ENE 311 Lecture 9. Junction Breakdown When a huge reverse voltage is applied to a p-n junction, the junction breaks down and conducts a very large current

ENE 311 Lecture 9

Junction Breakdown

• When a huge reverse voltage is applied to a- p n junction, the junction breaks down and c

onducts a very large current.

• Although, the breakdown process is not nat urally destructive, the maximum current mu

st be limited by an external circuit to avoid excessive junction heating.

• There are two mechanisms dealing with the breakdown: tunneling effect and avalanche

multiplication.

Tunneling Effect

• If a very high electric field is applied to a p-n junction in the reverse direction, a valence electron can make a transition from the valence band to the conduction band by penetrating through the energy bandgap called tunneling.

• The typical field for Si and GaAs is about 106 V/cm or higher.

Tunneling Effect

• To achieve such a high field, the doping con - - centration for both p and n regions must be

very high such as more than5 x 1017 cm-3.

• The breakdown voltage for Si and GaAs jun ctions about4Eg/e is the result of the tunneli

ng effect. With the breakdown voltage is mo re than6Eg/e , the breakdown mechanism is

the result of avalanche multiplication .• As the voltage is in between4Eg/e and6Eg/e

, the breakdown is due to a mix of both tunn eling effect and avalanche multiplication.

Avalanche Multiplication

• Let consider a p+-n one-sided abrupt junction with a doping concentration of ND 1017 cm-3 or less is under reverse bias.

• As an electron in the depletion region gains kinectic energy from a high electric field, the electron gains enough energy and acceleration in order to break the lattice bonds creating an electron-hole pair, when it collides with an atom.


• A new electron also rec eives a high kinetic ene

rgy from the electric fie ld to create another ele

- ctron hole pair. This co ntinue the process crea

- ting other electron hole pairs and is called avala nche multiplication.


• Assume that In0 is incident current to the depl etion region at x =0 .

• If the avalanche multiplication occurs, the ele ctron current In will increase with distance th rough the depletion region to reach a value o

f M n In0 at W , where Mn is the multiplication factor.

(1)0

( )nn

n

I WM

I


• The breakdown voltage VB for one-sided abrupt junctions can be found by

• The breakdown voltage for linearly graded junc

tions is expressed as

2

1

2 2c s c

B B

EW EV N

e

1/ 23 / 22 4 2

3 3c c s

B

EW EV

ea


T he critical field Ec can be calculated for Si and GaAs by using the plot in t he above figure.

Ex. Calculate the breakdown voltage for a Si o- - ne sided p+ n abrupt junction with ND =5 x

1016 cm-3

Ex. Calculate the breakdown voltage for a Si o- - ne sided p+ n abrupt junction with ND =5 x

1016 cm-3

Soln From the figure, at the given NB, Ec is about 5.7 x 105 V/cm.

21

214 5116

19

2

11.9 8.85 10 5.7 105 10

2 1.6 1021.4 V

s cB B

EV N

e



• -Assume the depletion layer reaches the n n+ interface prior to breakdown.

• By increase the reverse bias further, the de vice will break down.

• This is called the -punch through.


• The critical field Ec is the same as the previous ca se, but the breakdown voltage VB

-for this punch t hrough diode is

(4)

• - Punch through occurs when the doping concentr ation NB is considerably low as in a p+--n+ or p+-

-n+ diode, where -stands for a lightly doped p type and -stands for a lightly doped n type.

2B

B m m

V W W

V W W


Breakdown voltage for p+-π-n+ and p+- -v n+ junctions.

W is the thickness of the lightly doped region


• Ex. For a GaAs p+- - n one sided abrupt junctio n with ND =8 x 1014 cm-3 , calculate the deple

- tion width at breakdown. If the n type regio n of this structure is reduced to 20 μm, calc

ulate the breakdown voltage if r for GaAs is12.4.


Soln For the figure, we can find VB is about 500 V (VB >> Vbi)

14

19 14

3

2

2 8.85 10 12.4 500

1.6 10 8 10

2.93 10 29.3 m

s bi

B

V VW

eN


Soln - When the n type region is reduced to 20 μ - m, the punch through will occur first.

2

20 20500 2 449 V

29.3 29.3

B

B m m

B

V W W

V W W

V

Heterojunction

A heterojunction is defi ned as a junction forme

d by two semiconducto rs with different energy

bandgaps Eg , different dielectric permittivities

s , different work functi on es , and different ele

ctron affinities eχ.

Heterojunction

• The difference energy betw een two conduction band ed ges and between two valenc

e band edges are represent ed by EC and EV , respectiv

ely, as

where Eg is the difference e nergy bandgap of two semic

onductors.

2 1CE e

1 1 2 2V g g g CE E e E e E E

Heterojunction

• Generally, heterojunction has to be formed between semiconductors with closely matched lattice constants.

• For example, the AlxGa1-xAs material is the most important material for heterojunction.

• When x = 0, the bandgap of GaAs is 1.42 eV with a lattice constant of 5.6533 Å at 300 K.

• When x = 1, the bandgap of AlAs is 2.17 eV with a lattice constant of 5.6605 Å .

Heterojunction

• We clearly see that the lattice constant is almost constant as x increased. The total built-in potential Vbi can be expressed by

(7)

where N1 and N

2 are the doping concentrations i

n semiconductor1 and2 , respectively.

1 2bi b bV V V

2 21

1 1 2 2

1 12

1 1 2 2

bib

bib

N V VV

N N

N V VV

N N

Heterojunction

• The depletion widths x1 and x

2 can be found

by

(8)

1 2 21

1 1 1 2 2

1 2 12

2 1 1 2 2

2

2

bi

bi

N V Vx

eN N N

N V Vx

eN N N

Heterojunction

Ex. Consider an ideal abrupt heterojunction wi - th a built in potential of 1.6 V. The impurity

concentrations in semiconductor1 and2 ar e1 x 1016 donors/cm3 and3 x 1019 acceptors

/cm3 , and the dielectric constants are 12 an d 13 , respectively. Find the electrostatic pot

ential and depletion width in each material at thermal equilibrium.

Heterojunction

Soln

19

1 16 19

16

42 16 19

14 19

51 19 16 16 19

14 16

2

13 3 10 1.61.6 V

12 1 10 13 3 10

12 1 10 1.64.9 10 V

12 1 10 13 3 10

2 12 13 8.85 10 3 10 1.64.608 10 cm

1.6 10 1 10 12 10 13 3 10

2 12 13 8.85 10 1 10 1.

b

b

V

V

x

x

8

19 19 16 19

61.536 10 cm

1.6 10 3 10 12 10 13 3 10

Note: - Most of the built in potential is in the semic onductor with a lower doping concentration and

also its depletion width is much wider.

Metal-Semiconductor Junctions

• The MS junction is more likely known as the Sch- ottky barrier diode.

• Let’s consider metal ba nd and semiconductor b

and diagram before thecontact.


• When the metal and se miconductor are joined,

electrons from the semi conductor cross over to

the metal until the Fer mi level is aligned (Ther

mal equilibrium conditi on).

• This leaves ionized don ors as fixed positive ch

arges that produce an i nternal electric field as

- -the case of one sided p n junction.


• At equilibrium, equal number of electrons across the interface in opposite directions.

• Hence, no net transport of charge, electron current I - e equals to zero. The built in voltage Vbi = m - s .

• The barrier for electrons to flow from the metal to se miconductor is given by eb = e(m - χs) or it is called t

he barrier height of MS contact.


• When a voltage is applied, the barrier heigh - t remains fixed but the built in voltage chan

ges as increasing when reverse biased and decreasing when forward biased.

Metal-Semiconductor Junctions• Reverse bias

• Few electrons move across the interface from metal t o semiconductor due to a barrier, but it is harder for

electrons in the semiconductor to move to the metal.

• Hence, net electron transport is caused by electrons moving from metal to semiconductor. Electron curre

nt flows from right to left which is a small value.

Metal-Semiconductor Junctions• Forward bias

• Few electrons move across the interface from metal to semiconductor, but many electrons move across t he interface from semiconductor to metal due to the

reduced barrier.

• Therefore, net transport of charge flows from semic onductor to metal and electron current flows from le

ft to right.


• Under forward bias, the electrons emitted to the metal have greater energy than that of the metal electrons by about e(m - χs).

• These electrons are called hot-carrier since their equivalent temperature is higher than that of electrons in the metal.

• Therefore, sometimes, Schottky-diode is called “hot-carrier diode”.


• This leads to the thermionic emission with t hermionic current density under forward bia

s as

/ /** 2 m s F Fe V kT eV kT

F sJ A T e J e

/** 2

*** *

0

where saturation current

. effective Richardson's constant

m se kT

sJ A T e

mA A

m


• This behavior is referred to a rectification an d can be described by an ideal diode equati

on of

(10)

where V positive for forward bias and negative for reverse bias.

/ 1eV kTsJ J e

Metal-Semiconductor Junctions• - The space charge region width of Schottky diode is i

- - dentical to that of a one sided p n junction.

• Therefore, under reverse bias, they can contain the charges in their depletion region and this is called S chottky diode capacitance.


Ex. A Schottky junction is formed between Au - and n type semiconductor of ND = 1016 cm-3 . Area of junction = 10-3 cm2 and me

* = 0.92 m0

. Work function of gold is 4.77 eV and eχs = 4.05 eV. Find current at VF = 0.3 volts.


Soln

*** * 2

0

/** 2

/

4.77 4.05 / 0.02592 0.3 / 0.0259

2

3

120 0.92 110 A/ cm .K

1

110 300 e . 1

0.897 A/(cm )

10 0.897 0.897 mA

m s

e

e kTs

eV kTs

mA A

m

J A T e

J J e

e

I A J


Ex. - Si Schottky diode of 100 μm diameter has(1/C2 ) v.s. VR slope of3 x 1019 F-2V-1 . Given r

= 11.9 for Si. Find NB for this semiconductor.


Soln 2

2

4 -2 -1

2

22419 19 12

19 -3

21; [F/cm ]

2slope [cm F V ]

2

2

100 103 10 1.6 10 8.85 10 11.9

2

6.414 10 cm

bi Rj

j s B

s B

Bs

B

V V CC

C e N Area

e N

Nslope Area e

N

Ohmic contact

• This contact is defined as a junction that will not add a significant parasitic impedance to the structure on which it is used and will not sufficiently change the equilibrium-carrier densities within the semiconductor to affect the device characteristics.

• The I-V characteristic of ohmic contact is linear for an ideal case.

Ohmic contact

• A specific contact resistance RC is given by

(11)

• A good ohmic contact should have a small s pecific con tact resistance about 10-6 Ω.cm2 .

12

0

1.cm

VC

J

R V

Ohmic contact

• When the semiconduct or is heavily doped wit

h an impurity density o f 1019 cm-3 or higher, th e depletion layer of the

junction becomes very thin so that carriers ca

n tunnel instead of goi ng over the potential b

arrier.

Documents

ENE 311 Lecture 9. Junction Breakdown When a huge reverse voltage is applied to a p-n junction, the junction breaks down and conducts a very large current