farzana r zakieee 231: electronics i1 semiconductor diode instructor: farzana rahmat zaki senior...

Farzana R Zaki EEE 231: Electronics I 1

Semiconductor Diode

Instructor: Farzana Rahmat ZakiSenior Lecturer, EEEEastern University


Physical Operation of Diodes

Lecture – 1


Outline

• Basic Semiconductor Concepts

• Classification of Silicon

• Intrinsic Silicon

• Current flowing mechanism in Crystals: diffusion & drift


Basic Semiconductor Concepts

• p-n junction:

• Semiconductor diode is basically a p-n junction.

• p-n junction consists of p-type semiconductor material (such as Si) & n-type semiconductor material (such as Si).


Classification of Silicon

Silicon

Intrinsic Si(pure Si)

Extrinsic Si

p-type (B, Al, etc)

n-type (P, N, etc)


Intrinsic Silicon• Silicon has 4 valence

electrons.• Forms covalent bonds

with other atoms.• At low temperatures,

all electrons are in a covalent bond and thus no electrons are free to conduct electricity.


Thermal Ionization• At room temperature,

some electrons break free from their bonds (free electron)

• This process leaves a positively charged “hole” at the atom. – Note that the

overall charge of the crystal is still neutral


Recombination

– As the free electrons move randomly around the silicon, some will fill in some of the holes.

– This process is called recombination.

– Recombination is proportional to the number of free electrons and holes, which is, in turn, determined by the ionization rate.

– The ionization rate is a strong function of temperature.

– At thermal equilibrium, the recombination rate is equal to the ionization rate.


Equilibrium• At equilibrium, the ionization rate is equal to the recombination rate, and

there is a concentration of free electrons and an equal concentration of holes.• n = p = ni; where ni = intrinsic concentration

2 3

31

5

5.4 10 for silicon (material dependent parameter)

Bandgap energy 1.12 electron volts (eV) for silicon

Boltzmann's constant 8.62 10 eV K

At room temperature, this results in

GE kTi i

G

n p n n BT e

B

E

k

10 3

22 3

1.5 10 carriers/cm

The silicon crystal has 5 10 atoms/cm

This means that only one of every billion atoms is ionized!

Silicon is a semi-conductor (conductivity in between

conductors like metal and insu

lators like glass).


Summary of Intrinsic Silicon

-Intrinsic Si is a crystal with four valence electrons

– At room temperature, a number of these break free (ionize).

– Some of these recombine with the holes that are left behind

– The silicon reaches equilibrium when the recombination rate is the same as the ionization rate.


Current flowing mechanism in Crystals

• Two mechanisms for motion (current) in crystals– Diffusion

• Random motion of particles (electrons or holes) due to thermal excitation.

• Elements move from an area of higher concentration to an area of lower concentration.

– Drift• Movement of electrons and holes in response to an

electric field.


Diffusion mechanism• Associated with random motion due to thermal

agitation.• In a Si crystal with uniform concentrations of free

electrons and holes, this random motion does not result in a net flow of current.

• So, if by some mechanism, the concentration of free electrons is made higher in one part of piece of Si than in other, then electrons will diffuse from the high concentration region to the low concentration region. This process gives rise to a Diffusion current.


Diffusion mechanism-cont.

• Consider the bar of Si in figure in which the hole concentration profile has been created along the x-axis by some unspecified mechanism.

+ + + + + + + ++ + + + + + + ++ + + + + + + ++ + + + + + + +

+ + + + ++ + + +

x0

Slope = dP/dxH

ole

conce

ntra

tion, P


Diffusion mechanism-cont

• Hole concentration profile results a hole diffusion current in x direction, magnitude of current at any point is proportional to the slope of the curve at that point

JP = -q DP dP/dx ( hole diffusion)

Where JP = hole current density ( i.e. current per unit area of the plane perpendicular to x axis) in A/cm2

q = magnitude of electron charge = 1.6×10-19 C DP = Diffusion constant or diffusivity of holes


Diffusion mechanism-cont

For current diffusion

Jn = q Dn dn/dx (electron current diffusion)

Where Jn = electron current density ( i.e. current per unit area of the plane perpendicular to x axis) in A/cm2

q = magnitude of electron charge = 1.6×10-19 C Dn = Diffusion constant or diffusivity of electrons

For holes and electrons diffusing in intrinsic Si, typical values for diffusion constants

Dp = 12 cm2/s and Dn = 34 cm2/s


Drift• Assume that an electric field is applied across a piece

of silicon.• Free electrons and holes are accelerated by the

electric field and acquire a velocity component called drift velocity.

• If the field strength is denoted E (in V/cm), the positively charged holes will drift in the direction of E and acquire a velocity given by:

Where is the mobility of a hole.

drift p

p

v E


Mobility

• Mobility

2

2

Where is the mobility of a hole.

The same is true for electrons going the other way

Where is the mobility of an electron

typical values:

480 V/cm

1350 V/cm

Note that 2.5

drift p

p

drift n

n

p

n

n

v E

v E

p


Drift Current Density

• Drift Current Density

As the electrons and holes are going in opposite directions

p drift p

n drift n

drift p n

J qp E

J qn E

J q p n E


Summary of Current flowing mechanism

• Diffusion Current– Based on concentration differences, particles move

from higher to lower concentrations

• Drift Current– In the presence of an electric field, free electrons

move opposite the field, holes move in the direction of the field.


Question: Consider a Si crystal having a hole density P and a free electron density n. An electric field E is applied to the crystal. Find the expression of resistivity for that material.

Solution:

Electric field = E

Hole density = p

Electron density = n

Holes will drift in same direction as E and drift velocity for holes = μP E

Thus, for positive charge density ( qP C/cm3) moving in x direction with velocity (μP E), in 1s amount of charge flow across a plane A perpendicular to x-axis is IP = qPμPEA ( current flows for holes)

so, JP-drift = IP/A = qPμPE ---------- (1)


• For electrons, negative charge density ( -qn) moving in –x direction with drift velocity (-μnE), so In = qnμnEA

and Jn-drift = qnμnE --------------- (2)

So, total drift current density, Jdrift = Jp-drift + Jn-drift

= q(pμp + nμn )EUsing Ohm’s law, E = IR = I(ρl/A)

For unit length (l=1cm), E = I ρ/A = Jdrift ρ

so, ρ =E/Jdrift

ρ = 1 /q(pμp + nμn )

ρ = 1 /q(pμp + nμn )


Einstein Relationship

• Relation between carrier diffusivity and mobility is known as Einstein relationship. The relation is as follow—

T

DD Vp

p

n

n Where VT = thermal voltage = 25 mV at room temperature for intrinsic Si


Lecture 2

•Doping of Semiconductors: p-type & n-type•Ideal Diode & its characteristics•Practical Diode & its characteristics


Doping Semiconductors

• Doped Semiconductors– By adding an impurity, one kind of carrier

predominates– Doped silicon where the majority of charge

carriers are the negatively charged electrons is called n-type

– Doped silicon where the majority of charge carriers are the positively charged holes is called p-type


N-type• Doped with a pentavalent element such as phosphorus.– Five valence electrons

• four form covalent bonds• one becomes a free

electron– This is called a donor.

• No holes are created by this doping, thus there are free electrons without associated holes

• Electrons – majority charge carrier (concentration independent of T)

• Holes – minority charge carrier


N type (contd.)

• If ND = concentration of donor atoms• nno = concentration of free electrons in N-type Si in thermal

equilibrium

then, nn0 = ND

So, product of hole & electron concentration remains constant i.e,

nno × pn0 = ni2

Or, pn0 = ni

2 / ND

Here, ni is a function of temperature

pn0 is a function of temperature

nn0 is independent of temperature.


P-type• Doped with a trivalent element such as boron.– Three valence electrons

• three form covalent bonds• other bond has a hole.

– This is called an acceptor.• No electrons are freed by this

doping, thus there are holes without associated free electrons.

• Holes – majority charge carrier (concentration independent of T)

• Electrons – minority charge carrier


P type (contd.)

• If NA = concentration of acceptor atoms• Pp0 = concentration of holes in P-type Si in thermal

equilibrium

then ppo = NA

So, product of hole & electron concentration remains constant i.e,

pp0 × np0 = ni2

Or, np0 = ni

2 / NA

Here, ni is a function of temperature

np0 is a function of temperature

pp0 is independent of temperature.


Ideal Diode


Current-Voltage Characteristic of an Ideal diode

• Forward/Reverse Bias

OFF mode ON mode

i-v characteristic curve of an ideal diode


Ideal diode• Positive terminal of diode is called anode and negative

terminal is called cathode.• OFF mode: if a negative voltage is applied to diode, no

current flows and the diode behaves as Open Circuit. (v<0 and i=0). This mode is called Reverse-biased mode of operation and is said to be Cut Off.

Vanode<Vcathode => Reverse bias• ON mode: If a positive voltage is applied to diode, zero

voltage drop appears across the diode and the diode behaves as Short circuit. (i>0 and v=0). This mode is called Forward and a forward-conducting diode is said to be Turned on or simply on.

Vanode > Vcathode=> Forward bias


Problem 1

• What is the current through the diode and the voltage across the diode for the following two circuits?


Problem 2• What is the output voltage for the following

circuit? (a Rectifier)


Problem 3• For the following circuit, if is a sinusoid with 24-V peak amplitude,

find the fraction of each cycle during which the diode conducts. Find the peak value of the diode current and the maximum reverse-bias voltage that appears across the diode.

24sin 12

30 150, or 1/3 of a cycle

24 120.12A

100The maximum revers voltage is 24+12=36V

dI

Conduction angle:

Peak value of diode current:


Problem 4• Find I and V in the following circuits.


• Find the values of I and V in the following circuits.

5 / 2.5 2

0

I k mA

V V

0

5

I mA

V V

0

5

I mA

V V

2

0

I mA

V V


Practical Diode


Areas of Operation• There are 3 areas of

operation– The forward-bias region

• v > 0

– The reverse-bias region• v < 0

– The breakdown region• v < -Vzk


Areas Expanded: I-V characteristic curve for practical diode


Forward-Bias Region

• I is the forward-bias current• Occurs when v on the diode is positive.• the “cut-in” voltage is the voltage beneath which

the current is negligible small (generally around 0.5V)

• The current exponentially increases, and the voltage drop typically lies in a narrow range from .6V to .8V

/ 1Tv nVsi I e


Saturation Current

• The saturation current is directly proportional to the cross-sectional area of the diode.

• For “small-signal” diodes, the saturation current is on the order of 10e-15A.

• Strongly correlated to temperature– doubles for every 5˚C rise in temperature.

/

saturation curre

1

is nt, t scale currenthe or

Tv nVs

s

i I e

I


Thermal Voltage

/

thermal vol

1

is the and is given by

where

tage

Tv nVs

T

T

i I e

V

kTV

q

-23

-19

Boltzmann's constant = 1.38x10 joules/kelvin

Absolute temperature in kelvins = 273+

the magnitude of electronic charge = 1.60x10 coulomb

k

T C

q

at room temperature (20 C), the value of is 25.2mV

We generally use 25mVT

T

V

V


Fudge Factor

/ 1Tv nVsi I e

• n is a constant between 1 and 2 that represents variances in the material and physical structure of the diode.

• Diodes made using standard integrated circuit techniques exhibit an n close to 1.

• Diodes available as two-terminal devices generally exhibit an n closer to 2.

• We will use n=1 unless specified.


Voltage as a function of Current

• This exponential relationship for v holds over many decades of current (as many as seven decades).

/

/

1

for

ln

T

T

v nVs

s

v nVs

TS

i I e

i I

i I e

iv nV

I


Current effect on voltage

1

2

2 1

/1

/2

/2

1

2 22 1

1 1

ln 2.3 log

T

T

T

V nVs

V nVs

V V nV

T T

i I e

i I e

Ie

I

I IV V nV nV

I I


Problem 1

• Consider a silicon diode with n=1.5. Find the change in voltage if the current changes from 0.1mA to 10mA.

2 22 1

1 1

ln 2.3 log

102.3 1.5 log 172.5

1

T T

diff T

I IV V nV nV

I I

maV V mV

ma


Problem 2• A silicon junction diode with n=1 has

v=0.7V at i=1mA. Find the voltage drop at i=.1mA and i=10mA.

2 22 1

1 1

2

2

ln 2.3 log

.12.3 1 log .58 , .64

110

2.3 1 log .58 , .761

T T

diff T

diff T

I IV V nV nV

I I

maV V mV V V

mama

V V mV V Vma


Reverse-Bias Region

• In the reverse-bias region, the current is theoretically

• Real diodes often exhibit a much larger current due to leakages. However, the current is still quite small (nA range).

• There is also a slight increase with voltage for reverse-bias current.

si I


Breakdown Region

• When the voltage reaches a certain negative potential, the diode will begin conducting current. This “knee” is known as the breakdown voltage, Vzk.

• The Z stands for Zener and the K for knee.

• diodes that make use of the breakdown voltage and it’s near constant voltage/current relationship to be used in voltage regulation.


Thank You

farzana r zakieee 231: electronics i1 semiconductor diode instructor: farzana rahmat zaki senior...

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