Lecture 6Solid-State Diodes and Diode Circuits

1

Goals

• Understand diode structure and basic layout• Explore various diode models including the mathematical model, the

ideal diode model, and the constant voltage drop model• Define regions of operation of the diode (forward, reverse bias, and

reverse breakdown)• Apply the various types of models in circuit analysis

2

Diode Introduction

• A diode is formed by interfacing an n-type semiconductor with a p-type semiconductor.

• A pn junction is the interface between n and p regions.

Diode symbol

3

Diode Layout

4

Space Charge Region Formation at the pn Junction

5

Diode i-v Characteristics

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

6

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 due to its dependence on ni

2. The nonideality factor is typically close to 1, but approaches 2 for devices with high current densities. It is assumed to be 1 if the value of “n”

Diode Equation

iD IS expqvD

nkT

1

IS exp

vD

nVT

1

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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

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Diode Current for Reverse, Zero, and Forward Bias

• Reverse bias:

• Zero bias:

• Forward bias:

iD IS expvD

nVT

1

IS 0 1 IS

iD IS expvD

nVT

1

IS 1 1 0

iD IS expvD

nVT

1

IS exp

vD

nVT

9

Ideal Diode

A semiconductor diode consists of a PN junction and has (2) terminals, an anode(+) and a cathode(-).The diode is a unidirectional device; Current flows only from anode to cathodeForward bias : v>0, any current will flow (ideal diode acts as a short circuit)Reverse Bias: v < 0, i=0 (open circuit)

+-Anode

CathodeForward bias

Vanode > vcathode

Reverse biasVanode < vcathode

Break point

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Diode Circuit Analysis: Basics

V and R may represent Thevenin equivalent of a more complex 2-terminal network.Objective of diode circuit analysis is to find quiescent operating point for diode, consisting of dc current and voltage that define diode’s i-v characteristic.

Loop equation for given circuit is:

This is also called the load line for the diode. Solution to this equation can be found by:

1. Simplified analysis with ideal diode model.

2. Simplified analysis using constant voltage drop model.

3. Graphical analysis using load-line method.

4. Analysis with diode’s mathematical model.

DD VRIV

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1.Analysis using Ideal Model for Diode

If diode is forward-biased, voltage across diode is zero. If diode is reverse-biased, current through diode is zero.

vD =0 for iD >0 and vD =0 for vD < 0 Thus diode is assumed to be either on or off.

Analysis is conducted in following steps:

1. Select diode model.2. Identify anode and cathode of diode and label

vD and iD.3. Guess diode’s region of operation from circuit.4. Analyze circuit using diode model appropriate

for assumed operation region.5. Check results to check consistency with

assumptions.

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Analysis using Ideal Model for Diode: Example

Since source is forcing positive current through diode assume diode is on.

Q-point is(1 mA, 0V)

0

m1k10

V)010(

D

D

I

AI

our assumption is right. our assumption is right.

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

Q-point is (0, -10 V)

V10

01010 4

D

DD

V

IV

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Two-Diode Circuit Analysis

Analysis: Ideal diode model is chosen. Since 15V source is forcing positive current through D1 and D2 and -10V source is forcing positive current through D2, assume both diodes are on.

Since voltage at node D is zero due to short circuit 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 ?.

mA50.1k10

V)015(1

I mA00.2k5

V)10(02

DI

211 DI

DII mA50.025.1

1

DI

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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 in D2 but that in D1 is invalid, the second guess is D1 off and D2 on.

V67.17.16151

000,10151

mA67.115,000

V251

0)10(2

000,51

000,1015

ID

V

I

DII

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Break point analysis

VV

VKK

KV

VVx

in

in

3

).12

2(2

2

The break point is wherei=0 and v=0 transition from ‘ON” to “OFF”OR

It is the point after which the diode will start conducting

Range of V after which the diode will be “ON”

Vin

1K

2K

2V

Vx1K

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2.Analysis using Constant Voltage Drop Model for Diode

Analysis:

Since 10V source is forcing positive current through diode assume diode is on.

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

mA94.0k10

V)6.010(k10

V)10(

onD

VI

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3. Graphical analysis using load-line method.

Linear Circuit

Nonlinear element

+

-

iI and V must satisfy both the linear

And non-linear circuit equations

For the linear circuit

Vth(VOC)

R

Ino(ISC) R

i

+

i

Finding Thevenin or Norton equivalent of the linear circuit

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 resulting load line are plotted.Q-point is given by intersection of load line and diode characteristic:

Q-point = (0.95 mA, 0.6 V)

1010 4DD VI

mA1)k10/V10( DI

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4. Analysis using Mathematical Model for Diode

Ideal Diode

)1(

9.25 voltageThermalq

KT

current saturation ReverseI

)21(factor ideality n

eTemperaturT

)/.38x10Constant(1 sBoltzmann'K

)(1.6x10 chargeelectron q

)1(

s

23-

19-

T

D

D

nV

v

sD

T

o

o

nKT

qv

sD

eIi

mVV

n

K

KJ

C

eIi

Iterative Method based on Mathematical Model

)1( T

D

nV

v

sD eIi

Linear Circuit

Nonlinear element

+

-

iI and V must satisfy both the linear

And non-linear circuit equations

VOC=Vth

i

ISC =VOC/R

i

vva

ia

ib

Repeat e.

for solve circuit,linear theinto Plug d.

ln

find andequation diode theinto Put c.

b.

for valuea Assume a.

R

vViv

I

inVv

vi

iR

vVi

v

aocba

S

aTa

aa

aDoc

D

D

Linear circuit equation

Exponential diode equation

Example: Analysis using Mathematical Model for Diode

DD

DT

DSD

VV

VnV

VII

140exp101010

140exp101exp

134

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

=0.0025 VAnalysis:

is load line, given by a transcendental equation. A numerical answer can be found by using iterative method.

Use the technique in the previous slide

Q-point = ( 0.9426 mA, 0.5742 V)

Since, usually we don’t have accurate saturation current and significant tolerances for sources and passive components, we need answers precise up to only 2or 3 significant digits.

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Summary: Diode circuit linear models

a. Ideal diode

v

i

0 ,0

0 ,0

iv

iv

b. Constant voltage model

0 ,

0 ,

iVv

iVv

D

D

vVD

For Silicon diodes

0.65 < vD < 0.8

+

VD

i

ideal

ideal

ideal

VDO

rD

c. Piecewise linear model

vVDO

i1/rD

Summary: Diode circuit linear models