Chapter 3 Diode

7/22/2019 Chapter 3 Diode

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Diodes

1



Diode• Class of non-linear circuits

– having non-linear v-i Characteristics

• Uses

– Generation of :

• DC voltage from the ac power supply

• Different wave (square wave, pulse) form

generation – Protection Circuits

– Digital logic & memory circuits



Creating a Diode

• A diode allows current to flow in one

direction but not the other.

• When you put N-type and P-type silicon

together gives a diode its unique

properties.





Diode

Equivalent circuit in the reverse direction

Equivalent circuit in the forward direction.



Reverse Bias

• -ve voltage is applied to Anode• Current through diode = 0 (cut off

operation)

• Diode act as open circuit

Forward Bias+ve voltage applied to Anode

• Current flows through diode

• voltage Drop is zero (Turned on)

Operation



The two modes of operation of ideal diodes

Forward biased

Forward Current 10

mA

Reverse biasedReverse Voltage 10 V



x .

v D= 1.5v

i D=1. 5

1

= 1.5A

v D= 0

1.5v

1Ω

v D

+

−

i P v

D

+

−

i D= 0

1Ω

1.5v



Rectifier circuit

Input waveformEquivalent circuit when v i 0

Equivalent circuit when v i ≤ 0

Waveform across diode

Output waveform.



Exercise 3-3

i D=

10− 0

1k Ω = 10mA

v D=

1

t 2− t

1

∫t 1

t 2

vidt

v D=1

2π [∫0π

10 sinθ + ∫π

2π

0dt ]v D= −

1

2π ∣10cosθ ∣0π

=−

10

2π (− 1− 1)=10

π = 3.18V







Battery Charger

sin θ =1

2⇒θ = 30

0

Conduction Angle=

π −

2θ=

120

0

one− third of cycle

24sinθ = 12 V



Figure 3.6 Circuits for Example 3.2. Diodes are ideal , Find the value of I and V



Example 3.2.

Assumption Both Diodes are conducting



V = 0, V B= 0

I D2=10− 0

10k = 1mA

I 5k Ω= I D1+ I D2=0− (− 10 )

5k = 2 mA

From

above

equation

I D1

should

be1

mA It is not possible

Node A

Node B

Assumption Both Diodes are conducting

Not Possible

Thus assumption of both diode

conducting is wrong



Example 3.2(b). Assumption # 2Diodes 1 is not conducting Diodes 2 is conducting

I D2=10− (− 10)

15=

20

15= 1.33mA

V B= V A= 3.3 V , I D1= 0, I D2= 1.33mA

Assumption is correct

V A= 10− (1.33)(5k )= 3.3v

V B= (1.33)(10k )− 10= 3.3v



Figure E3.4

Diodes are ideal , Find the value of I and V

Fi E3 4



Figure E3.4


I= 2mA

V= 0V

I= 0A

V= 5VI= 0A

V= -5V

I= 2mA

V= 0V



Figure E3.4 Diodes are ideal , Find the value of I and V

I= 3mAV= 3V

I= 4mAV= 1V



Figure P3.2 Diodes are ideal , Find the value of I and V



Figure P3.2 Diodes are ideal , Find the value of I and V

Diode is conducting

I = 0.6 mA

V = -3V

Diode is cut-off

I = 0 mA

V = 3VDiode is conducting

I = 0.6 mA

V = 3V Diode is cut-off

I = 0 mA

V = -3V

P bl 3 3



D1 Cut-Off & D2 Conducting

I = 3mA

Problem 3-3

D1 Cut-Off & D2 Conducting

I = 1mA , V=1 V




Figure P3.4

In ideal diodes circuits, v1 is a 1-kHz, 10V peak sine wave.

Sketch the waveform of vo

i



In ideal diodes circuits, v1 is a 1-kHz, 10V peak sine wave.


Vp+ = 10V

Vp- = 0V

f = 1 K-Hz

Vp+ = 0V

Vp- = - 10V

f = 1 K-Hz

Vo = 0V



Figure P3.4 In ideal diodes circuits, v1 s a 1-kHz, 10V peak sine wave.


Vp+ = 10V

Vp- = 0V

f = 1 K-Hz

Vp+ = 10V

Vp- = -10V

f = 1 K-Hz

Vp+ = 10V

Vp- = 0V

f = 1 K-Hz





Figure P3 4



Figure P3.4

In ideal diodes circuits, v1 s a 1-kHz, 10V peak sine wave.


Vp+ = 0V

Vp- = -10V

f = 1 K-Hz

V0 = 0V Vp+ = 10V

Vp- = -5V

f = 1 K-Hz





Vp+ = 10V

Vp- = -5V

f = 1 K-Hz

P bl 3 4(k)



Problem 3-4(k)

vi

⇒10 V peak @ frequency 1000 H z

v i= 10sin2000 πt

For Vi > 0 V D1 is cutoff D2 is conducting v o =1V For Vi < 0 V is conducting D2 is cutoff v o =v i +1V

- 9 V

P bl 3 4(k)



Problem 3-4(k)



Figure P3.6

X = A . BX = A + B



Problem 3-4 (c)

vi⇒10Vpeak @frequency 1000 H z

vi= 10sin2000 πt

v o =zero

Problem 3 4(f)



Problem 3-4(f)

+ve Half Cycle with 10 V peak

at 1 KHz

kHz 10-V peak sine wave.



Problem 3-4(h)


vi= 10sin2000 πt v o =zero




vi= 10sin2000 πt

Problem 3.5

vi is 10 V peak sine wave and I = 100 mA current source. B is battery of

4.5 V . Sketch and label the iB

4.5 v

100 mA



Solution P3-5

vi> 4.5V ,D1 cutoff all current flows thru battery

Conduction angle

10sinθ = 4.5V⇒θ = sin− 1(0.45)= 26.70 ,153.30

Conduction angle= π − 2θ= 126.60

Fraction of cycle that i B of 100mA flows=126.6

360= 0.35

v i⇒10 Vpeak @ frequency 1000 H zv i= 10sin2000 πt B = 4 .5 V

vi< 4.5V , D

1 conducts D

2 cutoff

All current flows thru D1 , i

B= 0A4.5 v

100 mA



Problem 3-5

100 mA

4.5 v



Problem 3-5

4.5

10

100 mA

i Baverage=1

T ∫ i B dt =

1

T [100× 0.35 T ]= 35 mA

REVERSE PO ARITY



REVERSE POLARITY

PROTECTOR

REVERSE POLARITY



• The diode in this circuit protects a

radio or a recorder etc... In the event

that the battery or power source is

connected the wrong way round, thediode does not allow current to flow.

REVERSE POLARITY

PROTECTOR

Problem 3-9



D1& D2 ConductingI 1=1mA

I 3=0.5 mA

I 2 =0.5 mA

V= 0 V

D1=off, D2=OnI 1= I 3=0.66 mA

V = -1.7 V

Problem 3-9

I1

I3

2

I1

I3

2

P bl 3 10



Problem 3-10

D conductingI=0.225 mA

V=4.5V

D is not conducting

I=0A

V=-2V

P bl 3 16



Problem 3-16

V RED GREEN

3V On Off D1 conducts0 V Off Off

-3 V Off On D2 conducts

Quiz No 3 DE 28 EE A



Quiz No 3 DE 28 EE -A

Sketch vO if vi is 8 sin

Find out the conduction angle for the diode &

fraction of the cycle the diode is conducting

Solution Quiz No 3



Solution Quiz No 3

8V

I1

I2

8= 4I1− 2I2

− 2= − 2I1

+ 3I2

2I2= 2⇒ I 2= 1mA

Vo= 1× 1+ 2= 3V

vi/2

I

I =

8

2

− 2

2 = 1 mA

Vo= 1× 1+ 2= 3V

10-10-07

Conduction angle⇒2θ= 60o

4sinθ = 2⇒θ = 30

Fraction of Cycle the diode conducts= π − 2θ2π

= 13= 33



22-10-07



fraction of the cycle the diode is conducts

D1

never conducts

Vi<5V D2 is cut-off, Vo=5V

Vi>5V D2 is conducts

V omax= 5+

10− 5

2= 7.5V

+12 V

5

D1

D2


10sin θ = 5⇒θ = 30

Fraction of Cycle the diode conducts=π − 2θ

2π=

1

3= 33



Quiz No 3 DE 27 CE -B

D1

never conducts

Vi<5V D2 is cut-off, Vo=Vi

Vi>5V D2 is conducts



fraction of the cycle the diode is conducts

V omax= 5+

10− 5

2

= 7.5V


10sin θ = 5⇒θ = 30

Fraction of Cycle the diode conducts=π − 2θ

2π=

1

3= 33



Problem

• Assume the diodes are ideal,

sketch vo if the input is 10sin

(9)

• Find out the conduction angles for

Diode D1 & D2 (4) and the fractionof the cycle these diodes conduct. (2)

2< < 1



− 2< vi< 1⇒vo= vi

vi> 1V⇒

v0=

vi− 1

4 × 1+ 1v

ipeak = 10V ⇒v

opeak = 4.25V

vi<− 2V⇒vo=− 2V



− 2< vi< 1⇒vo= vi

vi

> 1V⇒v0

=v

i− 1

4× 1+ 1

vi= 2V⇒v

o= 1.25V

vi<− 2V⇒vo=− 2V v0=

vi− 1

4× 1+ 1V

-2V

Two-dimensional representation of the silicon crystal



p y

14 Electrons

Silicon and Germanium



Silicon Lattice



Silicon Lattice



At room temperature, some of the covalent bonds are broken by

thermal ionization.

Each broken bond gives rise to a free electron and a hole, both

of which become available for current conduction.

Intrinsic Semiconductor



Electrons and holes

Semiconductor Current



Semiconductor Current

The Doping of Semiconductors



.

The Doping of Semiconductors



Valence Electrons

N Type



N Type

P Type



P Type





p-n Junction

• P Junction

– Concentration of holes is high

– Majority charge carrier are hole

• N Junction

– Concentration of electron is high

– Majority charge carrier are electron



Diffusion Current ID

• Hole diffuse across the junction from

the p side to the n side & similarly

electron

• Two current components add together

to form the diffusion current with

direction from p to n side

Drift Current I



Drift Current Is

• Diffusion current due to majority carrier

diffusion

• A component due to minority carrier

drift exists across the junction



(a)The pn junction with no applied voltage (open-circuited

terminals).

(b) The potential distribution along an axis perpendicular to the

F d Bi d C d ti



Forward Biased Conduction



• The polarity of applied voltage which can't produce any current is called

Reverse Bias.• The polarity of applied voltage which causes charge to flow through the

diode is called Forward Bias.





Terminal Characteristics

of

a Junction Diode

The diode i –

v relationship



The diode i – v relationship with some scales

expanded and others compressed in order to reveal

details.

Terminal Characteristics of a



Terminal Characteristics of a

Junction Diode

• Forward Biased Region v > 0

• Reversed Biased Region v < 0

• Breakdown Region v < -V ZK

Forward Biased Region




• Is Saturation current – Scale Current

– Is is constant at a given temperature

– Is is directly proportional to Cross-Sectional region of the diode, Is doubles if cross-sectional area is double

– Is is 10-15 A for small size diode

– Doubles in value for every 10OC rise in temperature

i= I s(e

v

nV T − 1)




• Thermal Voltage VT

– VT = kT/q• K = Boltzmann‟s constant = 1.38 X 10-23 Joules/Kelvin

• T = Absolute Temperature in Kelvin (273 +Temp in Co)

• q = Magnitude of charge = 1.6 X 10-19 Coulombs

– VT @ 20oC is 25.2mV, ~ 25 mV

• n is 1 or 2 depending on the material andthe physical structure of the diode

– n = 1 for Germanium Diode & n=2 for Silicon


i= I s(ev

nV T −

1)




b Relationship of the current i to the voltage v

holds good over many decades of current

(seven decades, a factor of 107

i= I s

evnV

T

ln i= ln( I s e

v

nV T

)⇒v

nV T ln I s

v= nV T

lni

I s


i= I s

(ev

nV T − 1)

i >> Is




I 1= I s e

v1

nV T

I 2= I s e

v2

nV T

I 2

I 1

= e

(v2− v

1)nV

T

(v2− v

1)= nV T

ln I

2

I 1

⇒2. 3 nV T

log I

2

I 1





• for v drop changes by

for n = 1

for n = 2

0.8v≈ 0.7v

I 2 I

1

= 10

2.3nV T ≈ 60mV

120 mV

¿v< 0.5v⇒cut − in− voltage

v= 0.6v

Forward Biased Region(v2− v1)= 2 .3nV T log

I 2 I 1

Illustrating the temperature dependence of the diode forward characteristi



At a constant current, the voltage drop decreases by

approximately 2 mV for every 1C increase in

temperature.

Figure E3.9



If V=1V at 20o C, Find V at

400C and 00C

At 20o C Reverse current Is = 1V/1M Ω= 1μ A

Since the reverse leakage current doubles for every 100 C increase,

At 400 C I = 4*1 = 4 μ A V = 4 μ A * 1MΩ = 4.0 V

At 0 C I = ¼ μ A V = 0.25 V

Is

Forward biased Diode



Characteristics

Example 3.3• A silicon diode displays a forward

voltage of 0.7 V at a current of 1mA.

Find Is at n=1 & 2

i= I s e

v

nV T ⇒ I s= ie

−vnV

T

η= 1 I s= 10− 3

e

− 0 . 7

25× 10− 3

= 6 . 9× 10− 16

A

η= 2 I s= 10− 3 e

− 0. 7

2× 25× 10− 3

= 8 .3× 10− 10

A

Ex 3.7



Ex 3.7

Silicon Diode with n=1 has VD=0.7V @

i=1mA. Find voltage drop at i=0.1mA &10mA

i= I s e

vnV

T ⇒ I s= ie

− vnV

T

η= 1 I s= 10− 3 e

− 0 . 7

25× 10− 3

= 6 . 9× 10− 15

A

For i= 0 .1 mA⇒V 1= ηV T ln i I s

= 25× 10− 3 ln 10

− 4

6 .9× 10− 16 ==0.64 V

For i= 10 mA⇒V 1= ηV T lni

I s= 25× 10− 3 ln

10− 2

6 . 9× 10− 16= 0.76 V

Solution P3-18



(a )

η= 2, Diode current = i= 1000IS

i= I S e

v

nV T ⇒1000 I S = I s e

v

2× 25× 10− 3

v= 0.345 V

(a) At what forward voltage does a diode for which n=2 conduct acurrent equal to 1000Is?

(b) In term if Is what current flows in the same diode when its

forward voltage is 0.7 V

(b)v= 0.7V

i= I S

e

vnV

T = I se0 .7

0 .05 = 1 . 2× 106 I S

P bl 3 23



Problem 3-23

• The circuit shown utilizes threeidentical diodes having n=1 andIs= 10 -14 A. Find the value of the

current I required to obtain an

output voltage Vo=2 V. Assume n=1

• If a current of 1mA is drawn away

from the output terminal by a

load, what if the change in theoutput voltage. Assume n=1

Solution 3-23



(b) Load current = 1mA, therefore I

DY = 2.81mA

I DY

I DX

= e

(v DY

− v DX

)

0.025 = e

(v DY

− 2 /3 )

0.025

ΔvoY = vO2− v01= 22.8mV

Info available n= 1 , I S = 10− 14 A,V o= 2V

The voltage across each diode isvo

3= v

DX =

2

3

I DX

= I S e

v DX

ηV T = 10− 14

e

2

3

0.025 = 3.81mA

The circuit shown utilizes three identical diodes

having n=1 and Is= 10 -14 A. Find the value of

the current I required to obtain an output voltage

Vo=2 V.

If a current of 1mA is drawn away from the output terminal

by a load, what if the change in the output voltage.

P bl 3 25



Problem 3-25

• In the circuit shown,both diode haven=1, but D1 has 10times the junction

area of D2. Whatvalue of V results?

Solution 3-25(a)In the circuit shown, both diode

have n=1, but D1 has 10 times the junction area of D2. What value of



V 0= V D2− V D1= ηV T ln10 I D2

I D1

. . . . . . . . . . . . . . . . .2

I D1= 10 I S2 e

V D1

ηV T

I D2

I D1=

I S2 e

V D2

ηV T

10 I S2 e

V D1

ηV T

= 0. 1 e

V D2

− V D2

ηV

T . . . . . . . . . . . . . .1

I D1

= I S1

e

V D1

ηV T I D2= I S2 e

V D2

ηV T

I S1= 10 I S2

V 0= V D2− V D1= 0.025× ln80

2

= 92.2 mV

I 1= I D2+ I D1⇒ I D2= I 1− I D1..........3

⇒ I D1= 2mA , I D2= 10− 2= 8mA

V results?

solution 2-25 (b) To obtain a value of 50 mV, what current I2 id needed.



V o= 50mA , Find ID1 , I D2

I D2= 0.01− I D1

I D2

I D1

= 0 . 1e

V D2

− V D2

ηV T =

I D2

0.01− I D2

= 0 .1e2

I D2

= 4.25mA

I D1=(10− 4.25)= 5.75mA

P bl 3 26



Problem 3-26

• For the circuit shown,both diodes are

identical, conducting

10mA at 0.7 V and 100

mA at 0.8 V.

• Find „n‟

• Find the value of R for which V = 80 m V.

Solution 3-26 (a)



Diodes are identical therefore IS ,η ,V

T are same

For Diode 1 ⇒V D1

= 0.7V@I D1

= 10mA

For Diode 2 ⇒V D2= 0.8V@I

D2= 100mA

V D2− V

D1= ηV

T ln

I D2

I D1

0 .8− 0 . 7= η× 0.025× ln100

10

η= 1.739

Find η

V = V D2− V D1= ηV T ln

I D2

I D1

= 0.08= 1.737× 0.025× ln0.01− I D1

I D1

I D1= 1. 4 mA

R=80

1. 4

= 57.1Ω

d R if Vo=80mV

P bl 3 36



Problem 3.36

Assuming identical diodes for whichVD =0.7V @ ID=1mA. Find R if V0 = 3 V

V Dx=3

4= 0.75V

I DX = I S e

V DX

ηV T

I D2

I D1

=e

V D2

ηV T

e

V D1

ηV T

= e

(V D2

− V D2

)

ηV T

I D2= I D1¿e

(V D2

− V D2

)

ηV T = 1× e

(0.75− 0 .7 )

25× 10− 3

= 7.389mA

.75= 0 .7+ ηV T lnI D2

10− 3⇒ I D2= 7.389 mA

R=10− 3

7.389× 10− 3

= 947Ω



Modeling the Diode

Forward

Characteristics

A simple circuit used to illustrate the analysis of circuits in which

the diode is forward conducting.



g

I D=V DD− V D

R

I D= I

S e

V D

ηV T

Graphical analysis of the circuit using the exponential diode m



Iterative Analysis using theE ti l M d l



Exponential Model

Determined the diode current ID andDiode voltage VD with VDD =5V and R

=1000 ohms. Diode has a current of

1mA @ a VD of .7 V, and that its voltagedrop changes by 0.1 V for every decade

change in current.

Solution



F irst iteration V D= 0.7V

I D= I S e

V D

ηV T

= 4 .3 mA

V 2− V 1= 2.3ηVT log I 2 I 1

ΔV = 2.3VT = 0.1V For Every decade change in current

V 2= V 1+ 0.1log

4 . 3

1. 0 = 0.763 V

S econd iteration V D= 0.763V

I D= I S e

V D

ηV T

= 4.237 mA

V 2− V 1= 2.3ηVT ln I 2 I 1

V 2= 0.763+ 0.1log4.237

4 .3= 0.762 V

Solution

I D

= 4.237mA ,

V D= 0.762V



The Piecewise-Linear Model

Approximating the diode forward characteristicwith two straight lines: the piecewise-linear model.



The Piecewise Linear Model



The Piecewise-Linear Model

• Exponential curve is approx into twostraight lines

• Line No 1 with zero slope & Line 2 witha slope of 1/r d

• The voltage change of less than 50 mV is observed in case the current changefrom 0.1 mA to 10 mA.i D= 0 v D

= 0V

i D

=( v D− V D0 )

r D v

D

≥ V D0

Piecewise-linear model of the diode forward characteristic and its

equivalent circuit representation



equivalent circuit representation.

Piecewise-l inear model





The

Constant – VoltageDrop Model

Constant – Voltage DropModel



Model

• Forward conducting diode exhibitsa constant voltage drop VD

• The voltage change of less than 50mV is observed in case the currentchange from 0.1 mA to 10 mA.

• Model is used when

– Detailed information about diode

characteristics in not available

Constant-voltage-drop model



The constant-voltage-drop model of the diode forward

characteristics and its equivalent circuit representation



characteristics and its equivalent-circuit representation.

The Small – Signal Model




• A small ac signal is superimposed onthe DC components.

• First determined dc Operating Point

• Then small signal operation around the

operating point

– Small portion of the curve is approximated

as almost linear segment of the diode

characteristics.







Figure 3.17 Development of the diode small-signal model. Note that the numerical values shown are for a diode with n = 2.




In absence of signal

I D= I s e V DηV T

Once signal is applied

v D( t )= V D+ vd (t )

i D( t )= I

sev D

ηV T

i D( t )= I

se[V D+ v

d (t )]

ηV T

i D( t )= I

se

V D

ηV T × e

vd (t )

ηV T

i D(t )= I

De

vd (t )

ηV T

For very small signal vd

ηV T

<<1

i D(t )= I

D(1+

vd

ηV T

)

i D (t )= I D+ id (t )





i D( t )= I D(1+

vd

ηV T )

i D( t )= I D+ id ( t )

id ( t )= I D vd

ηV T

r d =

ηV T

I D


r d is inversely proportional to I D

Modeling the Diode Forward Characteristic





Table 3.1 (Continued)





Exp 3-6



+

-

VDID

+

-

vd

Exp 3 6V

DD= 10V,v

d = 1V peak amplitude @ 60 Hz

Diode has a current of 1mA @ a V D

of .7 V, n= 2

Find r d ,V D, vd ( t )

Solution



I D=V

DD

− V D

R = 10−

0.710 = 0.93mA

r d =

ηV T

I D

=2× 250.93

= 53.8Ω

Small signal

v dpeak = v speak

r d

R+ r d

= 5.35mV

Input variation of 10% resulted in output voltage

variation of 0.7+5.4mV(0.8%) Voltage regulation

Exercise 3-16



Exercise 3 16

• Design a circuit shown so that Vo=3vwhen IL =0 A and Vo changes by 40 mV

per 1mA of diode current.

• (a) Find the value of R• (b) The junction area of each diode

relative to a diode with ).7 V drop at

1mA current. Assume n=1

Excercise 3-16



r DT

= Δv

o

Δio

=0.04

10− 3

= 40Ω

r DX

= 40/4= 10Ω

I DX

=nV T

r DX

= 2 .5mA

R=15− 3

2.5m= 4.8K Ω

At dc Operating Point VDX= 3 /4= .75V

I D1= 1mA ,V D1 0.7V

I DX

I D1

= I

SX

I S1

e

V DX

− V 1

nV T

⇒ I SX

I S1

= 0.34

The diodes have the junction area 0 .34 times the diode

Why 4 diodes and not 5? Diodes will

not conduct at 0.6 V

Diode Forward Drop inVoltage Regulation



Voltage Regulation

• Small signal model is used.

• Voltage remains constant in spite of :

– Changes in load current

– Changes in the dc power supply voltages

• One diode provides constant voltage of

0.7 V and for greater voltages diodes

can be connected in series.

Example 3-7



• A string of three diodes is used toprovide a constant voltage of about 2.1

V. We want to calculate the percentage

change in this regulated voltagecaused by

• (a) a + 10 % change to the power supply

voltage• (b) Connection of a 1 K ohms load

resistance , Assume n=2

Solution Exp 3-7



Solution Exp 3 7

P 3-53• In a particular cct application, ten “20



In a particular cct application, ten 20

mA diodes” ( a 20 mA diode is a

diode that provides a 0.7 V drop

when the current thru it is 20 mA)

connected in parallel operate at a

total current of 0.1 A. For the diodes

closely matched, with n=1, what

current flows in each.

What is the corresponding small signal

resistance of each diode and of the

combination?

i Dx=0.1

10==0.01 A

r dx= nV T

I Dx

= 2 . 5Ω

req=2. 5

10= 0.25Ω



• If each of the 20 mA diode has aseries resistance of 0.2 ohm

associated with the wire bonds to

the junction. What is the equivalent

resistance of the 10 parallel

connected diodes?

What connection resistance would

single diode need in order to be totally

equivalent?

Re q=1

10(2. 5+ 0. 2)= 0.27Ω

The diode i –

v relationship





Reversed Biased Diode

Leakage current:



ea age cu e

• In the reverse direction there is asmall leakage cur rent up until the

reverse breakdown vo l tage is reached.

• This leakage is undesirable, obviously

the lower the better.

• Diodes are intended to operate below

their breakdown voltage.

The Reversed Biased Region



g

i= I s(e

v

nV T − 1

v is negative ∧ >>V T (25mV )

i= − I S

Current in reserved biased diode circuit is due to

leakage current & increases with increase in reverse

voltage

Leakage current is proportional to the junction area

& temperature but doubles for every 10oC rise in

Breakdown Region



• Once reverse voltage exceeds a threshold value of

diode VZK

, this voltage is called breakdown voltage.

VZK Z – Zener, K – Knee

• At breakdown knee reverse current increases rapidly

with associated small increase in voltage drop

• Diode breakdown is not destructive if power

dissipated by diode is limited by external circuitry.

• Vertical line for current gives property of voltage

regulation

The diode i –

v characteristic with the breakdown region

shown in some detail.





Zener Diode

Zener Diode

O ti i th R B kd



• Operation in the Reverse Breakdown

Region

• Very steep i-v cu rve at breakdown w ith

almos t constant vol tage drop region

• Used the design ing vo l tage regu lator

• Diode manu factu red to operate

speci f ical ly in the Breakdown region

called Zener or B reakdown

Zener Diode : Symbol



I Z

- V Z +

Zener Diode : Symbol

Model: Zener • Manufacturer specify Zener Voltage V at a



Manufacturer specify Zener Voltage Vz at aspecified Zener test current Iz, the Max.

power that the device can safely dissipate 0.5W @ 6.8 v at max 70mA

• r z Dynamic resistance of the Zener and is the

inverse of the slope of the almost linear i-vcurve at operating point Q

• Lower r z, the more constant Zener Voltage

• The most common range of zener voltage is 3.3 volts to 75 volts,

ΔV z = ΔI z r z

Model for the zener diode.



Model: Zener



V z = V zo+ r z I z

I z> I

zk

V z> V zo

es gn ng o e ener s unregulator



+

-

Vo

Zener regulator

Supply voltage includesa large ripple component

Vo is an output of the zener regulator

that is as constant as possible in spite of

the ripples in the supply voltage VS

and the variations in the load current

Voltage regulator performance can be measured

Line Regulation & Load Regulation

Line Regulation = ΔV o /ΔV s

Load Regulation = ΔVo /ΔIL

Expression of performance : Zener regulator



I

IL

+

-

Vo

• Only the first term on right hand side is desirable one

Second and third terms depend upon Supply

Voltage Vs and Load current IL

• Line Regulation =

• Load Regulation =

(V s

-V o)

R = (V

o-V

zo)

r z + I L

V o= V zo(R

R+ r z )+ V S (

r z

R+ r z ) - I L(r z ∣∣ R )

ΔV o

ΔV s/ =

r z

(r z + R )

ΔVo

ΔI L= - (r z ∣∣ R )

Expression of performance : Zener regulator



I

IL

+

-

Vo

• An important consideration for the design

is

• To ensure that current through the zener

diode never becomes too low i.e less

than IZK or Izmin

• Minimum zener current Izmin occurs when

• Supply Voltage Vs is at its minimumVSmin

• Load current IL is at its maximum

ILmax

• Above design can be made be selecting R =

(V s min – V ZO - r z I z min)

(I z min+ I Lmax )

where I Lmax=V Z

R L

The circuit with the zener diode replaced with its equivalent circuit mode

Example 3.8



Exp 3-8



Example 3-8



V =+10 v± 1v

R= 0.5k Ω

V z= 6 .8v I z= 5 mA

r z= 20Ω

I zk = 0 . 2 mA

I RL= 1 mA

a) Find No Load Line RegulationV o∧ ΔV o



V z= V

zo+ r

z I z

V zo= V z−I zr z=

6.8− 5

×20

×10

−3

=6.7v

Depending upon the manufacturer provide DataFirst calculate Vzo if Vz =6.8 V & Iz=5mA, r Z=20 ohm

Now connecting the Zener diode in the Cct as shownCalculate actual Iz and resulting Vo

Thus establishing operating Point



Line

Regulation

I z = V − V zo

R+ r z

= 10− 6 .7500+ 20

= 6.35mA

V o= V

zo+ I

z r z = 6 .7+ 6.35× 20× 10− 3= 6.827V ≈ 6.83V

ΔV o ΔV

=38.5

1= 3.85mv /v

Now carry out Small Signal Analysis

Suppress DC source and calculate resultant change in VoUse voltage divider rule

ΔV o= ΔV

+r z

R+ r z

=± 1× 20

520=± 38.5mv

n vO oa res s ance L connected & draws 1mA and load

regulation



g

1mA drawn by load would decrease

by same amount so



R L≈6.83 v

1mA= 6.83 k Ω

R L∣∣ R=

20× 6830

6850 = 19.94ΩV Z = V o= V ZO+ I Z r Z = 6 . 7+ 5.35× 20= 6.807 V

I Z =Vs− V Z

R+ R L∣∣r Z

=10− 6.807

500+ 19.94= 6.14 mA

ΔI Z

= 6.35− 6.14= . 21 mA= 210 μA

I z ΔV o= r z ΔI z = 20× −1mA=− 20mV

= ΔV o

ΔI z

=− 20 mV /mA

Checkexact Calculations

by same amount so

Load Regulation

c) for ΔV o R L= 2k Ω



I R L= V Z

R L= 3 . 4 mA

ΔI Z = − 3 . 4 mA

ΔV o= r Z ΔI Z = − 68 mV



• 1) Check

Zener at Breakdown region

V o=2000

2500× 10= 8v

2000

500

+10



10v

V o

10

500

6.63 v

19.8Ω

Δ

+

6.7v−

20

2k Ω

0.5k Ω

A

B



V oc= 6 .7× 20002020

= 6.63v

Req= 19.8Ω

A

B

d ) RL= 500Ω



Zener is not

operating

V o=10× 500

1000= 5v

zk

@V o<<Valignl ¿¿¿

¿5<<6.8v

¿¿

L

10 v

500

500

e) Min value of for which the diode still

operates in the breakdown region

R L



p g

I z

• at BreakdownRegion

+ 10± 1v

500 I z

R L6.7v

0.2 mA

I z= I

zk = 0 .2 mA

V z= V zk = 6.7vV

DD= 9vmin

I =9− 6 . 7

500= 4 . 6mA

I = I zk + I

RL

I RL

= 4 .6− 0. 2= 4 .4 mA

R L=

V zk

I RL

=6 . 7

4 .4m= 1.5k Ω

Problem D3.68Design a 7.5-V zener regulator circuit using a 7.5-V

ifi d t 12 A Th h



zener specified at 12mA. The zener has anincremental resistance r z = 30 Ω and a knee current

of 0.5mA. The regulator operates from a 10-V supplyand has a 1.2-kΩ load.

(a) What is the value of R you have chosen?

(a) What is the regulator output voltage when thesupply is 10% high? Is 10% low?

(a) What is the output voltage when both the supply is

10% high and the load is removed?

(a) What is the smallest possible load resistor that canbe used while the zener operates at a current nolower than the knee current while the supply is 10%

low?

Solution 3-68

r = 30Ω



r z

30Ω

I Zk = 0 .5 mAV

Z = 7.5V

I Z = 12 mA7 .5= V ZO+ 12× 30× 10

− 3

⇒V ZO= 7.14 V

I RL=7 .5

1 .2= 6.25 mA

Design a 7.5-V zener regulator circuit usinga 7.5-V zener specified at 12mA. The zener has an incremental resistance r z = 30 Ω anda knee current of 0.5mA. The regulator operates from a 10-V supply and has a 1.2-kΩ l d



Select I = 10mA

I RL=

7.5

1.2= 6.25 mA

So that I Z = 3.75mA

W hichis> I Zk

R=10− 7 .5

10= 250Ω

kΩ load.

(a) What is the value of R you have chosen?

(a) What is the regulator outputvoltage when the supply is 10%high? Is 10% low?

For ΔV +=± 1V



ΔV O=± 1×1.2//0.03

0.250+ (1.2//.03 ) =± 0.1V

ThusV O =+7.4V to + 7.6V

With V += 11V and I L= 0

V O= V ZO+11− V O

0.28X0. 03

⇒V O= 7.55V

(a) What is the output voltage whenboth the supply is 10% high andthe load is removed?

(a) What is the smallest possible load resistor that can beused while the zener operates at a current no lower thanthe knee current while the supply is 10% low? IZK=0.5mA,VZO=7.14 V



11V

V O

R9− 7.155

0.25

= 7.38mA

7.14+ 0.03 X0 .5

7.155V 0.5 mA

250Ω

R Lmin

R Lmin=7.155

7.38− 0 . 5

= 1.04k Ω

1

3

2



Rectifier Circuit Power Supply



• Power supply must supply dc voltage to be constantin spite of – variation is ac line voltage

– Variation in current drawn by load, that is variable loadresistance

Rectifier Circuits



• Filter – Smoothes out pulsating dc but still some time-

dependent components-(ripple) remain in the output

• Voltage Regulation – Reduces ripples

– Stabilizes magnitude of dc output against variation inload current

– Regulation by Zener Diode or Voltage regulator I.C

Half Wave Rectifier



Transfer characteristic of

the rectifier circuit

Input and output waveforms assuming that rD >

Full Wave Rectifier



Input and output waveforms.



Full Wave Rectifier



• Diode in Reverse biased stateAnode @ - Vs

Cathode @ + Vo

• PIV = 2Vs - VDO

Twice as in case of half wave rectifier

Bridge Rectifier



The bridge rectifier: (a) circuit; (b) input and output waveforms.



Bridge Rectifier



Bridge Rectifier



Bridge Rectifier



• Peak Inverse Voltage

– PIV => consider loop D3, R & D2

– VD3(res) = Vo + VD2

– Vo = Vs – 2VD

– PIV = Vs – 2VD + VD = Vs – VD

Half of PIV for Full wave Rectifier

D4

D1

D2 D3



Figure 3.28 (a) A simple circuit used to illustrate the

effect of a filter capacitor. (b) Input and output

waveforms assuming an ideal diode.



g

Peak detector with Load



Figure 3.29 Voltage and current waveforms in thepeak rectifier circuit with CR<<T .



Charge / Discharge Cycle



Peak detector with Load



i L=

V o

Ri D= i

C + i

L

i D= C

dV s

dt

+ i L

Figure 3.30 Waveforms in the full-wave peak

rectifier.



• When Vr is small

ea ec er : u puVoltage



When Vr is small

– Vo = Vpeak

– iL is almost constant

– DC components of iL

• Accurate value of output dc voltage

Average Value

i L=V P

R

V o= V P −1

2V r

Charge / Discharge Cyclev o= V P e

−t

CR



V o= V P − 12 V r

i L=V P

R

e−

T

CR= 1−T

CR

V o= V P − V r ≈ V P e

−T

CR

⇒V r = V P (1− e−

T

CR

)

V r = I L

fC

, provided V r <<V p

V r = V

P (1− 1+T

CR)

I L=V

P

R

V r =V

P

T

CR ⇒

V P

fCR

Peak Rectifier : Ripple Voltage



• During Discharge cycle

• At the end of discharge cycle

• Since CR >> T

e−

T

CR= 1−T

CR

V o= V P − V r ≈ V P e− T CR

⇒V r = V

P

(1− e−T

CR )

v o= V P e−

t

CR

−T

CR T−T

Peak Rectifier : Ripple Voltage



Ripple Voltage

V r = I L

fC

, provided V r <<V p

V r = V

P (1− 1+T

CR)

V r =V P

CR

I L= V P

R

e CR= 1−T

CRV r = V P (1− e

−CR

)

V r =

V P T

CR ⇒

V P

fCR

ea ec er : on uc onInterval



V P

(1−

(ωΔt )2

2 )= V P − V r

⇒ωΔt =√2V

r

V P

V P cos(ωΔt )= V P − V r

Hence ωΔt is small

Cos (wΔt )= 1−(wΔt )2

2 !+ .. .

When Vr<<Vp the conduction angle will be small



Average Diode Current –During

Conductioni = i + i



During Discharge Qlost =CV r

i D= i

C + i

L

i Dav

= iCav

+ I L

iCav

= i Dav

− I L

During Charge Qsupplied

= icav Δt

iCav

= i Dav

− I L

Qsupplied= Qsupplied⇒icav Δt = CV r

Qlost= Qsu lied⇒CV r= icav Δt

Average Diode Current –During

Conduction



i Dav

=2π

√2V r

V P

V P

R+ I

L⇒i

L(1+ π 2V P

V r )

V r

<<V P

⇒ iDav

>> I L

Qlost Qsu lied r cav

V r =V P T CR

=⇒CV r =V P T R

V P T

R= (i Dav

− I L) Δt

Δt = 12πf

2V r

V P

= T 2π

2Vr

V P

V P

T

R= (i Dav

− I L)(T

2π

2Vr

V P )

Deduction



• As waveform of is almost right angle r

triangleV

r << V

P

i D max= 2i Dav

Observations• Diode current flows for short interval

d t l i h th h l t b



and must replenish the charge lost by

the capacitor. Discharge interval is long

& discharge is through high resistance

• Maximum diode current

Assuming that i L is almost constant = I L &

CR >>T

r D<< R L

i D=CdV i

dt + i L

i D max

= i L(1+ 2π

2V p

V r )≈ 2i

Dav

Example N0 3-9Consider a peak rectifier fed by a 60 Hz

sinusoidal having a peak value of Vp =



sinusoidal having a peak value of Vp =

100 V. Let the load resistance R =10 kOhms.

(a) Find the value of the capacitance C

that will result in peak to peak ripple of 2 V

(b) Calculate the fraction of the cycle

during which the diode is conduction

(c) Calculate the average and peak value

of the diode current.

Example 3.9



• Find value of C for Vr =2V (peak to peak)

• Find fraction of cycles that diode

conducts

• => Diode conducts of cycle

0.2

2π× 100= 3.18

100Sin2Π60 t

10 k Ω

C =V P

V r fR=

100

2× 60× 104= 83 .3μF

ωΔt =2Vr

V P

= 0 .2radian

Solution Exp 3-9



• Find &

i Dav= I L(1+ π

2V P

V r )

I L=

V P

R=

100

10000= 10 mA

i Dav= 10(1+ π

√2× 100

2 )= 324 mA

imax= 2i Dav= 648 mA

i D max

i Dav

Full wave peak Detector



In full wave rectifier, the capacitor discharge for almost T/2 time interval.

that mean ripple frequency is twice the

input, soV r =V

P 2 fCR

imax

= 2i Dav

= I L(1+ 2π

V P

2Vr )

i Dav

= I L(1+ π

V P

2Vr )

Applications



• Peak Rectifier – Peak detector is used

for

– Detecting the peak of the an input signal

for signal processing systems

– Demodulator for amplitude modulated

(AM) signals

.



Precision Half Wave Rectifier Super Diode



• Normal Diodes VD= 0.7v are used for

rectifier of input of much larger

amplitude then VD

• For smaller signals detection,

demodulation or rectificationOperational Amplifiers (Op Amp) are

used

Wave form GenerationLimiting Clamping



• Limiter Circuit – Vo is limited between two levels – upper

(L+) and lower (L-) thresholds

Figure 3.33 Applying a sine wave to a limiter can

result in clipping off its two peaks.



Figure 3.34 Soft limiting.



Wave form GenerationLimiting / Clamping

• Double Limiter



– Clips off both negative & positive peaks• Single Limiter

• Clips off only one side of the input peak

• Application – Limits the inputs to operation Amplifier to a

limit lower than the breakdown voltage of

transistors of input stage of operational

Amplifier

– Half / Full Rectifier for Battery Charger

Figure 3.35 A variety of basic limiting circuits.



Figure E3.27



Solution Ex 3-27

(a ) − 5≤ vi≤ 5 ⇒ v = v i



v R=

[10

10+ 10 ](v i− 5)=

1

2(vi− 5)

v o= 5+ v R=1

2v i+ 2 . 5

(a ) 5≤ vi≤ 5 ⇒ vo v i

(b) V I ≥+ 5 Vo D2 Conduct, D1 cut-off

(c ) v i≤ − 5 V D1 Conducts & D2 is off

v R= [10

10+ 10 ](v i+ 5)=1

2(vi+ 5)

vo=

1

2

(v i+ 5)− 5=

1

2

(vi− 2 . 5)

D C Restorer • The output waveform will have its lower

peak “Clamped” to O V therefore



peak Clamped to O V therefore

known as “Clamped Capacitor”

• Output waveform will have a finite

average value & is entirely different and

unrelated to the average value of the

input waveform

Application



T X R

DCRestorers

− 4v

+ 6

− 4

+ 2v

0v

4v

Figure 3.36 The clamped capacitor or dc restorer

with a square-wave input and no load.



Figure 3.37 The clamped capacitor with a loadresistance R .



Figure 3.38 Voltage doubler: (a) circuit; (b)

waveform of the voltage across D 1.



Figure P3.97



Figure P3.98



Figure P3.102



Figure P3.103





Figure P3.105

+ 4 v i



Diode Off V0= V i+ V c

Diode On Vo= − 0.7v

D on

− 6

− 0

D off

V C

v o

The Voltage Doubler



C1

V P sinωt

+ V P -

+

V D1

−

D1

−

2V P

+

+ V P -

V P sinωt

C 1 D1⇒a Clamp circuit

DC Restorer

Special Diode TypeSchottky-Barrier Diode (SBD)

• Shottky-Barrier Diode is formed by



Shottky Barrier Diode is formed by

bringing metal into contact with amoderately doped „n‟ type

semiconductor material

• Resulting in flow of the conducting

current in one direction from metal

anode to the semiconductor cathodeand acts as an open circuit in the other

direction

Schottky-Barrier Diode (SBD)



• Gets two important properties

– SBD switches on-off faster due to current

conducts due to majority carrier b

(electrons)

– Forward voltage drop is lower then P-n

junction diode

Varactor



• Variable Capacitor – Depletion layer acts as junction

capacitance

– Depletion layer varies CapacitanceDepletion Region

Dielectric

← D→

Metallic Plate

Varactor • When a reverse vo ltage is applied to a p-n junction ,

the depletion region, is essentially devoid of carriers



e dep e o eg o , s esse a y de o d o ca e s

and behaves as the dielectric of a capacitor.

• The depletion region increases as reverse voltage

across it increases; and since capacitance variesinversely as dielectric thickness, the junction

capacitance will decrease as the voltage across the

p-n junction increases.

• By varying the reverse voltage across a p-n

junction the junction capacitance can be varied .

Semiconductor diodes• The tunnel diode, the current through the

device decreases as the voltage is increased



within a certain range; this property, knownas negative resistance, makes it useful as an

amplifier.

• Gunn diodes are negative-resistance diodes

that are the basis of some microwave

oscillators.

• Light-sensitive or photosensitive diodes can

be used to measure illumination; the voltage

d th d d th t f

SCR (Thyristor)• The Silicon Controlled Rectifier (SCR) is simply a

conventional rectifier controlled by a gate signal.



• A gate signal controls the rectifier conduction.

• The rectifier circuit (anode-cathode) has a low

forward resistance and a high reverse resistance.

• It is controlled from an off state (high resistance) to

the on state (low resistance) by a signal applied to

the third terminal, the gate.

• Most SCR applications are in power switching,

phase control, chopper, and inverter circuits.



Photodiode



If reversed biased PN junction isexposed to incident light – the photons

impacting the junction cause covalent –

bond to break thus give rise to current

known as a photocurrent & is

proportional to the intensity of incident

light.

Converts Light energy into a electrical

signals

Photodiode



• Photodiode are manufactured usingGallium Arsenide (GaAs)

• Photodiodes are important element of optoelectronics or photonics circuit

(Combination of Electronics & optics)

used for signal processing, storage &transmission

Photodiode : Applications



• Fiber optics Transmission of telephonic& TV signals

• Opto-storage are CD ROM computer disks

• Wide bandwidth & low signalattenuation.

Light Emitting Diode (LED)• Inverse of Photodiode



• Converts a forward biased current into light

• GaAs used for manufacturing LEDs

• Used as electronics displays

• Coherent light into a narrow bandwidth laser diodes

Fib O i & CD ROM

LED



Double heterostructure laser



Optoisolator

• LED & Photodiode



Electrical to light Light toelectrical

• Provides complete electrical isolation between

electrical circuits

• Reduces the effects of electrical interference on

signal being fixed within a system

• Reduces risk of shock

• Can be implemented over long distance fiber optics

Laser Pointer

.



Laser Microphone





Problem 3-103



Sketch and label the transfer Characteristics of the circuit shown

over a + 10 V range of the input signal.

All diodes are VD =0.7 V @ 1 mA with

n=1.

What are the slopes of the

characteristic at the extreme + 10 V

levels?

+1 V



-2 V-2 V

V i

V 0

0<V i< 1 Vo= 0

Problem 3-103





Ist Sessional• Q No 1 (12 Marks) In the circuit shown, input

voltage is a 1kHz, 10 V peak to peak sine wave. The

diode is an ideal diode.



d ode s a dea d ode

• (a) Sketch the waveform resulting at output

terminal vO.

• (b) What are its positive and negative peak

values?

• Q No 2 (15 Marks) A circuit utilizes three

Ist Sessional



Q ( )

identical diodes connected in series having

n=1 and IS= 10-14 A.

• (a) Find the value of current required to

obtain an output voltage of 2 V across thethree diodes combined.

• (b) If a current of 1 mA is drawn away

from the output terminal by a load

• (i) What is the change in output

voltage?

• (ii) What is the value of the load?

• Q No 3 (13 Marks) For the circuit shown,

Ist Sessional



Q ( ) ,

sketch the output for the sine wave input of

10 volts peak. Label the positive and

negative peak values assuming that CR

>>T.

Ist Sessional



• Q No 4 (10 Marks) 9.25 V zener diodeexhibits its nominal voltage at a test

current of 28 mA. At this current the

incremental resistance is specified as 7

ohms.

– (a) Find VZO of the zener model.

– (b) Find the zener voltage at a current

of 10 mA.

• Q No 5 (20 Marks) Consider a bridge

rectifier circuit with a filter capacitor C

Ist Sessional



placed across the load resistor R for the casein which the transformer secondary delivers

a sinusoid of 12 V (rms) having the 60 Hz

frequency and assuming VD = 0.8 V and a

load resistance of 100 ohms.

– Find the value of C that results in a ripple voltage

no larger than 1 V peak to peak.

– Find the diode conduction angle. – Find the load current.

– What is the average load current?

• Q No 6 (10 Marks) In a circuit shown, the

Ist Sessional



( ) ,

output voltage is 2.4 V. Assuming that the

diodes are identical and are having 0.7 V drop

at 1mA.

– (a) Find the current following through the resistor R.

– (b) What the value of resistor R.



Figure 3.31 The “superdiode” precision half -wave rectifier and its almost-ideal transfer characteristic. Note that when v I

> 0 and the diode conducts, the op amp supplies the load current, and the source is conveniently buffered, an added

advantage. Not shown are the op-amp power supplies.

Figure P3.82



Figure P3.91



Figure P3.92



Figure P3.93



Figure P3.105



Figure P3.105



Quiz DE28 EE -B

(10 ) 9 2



(10 Marks) 9.25 V zener diode exhibitsits nominal voltage at a test current of

28 mA. At this current the incremental

resistance is specified as 7 ohms. –(a) Find VZO of the zener model.

–(b) Find the zener voltage at a current

of 10 mA.

Quiz DE 28 EE -A

A di d h i l lt



A zener diode whose nominal voltageis 10 V at 10 mA has an incremental

resistance of 50 Ω.

(a) What is the value of VZO of the zener

model?

Documents

Chapter 3 Diode