Chap3 Diode Circuits

Microelectronic Circuit DesignMcGraw-Hill

Chapter 3Solid-State Diodes and Diode Circuits

Microelectronic Circuit DesignRichard C. JaegerTravis N. Blalock

Chap 3 -1


Chapter Goals

• Understand diode structure and basic layout• Develop electrostatics of the pn junction• Explore various diode models including the mathematical model, the

ideal diode model, and the constant voltage drop model• Understand the SPICE representation and model parameters for the

diode• Define regions of operation of the diode (forward, reverse bias, and

reverse breakdown)• Apply the various types of models in circuit analysis• Explore different types of diodes Discuss the dynamic switching

behavior of the pn junction diode• Explore diode applications• Practice simulating diode circuits using SPICE

Chap 3 -2


Diode Introduction

• A diode is formed byinterfacing an n-typesemiconductor with a p-typesemiconductor.

• A pn junction is theinterface between n and pregions.

Diode symbol

Chap 3 -3


pn Junction Electrostatics

Donor and acceptor concentrationon either side of the junction.Concentration gradients give riseto diffusion currents.

Chap 3 -4


Drift Currents

• Diffusion currents lead to localized charge density variations near thepn junction.

• Gauss’ law predicts an electric field due to the charge distribution:

• Assuming constant permittivity,

• Resulting electric field gives rise to a drift current. With no externalcircuit connections, drift and diffusion currents cancel. There is noactual current, since this would imply power dissipation, rather theelectric field cancels the diffusion current ‘tendency.’

Ec

s

E(x)1s

(x)dx

Chap 3 -5


Space Charge Region Formationat the pn Junction

Chap 3 -6


Potential across the Junction

Charge Density Electric Field Potential

j E(x)dx VT ln NA ND

ni2

, VT kT

q

Chap 3 -7


Width of Depletion Region

wd 0 (xn xp) 2s

q1

NA 1

ND

j

Combining the previous expressions, we can form an expressionfor the width of the space-charge region, or depletion region. Itis called the depletion region since the excess holes and electronsare depleted from the dopant atoms on either side of the junction.

Chap 3 -8


Width of Depletion Region (Example)

Problem: Find built-in potential and depletion-region width for given diode

Given data:On p-type side: NA=1017/cm3 on n-type side: ND=1020/cm3

Assumptions: Room-temperature operation with VT=0.025 V

Analysis:

m113.01120

jDNANqs

dw

V979.0

/cm10/cm10/cm10V)ln(0.025ln 620

320317

2

i

DATj n

NNV

Chap 3 -9


Diode Electric Field (Example)

• Problem: Find electric field and size of individual depletion layers oneither side of pn junction for given diode• Given data:On p-type side: NA=1017/cm3 on n-type side: ND=1020/cm3

from earlier example,

• Assumptions: Room-temperature operation• Analysis:

V979.0j m113.00 dw

m4-1013.1

1

0

110

AN

DNd

wnx

DNAN

pxANDN

nxpxnxdw

kV/cm173m113.0)V979.0(2

0j2

dwMAXE

m113.0

1

0

DN

ANd

wpx

Chap 3 -10


Internal Diode Currents

jnT qnnE qDn

nx

= 0

jpT qp pEqDp

px

= 0

Mathematically, for a diode with no external connections, the total currentexpressions developed in chapter 2 are equal to zero. The equations onlydictate that the total currents are zero. However, as mentioned earlier,since there is no power dissipation, we must assume that the field anddiffusion current tendencies cancel and the actual currents are zero.

When external bias voltage is applied to the diode, the above equations areno longer equated to zero.

Chap 3 -11


Diode Junction Potential for DifferentApplied Voltages

Chap 3 -12


Diode i-v Characteristics

Turn-on voltage marks point of significant current flow. Is is called thereverse saturation current.

Chap 3 -13


where IS = reverse saturation current (A)vD = voltage applied to diode (V)q = electronic charge (1.60 x 10-19 C)k = Boltzmann’s constant (1.38 x 10-23 J/K)T = absolute temperaturen = nonideality factor (dimensionless)VT = kT/q = thermal voltage (V) (25 mV at room temp.)

IS is typically between 10-18 and 10-9 A, and is strongly temperature dependent dueto its dependence on ni

2. The nonideality factor is typically close to 1, butapproaches 2 for devices with high current densities. It is assumed to be 1 in thistext.

Diode Equation

iD I S exp qv D

nkT

1

I S exp vD

nVT

1

Chap 3 -14


Diode Voltage and Current Calculations(Example)

Problem: Find diode voltage for diode with given specifications

Given data: IS=0.1 fA ID= 300 A

Assumptions: Room-temperature dc operation with VT=0.025 V

Analysis:

With IS=0.1 fA

With IS=10 fA

With ID= 1 mA, IS=0.1 fA

V718.0)A16-10A4-103V)ln(10025.0(11ln

SIDI

TnVDV

V603.0DV

V748.0DV

Chap 3 -15


Diode Current for Reverse, Zero, andForward Bias

• Reverse bias:

• Zero bias:

• Forward bias:

iD I S exp vD

nVT

1

IS 01 I S

iD IS exp vD

nVT

1

I S 11 0

iD I S exp vD

nVT

1

I S exp vD

nVT

Chap 3 -16


Semi-log Plot of Diode Current andCurrent for Three Different Values of IS

I S[ A]10IS [B ]100IS [C ]

Chap 3 -17


Diode Temperature Coefficient

Diode voltage under forward bias:

Taking the derivative with respect to temperature yields

Assuming iD >> IS, IS ni2, and VGO is the silicon bandgap energy at 0K.

For a typical silicon diode

vD VT ln iD

IS1

kTq

ln iD

IS1

kTq

ln iDIS

dvD

dT

kq

lniD

IS

kTq

1I S

dIS

dT

vD

TVT

1I S

dI S

dT

vD VGO 3VT

TV/K

dvD

dT

0.651.120.075 V300K

=-1.82 mV/K - 1.8 mV/C

Chap 3 -18


Reverse Bias

External reverse bias adds to the built-in potential of the pnjunction. The shaded regions below illustrate the increase in thecharacteristics of the space charge region due to an externallyapplied reverse bias, vD.

Chap 3 -19


Reverse Bias (cont.)

External reverse bias also increases the width of the depletionregion since the larger electric field must be supported byadditional charge.

wd (xn x p)2s

q1

NA

1ND

j vR

wherewd0 (xn xp) 2s

q1

NA 1

ND

j

wd wd 0 1vR

j

Chap 3 -20


Reverse Bias Saturation Current

We earlier assumed that reverse saturation current was constant.Since it results from thermal generation of electron-hole pairs inthe depletion region, it is dependent on the volume of the spacecharge region. It can be shown that the reverse saturation graduallyincreases with increased reverse bias.

I S I S 0 1v R

j

IS is approximately constant at IS0 under forward bias.

Chap 3 -21


Reverse Breakdown

Increased reverse biaseventually results in the diodeentering the breakdownregion, resulting in a sharpincrease in the diode current.The voltage at which thisoccurs is the breakdownvoltage, VZ.

2 V < VZ < 2000 V

Chap 3 -22


Reverse Breakdown Mechanisms

• Avalanche BreakdownSi diodes with VZ greater than about 5.6 volts breakdown according toan avalanche mechanism. As the electric field increases, acceleratedcarriers begin to collide with fixed atoms. As the reverse biasincreases, the energy of the accelerated carriers increases, eventuallyleading to ionization of the impacted ions. The new carriers alsoaccelerate and ionize other atoms. This process feeds on itself andleads to avalanche breakdown.

Chap 3 -23


Reverse Breakdown Mechanisms (cont.)

• Zener BreakdownZener breakdown occurs in heavily doped diodes. The heavy dopingresults in a very narrow depletion region at the diode junction.Reverse bias leads to carriers with sufficient energy to tunnel directlybetween conduction and valence bands moving across the junction.Once the tunneling threshold is reached, additional reverse bias leadsto a rapidly increasing reverse current.

• Breakdown Voltage Temperature CoefficientTemperature coefficient is a quick way to distinguish breakdownmechanisms. Avalanche breakdown voltage increases withtemperature, zener breakdown decreases with temperature.

For silicon diodes, zero temperature coefficient is achieved atapproximately 5.6 V.

Chap 3 -24


Breakdown Region Diode Model

In breakdown, the diode ismodeled with a voltage source,VZ, and a series resistance, RZ.RZ models the slope of the i-vcharacteristic.

Diodes designed to operate inreverse breakdown are calledZener diodes and use theindicated symbol.

Chap 3 -25


Reverse Bias Capacitance

Qn qN D xn A qN A N D

N A N D

wd A Coulombs

C j dQn

dvR

C j 0 A

1vR

j

F/cm2 where C j 0 s

wd 0

Changes in voltage lead to changes in depletion width and charge.This leads to a capacitance that we can calculate from the chargevoltage dependence.

Cj0 is the zero bias junction capacitance per unit area.

Chap 3 -26


Reverse Bias Capacitance (cont.)

Diodes can be designed with hyper-abrupt doping profiles thatoptimize the reverse-biased diode as a voltage controlled capacitor.

Circuit symbol for the variablecapacitance diode (varactor)

Chap 3 -27


Forward Bias Capacitance

CoulombsTDD iQ

F

T

TD

T

TSD

D

Dj V

iV

IidvdQ

C

In forward bias operation, additional charge is stored in neutralregion near edges of space charge region.

T is called diode transit time and depends on size and type ofdiode.

Additional diffusion capacitance, associated with forward regionof operation is proportional to current and becomes quite large athigh currents.

Chap 3 -28


Schottky Barrier Diode

One semiconductor region of the pnjunction diode is replaced by a non-ohmic rectifying metal contact.ASchottky contact is easily added ton-type silicon,metal region becomesanode. n+ region is added to ensurethat cathode contact is ohmic.

Schottky diode turns on at lowervoltage than pn junction diode, hassignificantly reduced internalcharge storage under forward bias.

Chap 3 -29


Diode Spice Model

Rs is inevitable series resistance of areal device structure. Currentcontrolled current source representsideal exponential behavior of diode.Capacitor specification includesdepletion-layer capacitance forreverse-bias region as well asdiffusion capacitance associated withjunction under forward bias.

Typical default values: Saturationcurrent= 10 fA, Rs = 0, Transittime= 0 second

Chap 3 -30


Diode Layout

Chap 3 -31


Diode Circuit Analysis: Basics

V and R may represent Theveninequivalent of a more complex 2-terminal network.Objective ofdiode circuit analysis is to findquiescent operating point fordiode, consisting of dc current andvoltage that define diode’s i-vcharacteristic.

Loop equation for given circuit is:

This is also called the load line forthe diode. Solution to this equationcan be found by:• Graphical analysis using load-linemethod.• Analysis with diode’s mathematicalmodel.• Simplified analysis with idealdiode model.• Simplified analysis using constantvoltage drop model.

DD VRIV

Chap 3 -32


Load-Line Analysis (Example)

Problem: Find Q-pointGiven data: V=10 V, R=10k.Analysis:

To define the load line we use,VD= 0VD= 5 V, ID =0.5 mA

These points and the resultingload line are plotted.Q-point isgiven by intersection of load lineand diode characteristic:

Q-point = (0.95 mA, 0.6 V)

1010 4DD VI

mA1)k10/V10( DI

Chap 3 -33


Analysis using Mathematical Model forDiode

DD

DT

DSD

VV

VnVV

II

140exp101010

140exp101exp

134

13

Problem: Find Q-point for given diodecharacteristic.Given data: IS =10-13 A, n=1, VT=0.0025 VAnalysis:

is load line, given by a transcendentalequation. A numerical answer can befound by using Newton’s iterativemethod.

•Make initial guess VD0 .

•Evaluate f and its derivative f’ forthis value of VD.•Calculate new guess for VD using

•Repeat steps 2 and 3 till convergence.

Using a spreadsheet we get :Q-point = ( 0.9426 mA, 0.5742 V)

Since, usually we don’t have accuratesaturation current and significanttolerances for sources and passivecomponents, we need answers preciseup to only 2or 3 significant digits. DD VVf 140exp101010 134

)(' 0

001

D

DDD Vf

VfVV

Chap 3 -34


Analysis using Ideal Model for Diode

If diode is forward-biased, voltage across diodeis zero. If diode is reverse-biased, currentthrough diode is zero.vD =0 for iD >0 and vD =0 for vD < 0Thus diode is assumed to be either on or off.Analysis is conducted in following steps:• Select diode model.• Identify anode and cathode of diode and labelvD and iD.• Guess diode’s region of operation from circuit.• Analyze circuit using diode model appropriatefor assumed operation region.• Check results to check consistency withassumptions.

Chap 3 -35


Analysis using Ideal Model for Diode:Example

Since source is forcing positive currentthrough diode assume diode is on.

Q-point is(1 mA, 0V)0

m1k10

V)010(

D

D

I

AI

our assumption is right.

Since source is forcing currentbackward through diode assume diodeis off. Hence ID =0 . Loop equation is:

Q-point is (0, -10 V)V10

01010 4

D

DD

V

IV

our assumption is right.

Chap 3 -36


Analysis using Constant Voltage DropModel for Diode

Analysis:

Since 10V source is forcing positive currentthrough diode assume diode is on.

vD = Von for iD >0 andvD = 0 for vD < Von.

mA94.0k10

V)6.010(k10

V)10(

on

D

VI

Chap 3 -37


Two-Diode Circuit Analysis

Analysis: Ideal diode model is chosen. Since15V source is forcing positive current throughD1 and D2 and -10V source is forcing positivecurrent through D2, assume both diodes are on.

Since voltage at node D is zero due to shortcircuit of ideal diode D1,

Q-points are (-0.5 mA, 0 V) and (2.0 mA, 0 V)

But, ID1 <0 is not allowed by diode, so try again.

mA50.1k10

V)015(1

I mA00.2k5

V)10(02

DI

211 DIDII mA50.025.11 DI

Chap 3 -38


Two-Diode Circuit Analysis (contd.)

Since current in D1 is zero, ID2 = I1,

Q-points are D1 : (0 mA, -1.67 V):off

D2 : (1.67 mA, 0 V) :on

Analysis: Since current inD2 but that in D1 is invalid,the second guess is D1 offand D2 on.

V67.17.16151000,10151

mA67.115,000

V251

0)10(2000,51000,1015

IDV

IDII

Chap 3 -39


Analysis of Diodes in ReverseBreakdown Operation

Choose 2 points (0V, -4 mA) and (-5 V, -3mA) to draw the load line.It intersects with i-vcharacteristic at Q-point (-2.9 mA, -5.2 V).

Using piecewise linear model:

DIDV 500020 Using load-line analysis:

0 DIZI

mA94.25100

V)520(

05510020

ZI

ZI

Since IZ >0 (ID <0), solution is consistentwith Zener breakdown assumption.

Chap 3 -40


Voltage Regulator using Zener Diode

Zener diode keeps voltageacross load resistor constant.For Zener breakdownoperation, IZ >0.

mA3k5

V)520(

RZ

VS

VSI

mA1k5V5

L

RZ

VLI

mA2 LISIZI

011

LRRZVRS

VZI

For proper regulation, Zener current must bepositive. If Zener current <0, Zener diode nolonger controls voltage across load resistorand voltage regulator is said to have“dropped out of regulation”.

min1

R

ZVSVR

LR

Chap 3 -41


Voltage Regulator using Zener Diode:Example (Including Zener Resistance)

Problem: Find output voltage andZener diode current for Zener dioderegulator.Given data: VS=20 V, R=5 k, RZ=0.1 kVZ=5 VAnalysis: Output voltage is afunction of current through Zenerdiode.

0mA9.1100V5V19.5

100V5

V19.5

05000100V5

5000V20

LVZI

LV

LVLVZV

Chap 3 -42


Line and Load Regulation

Line regulation characterizes how sensitive output voltage is to inputvoltage changes.

SdV

LdV

Line Regulation mV/V

For fixed load current, Line regulation =

Load regulation characterizes how sensitive output voltage is to changesin load current withdrawn from regulator.

Load Regulation Ohms

For changes in load current, Load regulation =

Load regulation is Thevenin equivalent resistance looking back intoregulator from load terminals.

LdI

LdV

ZRRZ

R

)( RZR

Chap 3 -43


Rectifier Circuits

• Basic rectifier converts an ac voltage to a pulsating dc voltage.

• A filter then eliminates ac components of the waveform toproduce a nearly constant dc voltage output.

• Rectifier circuits are used in virtually all electronic devices toconvert the 120 V-60 Hz ac power line source to the dc voltagesrequired for operation of the electronic device.

• In rectifier circuits, the diode state changes with time and agiven piecewise linear model is valid only for a certain timeinterval.

Chap 3 -44


Half-Wave Rectifier Circuit withResistive Load

For positive half-cycle of input, source forces positive current throughdiode, diode is on, vo = vs.

During negative half cycle, negative current can’t exist in diode, diode isoff, current in resistor is zero and vo =0 .

Chap 3 -45


Half-Wave Rectifier Circuit withResistive Load (contd.)

Using CVD model, during on state of diode vo=(VP sint)- Von. Output voltage is zero whendiode is off.

Often a step-up or step-down transformer is usedto convert 120 V-60 Hz voltage available frompower line to desired ac voltage level as shown.

Time-varying components in circuit output areremoved using filter capacitor.

Chap 3 -46


Peak Detector Circuit

As input voltage rises, diode is on andcapacitor (initially discharged) chargesup to input voltage minus the diodevoltage drop.

At peak of input, diode current tries toreverse, diode cuts off, capacitor has nodischarge path and retains constantvoltage providing constant output voltage

Vdc = VP - Von.

Chap 3 -47


Half-Wave Rectifier Circuit with RCLoad

As input voltage rises during first quartercycle, diode is on and capacitor (initiallydischarged) charges up to peak value of inputvoltage.

At peak of input, diode current tries to reverse,diode cuts off, capacitor dischargesexponentially through R. Discharge continuestill input voltage exceeds output voltage whichoccurs near peak of next cycle. Process thenrepeats once every cycle.

This circuit can be used to generate negativeoutput voltage if the top plate of capacitor isgrounded instead of bottom plate. In this case,Vdc = -(VP - Von)

Chap 3 -48


Half-Wave Rectifier Circuit with RCLoad (contd.)

Output voltage is not constant as in ideal peak detector, but has ripplevoltage Vr.

Diode conducts for a short time T called conduction interval duringeach cycle and its angular equivalent is called conduction angle θc.

PVrVTc

PVrV

PV

onVP

V

RCTT

CT

RonV

PV

TT

RCT

onVPVrV

2

21)(21

)(1)(

Chap 3 -49


Half-Wave Rectifier Analysis: Example

Problem: Find dc output voltage, output current, ripple voltage, conductioninterval, conduction angle.Given data: secondary voltage Vrms 12.6 (60 Hz), R= 15 , C= 25,000 F,Von = 1 VAnalysis:

Using discharge interval T=1/60 s,

A12.11516.8V

16.8VV)126.12(

RonVPV

dcI

onVPVdcV

ms769.012029.0

2

fccT

V747.0)(

CT

RonV

PV

rV

06.16rad290.02 PVrVTc

Chap 3 -50


Peak Diode Current

In rectifiers, nonzero current exists indiode for only a very small fraction ofperiod T, yet an almost constant dccurrent flows out of filter capacitor toload.

Total charge lost from capacitor in eachcycle is replenished by diode duringshort conduction interval causing highpeak diode currents. If repetitive currentpulse is modeled as triangle of height IPand width T,

using values from previous example.

A6.482 TT

dcIPI

Chap 3 -51


Surge Current

Besides peak diode currents, when power supply is turned on, there is aneven larger current through diode called surge current.

During first quarter cycle, current through diode is approximately

Peak values of this initial surge current occurs at t = 0+:

using values from previous example.

Actual values of surge current won’t be as large as predicted because ofneglected series resistance associated with rectifier diode as well astransformer.

tPCVtP

VdtdCtcitdi cossin)()(

A168 PCVSCI

Chap 3 -52


Peak Inverse Voltage Rating

Peak inverse voltage (PIV) rating of therectifier diode gives the breakdownvoltage.

When diode is off, reverse-bias acrossdiode is Vdc - vs. When vs reachesnegative peak,

PIV value corresponds to minimumvalue of Zener breakdown voltage forrectifier diode.

PVPVonVPVsvdcV 2)(minPIV

Chap 3 -53


Diode Power Dissipation

dcIonV

TTPI

onVdtT

tD

ionVT

dtT

tD

itD

vTDP

20

)(1

0)()(1

Average power dissipation in diode is given by

The simplification is done by assuming the triangular approximation ofdiode current and that voltage across diode is constant at Vdc.

Average power dissipation in the diode series resistance is given by

This power dissipation can be reduced by minimizing peak currentthrough use of minimum required size of filter capacitor or using full-wave rectifiers.

SRdcITT

TT

SRPIdtT

SRt

Di

TDP 2342

31

0)(21

Chap 3 -54


Full-Wave Rectifiers

Full-wave rectifiers cut capacitor dischargetime in half and require half the filtercapacitance to achieve given ripple voltage.All specifications are same as for half-waverectifiers.Reversing polarity of diodes gives a full-wave rectifier with negative output voltage.

Chap 3 -55


Full-Wave Bridge Rectification

Requirement for a center-tappedtransformer in the full-waverectifier is eliminated through useof 2 extra diodes.All otherspecifications are same as for ahalf-wave rectifier exceptPIV=VP.

Chap 3 -56


Rectifier Topology Comparison

• Filter capacitor is a major factor in determining cost, size and weight indesign of rectifiers.

• For given ripple voltage, full-wave rectifier requires half the filtercapacitance as that in half-wave rectifier. Reduced peak current canreduce heat dissipation in diodes. Benefits of full-wave rectificationoutweigh increased expenses and circuit complexity (a extra diode andcenter-tapped transformer).

• Bridge rectifier eliminates center-tapped transformer, PIV rating ofdiodes is reduced. Cost of extra diodes is negligible.

Chap 3 -57


Rectifier Design Analysis

Problem: Design rectifier with given specifications.Given data: Vdc =15 V, Vr < 0.15 V, Idc = 2 AAnalysis: Use full-wave bridge rectifier that needs smaller value of filtercapacitance, smaller diode PIV rating and no center-tapped transformer.

A7.940.352mss1/60

A22T2

ms352.017V

V)15.0(2120

121T

F111.0V15.0

1s120

1A22/

rms12VV2

2152

2

2

TdcIPI

PV

rVrV

Tdc

IC

onVdc

VP

VV

A711)17)(111.0(120surgeI PCV

V17PIV PV

Chap 3 -58


Three-terminal IC Voltage Regulators

• Uses feedback with high-gain amplifier to reduce ripple voltage atoutput.Bypass capacitors provide low-impedance path for high-frequencysignals to ensure proper operation of regulator.

• Provides excellent line and load regulation, maintaining constant voltageeven if output current changes by many orders of magnitude.

• Main design constraint is VREG which must not fall below a minimumspecified “dropout” voltage (a few volts).

Chap 3 -59


DC-to-DC converters: Boost Converter

When switch is closed, diode is off:

onTLSV

LidtonT

LS

VLionTLi )0(

0)0()(

When switch is open, diode is on:

offTLoVSV

onTLSV

LiTLi

)0()(

)0()( LiTLi

1sV

oV

where = Ton/T is duty cycleSince 0<<1, Vo > VS.

Chap 3 -60


DC-to-DC converters: Boost Converter(contd.)

Ripple current is given by or

In ideal converter, power delivered to input of converter is same as powerdelivered to load resistor.

or

Ripple voltage is given as

onTLS

VrI

RC

ToV

TonT

RC

ToV

RConToV

onTCoIrV

frI

SV

TonT

rI

TS

VonT

rIS

VL

oIoVSISV

1

oI

offT

ToI

SV

oVoISI

offTL

SVoV

rI

Chap 3 -61


DC-to-DC converters: Buck Converter

When switch is closed, diode is off:

onTLoVSV

LidtonT

LoV

SV

LionTLi

)0(0

)0()(

When switch is open, diode is on:

offTLoV

onTLoVSV

LiTLi

)0()(

)0()( LiTLi

sVTonT

sVoV

where is switch duty cycleSince 0<<1, Vo < VS.

Chap 3 -62


DC-to-DC converters: Buck Converter(contd.)

Ripple current is given by or

In ideal converter, power delivered to input of converter is same as powerdelivered to load resistor.

Ripple voltage is given as

offTLoV

rI

CQoff

TonT

onTdtriCrV

)2/(

2/

1

frI

oV

Toff

T

rI

ToVoffT

rIoV

L

oITonT

oI

SV

oVoISI

onTL

oVS

VrI

82221 TrIoff

TonTrI

Q

)1(8

2

8

LT

rVoV

rV

TrIC

Chap 3 -63


Clamping or DC-Restoring Circuit

After the initial transient lastingless than one cycle in bothcircuits, output waveform is anundistorted replica of input.

Both waveforms are clamped tozero. Their dc levels are said tobe restored by the clampingcircuit.

Clamping level can also beshifted away from zero byadding a voltage source in serieswith diode.

Chap 3 -64


Clipping or Limiting Circuits

Clipping circuits have dc path between input andoutput, whereas clamping circuits use capacitivecoupling between input and output.The voltage transfer characteristic shows thatgain is unity for vI < VC, and gain is zero for vI >VC.A second clipping level can also be set as shownor diodes can be used to control circuit gain byswitching resistors in and out of circuits.

Chap 3 -65


Dynamic Switching Behavior of Diode

Non-linear depletion-layer capacitanceof diode prevents voltage from changinginstantaneously and determines turn-onand recovery times. Both forward andreverse current overshoot final valuewhen diode switches on and off asshown. Storage time is given by:

R

IF

ITS 1ln

Chap 3 -66


Photo Diodes and Photodetectors

If depletion region of pn junction diode is illuminated withlight with sufficiently high frequency, photons can provideenough energy to cause electrons to jump the semiconductorbandgap to generate electron-hole pairs:

GEhchPE

h =Planck’s constant= 6.626e-34 J.sυ= frequency of optical illumination= wavelength of optical illuminationc =velocity of light=3e+8m/sPhoton-generated current can be used in photodetectorcircuits to generate output voltage

Diode is reverse-biased to enhance depletion-region widthand electric field.

Riov PH

Chap 3 -67


Solar Cells and Light-Emitting Diodes

In solar cell applications, opticalillumination is constant, dc current IPH isgenerated. Aim is to extract power fromcell, i-v characteristics are plotted in termsof cell current and cell voltage. For solarcell to supply power to external circuit,ICVC product must be positive and cellshould be operated near point of maximumoutput power Pmax.

Light-Emitting Diodes use recombinationprocess in forward-biased pn junctiondiode. When a hole and electronrecombine, energy equal to bandgap ofsemiconductor is released as a photon.

Chap 3 -68

Chap3 Diode Circuits

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