ELEKTRONIKOS PAGRINDAI 2008 - VGTU junctions.pdf · Breakdown in a pn junction ELEKTRONIKOS...

GRADED SEMICONDUCTORS, pn JUNCTIONS

Objectives:

Analysis of phenomena in graded semiconductors and pn junctions and their

properties.

Methods of calculation of pn junction parameters.

Content

1. Graded semiconductor

2. A pn junction in equilibrium

3. The volt-ampere characteristic

4. The depletion layer and the depletion layer capacitance

• Thickness of the depletion layer

• Barrier capacitance

5. The junction under forward bias

• The continuity equation

• Distribution of excess carriers

• The density of the excess carriers

• Diffusion capacitance

6. Breakdown in a pn junction

ELEKTRONIKOS PAGRINDAI 2008

VGTU EF ESK [email protected]

1

GRADED SEMICONDUCTORS



2

Graded semiconductor

The doping of a graded (or inhomogeneous)

semiconductor is non-uniform.

At room temperature impurity atoms in the

semiconductor are ionised. Thus, the electron and

hole densities vary with x. Then electrons diffuse in

the x direction.

Moving from their atoms electrons leave positive

donor ions.

As a result an electric field between positive and

negative charges appears.

... The total electron current and hole current must be zero in equilibrium..

0d

dqq nnn =+=

x

nDEnj µ

0d

dqq ppp =−=

x

pDEpj µ

x

xn

xn

T

x

xn

xn

DxE

d

)(d

)(

1

q

k

d

)(d

)(

1)(

n

n −=−=µ

x

xp

xp

T

x

xp

xp

DxE

d

)(d

)(

1

q

k

d

)(d

)(

1)(

p

p ==µ



3

)exp()( 0 bxnxn −=

bT

xEq

k)( =

... If the density of the majority carriers varies in an

exponential way, the electric field is uniform in the

semiconductor.

xxE d)(d −=ϕp

pT

n

nT d

q

kd

q

kd −==ϕ

∫∫∫ −==2

1

2

1

2

1

d

q

kd

q

kd

p

p

n

np

pT

n

nTϕ

ϕ

ϕ

1

2

1

21221 ln

q

kln

q

k

p

pT

n

nTU −==−= ϕϕ

... The potential difference between

two points depends only on the

carrier densities at these two points

and is independent on their

separation.

The concentration gradient causes diffusion of the carriers from the region of high

density to the region of low density (in the direction of the negative gradient).

As a result diffusion current occurs.

Diffusing carriers leave charged impurity ions and electric field appears within a

graded semiconductor.

When both a potential gradient and a concentration gradient exist simultaneously

within a semiconductor, the total current is the sum of the drift current and the

diffusion current. In equilibrium the total current is zero.




4

211221 qUWWW −=−= q/211221 WU −=−= ϕϕ

Because in the graded n-type semiconductor the

height of the potential barrier does not exceed ∆W,

the built-in potential in a graded specimen satisfies

the condition

q2k

WU

∆≤

1. In equilibrium the Fermi level is constant.

2. The Fermi level in an intrinsic semiconductor is

in the mid-gap position.

3. The Fermi level in the n-type semiconductor is

over the middle of the forbidden band.

4. The Fermi level in the p-type semiconductor is

below the middle of the forbidden band.

... The potential barriers appear in graded

semiconductors.




5

Electrons and holes diffusing in a graded

semiconductor meet the potential barrier that

appears due to the action of the internal electric

field. The carriers that have small energy are

reflected by the barrier.

Analyzing the movement of holes we must

remember once more that a hole is like a gas

bubble in liquid. Its energy is higher, if its

position in the valence band is lower. The

distance between the energy level of a hole and

the top of the valence band corresponds to the

kinetic energy of the hole.

... In equilibrium drift and diffusion currents

compensate each other.




6

Estimate the strength of the internal electric field in the graded p-type base of

the npn silicon transistor assuming that the base width is about 1 µm.




7

A pn junction in equilibrium



8


The region at which p-type and n-type semiconductors meet is called a pn

junction.

Let us consider a step (abrupt) silicon pn junction. Let us assume that the

junction is symmetrical. Impurity densities are 1016/cm3.

1. Carrier concentration gradients across a

pn junction cause holes and electrons to

diffuse. As a result of carrier diffusion,

carrier densities at the junction do not

change abruptly and are sufficiently less

than the majority carrier densities in the p

and n regions.

2. ... A transition or depletion layer with

high resistance exists between p and

n parts of the semiconductor. This layer

forms a p-to-n interface.



9

3. The majority holes diffusing out the p-

region leave behind negatively charged

acceptor atoms bound to the lattice.

Similarly electrons diffusing from the n-

region expose positively ionized donor

atoms. As a result a double space

charge layer builds up at the junction.

4. The double-space-charge layer

causes an electric field across the

junction. The field is directed from the

n-region to the p-region.




10

5. Because of the concentration gradient

and the electric field the drift and the

diffusion currents flow across the

junction.

6. Majority carriers tend to diffuse across the

junction. The internal electric field impedes

their diffusion. If the energy of the carrier is

not enough, then it is sent back. So

diffusion currents flow due to majority

carriers.

7. If a minority carrier approaches the pn

junction, the internal electric field

accelerates it and the carrier is moved to the

other side of the junction. So the drift

currents flow due to minority carriers.




11

8. ... Four components of the total current flow

across the junction:

• diffusion current due to holes,

• diffusion current due to electrons,

• drift current due to holes,

• drift current due to electrons,

0nnDn =+= Ejjj 0ppDp =+= Ejjj

0D =+= Ejjj

9. Holes diffusing from the p-type region

leave it negatively charged raising all

energy levels. Similarly electrons migrating

in the opposite direction cause all levels in

the n-type material to be lowered. In

thermal equilibrium the Fermi level is

continuous across the junction. As a

result a potential barrier arises in the

depletion layer. It is caused by the electric

field and diffusion potential.




12

10. ... The height of the potential barrier depends

on the contact, built-in, or diffusion potential.

kb qUW −=

∫∫−

∞

∞−

−≅−=−=n

p

d)(d)(pnb

d

d

xxExxEU ϕϕ

2

i

pn

n

p

p

nb

lnq

k...ln

q

k

...lnq

kd

q

k n

p

n

pnT

p

pT

n

nT

n

nTU

n

n

===

==== ∫

2

i

adb ln

q

k

n

NNTU =

npb WWWW ∆∆∆ −−=

If the impurity densities are higher, the distances

∆Wn and ∆Wp are less. Thus, the height of the

barrier may approach ∆W.

q/maxb WU ∆≅




13

Electrons moving from n to p side of the junction

meet a potential barrier. If the energy of an electron

is not sufficiently high, the barrier is high for such an

electron and the electron is reflected by the barrier.

Electrons with high energy can get over the barrier.

… Not all electrons with high energy reach other

side of the junction. If an electron collides with the

lattice and loses energy, it is directed by the internal

electric field back to the n side. We have a similar

situation considering holes moving from the p to n

side of the junction. Thus, the potential barrier

prevents the diffusion of the majority carriers

across the junction.

Any minority carrier in the vicinity of the junction

does not meet a barrier. Thus, minority carriers can

travel freely through the junction. If a carrier collides

with the lattice, it loses energy. Thus, we can

consider that minority carriers slide down

potential hills.




14

An abrupt pn junction is made in silicon. Impurity densities in the n and p

regions are: 1016 and 4·1018 cm-3. Find the built-in potential and the height of the

barrier in the junction region at T = 300 K.

A pn junction in equilibrium. Problem



15

I-U characteristics of pn junctions



16

Mathematical model of an ideal pn junction.

I-U characteristic of the ideal junction.

On what, how and why I-U characteristcs depend.

I-U characteristics of real pn junctions.

The volt-ampere characteristic of a pn junction

When external voltage is connected across a pn junction, the thermal equilibrium is

disturbed and a current flows across the junction.

The resistance of the space-charge region is much

higher than that of the neutral regions. So, the

external voltage drops across the space-charge

region. The magnitude of the conduction current

depends strongly on the polarity of the applied

voltage.... A negative voltage applied to the p side with respect to the

n side creates the field that has the same direction as the

built-in electric field. Then the applied (reverse) voltage

increases the height of the potential barrier.

The increased potential barrier may inhibit almost completely

the diffusion of the majority carriers across the junction. The

minority carriers can travel freely down potential hills. But

their densities are small. That is why junction current is small

and almost independent on the voltage. This current is called

reverse saturation current.

The relationship between the current and the applied

voltage is I-U, current-voltage, or the volt-ampere

characteristic of the junction.



17

If positive voltage is applied to the p side with

respect to the n side, the applied electric field

opposes the internal built-in field and reduces the

height of the potential barrier.

Under the forward bias conditions again the drift

currents are practically unaltered since minority

carriers that approach the junction are swept across

it. But the big change comes with the diffusion

currents. If the forward bias increases, the height of

the potential barrier decreases and the number of

the majority carriers that can get over the barrier

increases rapidly (…in an exponential way).

)k/qexp(sD TUjj =

The current across the junction consists of two (diffusion and drift) components.

.)k/qexp(sD EE jTUjjjj +=+= sEE jjjjjU −==+== ,0:0 s

[ ]1)k/qexp()k/qexp( sss −=−= TUjjTUjj [ ]1)k/qexp(s −== TUIjSI




18

[ ]1)k/qexp(s −== TUIjSI

At , the current across a reverse biased junction is small and

equal to the saturation current.

At U > 0,1 V: )k/qexp(s TUII ≅

At U = 0 V, the current I = 0.

V1.0and0 >< UU

The volt-ampere characteristic of the ideal pn junction and its initial part


… The current across a forward biased junction varies exponentially with

the forward voltage.



19

[ ]1)k/qexp(s −== TUIjSIN

TWSI

)k/exp(~s

∆−

… The reverse saturation current and the I-U characteristic are strongly

dependent on the forbidden gap energy, doping levels, junction

temperature, and, of course, junction area.

The saturation current approximately

doubles for each 10 0C increase in

temperature.

The forward voltage drop required to push a

given current through the diode decreases

as the temperature is increased.


SI ~



20

If the gap energy is greater, the

density of the minority carriers is less.

As a consequence the saturation

current also is less.

… The current through a silicon diode

is less than through the similar

germanium diode.

The current through the diode is also less, if the impurity densities are higher.


The theoretical pn junction equation provides only an approximate

introduction to the characteristic curves of the junction diodes.



21

It is necessary to consider the equivalent circuit of the junction (for DC) in order

to reveal reasons why the real characteristic differs from the ideal.

1FFFD RIUU +=

... The n-type and p-type regions of a junction diode have the bulk resistance. At

a given current the forward voltage of a diode consists of two terms. The voltage

drop across the bulk regions reduces the current through the junction.

… The cumulative effect is to shift the characteristic curve in the forward bias

region to higher voltage levels and to decrease the slope of the volt-ampere

characteristic at higher current levels (the characteristic can be approximated

by a straight line).

… No significant current flows through the diode until the forward bias is large

enough to overcome the barrier potential. This would be about 0.3 V for a

germanium diode and 0.7 V for a silicon diode.

U > 0:




22

U < 0:

The leakage current flows due to the leakage resistance. The reverse current of a

diode is given by

The generation current is due to the generation of intrinsic carriers in the depletion

region of the junction. If the reverse bias increases, the width of the depletion layer

also increases. This is accompanied by the growth of the generation current and

the total reverse current.

The reverse current of a diode exceeds the saturation current. It consists of the

drift current, generation current, leakage current and breakdown current.


2RRRD / RUII +=



23

1. How and how many times do the saturation currents of germanium and

silicon diodes change with temperature increasing from 20 to 100 oC?

2. There are germanium and silicon diodes having the same cross-section

area. How many times the saturation current of the germanium diode

exceeds the saturation current of the silicon diode at T = 300 K?

3. At the forward current of 5 mA the voltage drop on a pn junction is 0.7 V.

The resistance of the p and n regions is 20 Ω. Find the forward voltage of

the diode.




24

pn junction at reverse voltage



25

Processes at the junction.

Solution of Poison’s equation.

Thickness of the junction.

On what, how and why the thickness depends.

Charges in the junction area.

Barier capacitance.

On what, how and why the capacitance depends.


When the reverse-bias is applied, the n-type side is made positive and the p-

side is made negative. The applied voltage drops on the depletion layer.

The external voltage attracts majority carriers

away from the pn junction. Because the majority

carriers are moved away from the junction, the

thickness of the depletion region is increased.

It is possible to find the thickness of the junction solving Poisson’s equation.

ερϕϕϕ ),,(

2

2

2

2

2

2 zyx

zyx−=

∂

∂+

∂

∂+

∂

∂

ερϕ )(

d

)(d2

2 x

x

x−=

In the p region (at ):

ap qN−≅ρ

p0 dx >>

εϕ a

2

2 q

d

)(d N

x

x=



26

In the p region (at ):

.q ap N−≅ρ

p0 dx <<

εϕ a

2

2 q

d

)(d N

x

x=

1aq

d

)(dCx

N

x

x+=

εϕ

1aq

d

)(d)( Cx

N

x

xxE −−=−=

εϕ

pdx = pa

1

qd

NC

ε= )(

q

d

)(d)( p

a dxN

x

xxE −−=−=

εϕ

)(q

d

)(dp

a dxN

x

x−=

εϕ

2

2pa

2

)(q)( C

dxNx +

−=

εϕ

pdx =2

)(q)(

2pa

p

dxNx

−+=

εϕϕ

p2 ϕ=C




27

)(q

d

)(d)( p

a dxN

x

xxE −−=−=

εϕ

0n <<− xd )(q

d

)(d)( n

d dxN

x

xxE +=−=

εϕ

:0=x

nd

pa qq

dN

dN

εε= ndpa dNdN =

... According to this equation the ratio of

the thicknesses of the depletion layers

is in inverse ratio to the ratio of doping

densities.

... If impurity density is less, the thickness

of the depletion layer is greater.


p0 dx <<



28

p0 dx >>2

)(q)(

2pa

p

dxNx

−+=

εϕϕ 0n <<− xd

2

)(q)(

2nd

n

dxNx

+−=

εϕϕ

:0=x2

q

2

q 2nd

n

2pa

pdNdN

εϕ

εϕ −=+ Rkpn UU +=−ϕϕ

( ) .11

2

q

2

q

da

2n

2d

2nd

2paRkpn

+=+=+=−

NNdNdNdNUU

εεϕϕ

( )( )da

Rk

dn

/1/1q

21

NN

UU

Nd

++

=ε ( )

( )da

Rk

ap

/1/1q

21

NN

UU

Nd

++

=ε

( )

+

+=+=

da

Rkpn

11

q

2

NN

UUddd

ε

ndpa dNdN =

... The thickness of the depletion layer is dependent on the built-in

potential, reverse bias and doping densities.

If the reverse bias increases, the thickness of the depletion region also increases.

The thickness of the depletion layer is less, if the doping levels are higher.




29

pn junction at reverse voltage. Barrier capacitance

If we know dn and dp, we can determine the charges of impurity ions at sides of

the junction.

ndnnd q SdNVQ == ρ pappa q SdNVQ −== ρ

... Absolute values of the charges are equal: QQQ == ad

... Because the thickness dn and dp depend upon the reverse voltage, the charges

Q vary with the voltage. The variation of the charge with the voltage means that

the capacitance exists.

Rb dd UQC =

This capacitance is associated with the height of the potential layer between p

and n regions and is called the transition, depletion, barrier, or space-charge

capacitance.



30

... the junction can be modelled as a parallel-plate capacitor filled with dielectric

of permittivity ε and of thickness d.

( )( )adRb

b/1/12

q

NNUUS

d

SC

++==

εε

NNN == ad

( )Rb

b4

q

UU

NSC

+=

ε

The barrier capacitance is proportional to

the cross-area of the junction.

If the reverse bias increases, the thickness

of the depletion layer increases and the

depletion capacitance of the step junction

decreases.

The capacitance depends also on the

doping densities. If these densities are

higher, the depletion layer is thinner and

the capacitance is greater.




31

The characteristic Cb(UR) is called the voltage-capacitance characteristic.

( )N

UUA

C

Rb

2

b

1 +=

( )Rb

b4

q

UU

NSC

+=

ε

If the capacitance of the step junction is measured as a function of the reverse

bias voltage, the information can yield experimental values for the doping levels

(from the slope of the line) and also the built-in potential (from the intercept of the

line with the axis).

...In the case of the graded pn junction, linearly depends on UR.3

b−C

The depletion-layer capacitance is one of the factors that limit the high-frequency

operation of junction devices.

On the other hand, this voltage-dependent capacitance is deliberately exploited in

varactor diodes or varicaps.

... In the case of a step

junction the graph of Cb-2

versus UR is of the form of

straight line.




32

1. Find the thickness of a step germanium pn junction at

2. Impurity densities of a step pn junction are 1016 and 4·1018 cm-3. The cross-

section area of the junction is 0,1 mm2. How and how many times will the

barrier capacitance change with reverse voltage increasing from 1 to 5 V?

. TUNN K300 ,0 ,cm 104 ,cm 10 -318-316

ad ==⋅==

pn junction at reverse voltage. Barrier capacitance. Problems



33

• Physical processes at the junction.

• The continuity equation.

• Distribution of excess carriers.

• Density of the excess carriers (versus x and U).

I-U characteristic of the junction.

• Charges of the injected carriers.

• Diffusion capacitance.

• Influence of capacitances onto properties of devices.

pn junction under forward bias



34

pn junction under forward bias

The forward bias lowers the height of the

barrier at the junction. Then strong

diffusion currents can flow across the

junction.

By forward bias holes are injected from the

p-side across the depletion region to the n-

side of the junction. Similarly electrons are

injected through the junction and they find

themselves on the p-side of the junction.

Injected carriers become minority carriers.

The charge formed by the injected

carriers, attracts majority carriers. At the

same time the injected carriers diffuse

in the direction of the diminishing

density and recombine with majority

carriers.

Recombined carriers need to be replaced.

This causes the flow of electrons and

holes coming in from the ohmic contacts.



35

The continuity equation applied to semiconductors describes how the carrier

density in a given elemental volume of a crystal varies with time.

),( txn )d,( ttxn +

[ ] xtt

nxtxnttxn ddd),()d,(

∂∂

=−+

The number of electrons can change as a result of their

generation, recombination, diffusion and drift.

( ) txRG dd−txG dd txR dd− [ ] txx

JtxxJxJ ddd)d()(

∂∂

−=+−

txx

JRGtx

t

ndddd

∂∂

−−=∂∂

x

JRG

t

n

∂∂

−−=∂∂

q/njJ −=x

nDEnj

d

dqq nnn += µ

2

2

nnnd

d

x

nD

x

nE

x

EnRG

t

n

∂

∂+

∂∂

++−=∂∂

µµ

2

2

pppd

d

x

pD

x

pE

x

EpRG

t

p

∂

∂+

∂∂

−−−=∂∂

µµContinuity equations represent

mathematically the charge

conservation law..

pn junction under forward bias. The continuity equation



36

In the case when electrons, for example, are injected across a pn junction into the

p-region, the continuity equation can be simplified. Considering the static (time-

independent) field-free situation and neglecting generation … we obtain:

2

2

nnnd

d

x

nD

x

nE

x

EnRG

t

n

∂

∂+

∂∂

++−=∂∂

µµ

0d

d2

2

n =− Rx

nD

τ∆∆ /0 e

tnn −=τ∆

τ∆∆ τ nn

t

nR t −=−== − /0 e

d

)d(

.0d

)(d2

2

n =+τ∆∆ n

x

nD nn //

ee)(LxLx

BAxn += −∆ τnn DL =

Assuming that the density of the excess carriers at

large x must be zero due to recombination, we can get

that B = 0.

Assuming that the depletion layer is thin at forward

bias, we can simplify the model of the junction. Then

An =)0(∆ n/pp e)0()(

Lxnxn

−∆=∆

pn junction under forward bias. Distribution of injected electrons



37

n/pp e)0()(

Lxnxn

−∆=∆ ... The density of the excess carriers in the p region

decays exponentially.

?)0(p =∆nTW

nTWWWNnk/

nFnbcncpbe]k/)(exp[

−=−+−=

)(q kb UUW −=TUU

nnk/)(q

npke−−=

)k/qexp( kn0p TUnn −= TUnn k/q0pp e=

)1(e)0( k/q0p0ppp −=−= TUnnnn∆

n/k/q0pnpp e)1(e)/exp()0()(

LxTUnLxnxn−−=−= ∆∆

p/k/q0npnn e)1(e)/exp()0()(

LxTUpLxpxp −== ∆∆ τpp DL =

τnn DL =

Because of the concentration gradients, electron and hole diffusion currents

exist.

pn junction under forward bias. Distribution of injected electrons



38

p/k/q

p

0npppD e)1(e

q

d

dq)(

LxTU

L

pD

x

pDxj −−=−=

n/k/q

n

0pnnnD e)1(e

q

d

dq)(

LxTU

L

nD

x

nDxj

−−−==

)1(eq

)0( k/q

n

0pnnD −−= TU

L

nDj )1(e

q)0( k/q

p

0np

pD −−= TU

L

pDj

[ ]1)k/qexp(s −= TUII

+=

p

0np

n

0pns q

L

pD

L

nDSI

... We derived once more the current-voltage characteristic of the junction.

… Now we obtained the expression for the saturation current.

pn junction under forward bias. I-U characteristic of the junction

Multiplying the sum of the current densities by the junction area, we obtain the

current across the junction:



39

Space charges appear due to injected excess charge carriers.

∫∞

−=

0

pnp d)(q xxnSQ ∆

( )∫∞

−−−=0

/k/q0pnp de1q n xeSnQ

LxTU

( )1q k/qp0npn −= TUeLSpQ

nppp QQ −=

pnnn QQ −=

[ ]1)k/qexp()(q n0pp0npnpp −+=+= TULnLpSQQQ

... The charge is dependent on the forward bias applied to the junction. Again, if

the charge depends on voltage, the capacitance exists.

This capacitance is called the storage, or diffusion capacitance.

( )1q k/qn0pnp −−= TUeLSnQ

UQC ddd =

pn junction under forward bias. Diffusion capacitance



40

[ ]1)k/qexp()(q n0pp0npnpp −+=+= TULnLpSQQQ UQC ddd =

( ) )k/qexp(k

qn0pp0n

2

d TULnLpT

SC +=

s

s)k/qexp(I

IITU

+=

τ)(k

qsd II

TC +=

... The diffusion capacitance of a pn junction is directly proportional to the

forward current across the junction and carrier life-time.

The operation speed of semiconductor devices depends on the diffusion

capacitance. In order to increase the operation speed we must reduce the

diffusion capacitance.

The total capacitance of the junction consists of barrier and diffusion

capacitances.

At reverse voltage that is higher than 0.1 V, the diffusion current does not flow,

and the total capacitance is determined by the barrier capacitance.

If the forward current flows, the diffusion capacitance is sufficiently greater than

the barrier capacitance, and the total capacitance is determined the diffusion

capacitance.

pn junction under forward bias. Diffusion capacitance



41

1. A pn junction is forward biased. At the forward voltage of 0.8 V, the forward

current is 10 mA. Carrier life-time is 0.1 µs. T = 300 K. Find the diffusion

capacitance of the junction.

2. The saturation current of a silicon pn junction is 1 mA. Carrier lifetime is

1 µs. Find the diffusion capacitance of the junction at forward bias UF = 0.1

and 0.3 V.

pn junction under forward bias. Problems



42

Breakdowns in pn junctions



43

The reverse branch of I-U characteristic.

Electrical breakdown.

Avalanche breakdown.

Zener breakdown.

Application in semiconductor devices.

Breakdown voltage versus temperature.

Thermal breakdown.

Breakdowns in pn junctions

At some particular reverse bias voltage a sudden

increase in reverse current is observed. It is due to

some sort of breakdown. When breakdown occurs, the

diode current is limited mostly by the resistance of the

external circuit.

Some mechanisms of pn junction breakdowns (Zener,

avalanche, thermal, surface breakdowns) are possible.

The Zener breakdown, requires relatively heavily doped pn junctions. Because of

the high doping levels, the depletion layer is very thin and, as a consequence, a

strong field (106 V/cm, or greater) exists across it. Then electron-hole pairs are

created as the force due to the high electric field causes ionization of

semiconductor atoms.

Using the crystal lattice model we can suppose that the strong electric field pulls

out electrons from the bonds. As a result, the density of free electrons and holes

in the depletion layer and the current across the junction increase.



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Zener breakdown

According to the energy diagram, the Zener

breakdown is due to the tunnel current. Under the

reverse bias, electrons tunnel directly from the

valence band to the conduction band through the

potential barrier that has width a and height ∆W .

Because the tunnel current can flow only when the

depletion layer is very thin, the Zener breakdown

voltage is small (of up to about 5-7 V).

According to its mechanism, the Zener breakdown

sometimes is called the tunnel breakdown.

The Zener effect was discovered by the American physicist Clarence Melvin Zener.

Clarence Melvin Zener (December 1, 1905 - July 15, 1993) was the American

physicist who first described the electrical property exploited by the Zener diode,

which Bell Labs then named after him.



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The avalanche breakdown is caused by cumulative

multiplication of free charge carriers under the action

of an applied electric field.

… Minority carriers are accelerated in the depletion

layer by the electric field. Then some carriers can

gain enough energy to liberate new electron-hole

pairs by impact ionisation.

… Suppose that an electron moving in the p-region

approaches the depletion region. Then it is

accelerated by the electric field. Its kinetic energy

increases. If the electron having kinetic energy

higher than the gap energy comes into collision with

the lattice, it can cause ionization of a semiconductor

atom. Then additional electron-hole pair appears.

In turn the liberated carriers are accelerated by the

electric field and can generate further pairs.

Avalanche breakdown

As a consequence, the current across the junction increases.

The avalanche breakdown tends to dominate when doping densities at one or

both sides are only moderate and the depletion layer is thick. The avalanche

breakdown voltage is higher than 5-7 V.



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Thermal breakdown

These breakdowns can turn into thermal breakdown, if

the current increases. The dashed part of the current-

voltage characteristic corresponds to the thermal

breakdown.

The processes that take place during the Zener

breakdown and the avalanche breakdown are

reversible processes.

The thermal breakdown (or thermal runaway) is caused by thermal generation of

excess charge carriers due to the cumulative interaction between increasing

junction temperature and increasing power dissipation.

As the reverse current increases, the power dissipation in the junction region

increases. An increase of the power dissipation produces increase of the junction

temperature. An increase in the junction temperature causes the reverse current to

increase and the cycle repeats. All semiconductor devices have a maximum

operating junction temperature ranging typically from 125 to 200oC for silicon. Above

this temperature catastrophic irreversible failure occurs.



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… If the junction temperature is increased, the lattice vibrations are more intense

and valence electrons have additional energy. Consequently, increasing

temperature reduces the Zener breakdown voltage.

… At low impurity density, the electron-phonon interaction predominates. Then, if

temperature increases, the mean free path of carriers decreases. A carrier can

gain sufficient energy between two collisions, if it is accelerated by a more intense

field. Therefore the avalanche breakdown voltage increases with temperature.

The voltage in the breakdown region remains almost constant even if the reverse

current changes sufficiently. This effect is used in Zener diodes. Diodes that have

adequate power-dissipation capabilities to operate in the breakdown region are

commonly called Zener diodes independently of the breakdown mechanism.

These devices are employed as voltage regulators and in other applications in

which a constant voltage is required.

The reference voltage must be stable. Therefore it is important to know the

temperature effect upon the Zener and avalanche breakdown voltages.

TU

UU ∆

∆α

p

p=… At the Zener breakdown a Zener diode has a negative

temperature coefficient.

The avalanche breakdown voltage increases with temperature and

the temperature coefficient is positive.

Breakdowns in pn junctions. Zener diodes



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Problems

1. Find the conductivity of an intrinsic semiconductor assuming that the density of

intrinsic carriers is 5⋅1010 cm-3, mobility of electrons is 1400 cm2/(V·s), mobility

of holes is 500 cm2/(V·s). Determine also the conductivity of the semiconductor

after doping with donor impurities at their density of 3.5⋅1014 cm-3.

2. There are samples of copper, intrinsic silicon, silicon doped with phosphorus

(Nd = 10 17 cm-3). How and how many times will the mobility of charge carriers

and conductivity of the samples change with temperature, increasing from 300

to 350 K?

3. A silicon diode is operated at a constant forward voltage of 0.7 V. What is the

ratio of the maximum to minimum current in the diode over a temperature range

–55 to 1000C?

4. A pn silicon diode is used as a varactor. The doping densities on the two sides

of the junction are Na = 1015 cm-3 and Nd = 1017 cm-3, respectively. The diode

area is 0.1 mm2. Find the diode capacitance at the reverse voltages of 1 and

5 V.

5. The saturation current of a silicon pn junction is 1 µA. Carrier lifetime is 1 µs.

Find the diffusion capacitance of the junction at the forward bias UF = 0.1 and

0.3 V.



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