1

Prof. Sang-Jo Yoo

Chapter 3 Diodes

2

Prof. Sang-Jo Yoo

Contents

Ideal Diode

Terminal Characteristics of Junction Diodes

Physical Operation of Diodes

Analysis of Diode Circuits

Small-signal Model and its Application

Operation in the Reverse Breakdown Region-Zener Diodes

Rectifier Circuits

Limiting and Clamping Circuits

3

Prof. Sang-Jo Yoo

Introduction

The simplest and most fundamental nonlinear circuit element is the diode. It is a two terminal device like a resistor but the two terminals are not interchangeable.

We will start by describing an “ideal” diode and then look at how closely a real diode approximates the ideal situation.

One of the most common uses of diode is in rectifier circuits(conversion of ac signals to dc).

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Prof. Sang-Jo Yoo

3.1 The Ideal Diode

The quantity VA is referred to as the Applied voltage.

The i-v characteristic for the ideal diode passes no current when the applied voltage is negative, and when the applied voltage is positive the diode is a perfect short circuit (zero resistance).

p n

−+ Av

anode cathode

i

i

Av

Forward BiasReverse Bias

Reverse Bias Circuit Model Forward Bias Circuit Model

00 =⇒< ivA

i −+ Av

00 =⇒> Avi

i −+ Av

“Cut off” “ON”The external circuit must limit the currentunder Forward Bias conditions since thediode will have no resistance

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Prof. Sang-Jo Yoo

The diode is polarity dependent!

Forward Bias Current Limit Example(resistor limits the current)

Ωk1

V10+

Ωk1

V10+

Ωk1

−

+V0

V10+

−

+V10 Ωk1

−

+V10

V10+

−

+V0

Short circuit Open circuit

Reverse Bias Current Limit Example(diode, in cut off, limits the current)

mA1

mA0PN

NP

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Prof. Sang-Jo Yoo

A Simple Application: The Rectifier

−

+

OvDi

−+ Dv

Iv R

IvPv

t

−=

+

IO vvDi

−=+ 0Dv

Iv R−

=+

0Ov0=Di

−=+ ID vv

Iv R

OvPv

t

The negative half-cycleis blocked

The positive half-cycleis transmitted

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Prof. Sang-Jo Yoo

Battery Charger Rectification Example

Example 3.1 The circuit below is used to charge a 12V battery, where vS is a sinusoid with a 24V peak amplitude. Find the fraction of each cycle during which the diode conducts, also find the peak value of the diode current and the maximum reverse-bias voltage that appears across the diode.

−

+V12

Di

Sv

Ω= 100R

( )cycle theof thirdoneor 1202

605.0cos 1

o

o

=

== −

θ

θ

AVV

I Pd 12.0

100

1224=

Ω−

= VVVV Pd 362412max =−−+=

t

V12V24

θ212cos24 =θ

Di

Sv

Di

V24−

maxdV

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Prof. Sang-Jo Yoo

Another Application: Diode Logic Gates

Diodes and resistors can be used to implement digital logic functions : 0V is a Low and +5V is a high

R

Av

Bv

Cv Qv

Av Bv Cv Qv

0 0 0 00 0 5 50 5 0 50 5 5 55 0 0 55 0 5 55 5 0 55 5 5 5

Inputs OutputOR Gate

R

Av

Bv

Cv

Qv

V5+Av Bv Cv Qv

0 0 0 00 0 5 00 5 0 00 5 5 05 0 0 05 0 5 05 5 0 05 5 5 5

Inputs Output

AND Gate

9

Prof. Sang-Jo Yoo

Simple DC Analysis of Ideal Diode CircuitsHow do we know which diodes are conducting and which are not?

It might be hard to tell, so we make an assumption, then proceed with your analysis and then check to see if everything is consistent with your initial assumption. If things are not consistent then the assumption was invalid.

For now, lets assume that both diodes are conducting

If D1 is on VB=0 and the output V=0 also.We can now find the current through D2

Ωk5

I1D

V10−

−

+V

2D

V10+

Ωk102DI

B

mAVV

ID 1000,10

0102 =

Ω−

=

We can write a node equation at node B,looking at the sum of the currents

( )

mAI

VVmAI

III kD

1

000,5

1001

52

=Ω

−−=+

=+

Therefore D1 is on as assumed

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Prof. Sang-Jo Yoo

Another CircuitThis is the same circuit as the previous one except that the values of the two resistors have been exchanged.

Again assume that both diodes are on, do the analysis and check the results.

Again VB=0 and V=0.

Ωk5

I1D

V10−

−

+V

2D

V10+

Ωk10

2DI

B

mAVV

ID 2000,5

0102 =

Ω−

=

( )

mAI

VVmAI

III kD

1

000,10

1002

102

−=Ω

−−=+

=+

Not possible, thereforeassumption was wrong

( )mA

VVID 33.1

000,5000,10

10102 =

Ω+Ω−−

=

Now assume D1 is off and D2 is on

Now solving for VB we get 3.33V and I=0 since D1 is off

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Prof. Sang-Jo Yoo

3.2 Diode Terminal Characteristics

An Analog sweep has been converted to Digital (discrete values)

- 4mA

- 3mA

- 2mA

- 1mA

1mA

2mA

3mA

4mA

5mA

(+0.2 volt increments)

∆Ι

∆VRd =

∆V

∆Ι slope = ∆Ι∆V

1slope=

Va

I

1.0slightly negative

ForwardBias Va > 0

ReverseBias Va < 0

Rd is the dynamic (changing) resistance

BreakdownVoltage

Turn-onVoltage

from -6 to -hundreds of volts

Rd

Va

The higher the doping levels of then and p sides of the diode, the lowerthe breakdown voltage.

“on” R is low

“Off”R is high

breakdown

The turn-on voltge is a function of the semiconductor used.~ 0.7V for Si and ~ 1.7V for GaAs

-VZK

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Prof. Sang-Jo Yoo

The Forward Bias Region

Forward-bias is entered when va>0

The i-v characteristic is closely approximated by Is, saturation current or scale current, is a constant for a given diode at a given temperature, and is directly proportional to the cross-sectional area of the diode

VT, thermal voltage, is a constant given byK = Boltzman’s constant = 1.38 x 10-23 joules/kelvinT = the absolute temperature in kelvins = 273 + temp in Cq = the magnitude of electronic charge = 1.60 x 10-19 coulomb

For appreciable current i, i >>IS, current can be approximated by

or alternatively

Va

I

1.0

p n−+ Av

iq

kTV T

=

eI Vi Tnv

S≈

IVS

T

inv ln=

°

⎟⎠⎞⎜

⎝⎛ −= 1eI V Tn

v

Si

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Prof. Sang-Jo Yoo

The Reverse Bias Region

Reverse-bias is entered when va < 0 and the diode current becomes

Real diodes exhibit reverse currents that are much larger than IS. For instance, IS for a small signal diode is on the order of 10-14 to 10-15 A, while the reverse current could be on the order of 1 nA (10-9 A).

A large part of the reverse current is due to leakage effects, which are proportional to the junction area.

I Si −≈

I

Va

VZK

breakdown voltage

reverse-bias region

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Prof. Sang-Jo Yoo

The Breakdown Region

The breakdown region is entered when the magnitude of the reverse voltage exceeds the breakdown voltage, a threshold value specific to the particular diode. The value corresponds to the “knee” of the i-v curve and is denoted VZK. Z stands for Zener, which will be discussed later, and K stands for knee.

In the breakdown region, the reverse current increases rapidly, with the associated increase in voltage drop being very small.

I

Va

VZK

breakdown voltage

breakdown region

reverse-bias region

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Prof. Sang-Jo Yoo

3.3 Physical Operation of DiodesTetrahedron Covalent Bonds in a Semiconductor

atom

valenceelectrons

covalent bonds

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Prof. Sang-Jo Yoo

Semiconductors (cont.)Bonds, Holes, and Electrons in Intrinsic Silicon

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Prof. Sang-Jo Yoo

Doped SemiconductorsBonds, Holes, and Electrons in Doped Silicon

donor accept

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Prof. Sang-Jo Yoo

The Diode

p

n

B A SiO2Al

Cross-section of pn -junction in an IC process

A

B

diode symbol

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Prof. Sang-Jo Yoo

Carrier Motion

Carriers move due two two different mechanisms

Carrier drift in response to an electric field

Carriers diffuse from areas of high concentration to areas of lower concentration

Since both carrier types (electrons and holes) can be present and there are two mechanisms for each carrier there are four components to the overall current, as shown below

J= current density (A/cm2)

J J J q pE qD p

J J J q nE qD n

p p drift p diffusion p p

n n drift n diffusion n n

= + = − ∇

= + = + ∇

| |

| |

µ

µ

r

b br

drift diffusion

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Prof. Sang-Jo Yoo

Diffusion Current

Carriers move from areas of high concentration to low concentration

dx

dnqDJ nn =

negative is dx

dn

electron conc.

electron motion

current flowdx

dpqDJ pp −=hole conc.

hole motion

current flow

negative is dx

dp

++++++

++

+++ +

++

diffusion constantDp=12 cm2/s Dn=34 cm2/s

21

Prof. Sang-Jo Yoo

Carrier Drift

Definition - Drift is the motion of a charged particle in response to an applied electric field.

Holes are accelerated in the direction of the applied field

Electrons move in a direction opposite to the applied field

Definition - Current, is the charge per unit time crossing an arbitrarily chosen plane of observation oriented normal to the direction of current flow.

e-

Er

vth

vdrift

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Prof. Sang-Jo Yoo

Carrier Drift, continued

Drift current density (divided by area A) for hole

Total drift current

Einstein relationship :

VT=thermal voltage (25mV)

EqpJ

Ev

pdriftp

ppdrift

µ

µµ

=

=

−

srength field electric :E hole, ofmobility :

EqnJ ndriftn µ=−

( )EnpqJ npdrift µµ +=

Tp

p

n

n VDD

==µµ

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Prof. Sang-Jo Yoo

Doped Semiconductors

N-type : by introducing a small number of impurity atoms (phosphorus) donor

P-type : by introducing a small number of impurity atoms (boron) acceptor

D

in

inn

Dn

N

np

npn

Nn

2

0

200

0

≈

=

≈

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Prof. Sang-Jo Yoo

Abrupt Junction Formation

Junction Formation

Carrier Concentrations

pp ~ Na

np0 ~ (ni )2 / Na

nn ~ Nd

pn0 ~ (ni )2 / Nd

P N

X0

pp

np0

nn

pn0

The Depletion Region

represents an immobile donor impurity (i.e. P+ )

represents an immobile acceptor impurity (i.e. B- )

- represents a mobile electron

+ represents a mobile hole

+-

- -- - - - -

- - - - -+++ +

++

+++ ++

+- + +

+ + +

++ +

+ + ++

+ + +++

++

+

-

- -

- --

- - - - -- - - - -

n typep type

x

+

+

+ ++

++

+

+

--

--

-

---

-

DepletionRegion

25

Prof. Sang-Jo Yoo

The Depletion or Space Charge Region

Maximum Field (Emax )

xn

+qNd Q n = qNdxnx

Q p = -qNaxp

Electric field (x)

-qNa

+

-

xp

Abrupt depletion approximation

xn

Electrostaticpotential V(x) (Volts)

xp

Vbi

charge density

xp

xn xd = xn + xp

CArea

xr Si O

d

=ε ε

2 0

(Coulombs/cm-3)

(Volts/cm)

( )∇ • =E le c tr ic f ie ldK s

_ρε 0

d E le c tr ic f ie ld

d x K s

( _ )=

ρε 0

( )ρ = − + −q p n N ND A

P o te n tia l E le c tr ic fie ld

E le c tr ic f ie ld

= −

∝

∫∫

_

_ ρ

-+

++

++

electron diffusion

hole diffusion

hole drift

electron drift

---

--

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Prof. Sang-Jo Yoo

Reverse Bias

x

pn0

np0

-W1 W20n-regionp-region

diffusion

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Prof. Sang-Jo Yoo

Forward Bias

x

pn0

np0

-W1 W20p n

(W2)

n-regionp-region

Lp

diffusion

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Prof. Sang-Jo Yoo

3.4 Analysis of Forward Biased Diode Circuits

We have already looked at the ideal diode model for forward bias (short circuit). In this section we will work with a detailed model and then explore simplifying assumptions that allows us to work back towards our ideal case.

We want to determine the exact current through the circuit, ID and the exact voltage dropped across the diode VD.

If we assume that the voltage source VDD is greater than ~0.5 volts the diode will obviously be in the forward mode of operation and the current through the diode will be given by the following equation

T

D

nV

V

SD eII =

R

VVI DDD

D

−=

+VD

-

ID

VDD +

R

29

Prof. Sang-Jo Yoo

Graphical (Load Line) Analysis continued

We can determine this unknown voltage (operating point) by superimposing the graphs of the expressed for diode current.

i (mA)

v (V)

00

R

VDD

DDV

T

D

nV

V

SD eII =

R

VVI DDD

D

−=

QI D

V D

Rslope

1−=

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Prof. Sang-Jo Yoo

Iterative Analysis

Example 3.4 Assume that the resistor in our graphical analysis circuit is 1kΩ and VDD is 5V

The diode has a current of 1mA if it is at a voltage of 0.7 volts and the voltage drops by 0.1 volt for every decade decrease in current.

Find the current through the circuit and the exact voltage across the diode.

We can start by assuming we have set up the conditions so that the voltage across the diode is 0.7 volts, we do this so that we can do some calculations about our diode that we can use later to zero in on our actual conditions

mAR

VVI DDD

D 3.41000

7.05=

−=

−=

31

Prof. Sang-Jo Yoo

Iterative Analysis

This current is larger than the 1mA current at 0.7 volts so we conclude that the actual diode voltage will be larger than 0.7 volts.

Since the relationship between the current and the voltage is exponential we can adjust our voltage estimate slightly using anequation we derived earlier relating the voltage change to the current ratio, namely

Now using this value in our original equation we get

( )

74.1 then1.03.2 If :note 763.0

0.001

0.0043log1.0 so log3.2

2

121

212

===

⎟⎠⎞

⎜⎝⎛+=⎟⎟

⎠

⎞⎜⎜⎝

⎛=−

nnVVV

VVI

InVVV

T

T

mAR

VVI DDD

D 237.41000

763.05=

−=

−= ( ) VV 762.0

0.0043

0.004237log1.0763.02 =⎟

⎠⎞

⎜⎝⎛+=

Converged toID=4.237mAVD=0.762V

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Prof. Sang-Jo Yoo

A graphical view of the iterative analysis

i (mA)

v (V)

00

R

VDD

DDV0.7V

1

1.0 mA

0.763V

1

2

4.3 mA2

0.762V 3START

4.237 mA

3END

33

Prof. Sang-Jo Yoo

Approximating the diode forward characteristic with two straight lines

The analysis of a diode circuit can be greatly simplified by approximating the exponential i-v curve with two straight lines. One line, A, has a zero slope and the second line, B, has a slope of 1/rD

The piecewise-linear model is described as follows:

( ) ,

,0

VD0D0

0

≥−=

≤=

vrVviVvi

DDDD

DDD

−

+

vD

rD

ideal

iD

V D0

iD (mA)

vD (V)0

B, slope =rD

1

A, slope = 0

V D0

iD (mA)

vD (V)0

B, slope =rD

1B, slope =

rD

1

A, slope = 0

V D0

34

Prof. Sang-Jo Yoo

Constant-Voltage Drop Model

We assume that a forward-conducting diode exhibits a constant voltage drop, VD, which is approximately 0.7 V.

This model is used in the initial phases of analysis and design to give a rough estimate of circuit behavior.

−

+

vD

ideal

iD

V 7.0=V D

iD (mA)

vD (V)0

A, horizontal

V D

B, vertical

iD (mA)

vD (V)0

A, horizontal

V D

B, vertical

35

Prof. Sang-Jo Yoo

3.5 DC Forward Bias with an ac small signal

The DC bias level determines the ac parameters

By restricting the input signal swing to small values we can “linearize” the characteristic like we did for amplifier transfer characteristics

+vD(t)

-

vd(t)

iD(t)

VD +DC

ac

(DC+ac)

(DC+ac)VD=0.7

VD0 vd (t)

iD (mA)

vD (V)

t

ID

Bias Point - Q

tangent at Q

dr

1slope =

id (t)

0.5500

1.0

0.750.65

36

Prof. Sang-Jo Yoo

Small Signal Analysis

If we set the ac signal to zero, the current through the diode due to the DC bias is given by

When we add in the ac small signal to the DC voltage bias the total signal is

The total (DC +ac) instantaneous current is

Which we can re-arrange to get

Substituting in the DC equation from above, we get

T

D

nV

V

SD eII =

T

D

nV

v

SD eIti =)(

)()( tvVtv dDD +=

T

dD

nV

vV

SD eIti+

=)(

T

d

T

D

nV

v

nV

V

SD eeIti =)(

T

d

nV

v

DD eIti =)(

1<<T

d

nV

v

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

T

dDD nV

vIti 1)(

T

dDDD nV

vIIti +=)(

dDD iIi +=

If we keep the amplitude of the ac signal small, such that

We can expand the exponential in an infinite series, but we find that a sufficiently accurate expression can be found using only the first two terms.

This IS the small signal approximation, valid for amplitudes less than about 10mV

We find that the total current is made up of a DC component and an ac component that is directly proportional to the small signal voltage AND the DC bias level

T

dDd nV

vIi =Where

37

Prof. Sang-Jo Yoo

Small Signal Resistance (incremental resistance)

On the previous page we found

And since,

The ac small-signal resistance is inversely proportional to the DC bias current ID

In the graphical representation we find that about the Q point

DD IiD

D

d

vi

r

=⎥⎦

⎤⎢⎣

⎡∂∂

=1

( )0

1DD

dD Vv

ri −=

D

T

dd

T

Dd

d

d

I

nV

gr

nV

Ig

v

i==∴==

1

T

dDd nV

vIi =

iD (mA)

vD (V)

ID

Bias Point - Q

tangent at Qdr

1slope =

00

VD0

The equation of the tangent line is given by:

38

Prof. Sang-Jo Yoo

The Equivalent Circuit Model for the Diode

The equation of the tangent line is a model of the diode operation for small signal changes about the bias DC point (Q point)

The total model has the components shown below

The incremental voltage

across the diode is

( )0

1DD

dD Vv

ri −=

( )( )

ddDD

dddDDD

ddDDD

dDDD

riVv

rirIVv

riIVv

riVv

+=++=

++=+=

0

0

0

+vD

-

iD

rd

VD0

ideal

ddd riv =vD (V)

ID

Bias Point - Q

tangent at Q

dr

1slope =

00

VD0 VD

di

dv

39

Prof. Sang-Jo Yoo

Application of the Diode Small-Signal Model

Combined DC and ac voltage input causing a DC and ac current. We can analyze the response of the circuit.

+vD=VD+vd

-vs

iD=ID+id

VDD +

R

( ) ( )( ) ( )

( )ddDDsDD

dddDDDsDD

ddDDdDsDD

dDDDsDD

rRiVRIvV

rRirIVRIvV

riIVRiIvV

riVRivV

+++=+++++=+

++++=+++=+

0

0

0

+

vD=VD+vd

-

vs

iD=ID+id

VDD

R

rd

VD0

ideal( )dds rRiv +=

40

Prof. Sang-Jo Yoo

Application of the Diode Small-Signal Model continued

The small-signal analysis is done by eliminating all DC sources and replacing the diode with the small-signal resistance.

+

VD

-

ID

VDD

R

rd

VD0

ideal

DDDD VRIV +=

Circuit for DC Analysis

+

vd

-

vs

id

R

rd

( )dds rRiv +=

Circuit for small-signal Analysis

d

dsd rR

rvv

+=

41

Prof. Sang-Jo Yoo

Power Supply Ripple Example 3.6

Example 3.6 The power supply has a 10V DC value and a 1V peak sinusoidal ripple at a frequency of 60 Hz.

Calculate the dc voltage across the diode and the magnitude of the sine-wave signal appearing across it

Assume the diode has a 0.7V drop at a current of 1mA and that the ideality factor n=2

Calculate the dc diode current by assuming VD=0.7V

Since this value is close to 1mA the diode voltage will be closeto the assumed value of 0.7V. At this DC operating point we can calculate the incremental (dynamic) resistance rd as follows

The peak-to-peak small signal voltage across the diode can be found using the ac model and the voltage divider rule

+vd

-

+V=10V+ripple

R=10kΩ

mAID 93.0000,10

7.010=

−=

( )Ω=== − 8.53

1093.0

025.023xI

nVr

D

Td

( ) mVrR

rpeaktopeakv

d

dd 7.10

8.53000,10

8.5322 =

+=

+=−−

+V=10V+ripple

R=10kΩ

rd=53.8Ω+vd

-

42

Prof. Sang-Jo Yoo

Voltage Regulation Using Diode Forward Voltage Drops

A string of three diodes is used to provide a constant voltage of about 2.1V

We want to see

a) how much of a fluctuation (percentage change in regulation) there is in the output for a 1V (10%) change in the power supply voltage

b) percentage change in regulation when there is a 1kΩload resistance. Assume n=2

With no load the nominal dc current is given by

Thus the dynamic resistance of each diode is

The total resistance of the diodes will be 3rd or 18.9Ω

Using voltage division on the 1V change (10%) we get

+

vo

-

10V + 1V

R=1kΩ

RL=1kΩ

mAID 9.7000,1

1.210=

−=

( )Ω=== − 3.6

109.7

025.023xI

nVr

D

Td

( )( ) mV

rR

rv

d

do 1.37

9.18000,1

9.182

3

32 =

+=

+=∆

43

Prof. Sang-Jo Yoo

Diode Model for High Frequencies

Charge storage effects were modeled by two capacitances

The diode depletion layer capacitance (Cj)

The forward biased diffusion capacitance (Cd)

Cjrd Cd

0for 2

0for

1

, :Point Bias DC

0

0

0

>=

<

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

=

=

Djj

Dm

D

jj

DT

Td

D

Td

DD

VCC

V

VV

CC

IV

C

I

nVr

VI

τ

44

Prof. Sang-Jo Yoo

3.6 Diode Characteristic in the Reverse Breakdown Region - Zener Diodes

If the Zener diode is biased in the reverse breakdown region of operation the current can fluctuate wildly about the Q point and the voltage across the diode will remain relatively unchanged

The incremental (dynamic) resistance in reverse breakdown is given by rZ

+VZ

-

IZ

Circuit symbolfor a Zener diode

i

v

IZT

∆V

VZ0

Bias Point - QZr

1slope =

VZKnee VZ

∆Ι

∆V=∆Ι(rΖ)

IZKnee

Test current

45

Prof. Sang-Jo Yoo

The Reverse Bias Zener Model

We can model the zener diode in the breakdown region as straight line having an x (voltage) intercept at VZ0

and a slope of 1/rZ.

The reverse breakdown characteristic of a Zener diode is very steep (low resistance). For a very small change in voltage biased in the breakdown region the current changes significantly.

+VZ

-

IZ

rz

VZ0

ZzZZ IrVV += 0

46

Prof. Sang-Jo Yoo

A Shunt Regulator

Load

+VO

-

IL

IZ

I

R

Zenerregulator

Load

+VO

-

IL

IZ

I

R

Zenerregulator

i

v

i

v

The zener diode can be used to absorb or buffer a load from large current changes, I.e. keep the voltage across the load approximately constant

47

Prof. Sang-Jo Yoo

Zener Voltage Regulation

Example 3.8 A 6.8 V Zener diode in the circuit shown below is specified to have VZ = 6.8V at IZ = 5mA and rZ=20Ω, and IZK = 0.2mA.

The supply voltage is nominally +10V but can vary by plus or minus 1 V.

(a) Find the output VO with no load and V+ at 10V

(b) Find the value of VO resulting from the +/- 1 V change in V+

(c) Find the change in VO resulting from connecting a load resistance RL= 2 kΩ(d) Find the value of VO when RL =0.5 kΩ(e) What is the minimum value of RL for which the diode still operates in the breakdown region.

We can start by determining the value of VZ0. VZ0 is the x-axis intercept of the line tangent to the characteristic at the reverse bias operating point

+

vo

-

V+ (10V + 1V)

R = 0.5 kΩ

RL

6.8Vzener

( ) VVV

IrVVIrVV

ZZ

ZzZZZzZZ

7.6 005.0208.6

00

00

=−=−=+=

48

Prof. Sang-Jo Yoo

Zener example continued

With no load connected, the current through the zener diode is given by

We can now find V0, the voltage at the operating point current

Now, for a +/- 1V change in V+ can be found from

When a load resistance of 2kΩ is connected, the load current will be approximately 6.8V/2000Ω or 3.4mA. This current will not be flowing through the zener diode if it is flowing through the load so the change in the zener current is -3.4mA. The corresponding change in the zener voltage (which is also the output voltage) is,

A more accurate result

mArR

VVII

Z

ZZ 35.6

20500

7.6100 =+−

=+−

==+

( ) VrIVV ZZZ 83.602.035.67.600 =+=+=

( ) mVrR

rVV

Z

Z 5.3820500

2010 ±=

+±=

+∆=∆ +

( ) mVmAIrV ZZ 684.3200 −=−=∆=∆

mVV 700 −=∆

49

Prof. Sang-Jo Yoo

Zener Example continued

If we change the load resistance to 500Ω the load current would increased to 6.8V/500Ω = 13.6mA,

Not possible because the current I supplied through R s only 6.4 mA.

Cut off

For the zener to be at the edge of the breakdown region, the current has to be IZ=IZK=0.2mA and VZ=VZK=6.7V. At this point the current supplied through the resistor R is (9-6.7)/500 or 4.6 mA.

The load current would be 4.6mA - 0.2mA = 4.4mA. We can now find the value of RL for to cause this

mAI

VRR

RVV

L

L

LO

10500500

10

voltagebreakdownzener thelower than is which

5500500

50010

=+

=

=+

=+

= +

Ω== 500,10044.0

7.6LR

50

Prof. Sang-Jo Yoo

Shunt Regulator

S

O

V

V

∆∆

←

L

O

I

V

∆∆

←

VO

VS VSmax

VSmin

tt

Load+VO

-

VS

IL

IZ

I

R

Zenerregulator

reduced ripple

( )RrIrR

rV

rR

RVV zL

z

zS

zZO −

++

+= 0

z

z

rR

r

+=Regulation Line

RrZ−=Regulation Load

maxmin

min0minmax

LZ

ZZZS

II

IrVVR

+−−

=

rz

VZ0+VO

-

VS

IL

IZ

I

R

↑↑=>↓=>↓=>↑=>↓=> gulationLoadrIIRgulation zZ ReRe

51

Prof. Sang-Jo Yoo

Example 3.9 Design of a Zener Shunt Regulator

It is required to design a zener shunt regulator to provide a voltage of approximately 7.5 volts. The original supply varies between 15 and 25 volts and the load current varies between 0 and 15 mA. The zener diode we have available has a VZ of 7.5 V at a current of 20 mA and itsrZ is 10 Ω.

Find the required value of R and determine the line and load regulation. Also determine the percentage change in VO corresponding to the full change in VS and IL.

)20(105.7 00 mAVVIrVV ZZZZZ Ω+==+= VVZ 3.70 =

maxmin

min0min

LZ

ZZZS

II

IrVVR

+−−

= Ω=+

Ω−−= 383

155

)5(103.715

mAmA

mAVVR designing for Izmin=(1/3)ILmax

VmVRr

r

Z

Z /4.2538310

10Regulation Line =

+=

+=

mAmVRrZ /7.9)383//10()//(Regulation Load −=−=−=

%4.3or 254.010/4.25 - Vin Change Full S VVmVVo =×=∆

%2or 15.015/7.9 - Vin Change Full S −−=×−=∆ VmAmAmVVo

52

Prof. Sang-Jo Yoo

3.7 Rectifier Circuits

Dioderectifier Filter

VoltageRegulator

Load

+

-

VO

IL

t t t t t

Powertransformer

+

-

ac line

120V (rms)

60 Hz

+

-

vO

1

2

N

N

53

Prof. Sang-Jo Yoo

Half-Wave Rectifier

D

Rvs

+

-

vo

+

-

Ideal

Rvs+- vo

+

-

VD0 rD

vo

vSVD00

DrR

RSlope

+=

VS

VD0vS

t

v

vo

VD0,0=ov 0Ds Vv <

,0D

DSD

o rR

RVv

rR

Rv

+−

+=

0Ds Vv ≥RrD << 0DSO Vvv −≅

SVPIV = PIV: peak inverse voltageWithstand without breakdown

54

Prof. Sang-Jo Yoo

Full-Wave Rectifier with Center Tapped Transformer

vo

vSVD00

DrR

RSlope

+=

VS

VD0vS

t

v

vo

02 DS VVPIV −=

D1

Rvs

+

-vo

+

-

D2vs

+

-

-VD0

-vS

Vs: positive case

55

Prof. Sang-Jo Yoo

Full-Wave Bridge Rectifier

VS

2VD0vS

t

v

vo

0002 DSDDS VVVVVPIV −=+−=

-vS

D1D4

Rvs

+

vo+ -

D2

- D3

56

Prof. Sang-Jo Yoo

Rectifier with a Filter Capacitor

To reduce the variation of the output voltage : filter capacitor

RC

t

e−

LI

LCD

oL

idt

dvCiii

Rvi

+=+=

= /

57

Prof. Sang-Jo Yoo

Rectifier with a Filter Capacitor

Vr=peak-to-peak ripple voltage

CR>>T, then Vr is small.

( )

fCR

V

CR

TVV

CRTVeVVV

VVV

RVIVv

ppr

pCRT

prp

rpo

pLpo

==

>>−≈≈−

−=

=→≈

− TCR if /1

interval discarge of end at the

output average accurate 2

1

58

Prof. Sang-Jo Yoo

3.8 Limiting and Clamping Circuit

Limiters are sometimes referred as clippers

59

Prof. Sang-Jo Yoo

Limiting and Clamping Circuit

The clamped capacitor or dc restorer

Voltage doubler

Schottky-Barrier Diode (SBD)

Varactors

Photodiodes

Light-Emitting Diode (LED)