ESS2220-10to11

Chapter 10Chapter 10 Diodes Diodes

1. Understand diode operation and select diodes for various applications.

2. Analyze nonlinear circuits using the graphical load-line technique.

3. Analyze and design simple voltage-regulator circuits.

4. Solve circuits using the ideal-diode model andpiecewise-linear models.

5. Understand various rectifier and wave-shaping circuits.

6. Understand small-signal equivalent circuits.

2

10. Diodes – Basic Diode Concepts

10.1 Basic Diode Concepts

10.1.1 Intrinsic Semiconductors

* Energy Diagrams – Insulator, Semiconductor, and Conductor

the energy diagram for the three types of solids

3



* Intrinsic (pure) Si Semiconductor:

Thermal Excitation, Electron-Hole Pair, Recombination,

and Equilibrium

)cm 105~ is

density atom crystal Si:Note (

K 300 at crystal Siintrinsic for

cm 101.5pn

density hole density electron

: reached is

ionrecombinat and excitation

between mequilibriu When

3-22

3-10ii

4



*Apply a voltage across

a piece of Si:

electron current

and hole current

5


10.1.2 N- and P- Type Semiconductors

* Doping: adding of impurities (i.e., dopants) to the intrinsic semi-conductor material.

* N-type: adding Group V dopant (or donor) such as As, P, Sb,…

carrier cahage minor the hole

carrier charge major the electron

call We

pp ,nNn

nconceratio donor the Nn

material type-n In

101.5pnpn

300K at SiFor

ctor semicondua for constantpn

iid

d

2102i

2i

6


10.1.2 N- and P- Type Semiconductors

* Doping: adding of impurities (i.e., dopants) to the intrinsic semi-conductor material.

* P-type: adding Group III dopant (or acceptor) such as Al, B, Ga,…

carrier cahage minor the electron

carrier charge major the hole

call We

nn ,pNp

nconceratio acceptor the Np

material type-p In

101.5pnpn

300K at SiFor

ctor semicondua for constantpn

iia

a

2102i

2i

7


10.1.3 The PN-Junction

* The interface in-between p-type and n-type material is called a

pn-junction.

. V ,T as :300K at

Ge for 0.3V and Sifor 0.7V6.0V potential barrier The

B

B

8


10.1.4 Biasing the PN-Junction

* There is no movement of charge through a pn-junction at equilibrium.

* The pn-junction form a diode which allows current in only one direction and prevent the current in the other direction as determined by the bias.

9



*Forward Bias: dc voltage positive terminal connected to the p region and negative to the n region. It is the condition that permits current through the pn-junction of a diode.

10


*Forward Bias: dc voltage positive terminal connected to the p region and negative to the n region. It is the condition that permits current through the pn-junction of a diode.

11



*Forward Bias:

12


*Reverse Bias: dc voltage negative terminal connected to the p region and positive to the n region. Depletion region widens until its potential difference equals the bias voltage, majority-carrier current ceases.

13


*Reverse Bias:

majority-carrier current ceases.

* However, there is still a very small current produced by minority carriers.

14



* Reverse Breakdown: As reverse voltage reach certain value, avalanche occurs and generates large current.

15


10.1.5 The Diode Characteristic I-V Curve

16


10.1.6 Shockley Equation

* The Shockley equation is a theoretical result under certain simplification:

0v whenapplicable not is equation This

Vn

vexpIi 0.1V,v when

C101.60q constant, sBoltzman' the is k

voltage thermal the is 300K at 0.026V q

TkV

t,coefficien emission the is 2 to 1 n current,

n saturatio(reverse) the is 300K at A10I where

1Vn

vexpIi

D

T

DsDD

19-

T

14-s

T

DsD

17

10. Diodes – Load-Line Analysis of Diode Circuits

10.2 Load-Line Analysis of Diode Circuit

equations. KVL or KCL writeto difficult is It

1Vn

vexpIi: diode a is there whenbut

,...dt

diLv ,

dt

dvCi iR,v use can We

T

DsD

Analysis" Line-Load"

the perform can we

given, is diode the

of curve V-I the If

viRV

:gives KVL

shown,circuit the For

DDSS

18


Example 10.1- Load-Line Analysis

point)-(Q point operating the at

voltage and current diode the :Find

diode the of curve V-I the

,Ω1kR 2V,V :Given


SS

mA 1.3i V, 0.70V

point operating the at

analysis line-load perform

vi 10002

i.e., ,viRV

DQDQ

DD

DDSS

19


Example 10.2 - Load-Line Analysis


voltage and current diode the :Find

diode the of curve V-I the

,k 10R V, 10Vss :Given


mA 0.93i V, 0.68V


analysis line-load perform

vi 10k10

i.e., ,viRV

DQDQ

DD

DDSS

20

10. Diodes – Zener Diode Voltage-Regulator Circuits

10.3 Zener-Diode Voltage-Regulator Circuits

10.3.1 The Zener Diode

* Zener diode is designed for operation in the reverse-breakdown region.

* The breakdown voltage is controlled by the doping level (-1.8 V to -200 V).

* The major application of Zener diode is to provide an output reference that is stable despite changes in input voltage – power supplies, voltmeter,…

21

10. Diodes – Zener-Diode Voltage-Regulator Circuits

10.3.2 Zener-Diode Voltage-Regulator Circuits

* Sometimes, a circuit that produces constant output voltage while operating from a variable supply voltage is needed. Such circuits are called voltage regulator.

* The Zener diode has a breakdown voltage equal to the desired output voltage.

* The resistor limits the diode current to a safe value so that Zener diode does not overheat.

22


Example 10.3 – Zener-Diode Voltage-Regulator Circuits

Actual Zener diode

performs much better!

V 20V

and V 15V for voltage output the :Find

1kR curve, V-I diode Zenerthe :Given

SS

SS

o

SSo

SSo

DDSS

v in change 0.5V

input in change 5V

V 20V for V 10.5v

V 15V for V 10.0v

:have wepoint-Q the From

0viRV

:line load the gives KVL

23


10.3.3 Load-Line Analysis of Complex Circuits

* Use the Thevenin Equivalent

24


Example 10.4 – Zener-Diode Voltage-Regulator with a Load

SL

LSS

I currents sourceand v voltage load the :Find

6kR ,1.2kR 24V,V curve, V-I diode Zener:Given

mA 11.67 )/Rv-(VI

V 10.0-vv

0viRV

k1RR

RRR ,V20

RR

RVV Equivalent Thevenin Applying

LSSS

DL

DDTT

L

LT

L

LSST

25


Quiz – Exercise 10.5

100mAi and 20mA,i 0,i for v voltage output the :Find

shown.as curve V-I doide Zenerthe and circuit the :Given

LLLo

26

10. Diodes – Ideal-Diode Model

10.4 Ideal-Diode Model

* Graphical load-line analysis is too cumbersome for complex circuits,

* We may apply “Ideal-Diode Model” to simplify the analysis:

(1) in forward direction: short-circuit assumption, zero voltage drop;

(2) in reverse direction: open-circuit assumption.

* The ideal-diode model can be used when the forward voltage drop and reverse currents are negligible.

27


10.4 Ideal-Diode Model

* In analysis of a circuit containing diodes, we may not know in advance which diodes are on and which are off.

* What we do is first to make a guess on the state of the diodes in the circuit:

YES" ALL" until iterates

guess.... another make YES ALL not

BINGO! YES ALL

negative is v if check :diodes" off assumed" For (2)

positive; is i if check :diodes" on assumed" (1)For

D

D

28


Example 10.5 – Analysis by Assumed Diode States

on D off,D

assume (1)

21

on D and off isD assumingby circuit the Analysis 21

OK! not 7Vv

OK! 0.5mAi

D1

D2

off D on, D

assume (2)

21

OK! V -3v

OK! mA 1i

D2

D1

29


Quiz – Exercise 10.8c

* Find the diode states by using ideal-diode model. Starting by assuming both diodes are on.

on D

on D

assume (1)

4

3

OK mA, 6.7i

OK not mA, -1.7i

4 D

3 D

on D and off D assume (2) 43

OK V, -5v

OK mA, 5i

3 D

4 D

30

10. Diodes – Piecewise-Linear Diode Models

10.5 Piecewise-Linear Diode Models

10.5.1 Modified Ideal-Diode Model

* This modified ideal-diode model is usually accurate enough in

most of the circuit analysis.

31

10. Diodes – Piecewise-Linear Diode Models

10.5.2 Piecewise-Linear Diode Models

aa ViRv

32

10. Diodes – Rectifier Circuits

10.6 Rectifier Circuits

* Rectifiers convert ac power to dc power.

* Rectifiers form the basis for electronic power suppliers and battery charging circuits.

10.6.1 Half-Wave Rectifier

33


* Battery-Charging Circuit

* The current flows only in the direction that charges the battery.

34


* Half-Wave Rectifier with Smoothing Capacitor

* To place a large capacitance across the output terminals:

35


10.6.2 Full-Wave Rectifier Circuits

* Center-Tapped Full-Wave Rectifier – two half-wave rectifier with out-of-phase source voltages and a common ground.

* When upper source supplies “+” voltage to diode A,

the lower source supplies “-” voltage to diode B;

and vice versa.

* We can also smooth the output by using a large capacitance.

36


10.6.2 Full-Wave Rectifier Circuits

* The Diode-Bridge Full-Wave Rectifier:

A,B C,D

37

10. Diodes – Wave-Shaping Circuits

10.7 Wave-Shaping Circuits

10.7.1 Clipper Circuits

* A portion of an input signal waveform is “clipped” off.

38

10. Diodes – Wave-Shaping Circuits

10.7 Wave-Shaping Circuits

10.7.2 Clamper Circuits

* Clamp circuits are used to add a dc component to an ac input waveform so that the positive (or negative) peaks are “clamped” to a specified voltage value.

39

10. Diodes – Linear Small-Signal Equivalent Circuits

10.8 Linear Small-Signal Equivalent Circuits

* In most of the electronic circuits, dc supply voltages are used to bias a nonlinear device at an operating point and a small signal is injected into the circuits.

* We often split the analysis of such circuit into two parts:

(1) Analyze the dc circuit to find operating point,

(2) Analyze the small signal ( by using the “linear small-

signal equivalent circuit”.)

40



* A diode in linear small-signal equivalent

circuit is simplified to a resistor.

* We first determine the operating point

(or the “quiescent point” or Q point) by

dc bias.

* When small ac signal injects, it swings the Q point slightly up and down.

* If the signal is small enough, the characteristic is straight.

voltage diode in change smallthe is v

current diode in change smallthe is i

vvd

idi

D

D

D

QD

DD

QD

D

vd

id

41



QD

D

vd

id

QD

Td

d

dd

ddDD

d

DDD

QD

DD

1

QD

Dd

I

Vnr :have we

equation, Shockley the applyingby e,Furthermor

r

vi

: signalsac for have wechanges, small

denoting v and iby v and i Replace

r

vi v

vd

idi

:have willWe vd

idr

:as diode the of resistance dynamic the Define

42



* By using these two equations, we can treat diode simply as a linear resistor in small ac signal analysis.

* Note: An ac voltage of fixed amplitude produces different ac current change at different Q point.

QD

Td

d

dd I

Vnr ,

r

vi

43

10. Diodes – Linear Small-Signal Equivalent Circuits10.8 Linear Small-Signal Equivalent Circuits

current. andvoltage diode ousinstantane

totalthe represent i and v (3)

signals.sc smallthe represent i and v (2)

point.Q the at

signalsdc the represent I and V (1)

DD

dd

DQDQ

dDQD

dDQD

vVv

iIi

QD

Td

d

dd I

Vnr ,

r

vi

44


Voltage-Controlled Attenuator

* The function of this circuit is to produce an output signal that is a variable fraction of the ac input signal.

* Two large coupling capacitors: behave like short circuit for ac signal and open circuit for dc, thus the Q point of the diode is unaffected by the ac input and the load.

Cj

1ZC

45


Voltage-Controlled Attenuator

QD

TddDQ I

Vnr :diode the of r the then ,I determine

point, Q diode the find to analysis dcapply First

1RR

R

v

vA :divider voltage on based ,

r1R1R1

1R

signal.)ac for circuit shorta to equivalent is sourcevoltage dc the

voltage, ac no but current of component ac an has sourcevoltage dc the :(note

: analysis signalac smallperform weNext,

p

p

in

ov

dLCp

46


Exercise 10.20 - Voltage-Controlled Attenuator

0.026VV withI

Vnr ,

R

0.6-VI

point,Q diode the find to analysisdc apply First

10.6V and 1.6V for A and

0.6VVassuming values point-Qthe :Find

300Kat 1ndiode ,Ω2kRR ,Ω100R :Given

TQD

Td

C

CDQ

Cv

f

LC

p

p

in

ov

dLCp RR

R

v

vA ,

r1R1R1

1R

: analysis signalac small perform we Next,

Chapter 11: Amplifiers:Chapter 11: Amplifiers:Specifications and External CharacteristicsSpecifications and External Characteristics

48

11. Amplifiers – Basic Amplifier Concepts

11.1 Basic Amplifier Concepts

11.1.1 For Starting

* Ideally, an amplifier produces an output signal with identical wave-shape as the input signal but with a larger amplitude.

GainVoltage the is A

tvAtv

v

ivo

49


11.1.1 For Starting

* inverting and

non-inverting amplifiers

50


11.1.1 For Starting

* Often, one of the amplifier input terminals and one of the output terminals are connected to a common ground.

* The common ground serve as the return path for signal and the dc power supply currents.

51


11.1.1 For Starting

* Another example for common ground

52


11.1.2 Voltage-Amplifier Model

* Amplification can be modeled by a controlled source.

Amplifier

)A tham smaller is gain realthe :(note

gainvoltage circuit-openthe : /vvA

impedance) (or resistance outputthe is

terminals, outputthe withseries in :R

terminals. inputthe intolooking when

seen resistance equivalentthe is ,impedance) (or resistance inputthe :R

vo

iocvo

o

i

53


11.1.3 Current and Voltage Gains

load.the through

flowing currentthe is i

amplifier;the of

terminals inputthe into

delivered currentthe is i

o

i

vov

i

ov

L

iv

ii

Lo

i

oi

i

oii

A gainvoltage circuit-openthe than smallerusually is A gainvoltage The

gainvoltage the is v

vA where ,

R

RA

Rv

Rv

i

iA e,Furthermor

i

iA :currents input and output between ratiothe is A gain currentThe

54


11.1.3 Power Gain

L

i2viv

i rmsi rms

o rmso rms

i

o

rmsrms

i

o

R

RAAA

IV

IV

P

PG :havewe

,I and V values rmsthe of productthe is poweraverage the Since

P

PG :power inputthe to power outputthe of ratiothe is gain powerThe

55


Example 11.1 Calculating Amplifier Performance

gain powerthe and

gain currentthe find Also,

/VVA and /VVA

gainsvoltage the Find

iovsovs

12iv

9

L

ivi

ov,vsv

si

iv

isii

o

s

osv

4

Lo

Lov

i

LoLivo

i

ov

1016AAG ,102R

RAA

AAA effect,loading the todue :Note

5333RR

RA

)/RR(RV

V

V

VA

800082

810

RR

RA

V

)R/(RRVA

V

VA

56

11. Amplifiers – Cascade Amplifiers

11.2 Cascade Amplifiers

212i1ii

2ov1vov2i

1o2ov

1i

2i2ov

1i

2coov

2v1vv2i

2o

1i

1o

01

2o

1i

1o

1i

2ov

GGG , AAA addition, In

AAA v

vA

v

vA

v

vA

AAA v

v

v

v

v

v

v

v

v

vA

57


Example 11.2 – Calculating Performance of Cascade Amplifier

Find the current gain, voltage gain, and power gain

1121

42i2v2

71i1v1

62i1ii

L

2i2v2i

5

2i

1i1v1i

2v1vv2oL

L2vo2v

1o2i

2i1vo1v

10625.5GGG

1075.3AAG ,105.1AAG

1075AAA 750R

RAA ,10

R

RAA

7500AAA 50RR

RAA ,150

RR

RAA

1L2i RR

58


Example 11.3– Simplified Model for Amplifier Cascade

32ov1vov

2ov

1o2i

2i1ov1v

1015AAA

100A

150RR

RAA

59

11. Amplifiers – Power Supplies and Efficiency

11. 3 Power Supplies and Efficiency

* The power gain of an amplifier is usually large, the additional power is taken from the power supply.

100%P

Pη

:amplifier an of efficiencyThe

carcuits internal

in dissipated powerthe is P

signal output of powerthe is P

signal input of powerthe is P

PPPP

:energy of onConservati

IVIVP

is amplifierthe to supplied

poweraverage totalThe

s

o

d

o

i

dosi

BBBAAAs

60

11. Amplifiers – Power Supplies and Efficiency

Example 11.4 Amplifier Efficiency

%6.35P

Pη

W5.14PPPP

W5.22IVIVP

W8R

VP

rmsV8RR

RVAV

pW10W10R

VP

s

o

oisd

BBBAAAs

L

2o

o

oL

Livoo

11

i

2i

i

61

11. Amplifiers – Additional Amplifier Models

11.4 Additional Amplifier Models

Current-Amplifier Model

models. twothe for

samethe are R and R

R

RA

i

iA

:gain current circuit shortthe

R

vAiand

R

vi

approach Norton vs. Thevenin

oi

o

ivo

i

oscisc

o

ivoosc

i

ii

62


Example 11.5 – Transform Voltage-Amplifier to

Current-Amplifier Model

3

o

ivo

i

oscisc

o

ivoosc

i

ii

10R

RA

i

iA

R

vAiand

R

vi

63


11.4 Additional Amplifier Models

Transconductance-Amplifier Model

Transresistance-Amplifier Model

i

oscmsc v

iG

i

oscmsc i

vR

64

11. Amplifiers – Ideal Amplifiers

11.6 Ideal Amplifiers

Gain!Voltage Maximum

max vAv 0, R

max. vv , R

:AmplifierVoltage

ivooo

sii

Gain! Current Maximum

max. iAi , R

max. ii 0, R

:Amplifier Current

iscioo

sciii

65

11. Amplifiers – Frequency Response

11.7 Frequency Response

* The gain of an amplifier is a function of frequency.

* Both amplitude and phase are affected.

Example 11.8

i

ov V

VA

dB40)100log(20Alog20A

shift.phase 45a isthere :Note

45100301.0

1510

V

VA 1510V , 301.0V

dB inamplitude the express and gainvoltage complex the Find

15tπ2000cos10tv :is outpotthe

30tπ2000cos1.0tv:is amplifier certaina for inputThe

vdBv

i

ovoi

0

i

66


Gain as a Function of Frequency

67


AC Coupled Amplifiers

The gain drops to zero at dc (low frequency).

68


DC Coupled Amplifiers

Amplifiers that are realized as integrated circuits are often dc coupled, because capacitors or transformers can not be fabricated in integrated form.

69


High-Frequency Drop Off

* The small amount of capacitance in parallel or inductances in series with the signal path in the amplifier circuit will cause the gain of the amplifier to drop at high frequencies.

f as short fC,π1/j2 f as open fL,πj2

70


Half-Power Frequencies and Bandwidth

* Bandwidth is the distance between the half-power frequencies.

* Half-power frequencies:

* Wideband (Baseband) Amp.

* Narrowband (Bandpass) Amp.

dB -3.01)220log(1/

A from 3dB 2/AA midmid

71

11. Amplifiers – Linear Waveform Distortion

11.8 Linear Waveform Distortion

* Distortion may occur even though the amplifier is linear (i.e., obeys superposition principle).

Amplitude Distortion

If a signal contains components of various frequencies, the output waveform may be distorted due to the frequency response of the amplifier gain.

72


Phase Distortion

* If the phase shift of an amplifier is not proportional to the frequency, phase distortion may occur.

:gainsthe having samplifiierthree and

tπ6000costπ2000cos3tv

:like input anhave Assume we

i

45tπ6000cos1045tπ2000cos30tv

135tπ6000cos1045tπ2000cos30tv

tπ6000cos10tπ2000cos30tv

:like look wouldoutputThe

oC

oB

oA

73


11.8 Linear Waveform Distortion

* In order to avoid distortion:

(1) the magnitude of the gain must be constant against frequency.

(2) The phase response must be proportional to the frequency.

delay!time same The

1:1

)ω/ω(Tπ2

)ω/ω(θ:T

π2

θ

Tπ2

θ:T

π2

θtΔ:tΔ

T:Tω:ωθ:θ

121211

11

22

11

21

122121

74

11. Amplifiers – Pulse Response

11.9 Pulse Response

75

11. Amplifiers – Nonlinear Distortion

11.10 Nonlinear Distortion

76

11. Amplifiers – Differential Amplifiers

11.11 Differential Amplifiers

* A Differential amplifier has two input sources, the output voltage is proportional to the difference between the two input voltages.

Non-inverting input

Inverting input

tvAtvA

tvtvAtv

2id1id

2i1ido

iddo

d

2i1iid

vAv

:as amplifier ildifferentaa

of outputthe write canWe

gain ildifferentathe A and

vvv

:signal aldifferentithe Define

77

11. Amplifiers – Differential AmplifiersElectrocardiogram

* The desired waveform is given by the difference between the potentials measured by the two electrodes, i.e., the output of an ideal differential amplifier.

* While both electrodes (also act like antennas) pick a common-mode signal (noise) from the 60 Hz power line:

)v-(vAvAv i2i1diddo

)φcos(377tVv ),φcos(377tVv nni2nni1

78

11. Amplifiers – Differential AmplifiersElectrocardiogram

* Ideally, the common-mode signal was nullified by the differential amplifier.

* However, real differential amplifier responds to both differential-mode and common-mode signals:

Common-Mode Rejection Ratio (CMRR):

* At 60 Hz, CMRR of 120 dB is considered good.

)v-(vAvAv i2i1diddo

)φcos(377tVv

),φcos(377tVv

nni2

nni1

signalmode -commonthe v and signalmode -aldifferentithe is v

gain,mode -commonthe is A where vAvAv

icmid

cmicmcmiddo

cm

d

A

Alog20CMRR

79

11. Amplifiers – Differential Amplifiers

Example 11.12 Determination of Minimum CMRR Specification

* Find the minimum CMRR for an electrocardiogram amplifier if the differential gain is 1000, the desired differential input signal has a peak amplitude of 1 mV, the common-mode signal is 100 V-peak 60 Hz sine wave, and it is desired that output contain a peak common-mode signal that is 1% or less of the peak output caused by the differential signal.

dB140 CMRR :Ans

dB14010

1000log20

A

Alog20CMRR

10V100

V01.0A V 0.011%V 1 v peak

V 1v 1000 A mV, 1v peak

4cm

d

4cmcmo

doddi

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