

vs

+

-voutRL

in)oc(Lo

Lo R

vAZ

Rv vut +=

)oc(Lo

L

in

ov

utv A

RZR

vvA

+==

30

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

==iS

i

Lo

L)oc(

s

o

ZRZ

RZRA

vvA v

utvs

-

12/15/2009

16

1.4 Current Gain of a Voltage AmplifierBy inspection of the output circuit:


in)oc(o RZ

vAi v

ut =Lo

o RZut +

iinZivin =

iinLo

)oc(o Zi

RZA

i vut +=

R

31

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

==Lo

i

iS

S)oc(

s

o

RZZ

ZRRA

iiA v

utis

sis

sin i

ZRRi+

=

1.5 DecibelsTo simplify the analysis of electronic amplifiers and systems, a logarithmic measure called decibel is used.


v10dBlog20 AAv =

i10dBlog20 AAi =

Voltage Gain

Current Gain

32

iv10dBlog10 AAAP =

dB

Power Gain

Gain

12/15/2009

17

Equations to Remember


v(oc)v RZRA

vvA ⎟⎟

⎠

⎞⎜⎜⎝

⎛+

==Lo

L

in

out

v(oc)i

vv(oc)vs

RRZi

RZZA

iiA

ZRZA

ZRZ

RZRA

vvA

⎞⎛⎞⎛⎞⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+

==

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

==

Lo

i

in

out

is

i

is

i

Lo

L

s

out

33

ivp

iv(oc)is

AAppA

ZRRA

ZRR

RZZA

iiA

==

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

==

in

out

is

s

is

s

Lo

i

s

out

Exercise:Determine the values of :(a) Av

(b) A


(b) Avs, (c) vi and vs if vL = 4.11 vac

(Ans: 363.6, 342.2, 11.3 mVac, 12 mVac)

Ro300 Ω

RS50 Ω

34

Ri800 Ω

RL3 kΩvS

+vi_

+vL_

AV(OC)vi

AV(OC)= 400

12/15/2009

18

1.6 BJT Transistor Modeling

A model is an equivalent circuit that represents the AC characteristics of the transistor.


A model uses circuit elements that approximate the behavior of the transistor.

There are two models commonly used in small i l AC l i f t i t

35

signal AC analysis of a transistor:re modelHybrid-π equivalent model (will be used exclusively)

The re Transistor ModelBJTs are basically current-controlled devices, therefore the re model uses a diode and a current source to duplicate the behavior of the transistor.One disadvantage to this model is its sensitivity to the DC


One disadvantage to this model is its sensitivity to the DC level. This model is designed for specific circuit conditions.

The Hybrid-π ModelThe hybrid-π model is most useful for analysis of high-frequency transistor applications.

36

frequency transistor applications.At lower frequencies the hybrid-π model closely approximate the re parameters, and can be replaced by them.

12/15/2009

19


1.7 Fundamental Voltage AmplifiersThe procedure for the analysis of any amplifier is as follows:

1. Perform a dc analysis to find the Q-point.2. Determine the small-signal (device) parameters

at that Q-point.3. Draw the ac equivalent circuit and determine Zi,

Zo and Av(oc).U th ti d l d f th l

37

4. Use the equations developed for the general amplifier model to find Av, Avs, Ai, Ais and Ap.


1.8 BJT Hybrid-π Model (at low & mid freq.)C

iC2N4124 +

B CiC

B

E

+

-

v

gmv

-

v r ro

E

Ao

200I

VIVr ==r β

=π V26CC

mI

VIg ==

mV26==q

kTVT

38

CCo IImgπ mV26T

m Vg

gm - Transconductance

VA - Early voltage

VT - Voltage equivalent of temperature

k – Boltzman Constant (1.38x10-23 J/K)

q – Electron Charge (1.6x10-19 C)

T = 273O + TC Kelvin

12/15/2009

20


1.9 Common-Emitter Fixed-Bias ConfigurationThe input is applied to the baseThe output is from theThe output is from the collectorHigh input impedanceLow output impedanceHigh voltage and current gainPh hift b t

39

Phase shift between input and output is 180°


Common-Emitter Fixed-Bias Configuration

40

12/15/2009

21


Common-Emitter Fixed-Bias Configuration

Input impedance:

= rRZ

v)R(rvg)R(rvgv CoimComo −=−= π

Output impedance:

Voltage gain:

π= rRZ Bi

CCo

Co

10Rro

O

RZ

rRZ

≥≅

=

41

)R(rgvvA Com

i

o)oc(v −==

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=⎟⎟⎠

⎞⎜⎜⎝

⎛+

=Lo

iCom

Lo

iv(oc)i RZ

Z)R(rgRZ

ZAA

Current gain from voltage gain:


Example:Determine the values of Zi, Zo, Av(oc), Av, Avs, Ai, Aisand Ap for the amplifier circuit below. Given VA=200V and IC = 2.37mA

VCC20V

RC4.3k

=120

R18.2k

C1

C2

600

RS

42

RE1k

R21.5k

C3

RL10k

600

vsvoutvin

Zi Zo

12/15/2009

22


373)09.4)(mS2.91(-Comin

out)oc(v −=Ω=−== k)R(rg

vvA

Input impedance:

Output impedance:

Voltage gain:

Ω=ΩΩΩ== π 64732.15.12.821i kkkrRRZ

Ω=ΩΩ== kkkRrZ 09.43.44.84Co O

43

in

5.137647600

647)265(is

iv

s

outvs −=⎟

⎠⎞

⎜⎝⎛

Ω+ΩΩ

−=⎟⎟⎠

⎞⎜⎜⎝

⎛+

==ZR

ZAvvA

2651009.4

10)373(Lo

L)oc(v

in

outv −=⎟

⎠⎞

⎜⎝⎛

Ω+ΩΩ

−=⎟⎟⎠

⎞⎜⎜⎝

⎛+

==kk

kRZ

RAvvA


13.1710094

647)373(iv(oc)

outi −=

⎠⎞

⎜⎝⎛

ΩΩΩ

−=⎟⎠

⎞⎜⎜⎝

⎛==

kkRZZA

iiA

Current gain:

10094Lo(oc)

in ⎠⎝ Ω+Ω⎠⎜⎝ + kk.RZi

24.8647600

600)13.17(is

si

s

outis −=⎟

⎠⎞

⎜⎝⎛

Ω+ΩΩ

−=⎟⎟⎠

⎞⎜⎜⎝

⎛+

==ZR

RAiiA

Power gain:

44

g

45.4539)13.17)(265(ivin

outp =−−=== AA

ppA

12/15/2009

23


45

Draw a small signal AC equivalent circuit


1.10 Common-Emitter (Unbypassed RE)The removal of the bypass capacitor results in an increase in the amplifier’s input impedance, a reduction in its voltage gain and an increase in its output impedance looking into the collector.

46

12/15/2009

24


Eb

bBi

)R1(βrZZRZ

++=

=

π

Input impedance:

Output impedance (ignore ro):

Co RZ =

Voltage gain:

bb

Cm

i

ov(oc) E++=

π−==π 1)R(βrZZi

RvgvvA

b

Current gain:

BBmoi ZR

RZR

RrgiiA

+β

−=+

−== π

47

( )

1E

C

i

ov(oc)

Em

Cm

Emb

Cbm

bbi

Em >>

ππ

π

−≅=

+−=

+−=

RgRR

vvA

Rg1Rg

RrgriRrig


bBbBi ZRZRi ++

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

=CLo

i RZA

RZZAA i

)oc(vi

)oc(v


To determine Zo using imaginary voltage source

i iTESTgmv

ZTEST

VTEST

iTEST

RE

(iTEST – gmvro

vr

RTH’(RS//RB)

48

ETESTOmOTEST

ETESTOmTESTTEST

TEST

TESTTEST

)(RirvgriRirvgiv

ivZ

+−≈

+−≈

=

π

π

12/15/2009

25


TESTE

TEST'THE

E

)(

irRirv

iRrR

Ri

×−

=−=

×++

=

π

π

E'THE

EOmOTEST

ETESTTEST'THE

EOmOTESTTEST

TEST'THE

RRrRrRrgri

RiiRrR

rRrgriv

iRrR

irv

⎥⎦

⎤⎢⎣

⎡+

+++=

+×⎥⎦

⎤⎢⎣

⎡++

−−=

×++

π

π

π

π

πππ

49

CCTESTO

E'THE

EOmO

TEST

TESTTEST

)( RRZZ

RRrRrRrgr

ivZ

≈=∴

+++

+==∴

⎦⎣

π

π


Example:Determine the values of Zi, Zo, Av(oc), Av, Avs, Ai, Aisand Ap for the amplifier circuit shown below. Given VA= ∞ and IC = 2.37mA

VCC20V

RC4.3k

=120

R18.2k

C1

C2

600

RS

50

RE1k

R21.5k

RL10kvs

voutvin

Zi Zo

12/15/2009

26


Input impedance:

kkkkZRRZ 261321225128

Ω=Ω+Ω=++= π kkk1)R(βrZ 32.122)1)(121(32.1Eb

51

25.4)1)(mS2.91(1)09.4)(mS2.91(-

1 Em

Cm

in

out)oc(v −=

Ω+Ω

=+

−==kk

RgRg

vvA

Output impedance:

Voltage gain:

Ω=ΩΩΩ== kkkkZRRZ 26.132.1225.12.8b21i

Ω== kRZ 3.4Co


01.2261600

26.1)97.2(iv

outvs −=⎟

⎠⎞

⎜⎝⎛

Ω+ΩΩ

−=⎟⎠

⎞⎜⎜⎝

⎛+

==k

kZR

ZAvvA

97.2103.4

10)25.4(Lo

L)oc(v

in

outv −=⎟

⎠⎞

⎜⎝⎛

Ω+ΩΩ

−=⎟⎟⎠

⎞⎜⎜⎝

⎛+

==kk

kRZ

RAvvA

37.01034

26.1)25.4(Lo

iv(oc)

in

outi −=⎟

⎠⎞

⎜⎝⎛

Ω+ΩΩ

−=⎟⎟⎠

⎞⎜⎜⎝

⎛+

==kk.

kRZ

ZAiiA

Current gain:

120600)370(sout =⎞⎜⎛ Ω

=⎞

⎜⎜⎛

==RAiA

26.1600iss ⎠⎝ Ω+Ω⎠⎝ + kZRv

52

12.026.1600

)37.0(is

is

is −=⎠

⎜⎝ Ω+Ω

−=⎠

⎜⎜⎝ +

==kZR

Ai

A

1.1)37.0)(97.2(ivin

outp =−−=== AA

ppAPower gain:

12/15/2009

27


1.11 Emitter Follower ConfigurationThis is also known as the common-collector configuration.The input is applied to the base and the outputThe input is applied to the base and the output is taken from the emitter.There is no phase shift between input and output.

53


r gmv

Eb

bBi

)R1(βrZZRZ

++=

=

π

Input impedance:

Ie gmVβ+=== π

1rRZR

ivZ EeE

o

oo

Output impedance:

Voltage gain:

o)oc(v +

== Em

Rg1Rg

vvA

Current gain:

BBmoi ZR

βRZRRrg

iiA

+=

+== π

54

1i

o)oc(v

i

>>≅=

+

EmRg

Em

1vvA

Rg1v


bBbBi ZRZRi ++

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=Lo

i RZZAA i

)oc(v

12/15/2009

28


To determine Zo using imaginary voltage source

×+

=π

ππ

TESTTESTTEST

TEST'TH

vrgvv

vRr

rv

×⎥⎦

⎤⎢⎣

⎡++

=

×⎥⎦

⎤⎢⎣

⎡+

++

=

++

+=+

+=

π

π

π

π

π

π

π

ππ

π

1

r1

TEST'TH

m

TEST'TH

m

'THTEST

'TH

TESTm

'TH

TESTm

'TH

TESTTEST

RR

vRr

rg

vRrg

Rri

Rrvrg

Rrvvg

Rrvi

gmv

i

RTH’(RS//RB) rv

i

55

⎟⎟⎠

⎞⎜⎜⎝

⎛β+

+==∴

β++

=++

==∴

π

π

π

π

1)(

)1(1

'THEETESTO

'TH

m

'TH

TEST

TESTTEST

RrRRZZ

Rrrg

RrivZ

VTEST

ZTEST

iTEST


Example:Determine the values of Zi, Zo, Av(oc), Av, Avs, Ai, Aisand Ap for the amplifier circuit below. Given VA→∞and IC = 2.37mA

56

12/15/2009

29


R1//R2

r

600

RS

vsvin

v gmv

Input impedance:

kkkkZRRZ 251321115128

( ) Ω=ΩΩ+Ω=++= π kkkkRRrZ 32.111)101)(121(32.1)1(β LEb

vout

Zi Zo

RLRE

Zb

57

Output impedance:

Ω=ΩΩΩ== kkkkZRRZ 25.132.1115.12.8b21i

( )( )

Ω=ΩΩΩ+Ω

Ω=+β+

= π 07.14121

60012.832.11

1TH

Eo

.5kkkkRrRZ


988.010)989.0(L)oc(v

outv =⎞⎜

⎛ Ω=

⎞⎜⎜⎛

==kRAvA

989.0)1)(mS2.91(1

)1)(mS2.91(1 Em

Em)oc(v =

Ω+Ω

=+

=k

kRg

RgA

Voltage gain:

123.01007.14

25.1)989.0(Lo

iv(oc)

in

outi =⎟

⎠⎞

⎜⎝⎛

Ω+ΩΩ

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

==k

kRZ

ZAiiA

Current gain:

668.025.1600

25.1)988.0(is

iv

s

outvs =⎟

⎠⎞

⎜⎝⎛

Ω+ΩΩ

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

==k

kZR

ZAvvA

988.01007.14

)989.0(Lo

)oc(vin

v⎠

⎜⎝ Ω+Ω⎠

⎜⎜⎝ + kRZ

Av

A

58

00Loin ⎠⎝⎠⎝

04.025.1600

600)123.0(is

si

s

outis =⎟

⎠⎞

⎜⎝⎛

Ω+ΩΩ

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

==kZR

RAiiA

122.0)123.0)(988.0(ivin

outp ==== AA

ppAPower gain:

12/15/2009

30


1.12 Common-Base ConfigurationThe input is applied to the emitter.The output is taken from the collector.Low input impedance.High output impedance.Current gain less than unity.Very high voltage gain.No phase shift between input and output.

59


Inp t impedancer vout

vin v RC

gmv

RE

)e

eEi

1(βrZ

ZRZ

+=

=

π

RZOutput impedance:

Input impedance:

Current gain:

βZRRrg

iiA ≅

+== π Emo

i

Zi ZoZe

60

Co RZ =

Voltage gain:

CmRgvvA ==

i

o)oc(v


ZRi + eEi

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=Lo

i RZZAA i

)oc(v

12/15/2009

31


Example:Determine the values of Zi, Zo, Av(oc), Av, Avs, Ai, Aisand Ap for the amplifier circuit below. Given VA→∞and IC = 2.37mA

VCC20V

=120C1 C2

600

RS

VCC20V

RC4.3k

61

20VRE1k R2

1.5k

R18.2k

RL10k

vs voutvin

Zi Zo

C3


Input impedance:

Output impedance:

( )Ω=ΩΩ==

Ω=Ω

=+β

= π

79.1091.101

91.1012132.1

1

eEi

e

kZRZ

krZ

Ω== kRZ 3.4Co

62

392)3.4)(mS2.91(Cmin

out)oc(v =Ω=== kRg

vvA

p p

Voltage gain:

Co

274103.4

10)392(Lo

L)oc(v

in

outv =⎟

⎠⎞

⎜⎝⎛

Ω+ΩΩ

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

==kk

kRZ

RAvvA

12/15/2009

32


Current gain:

84.479.10600

79.10)274(is

iv

s

outvs =⎟

⎠⎞

⎜⎝⎛

Ω+ΩΩ

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

==ZR

ZAvvA

296.01034

79.10)392(Lo

iv(oc)

in

outi =⎟

⎠⎞

⎜⎝⎛

Ω+ΩΩ

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

==kk.RZ

ZAiiA

Current gain:

291.079.10600

600)296.0(is

si

s

outis =⎟

⎠⎞

⎜⎝⎛

Ω+ΩΩ

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

==ZR

RAiiA

63

1.81)296.0)(274(ivin

outp ==== AA

ppA

Power gain:


C-E C-C C-B

VOLTAGEGAIN A

Large(180o h hift)

≈1(i h )

Large(i h )GAIN, Av (180o phase shift) (in phase) (in phase)

CURRENTGAIN, Ai

Large Large ≈1

POWERGAIN, AP

Large Small Medium

64

INPUTIMPEDANCE, Zi

Medium High Low

OUTPUTIMPEDANCE, Zo

Medium Low High

12/15/2009

33


Exercise 1For the circuit shown below, if β=100, VEB = 0.7V and VA→∞,

determine:

(a) ICQ and VCEQ(a) ICQ and VCEQ.(b) draw the ac equivalent circuit at middle frequency.(c) impedances Zi and ZO.(d) gains AVS = vo/vs dan Ais = iL/is at middle frequency.

+4V

RE5kΩ C3

10µF

65-6V

RC4kΩ

RL4kΩ

RB5kΩ

RS1kΩ

C10.1µF

C20.1µFvs

+vo-Zi

Zo

is

iL


Exercise 2Draw the small signal ac equivalent circuit. If +VCC = 12V and β = 80, calculate the values of :(a) Input and output impedances( ) p p p(b) Voltage gain AV and AVS

66

12/15/2009

34


Exercise 3Draw the small signal ac equivalent circuit. If β = 100,calculate the values of :(a) Input and output impedances(a) Input and output impedances(b) Ratio of vL/ii

R168K

RC2.7K

-10V

ii C1

C

C3

67Zi

R215K

Vi

+12V

C2

RE1.5K

RL1K

+

-

VL

ZO


Exercise 4Draw the small signal AC equivalent circuit. If β = 100, calculate the values of :(a) Input and output impedances( ) p p p(b) Voltage gain, AV in decibel [dB]

68

12/15/2009

35


Given β = hfe = 200, VBE = 0.6 V, VCEQ = 10 V, VA→∞, Av(dB)

f

+ 15 V

Final Exam Paper 1998/99 semester 1

at middle frequency = 14dB.(a) Draw the mid-frequency AC equivalent circuit.(b) Determine RE1 5 K

10 K 1 KR1

R2

RC

RE1

iL

vo

vi

C1 C2

C3

RL1 K

69

and RE2.(c) Determine the amplifier input impedance, Zi.(d) Ratio of iL/vi.

RE2

vi

Zi

RE1 = 56.67Ω; RE2 = 7.27kΩ; Zi = 2.86kΩ; iL/vi = 4.99mS


chapter 1: (continue)chapter 1: (continue)Amplifier Frequency ResponseAmplifier Frequency Response

70

12/15/2009

36

SEE 2253 FREQUENCY RESPONSE

1.13 General Frequency ConsiderationsThe frequency response of an amplifier refers to the frequency range in which the amplifier will operate with negligible effects from capacitors and capacitance in devices. This range of frequencies can be called the mid-range.

At frequencies above and below the mid-range, capacitance and any inductance will affect the gain of the amplifier.At lo freq encies the co pling and b pass capacitors

71

At low frequencies the coupling and bypass capacitors lower the gain.At high frequencies stray capacitances associated with the active device lower the gain.Also, cascading amplifiers limits the gain at high and low frequencies.


1.14 Effect of Capacitors on Frequency Response

72

At Low Frequency At High Frequency

12/15/2009

37

1.15 Bode PlotA Bode plot indicates the frequency


response of an amplifier.The horizontal scale indicates the frequency (in Hz) and the

73

vertical scale indicates the gain (in dB).

1.16 Cutoff FrequenciesThe mid-range frequency of an amplifier is called the bandwidth of the


bandwidth of the amplifier.The bandwidth is defined by the lower and upper cutoff frequencies.Cutoff—frequency at which the gain has

74

which the gain has dropped by:

0.5 power0.707 voltage-3dB

12/15/2009

38

1.17 Semi-Log GraphThe use of log scales can significantly expand therange of variation of a particular variable on a graph.


1 cycle

75

1 2 3 4 65 7 8 9

≈ 30%log10 2 = 0.3010

≈ 48%log10 3 = 0.4771 log10 4 = 0.6021 (≈ 60%)

log10 5 = 0.6999

log10 6 = 0.7781

log10 7 = 0.8451

log10 8 = 0.9031

log10 9 = 0.954311 2 3 4 65 7 8 9

1.18 Low Frequency ResponseAt low frequencies coupling capacitors (CS, CC) and bypass capacitors (CE) will have capacitive reactances (XC) that affect the circuit impedances.


76

12/15/2009

39

Low Frequency ResponseCommon-Emitter ac equivalent circuit at low and middle frequency.


RC

RL

R

(R1//R

2)

RBB

RS

vs C

CcCs

rπ g

mvπ

+vπ-+

vi

-

77

RE

s CE

CouplingCoupling CapacitorCapacitor —— CCSSThe cutoff frequency due to CS can be calculated by:

rRR Z )CZ(R

fπ

=+

= 21iL and2

1S


)CZ(R π+π sis2S

+

CS

RS

78

RL

(R1//R2)RBB rπ gmvπ

+vπ−

vi

−

RC

S

vS

Zi

12/15/2009

40

CouplingCoupling CapacitorCapacitor —— CCCCThe cutoff frequency due to CC can be calculated with:

RZf and1


RZ )CZ(R

f CoCoL

L and2C

=+π

=

++

C2

RS

79

RL

(R1//R2)RBB rπ

gmvπ+vπ−

vi

−

RC

vS

ZO

BypassBypass CapacitorCapacitor —— CCEEThe cutoff frequency due to CE can be calculated with:

Rr1 ⎞⎜⎛ +


BRRR CR

f BS'TH'TH

EeqEeq

L where1

RrRRand2

1E

=⎟⎠

⎞⎜⎜⎝

⎛β+

+=

π= π

RL(R //R )RBB

rπ gmvπ

+vπ−+

viRC

RS

80

L(R1//R2)vi

−

C

vS

RE CE

Req

12/15/2009

41

1.19 Bode Plot of Low Frequency ResponseThe Bode plot not only indicates the cutoff frequencies of the various capacitors but it also indicates the amount of attenuation (loss in gain) at these frequencies.The amount of attenuation is sometimes referred to as rollroll offoff


The amount of attenuation is sometimes referred to as rollroll--offoff.The roll-off is described as dB loss-per-octave or dB loss-per-decade.

81

1.20 –dB/Decade--dB/decadedB/decade refers to the attenuation for every 10-fold change in frequency.For attenuations at the low-frequency end, it refers to the loss in gain from the lower cutoff frequency to a frequency that is one


gain from the lower cutoff frequency to a frequency that is one-tenth the cutoff value.

In this examplefLS = 9kHz gain is 0dBfLS/10 = 0 9kHz

82

fLS/10 0.9kHz gain is –20dBThus the roll-off is 20dB/decadeThe gain decreases by –20dB/decade

12/15/2009

42

1.21 –dB/Octave--dB/octavedB/octave refers to the attenuation for every 2-fold change in frequency.For attenuations at the low-frequency end, it refers to the loss in gain from the lower cutoff frequency to a frequency one half the


gain from the lower cutoff frequency to a frequency one-half the cutoff value.

In this examplefLS = 9kHz gain is 0dBfLS/2 = 4.5kHz gain is –6dBTherefore the roll off

83

Therefore the roll-off is 6dB/octaveThis is a little difficult to see on this graph because the horizontal scale is a logarithmic scale.

1.22 Bode Plot of Low Frequency ResponseThe Bode plot indicates that each capacitor may have a different cutoff frequency. The capacitor that has the highest lower cutoff


p gfrequency (fL) is closest to the actual cutoff frequency of the amplifier.

84

12/15/2009

43

ExampleDetermine the values of cutoff frequency due to CS,CC and CE for the amplifier circuit shown below. Given gm=50mS and VA = ∞


85


RRBB

rπ gmvπ

+vπ−+

R

RS

CS CC

RL(R1//R2)vi

−

RC

vS

RE

Req

CE

ZOZ i

86

Ω=ΩΩΩ== π 76.4411k6k20k100rRRZ 21i

( ) ( )( ) HzF4kCZR

fSiS

LS 35.776.441112

12

1=

µΩ+Ωπ=

+π=

12/15/2009

44


RL(R1//R2)RBB

rπ gmvπ

+vπ−+

viRC

RS

CS CC

1 2

−vS

RE

Req

CE

ZOZ i Ω== k5CO RZ

( ) ( )( ) HzF10kkCRZ

fLC 23521

21

=ΩΩ

==

87

( ) ( )( )F10kkCRZ CLO 3522 µΩ+Ωπ+π

[ ] [ ]Ω=Ω⎟

⎠⎞

⎜⎝⎛

+Ω+ΩΩΩ

=⎟⎟⎠

⎞⎜⎜⎝

⎛+β

+= π 96.225

1300620//1

1kkk//100kkR

rRRR E

BBSeq

( )( ) HzF70CR

fEeq

LE 02.9996.222

12

1=

µΩπ=

π=


0 0.2 0.75 2 3.75 7.5 10 100 [Hz]fLEAV [dB]

actual

decade-3

- 6

fLSfLC

log f

Sketch of the overall

plot

roll-off:20dB/decade

or 6dB/octave

20dB

octave

- 15

- 9

- 12

- 21

- 18

- 24

88

Sketch of the overalllow frequency response

12dBroll-off:

40dB/decade or

12dB/octave

- 27

- 30

- 36

- 33

- 39

12/15/2009

45


AV [dB]

f [Hz]

20dB/decade

fLC1fLC2 fLCE

20dB/decade@ 6dB/octave


89

Bode plot showing the differences in roll-off due to severalcutoff frequencies.



Exercise: (Test 1 2005/06 sem1)Determine the lower cutoff frequency, fL for the amplifier circuit if ICQ = 3.13mA.

260.32 Hz

Given: β = 100, VCC = -20 VR1 =56 kΩ, R2 = 15 kΩRC = 2.2 kΩ, RL = 2 kΩRE = 1 kΩ, RS = 500 Ω

90

E SC1 = 1.0 µF, C2 = 1.5 µFVT = 26 mV, VA = ∞VEB = 0.7 V

12/15/2009

46


1.23 Miller Effect Capacitance:Any p-n junction can develop capacitance.In a BJT amplifier, this capacitance becomes noticeable between:noticeable between:

The base-collector junction at high frequencies in common emitter BJT amplifier configurations.The gate-drain junction at high frequencies in common source MOSFET amplifier configurations.

91

It is called the Miller Capacitance, and it affects the input and output circuits.


Miller Input Capacitance (CMi)

ZMo

RL

If +

vO-i

OV V

VA =

+

vi

-

ZMi

If

92

Note that the amount of Miller capacitance isdependent on inter-electrode capacitance frominput to output (Cf) and the gain (AV).

12/15/2009

47


Miller Input Capacitance (CMi)

Oif V-VV =Voltage across Z :

VAV iVO VAV =But

)A(1-VV Vif =Therefore

( )Z

AVZVI vif

f−

==1

Thus the current through Z :

93

( ) fVMi CA1C −=

( )Z

AVV

IVZ

vi

i

f

iMi −

== 1 vMi A1

ZZ−

=or


Miller Output Capacitance (CMo):If the gain (AV) is considerably greater than 1,then:

ZMo

RL

If +

vO

-i

O

VV

VA =

+

vi

-

ZMi

If

VVZ oo ⎞⎜⎛ AZZ v

94

( )Z

AV-I-Z

vi

o

f

oMo −

== 1 or ⎟⎠

⎞⎜⎜⎝

⎛−

=1A

AZZv

vMo

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

V

VfMo A

1ACC 1>>≅VAfMo CC

12/15/2009

48


1.24 High-Frequency ResponseCapacitances that affect the high-frequencyresponse are:

Junction capacitancesCbe, Cbc, Cce

Wiring capacitancesCwi, Cwo

95


High-Frequency Cutoff—Input Network (fHi)

RCCwirπ Cπ CMiCMoCwo CceRLRBB

RS

Cwi

gm vππ π Mi Mowo ce

RTHi

LBB

vS

Ci CoRTHo

vCi gm vπ CoRTHo

RTHi

Hi1f =

96

vS

iThiHi CR2f

π=

isThii21sThi ZRRRRRRR == @

bcvbeWiMibeWii )CA(1CCCCCC −++=++=

12/15/2009

49


High-Frequency Cutoff—Output Network (fHo)

RCCwirπ Cπ CMiCMoCwo CceRLRBB

RS

Cwi

gm vππ π Mi Mowo ce

RTHi

LBB

vS

Ci CoRTHo

vCi gm vπ CoRTHo

RTHi

oThoHo CR2

1fπ

=

97

vS

LCLoTho RRRZR ==

bcceWoMoceWoo CCCCCCC ++=++=


Example:Determine the values of higher cutoff frequency for the amplifier circuit below . Given gm=50mS Cwo=8pF, Cwi=6pF, Cce=1pF, Cπ=Cbe=100pF, Cµ=Cbc=3pF and VVA = ∞

98

12/15/2009

50


vS

+vi-

gm

vπ

RSR

BB rπ RC

RL

+vπ-

+vo-

( ) ( )( ) 603k2k50mSRRgVVA LCm

i

OV −=−=−==

Voltage Gain:

Miller Capacitances:

99

3.05pF60

1603pFA

1ACCV

VµMo =⎟

⎠⎞

⎜⎝⎛

−−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

( ) 183pF(-60)]3pF[1-A1CC VµMi ==−=

Miller Capacitances:


vS

+vi-

gm vπ

RSRBB rπ RC

RL

+vπ-

+vo-

At Input:

620.7ΩrRRRR 21STHi == π

( )( ) 887.239kHz620.7Ω289pF2π

1RC2

1fTHii

Hi ===π

289pF183pF100pF6pFCCCC Miwii =++=++= π

100

At Output:( )( )pTHii

1.2kΩRRR LCTHo ==

12.05pF3.05pF1pF8pFCCCC Mocewoo =++=++=

( )( ) 11.05MHz1.2kΩ12.05pF2

1RC2

1fTHoo

Ho ===ππ

12/15/2009

51


There are two cutoff frequencies i.e fHi = 859.11kHz and fHo = 11.01MHz.Therefore the lowest cutoff frequency (fH) that dominates the overall frequency response of the

lifi i f f 859 11kHamplifier is fH = fHi = 859.11kHz.

Actual Plot

Roll-off:20dB/decade or

6dB/octave

Bode Plot

AV[dB] 887kf [Hz]

110M2M11M8.87M0

-3

101

Roll-off:40dB/decade or

12dB/octave

6dB/octave

octave

12dB

-20


1.25 Short Circuit Current Gain For C-E

RS

Cµ = Cbc

++ +RSRBB rπ RC RLgm vπ

vS

Cbe = Cπ

+vπ−

+vi−

+vo−

CE with shorted output

102

µMVMi Cf2

1Cf21orZ

A1ZZ

iππ

==−

=

From Miller Theorem:

12/15/2009

52


bcµM CCCi

==

0AZZ V =⎞

⎜⎜⎛

= 01A

ZZV

Mo =⎠

⎜⎜⎝ −

=

RSRBB rπ gm vπ

vS

Cbe = Cπ

+vπ−

+vi−

+vo−

CMi = Cµ

ioi1ii

103

Z1

vS

Simplified equivalent circuit


Since RBB >> rπ therefore i1≈ ii

Current gain:1

O

i

Oi i

iiiA ≈=

1i ii

πVgi mO =

ππ

π

ππ r)C(Cjω1

rCjω1

Cjω1rZ

µµ1 ++

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎞⎛

104

⎟⎟⎠

⎞⎜⎜⎝

⎛ ++==

π

πππ

π

r)C(Crjω1

VZVi µ

11

)C(Crjω1rg

ii

iiA

µ

m

1

O

i

Oi ++

=≈=ππ

π

12/15/2009

53


)C(Crf2j1)C(Crjω1Ai π

ββ++

=++

=∴

ππ rgβorgβr mm

==Since

)C(Crf2j1)C(Crjω1 µππµππ π ++++

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

β

i

ffj1

βACurrent gain (Ai) as a function of frequency:

g1f mβ ==Where:

105

β)C(C2r)C(C2βµππµπ ππ ++

⇒ This shows that the hfe parameter (or β) of a

transistor varies with frequency.

20 log β

Ai [dB]

3dB


fTfβ0

f

Note that at f = f the gain is

106

Note that at f = fT the gain is 0 dB or |Ai= 1. Thus

2

β

T

i

ff1

β1A

⎟⎠⎞

⎜⎝⎛+

==

12/15/2009

54

Note that fT >> fβ. Therefore (fT/fβ)2 >> 1.Current gain become:


βT2i βf f or fββ1A ==== β

β

T2

β

T ff

ff1 ⎟

⎠⎞

⎜⎝⎛+

)C(C2g

)C(Cr2β f m

Tµππ +

=+π

=µππ

107

fT = βfβ shows that fT is a current gain bandwidth product.This is because, the bandwidth, BW = fH - fL ≈ fH = fβand β is the current gain at middle frequency(|Ai|= β).


For common-base amplifier (CB), the following equation can be used to relate between fβ and fα:

1)(βff or )(1-ff ββ +=α= αα )(β)( ββ αα

fα is the higher cutoff frequency for common-baseConfiguration.

But Tββ fβf1)(βff =≈+=α

108

And α ≈ 1 or 0 dB

12/15/2009

55


Note that fT is closer to fα thus fα ≈ fT can be

dused.

109

β = hfe and α = hfb versus frequency at high frequency.


Full Frequency Response of a BJT Amplifier

110

Note the highest lower cutoff frequency (fL) and the lowest upper cutoff frequency (fH) are closest to the actual response of the amplifier.

12/15/2009

56


1.26 Common-Base Frequency Response

111


RE rπ RC RL

RS+vO

+vi

i i ie

ib

ioC1 C2

+vπ

gmvπ

CB equivalent circuit at

lowvS

--

ZiZO

-

⎞⎜⎛ r

low frequency

lower cutoff frequency due to coupling capacitors C1 and C2 are:

112

⎟⎠

⎞⎜⎜⎝

⎛+

=1β

rRZ Eiπ

CO RZ = ( ) 1iSL1 CZR2

1f+

=π ( ) 2Lo

L2 CRZ21f+

=π

capacitors C1 and C2 are:

12/15/2009

57


At high frequencies, the analysis becomes simplercompared to CE because of small value of Cce between input and output which can be neglected. Thus Miller theorem will not be used.

113

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=1β

rRRR ESTHiπ

CLTHo RRR =

bewii CCC += bcwoo CCC +=

iTHiHi CR2

1fπ

=oTHo

Ho CR21f

π=

The dominanthigher cutofffrequency is thelowest betweenthe two.


1.27 Common-Collector Frequency Response

R B

+V CC

C1

RS

B

RE RLv S

C1

C2

RS

RB

RE RL

vS

C2

gmvπrπ

ZZ

114

ZoZi

CC equivalent circuit at low frequency.

12/15/2009

58


( )( )[ ]1βRRrRZ LEBi ++= π

( ) rRR ⎞⎜⎛ +( )

EBS

0 R1β

rRRZ ⎟

⎠

⎞⎜⎜⎝

⎛+

+= π

( ) 1iSL1 CZR2

1f+

=π ( ) 2Lo

L2 CRZ21f+

=π

115

The dominant cutoff frequency is the highest between fL1and fL2.


For high frequency analysis, there are two methods for finding the cutoff frequencies.First method is using Miller Theorem, since the parallel combination of capacitor Cbe and resistor rπ

b t i t d t t f th lifi

RS

R R

Cbe

C

rπR CCwi Cbc

appear between input and output of the amplifier.

116

RB RL

vS

Cce

gmvπ

RE Cwowi bc

CC equivalent circuit at high frequency

12/15/2009

59


RS

RB RLvS

CMi CceRECwoCwi CbcRMi

CMoRMoS gmvπ

RTHi RTHoCi Co

Equivalent circuit after using Miller Theorem

⎞⎜⎛ 1A

117

( )vbeMi A1CC −=

vMi A1

rR−

= π

⎟⎠

⎞⎜⎜⎝

⎛ −=

v

vbeMo A

1ACC

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=1A

ArRv

vMo π


MiBSTHi RRRR =

Mibcwii CCCC ++=1

MoLETHo RRRR =

Mocewoo CCCC ++=

1

Second method is using open-circuit time constants(Thevenin equivalent resistance looking from each capacitor).

iTHiHi CR2

1fπ

=oTHo

Ho CR21f

π=

118

capacitor).

For this example, there are only three Thevenin equivalent resistance that we need to find, i.e Cbe, Ci= Cwi//Cbc and Co= Cwo//Cce.

12/15/2009

60


To find the equivalent resistance looking from capacitor Ci, other capacitors must open circuit.

The equivalent circuit can be redrawn as shownThe equivalent circuit can be redrawn as shown below.

RS

Cbe

r

119

RS

RB

RL

vS

RE

Cwi

+ Cbc

gm

v π

rπC

ce+ C

wo


( )( )[ ]1βRRrRRR LEBSTH1 ++= π

( ) ( )[ ]RRRR 1

( )bcwiTH1H1 CCR2

1f+

=π

( ) ( )[ ]( ) ( )( )1βRRRRr

RRRRrR

LEBS

LEBSTH2 +++

+=

π

π

( )⎟⎟⎠

⎞⎜⎜⎝

⎛+

+=

1βRRr

RRR BSLETH3

π

beTH2H2 CR2

1fπ

=

( )cewoTH3H3 CCR2

1f+

=π

120

The dominant cutoff frequency:

H3H2H1

H

f1

f1

f1

1f++

=

12/15/2009

61


Final Exam Paper 2005/06 semester 2Determine the higher cutoff frequency, fH if β = 100, fT = 450 MHz, Cbc = Cµ = 2 pF, ICQ = 0.97 mA and VA = ∞. 1.21 MHz

121


Final Exam Paper 2006/07 semester 2Determine fL due to coupling capacitors and draw the Bode plot of the frequency response. Given β = 200, ICQ = 5.515 mA and VA = ∞. 26.37 Hz; 1.52kHz

122

12/15/2009

62


T H E T H E EE N DN D

123

chapter 1 ac analysis skee 2253

Education