PN Junction Diode - lark.tu-sofia.bg

EE141© 2009, Associate Professor PhD. T.VasilevaProf. Tania Vasileva, Ph.D1

PN Junction Diode

Semiconductor

Elements


Diode Cases


Basic Diode Feature

The essential feature of a diode is that the magnitude of the current greatly

depends on the polarity of applied voltage.

Basically the diode conducts current in only one direction. It is applied in

rectifiers for converting an alternating current into a direct current.


PN Junction Diode Structure

An ideal diode can conduct current in only one direction.

Diode specific electrical properties depends on it’s structure.


Junction Diode Schematic Symbol

A diode is a nonlinear semiconductor device with two electrodes:

an anode and a cathode.

An arrow points from the anode to the cathode to remind that the

current will only flow easily in this direction.


Diode's Mode of Operation

A diode's mode of operation depends on the various conditions possible

at the pn junction. They are caused by the polarity of the applied voltage.


Unbiased Diode

Equal number

of plus and

minus charges

Electrically

neutral as

a whole


PN Junction Formation

Unlike the free carriers, the ions will never move. They remains fixed due

to their covalent bonds in the crystal structure.

Majority carriers – free electrons and holes, diffuse (spread) across the

junction because difference in concentration in both side of the crystal.


Depletion Layer

When an electron leaves the n side, it will leave behind an uncompensated

positive ion. A positive space charge is created to the right of the junction in

the n-region.

Similarly a negative space charge will be created to the left of the junction

in the p-region.

The depletion layer is almost completely depleted of mobile charge

carriers and has very high resistance.


Barrier Potential & Electric Field

The uncompensated positively- and negatively-charged ions that are within

the depletion layer generate an electric field Eo the barrier potential Uo.

At room temperature (25 oC) the barrier potential has for Si diodes a

voltage of approximately 0.7V and for Ge diodes a voltage of about 0.3V.

ID

IE

Majority carriers

Minority carriers

Diffusion current ID

Drift current IE


Forward Bias

The barrier potential will reduce to Uo - Us and the electric field to E < Eo.

A forward current is always composed of majority carriers whose energies

are sufficient to overcome the barrier potential.

Current will easily flow in forward-biased diode.


Reverse Bias

The barrier potential will increase to Uo + Us and the electric field to E > Eo.

The diffusion of majority carriers across the junction has been greatly diminished.

A very small reverse current IR, composed of thermally produced minority carriers

still cross the junction. This makes the reverse current IR independent from the

direction of the voltage polarity and of the barrier potential.


VA Characteristic

A diode is a nonlinear device. It conducts in only one direction.

An ideal diode functions like a switch – open and closed.

In forward-biased diode, the current increases very rapidly with U. The

reverse current is very small.


Ideal Diode Equation

)1( T

U

s eII

q

kTT

11600

)(KTT 0258.0T

- Thermal potential

V for T = 25 oC

Is - Saturation current

ss SJI S - Junction area

Js - Current density

1ln

S

TI

IU

T


Ideal Diode Equation

T

U

seII

sII


Saturation Current

),,( STWfIs

Is doubles with every 10oC increase.

Because the reverse current is caused by

thermally created minority carriers, it will

also be highly sensitive to temperature

changes.

There are fewer minority carriers

in Si diodes than in Ge diodes.

A Si diode has a much smaller

Is than a Ge diode.


VA Characteristics of Si and Ge Diode

IF , mA

UF , V

IR , (µA or nA)

UR , V


Real Diode – Forward Region

R

U

d

+_

p n

+_

IF , mA

UF, V

rE rB

UB

UPN

rBI

UB

UPNI.rB

U = UPN + IrB

Ohmic

region

Е

Recombination current, base ohmic resistance rB, rB = f (I) are considered.

)( T

B

m

IrU

S eII


Real Diode – Reverse Region

UR , V

I leakage

IS

IGIR

Breakdown

region

Si ~ IG

Ge ~ IS

UR, V

IR, nA

BRутGSR IIIII

Generation current, leakage current and breakdown are considered.

IG ≈√U

IG >> IS

IS >> IG

Si diode IR = IS + Ilekage (IG >> IS ),

Ge diode IR = IG + Ileakage (IS >> IG)

IR, nA


Forward Voltage and Temperature

constIT

U

dT

dUTKU F

TKUF ≈ - 2 mV/ oC


Diode Ratings

Max junction temperature Tjmax < Ti, where n=p=ni

Max power Pjmax

th

ai

R

TTP

max

max

th

aj

R

TTP

P = UI Power dissipated in diode

Power conducted

into the ambient

surroundings

th

aj

R

TTUI

maxmax

11j

th

a

th

TR

TR

P

EE141© 2009, Associate Professor PhD. T.VasilevaProf. Tania Vasileva, Ph.D

Maximum Power & Temperature

maxmax

11j

th

a

th

TR

TR

P

Ta,oC

Pjmax, W

Tjmax25oC

Ta,oC

Pjmax, W

Tjmax25oC


Maх Power & Maх Current

th

aj

FFR

TTPIU

max

maxmax

IFmax, A

Ta,oC

Tjmax

Ta,oC

Pjmax, W

Tjmax25oC

maxmax

11j

th

a

th

TR

TR

P


Current-Limiting Resistor

The current-limiting resistor has the function to keep the diode current

smaller than the maximum rating.

maxFF I

R

UEI


Heat Removal

jcca thth RR cajc ththth RRR

hachjc thththth RRRR

Thermal energy can be easily reduced through conduction and radiation from

the device's case.

Heat sink

Rth indicates efficiency in removing

heat from the transistor in units oK/W.

The less thermal resistance the higher power rating.


Reverse Breakdown

At a reverse breakdown voltage UBR, the current will rapidly increase with

only small changes in the voltage.

Electrical breakdown

Thermal breakdownUBR

Avalanche breakdown

Zener breakdown


Avalanche Breakdown

n

BR

RR

BR

U

UI

IM

1

1

Minority carriers can be accelerated by the field and

generate another two electron-hole pairs on collisions

with crystal atoms. The process may continue, causing

the increase of the current.

When the temperature increases, the crystal atoms

start moving randomly in the lattice, preventing in this

way acceleration of carriers, so they can not reach

enough energy when to generate new electron-hole

pairs by hitting atoms. For this reason the breakdown

voltage increases.

UBR >7V

Avalanche breakdown occurs in wide depletion

layers with a reverse bias higher than 7V.


Zener Breakdown

When an electric field of the barrier potential is large enough to break the

covalent bonds additional free electrons and holes are created. Such effect is

called Zener effect or Zener breakdown. This requires high electric fields on

the order of 300 000 V/cm.

When the temperature increases the process of breaking up covalent bonds is

facilitated and the breakdown voltage decreases.

Zener breakdown occurs in very narrow depletion layers with a reverse

bias less than 5V.

Covalent

bond

UBR < 5V


Thermal Breakdown

th

aj

RRR

TTIU

UR,V

IR, mA

UBR1 UBR2

T2T1

r < 0

T2 > T1

UR,V

IR, mA

Avalanche Thermal

Zener

Surface

This breakdown occurs if


Load Line and Operating Point

E = U + IR

R

EU

RI

1

E1

E1

R

Е2E

E

R1

E2

R

E2 > E1E

R2

R2 > R1

Q

Q

QI

UR R

E

E

EQ

Kirchhoff law

UF,VUF,V

IF, mAIF, mA

If I = 0, U = ELoaf line

End points


Equivalent Circuits

Threshold model


Examples

1k

12mA

12V

11.3 mA

The diode is forward connected with Uo = 0.7V

The voltage over resistor UR се is calculated from Kirchhoff's law

UR = V1 – Uo = 12 – 0.7 = 11.3V

Calculation current and voltages in the circuit by using diode threshold model.

Diode

Resistor

U, V I, mA

0.7

11.3

11.3

11.3

The current is the same through all

elements in serial circuits

Current through R1 from Ohm law is:

Operating

point12V

1k


Examples

Diode

Resistor

U, V I, mA

3

0

0

0

The diode is revers connected, I=0.

From Ohm law UR = I.R1 = 0

Voltage oved diode is calculated from Kirchhoff's law

UD = V1 – I.R1 = 3-0 = 3V

=13 mA

Operating

point

Calculation current and voltages in the circuit by using diode threshold model.


What will be the reading of the voltmeter (UD) if the

diode is made of Si? What current flows through the

diode? Build the load line and mark the operating

point. Use a threshold model of the diode.

What will the voltmeter read if the diode is made of

Ge? What current flows through the diode?

Build the load line and mark the operating point. Use

a threshold model of the diode.

Calculate the power dissipated on the Si diode in the

circuit.

Examples – self learning

PD = I.UD I = ? Use the diode threshold voltage.

Ge


Diode Testing

PN Junction Diode - lark.tu-sofia.bg

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