Semiconductor Devices Module 2 Notes

8/10/2019 Semiconductor Devices Module 2 Notes

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109. Semiconductor Devices Module 2

Department of ECE, VKCET Page3

PN junction under thermal equilibrium:

How p-n junction formed?

Consider a p-type material of doping concentration Na and n-

type material of doping concentration Nd brought together to

form a contact at room temperature, where Na> Nd.

The carrier concentration, diffusion and drift of carriers and

formation of p-n junction are as shown below:

Due to the gradient of carrier concentration in both sides, holes diffuse from p-side to n-

side and electron diffuse from n-side to p-side.

The diffused carriers recombine and results uncompensated acceptor ions left behind in

the p-side and donor ions left behind n-side as shown in figure b. Then forms

uncompensated negative ions on p-side and positive ions on n-side.

The region of immobile ions between p-type and n-type material is called depletion

region or space charge region.

Due to the presence of ionic charge between p-n junction results a potential difference

between the junction and also develops an electric field.

Question:

Part B

What is meant by depletion

region of a p-n junction?How is it formed?


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This electric field causes drift minority carriers across the junction and also reduces the

diffusion of the carrier due to its gradient.

The properties of the depletion layer are:

1. This region is depleted of mobile charges electrons and

holes.

2.

The p-side of depletion layer is negatively charged and n-

side is positively charged.

3. Number ions on p-side are equal to the number of ions on n-

side. Thus the electrical charge of depletion region is neutral.

4. There is a potential difference exists between two edges of depletion layer, because of

the opposite charges of immobile ions.

5. The width of the depletion layer decrease with increase in doping concentration.

6. If doping concentration are equal in p and n sides, the width of the depletion region

are equal on both sides.

7.

If one side is lightly doped and other side is heavily doped, the lightly doped side willhave more width and heavily doped side will have less width.

The direction of current flow in p-n junction under equilibrium is shown below.

Under thermal equilibrium the net current across the junction must be zero. Thus the

electron and hole current are zero and hence the net current across the junction is zero.

Question:

Part BList the properties of

depletion region of a p-n

junction.


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Equilibrium energy band diagram of p-n junction:

Consider the following principles to draw the energy band diagram under thermal

equilibrium:

1. There is no gradient for Fermi-level EF,

i.e.

dE F

dx = 02. Depletion layer is depleted of mobile charge carriers and the region outside the layer

is neutral.

3. The electric field in the neutral region is zero and Fermi-levels in the neutral regions

depend on doping concentration.

4. The energy bands bends upward in the direction of the electric field or the direction

of energy barrier is opposite to the direction of potential barrier.

The steps to draw the energy band diagram:

1. Draw the equilibrium Fermi-level EF.

2.

Mark the depletion layer.3. Draw the valance band edge on the p-side EVp and

conduction band edge on the n-side ECnrelative to EF.

4. Draw the other edges of the bands ECp and EVn, keeping

constant band gap on both sides.

5. Connect ECpto ECnand EVpto EVnwhich completes energy band diagram.

Isolated p-n materials and p-n junction under thermal equilibrium are shown below:

Question:

Part B

Draw the equilibrium energyband diagram of a p-n

junction.


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Derivative information from energy band diagram

There is an energy barrier for the movement of electron from conduction band on n-side

to conduction band on p-side and is ECp - ECn. Like that the energy barrier for the

movement of hole from the p-side valance band to n-side valance band is EVpEVn.

The gradient of all energies are same, then

= = The potential across the junction is called built-in potential or contact potential or barrierpotential or diffusion potential V0.

It is called built-in potential, because it is built-in the

semiconductor and not an external one.

This potential develops due to the physical contact of

semiconductor, so it is called contact potential.

This potential acts as a barrier for majority carriers, hence named

as barrier potential. This potential is due to the diffusion of

majority carriers, so it is also called as diffusion potential.

The built-in potential across the junction is given by0 = ---(1)= +

=

= ---(2)where and are called Fermi-potentials on the p-side and n-side respectively. The Fermi-potentials increase with increase in doping, therefore the built-in potential also

increases with increase in doping.

Equilibrium energy band confirms that an electric field exists in the depletion layer which

is directed from n-side to p-side.

Question:

Part A

Define built-in potential.

Why it is called diffusionpotential?

Question:

Part A

Define Fermi-potentials.How it is related to built-in

potential in p-n junction?


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Relation between doping concentration and depletion layer width

For space charge neutrality, the total charges on both sides are equal, so the net charge is

zero.

i.e.

=

---(1)

where and are total charge in n-side and p-side respectively. Then = 0 = 0 where A is area of cross section.

Using (1), 0 = 0

=

---(2)

= ---(3)And 0 = 0 + 0

= 0 + 0=

0

1 +

0 = 0 + = +---(4)Similarly

=

+

---(5)


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Problem:

For Si p-n junction at 300K, Na= 1018

cm-3

on p-side and Nd= 1014

cm-3

on n-side. Find the ratio

of depletion layer.

Solution:

= = = Built-in potential/Contact potential:

Under thermal equilibrium = = 0 Consider hole current

= + = 0---(1) We know that there is a diffusion and drift of carrier and isfunction of distance. Then = 0() = 0() Using (1)

+

= 0

0 0(

)

= 00 = 0() = 10 0() Applying Einsteins relation,

=

1

0

0()

---(2)

By definition

= Using (2) = 1

00()

Question:

Part B

Derive the expression for

built-in potential of an abrupp-n junction.


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= 100()

Integrating both sides

=

1

00(

)---(3)

The LHS of (3) is the integration of potential distribution and RHS is integration of hole

distribution.

Apply limit to (3)

00 = 10 0()

00

00

= [(0)]00

00 = ln0 ln000 = ln0 ln0

= ln 00

Then built-in potential

0 = 0 0 = ---(4)We have = = Substituting this to (4)

= = ---(5)


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= 3.88 1016(3 2 )(7000 ) cm-3= 3.88 1016(4503 2 )(7000 450 )= 6.5 1013cm-3

0 = ln 2 = 0.039 ln1017 1015(6.5 1013)

2

= 0.393Problem:

An abrupt p-n junction made up of Si has Na= 1018

cm-3

on the p-side and Nd = 1015

cm-3

on the n-

side at 300K.

a) Find position of Fermi-levels on p-side and n-sides

b) Draw the equilibrium energy band diagram and determine the Vo, from the diagram.c) Compare V0obtained from energy band diagram with the calculated value using expression

for V0

Solution:

a) Fermi-level on p-side

We have 0 = () 0

=

()

= ln 0 = 0.026 ln = 0.026 ln 10181.5 1010

= 0.468Similarly

= ln0 = 0.026 ln = 0.026 ln 10151.5 1010

= 0.289


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b) Equilibrium energy band diagram:

Built-in potential0 = + = 0.468 + 0.289 0 = 0.757 = 0.757c) Built-in potential by expression

0 =

ln

2

= 0.026 ln1018 1015(1.5 1010)2 = 0.757


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Electric field:

The electric field distribution of depletion layer can be obtained

by Poissons equation. It states that

(

)

=

---(1)

i.e. the gradient in electric field is ratio of charge density andpermittivity = 0 Then for semiconductor() = + + Apply Poissons equation on p-side of the depletion layer,

(

)

= ---(1)For 0 < < 0Integrating both sides = +

=

+

---(2)

To find the constant C, apply the condition,

at = 0, = 0Then (2) becomes (0) + = 0 =

0

Substituting value of C to (2)

= + 0 = ( + ), for < < 0---(3) Apply Poissons equation on n-side()

=

+---(4)

Question:

Part B

Derive the expression for

maximum electric fieldacross an abrupt p-n junction


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For 0< < 0Integrating both sides

=

+

= + ---(5)To find the constant C, apply the condition,at = 0, = 0Then (5) becomes

0 + = 0

= 0Substituting value of C to (5) = + 0 = ( ), for 0< < ---(6) In depletion region, electric field is maximum at x = 0. Apply this condition to (3) and

(6),From (3) = 0 + 0 = 0From (6) =

0 0 =

0

Maximum electric field intensity is denoted as

= = ---(7) Substituting (7) into (3) and (6)From (3) = 0 0 + 1

Question:

Part A

Express electric field as a

function of distance in the

depletion region of an abrupt p-njunction.


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= 0 1 + 0

= + for < < 0 ---(8)From (6) = 0 0 + 1

=

0

1

0

= for 0< < ---(9)

The equations (8) and (9) shows that the electric field inside the depletion layer is linearly

related to distance x and its shape is like triangle with maximum value at x=0.

Potential distribution:

Assume potential at p-side or

=

0is zero.

By definition, potential on the p-side of the depletion layer = for 0 < < 0---(1)We have = 0 1 + 0Substituting to (1)

= 0 1 + 0 = 0 + 220 + ---(2)

Question:

Part B

Express the potential in a p-n

junction as a function ofdistance.


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To find constant C, apply the condition = 0, = 0Then (2) becomes

0 =

0

0 +

0

2

20+

= 0 0 + 02 = 0 20 + 02

= 0 02 Substituting the value of C to (2)

= 0 + 2200 02 = + + for < < 0---(3)

Similarly on the n-side of the depletion layer, for 0< < 0

=

---(4)

We have = 0 1 0Substituting to (4) = 0 1 0

=

0

2

20+

---(5)

To find the value of C:

The potential across depletion layer is continuous and at x = 0, Vp= Vn.

Apply this condition to (3) and (5).

Then = 00 0 +


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= 0 0 + 0 + 02

=

0

02

Substituting this to (5) = 0 220 0 02 = + for 0 Nd.

Then depletion width

= +

= +

=

+

Reverse saturation current = + Since


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Electrical breakdown of p-n junction:

When diode is reverse biased, according to ideal

concept a small reverse saturation current IS willflow and is independent to applied biasing voltage.

When reverse bias voltage increases to a particular

value, the current abruptly increases. This

phenomenon is called breakdown of p-n junction

and the reverse voltage for breakdown is called breakdown voltage Vbr.

Breakdown is an electrical phenomenon, it is reversible and there is no

mechanical damage for the diode.

The I-V characteristics of diode with breakdown is shown below:

There are two mechanisms causes breakdown: 1) Zener breakdown 2) Avalanche

break down.

A diode operating under Zener breakdown, Avalanche breakdown or mechanisms

is called Zener diode.

Question:

Part A

What is meant bybreakdown in p-n junction?

(April 2014)Does breakdown damage a p-n

junction? Why?


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Zener breakdown:

It is the breakdown occurs in heavily doped p-n

junction.

This breakdown is due to tunneling of electrons

from p-side to n-side.

Consider a heavily doped p-n junction, when it is reverse biased the energy bands

get crossed at relatively crossed. i.e. the filled states on p-side (valance band)

comes opposite to vacant states on n-side (conduction band). This is shown

below:

If the barrier separation between the valance band on p-side and conduction band

on n-side is too narrow, tunneling of electrons from p-side to n-side takes place.

This tunneling of electrons causes a reverse current from n to p.

Question:

Part B

Explain different

breakdown mechanisms

in p-n junctions. (April

2014)

Question:

Part A

What is zener breakdown?


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For tunneling under low reverse bias voltage, the width of the depletion region W

should be as low as possible. This will happen only for heavily doped p and n

regions.

Another reason for tunneling is due to the heavy electric field in depletion region

under reverse bias. This electric field increases the slope of the energy bands and

accelerates holes from n-side to p-side.

Typical electrical field for zener effect is 105V/cm.

Avalanche breakdown:

It is the breakdown occur in lightly doped p-n

junctions, where tunneling is negligible.

This breakdown is due to the ionizing collision of

carriers in depletion region.

When reverse bias voltage increases, an electron entering in depletion region

gains sufficient kinetic energy. This causes an ionizing collision and generates anEHP.

Then the original and generated electrons move

towards n-side and holes to p-side.

If the W is large and electric field is high, carrier

multiplication process continues in a cumulative

way as shown below:

This process is called Avalanche multiplication.

The electron multiplication factor in depletion region due to the Avalanche

multiplication M is given by = = Where the value n varies between 3 and 6 depending on the material.

Question:

Part A

What is Avalanche breakdown?

Question:

Part B

What is meant by Avalanche

multiplication? How is

multiplication factor related to

reverse bias voltage?


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Critical electric field under breakdown:

The maximum electric field in depletion region is = ---(1)= +

= + +

= + + = +---(2)

When reverse bias voltage Va = -VR = Vbr breakdown voltage, at which

maximum electric field is referred as critical electrical field

.

Using (2) = + +---(3) We know that

Using (3)

=

+

= + = + ---(4) This shows the relationship between breakdown voltage and critical electric field.

Question:

Part A

What is meant by critical field?

Question:

Part B

Derive an expression forbreakdown voltage of p-n

junction.


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Problem:

A Si abrupt p-n junction has Na= 1018

cm-3

on the p-side and Nd= 1016

cm-3

on the n-

side. Determine the breakdown voltage if critical electric field is 3 x 105 V/cm.

Determine the avalanche multiplication factor at reverse voltages of 10, 20, 29, 29.2

and 29.6V. Assume n = 3.

Solution: = + =. . . + = .

=

,

=

=

.

=

.

= , = . = . = , = . = . = ., = .. = . = ., = .. = .(This shows that M suddenly increase near the Vbr)


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Comparison between Zener breakdown and Avalanche breakdown:

Zener breakdown Avalanche breakdown

Breakdown due to

tunneling of

electrons.

Breakdown due to

Avalanche

multiplication.Characteristics near

the breakdown is

smooth, because oftunneling.

Characteristics near the

breakdown is sharp,

because of Avalanchemultiplication.

Occur in heavily

doped p-n junction

Occur in lightly doped

p-n junction

Breakdown voltagedecreases with

increase in

temperature.

Breakdown voltageincreases with increase

in temperature.

Vbris less than 4Vg,

where = Vbrgreater than 8VgApplications of diodes:

Diodes:

1. RectifierConverting AC to DC.

2. IsolationIsolating signals from power supply.

3. Wave shaping.

4. Temperature sensor.

Zener diodes:

1. Voltage regulators

2. Reference voltage source.

Question:

Part A

What are the differences

between Zener breakdown andAvalanche breakdown?

Question:

Part A

What are the applications of

diode breakdown?


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Reverse recovery transient:

Most of the switching applications require to turn

ON (forward biased) and OFF (reverse biased)

diode continuously.

When diode is switched between ON and OFF

states, a reverse current which is greater than

reverse saturation current can flow through the

diode.

Consider a p+-n diode suddenly changes from +V toV as shown below:

During t < 0, diode is forward biased by +V and current through the diode is = + (where forward resistance of the diode is negligible than R). At t = 0, the diode is reverse biased by V and the initial current must be reverse

and is = . This current is due to the stored charge and junction voltage, bothchange with time. At t = 0, the current is reversed and the voltage drop across the diode remains

same as in forward biased.

This is due to the excess minority carrier distribution as shown below:

Question:

Part B

What is meant by stored delay

time? How it is related to thecurrent through the diode? How

it is related to carrier life time?


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When stored charge decayed, the current i(t) remains until stored charge is

zero and isve. The current flow and voltage drop across the diode is shownbelow:

The time required for stored charge becomes zero is called storage delay time t sd.

It is given by = +---(1)The storage delay time is measured by = + ---(2)


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Storage capacitance, Cs:

Under forward bias the capacitance due to stored

charge dominates the depletion layer capacitance.

This capacitance is referred as storage capacitance

or diffusion capacitance.

The stored charge = ---(1)But the forward current = ---(2)Then

=

= ---(3)We have = ( )---(3)For forward bias = Then

=

---(4)

The capacitance due to small changes in stored charge is = =

=

---(5) Using (1), (5) becomes

=

= ---(6)

Question:

Part B

What is storage capacitance or

diffusion capacitance? How it isrelated to forward current?

Question:

Part B

Derive the expression forconductance of diode. How

does it vary with variation in

forward bias?


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Similarly the ac conductance is given by = Using (2) and (3)

= = = =

=

= ---(7)Then the ac resistance of the diode = And = .

The current flow in diode is given by

=

(

)

+

(

)

This gives that the storage capacitance introduces a serious limitation to high

frequency switching circuits.

Small signal equivalent circuit:

The small signal equivalent circuit of diode under

reverse bias is shown below:

Where

Cjis junction capacitance,

rRis dynamic reverse resistance, typically it is high value

Rsis resistance of p and n regions, typically it is low value

Question:

Part B

Draw the small signal

equivalent circuits of an

abrupt p-n junctionunder forward bias and

under reverse bias.

(April 2014)


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The small signal equivalent circuit of diode under forward bias is shown below:

Where

Cs is storage capacitance and its value is lesser than Cj, so this capacitance

dominates.

rFis dynamic forward resistance. We have

=

(

)

= ---(1) = From (1) =

=

Then = ---(2)Problem:

For a p+-n Si diode at 300K, Nd= 10

15cm

-3on the n-side, cross sectional area is 10

-3cm

2

on n-side and minority carrier life time is 0.1s. If the diffusion constant is 16cm2/s,

calculate dynamic forward resistance and storage capacitance at forward-bias of 0.6V.

Assume

= 1.

Solution:

We have = and =


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For p+-n junction = ( )

If forward biased

=

= Given = / = . = Then = = . = . = = . = .

=

= .

.

.

= . = . . (.) = . = = .. = . = = . . . = .


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Problem:

A Si p-n junction at 300K has Na= 1016

cm-3

, Nd= 1015

cm-3

, n= p= 0.1s, A = 10-3

cm2.

Determine:

a) Junction capacitance at zero bias Cj0

b) Junction capacitance at Va= -5V

c) Storage capacitance at Va= 0.5V

Assume p = 480 cm2/V-s and n = 1300 cm

2/V-s

Solution:

a) =

=

+

= = . . = . = . .. . +

=

.

= = . . . = . = .

b) =

=

=. + . = .

= .


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c) =

=

(

)

= + = , = = . = ./ = , = = . = ./

=

=

.

.

=

.

= = . . = . = = . = = . = + = . .. .

+.. . = .

=

=

.

. .

= . = = . . . = . = .


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Metal Semiconductor Contact:

The junction between a metal and semiconductor may behave like a diode or an

ohmic contact.

The diode made up of metal semiconductor contact is called Schottky diode. This

type of diodes are mainly used for fastest switching applications.

The behavior of metal semiconductor contacts depends on the work functions of

metal mand semiconductor s.

The work function is the difference between vacuum level energy and Fermi-level

energy. Vacuum level energy is the energy required to remove an electron at the

Fermi-level to the vacuum outside the metal.

According the relative work functions between metal and semiconductor, the contact

may be rectifying or ohmic and are as follows:

Type of contactRelative work functions

m> s m< s

Metal n-type semiconductor Rectifying OhmicMetal p-type semiconductor Ohmic Rectifying

Metal n-type semiconductor Schottky contact:

Consider the energy band diagram of metal n-type

semiconductor during isolated and equilibrium states:

Question:

Part B

Draw the energy band diagram

of metal n-type semiconductor

with m> sunderequilibrium and bias.


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= = . . =

.

= ( ) = . .. = . = = . .

.

.

= . = . = = . . = ./Depletion layer capacitance of metal semiconductor contact:

It is similar to that of p+-n type diode.

Question:

Part A

Derive the expression fordepletion layer capacitance of

Schottky diode.


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The main difference between Si p-n junction diode and Si Schottky diode is: Schottky

diode has high forward current as well as reverse saturation current. This is shown

below:

Comparison between Schottky diode and p-n junction diode:

Schottky diode p-n junction diode

High reverse saturation

current

Low reverse saturation

current

Lower forward voltage

drop

Higher forward voltage

drop

Used as low-voltage high-

current rectifier

Used as high-voltage

high/low current rectifier

There is no minority

carrier diffusion, hence nostorage capacitance.

Have storage capacitance

Reverse recovery time is

decided only by the

junction capacitance,hence suitable for high

speed switching

Reverse recovery time is

decided by junctioncapacitance as well as

storage capacitance,

hence not suitable forhigh speed applications.

Question:

Part ADraw the V-I

characteristics of Si p-n

junction and Si

Schotkky diode. (April

2014)

Question:

Part A

What are the advantages of

Schottky diode over abrupt p-njunction diode?


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Problem

Question:

Part B

A metal with work function of 4.3eV is deposited on n-type Si.

Determine the doping density required at 300K, so that there is no

space charge region at equilibrium. Electron affinity of Si is

4.15eV. (April 2014)Solution:

Given = . = .If there is no space charge region, V0= 0V.

We have

= ---(1) = ( ) = ( ) = + ( ) = = . . = .---(2) = For Si, Eg= 1.1eV, then

=

.

=

.

= (. + )Substitute this into (2), = .. + = . = .. = .Using (1) = = . . . = . Problem

A Schottky barrier diode is formed by depositing tungsten on n-type Si. Determine at 300K,if Nd= 10

15cm

-3, = .and electron affinity = 4.15eV

a) V0b) W0c) Solution:

a) = ( ) =?


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= = .---(1) =

=

= = = . . = . = .Substitute to (1)

= . + . = . = .. = . = = . .. = . = .b)

=

= . .. . = =

=

.

= . /


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Problem

A Schottky diode between tungsten and Si doped with 1015

As atoms /cm3has a junction

area of 10-3

cm2. R

*= 110 A/K

2cm

2, = .at 300V.

a) Determine the current through the diode at 300K for forward bias of 0.3V.

b) Consider a p

+

-n junction diode with equal doping n-side. What is the current at thesame forward bias? Take Dp= 12 cm

2/s and p= 1s.

c) What is the forward voltage required for the same forward current as that in part (a)?

Solution:

a) = =

. .

. .

=

.

=

.

b)

For p+-n junction diode = = = = .

= = .

= . = . . . . . = .

c). = . . .

.

=

.

= . . = . = . = . = . . = .(This shows that p+-n junction diode require more biasing voltage than Schottky diode to get

same current)


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Ohmic contact:

These are for making contact with semiconductor.

The energy band diagram of metal n-type

semiconductor ohmic contact under isolated, equilibrium, forward biased and reverse

biased conditions are shown below:

Under isolate condition, Fermi level of metal is greater than semiconductor, because

m< s

At equilibrium, EFsmoves closer to EFmand there is no barrier for the movement of

electron from metal to semiconductor or vice versa.

The electrons move from metal to semiconductor and accumulate near the interface.

There is no depletion layer near the junction. During +ve bias between metal and semiconductor, the potential and electric field

across the semiconductor bend upwards. Then the carriers from both directions cross

the junction.

During ve bias between metal and semiconductor, the potential and electric field

bend downwards and carriers flow in both direction.

Question:

Part A

What is ohmic contact?

Semiconductor Devices Module 2 Notes

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