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Microelectronic Circuits, Kyung Hee Univ. Spring, 2016 1 1.10.1. Physical Structure pn junction structure p-type semiconductor n-type semiconductor metal contact for connection Figure 1.35: Simplified physical structure of the pn junction. (Actual geometries are given in Appendix A.) As the pn junction implements the junction diode, its terminals are labeled anode and cathode.

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1.10.1. Physical Structure• pn junction structure

• p-type semiconductor• n-type semiconductor• metal contact for connection

Figure 1.35: Simplified physical structure of the pnjunction. (Actual geometries are given in Appendix A.) As

the pn junction implements the junction diode, its terminals are labeled anode and cathode.

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1.10.2 Operation with Open-Circuit• Q: What is state of pn junction with open-circuit terminals?• A:

• p-type material contains majority of holes• these holes are neutralized by equal amount of bound

negative charge• n-type material contains majority of free electrons

• these electrons are neutralized by equal amount of bound positive charge

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• Bound charge• charge of opposite polarity to free electrons / holes of a

given material• neutralizes the electrical charge of these majority carriers• does not affect concentration gradients

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• Q: What happens when a pn-junction is newly formed – aka. when the p-type and n-type semiconductors first touch one another?

• A: See following slides…

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Step #1: The p-type and n-type semiconductors are joined at the junction

Figure: The pn junction with no applied voltage (open-circuited terminals).

n-type semiconductor filled with free electrons

p-type semiconductor filled with holes junction

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Step #1A: Bound charges are attracted (from environment) by free electrons and holes in the p-type and n-type semiconductors, respectively. They remain weakly “bound” to these majority carriers; however, they do not recombine.

Figure: The pn junction with no applied voltage (open-circuited terminals).

positive bound charges

negative bound charges

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Step #2: Diffusion begins. Those free electrons and holes which are closest to the junction will recombine and, essentially, eliminate one another.

Figure: The pn junction with no applied voltage (open-circuited terminals).

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Step #3: The depletion region begins to form – as diffusion occurs and free electrons recombine with holes.

The depletion region is filled with “uncovered” bound charges – who have lost the majority carriers to which they were linked.

Figure: The pn junction with no applied voltage (open-circuited terminals).

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• Q: Why does diffusion occur even when bound charges neutralize the electrical attraction of majority carriers to one another?

• A: Diffusion current, as shown in (1.43) and (1.44), is effected by a gradient in concentration of majority carriers – not an electrical attraction of these particles to one another.

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Step #4: The “uncovered” bound charges effect a voltage differential across the depletion region. The magnitude of this barrier voltage (V0) differential grows, as diffusion continues.

volta

ge p

oten

tial

location (x)

barrier voltage (Vo)

No voltage differential exists across regions of the pn-junction outside of the depletion region because of the neutralizing effect of

positive and negative bound charges.

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Step #5: The barrier voltage (V0) is an electric field whose polarity opposes the direction of diffusion current (ID). As the magnitude of V0increases, the magnitude of ID decreases.

Figure: The pn junction with no applied voltage (open-circuited terminals).

diffusion current (ID)

drift current (IS)

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Step #6: Equilibrium is reached, and diffusion ceases, once the magnitudes of diffusion and drift currents equal one another – resulting in no net flow.

diffusion current (ID)

drift current (IS)

Once equilibrium is achieved, no net current flow exists (Inet = ID – IS) within the pn-junction while under open-circuit condition.

p-type n-typedepletion region

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The Drift Current IS and Equilibrium

• In addition to majority-carrier diffusion current (ID), a component of current due to minority carrier drift exists (IS).

• Specifically, some of the thermally generated electrons and holes in the p-type and n-type materials move toward and reach the edge of the depletion region.

• There, they experience the electric field (V0) in the depletion region and are swept across it.

• Unlike diffusion current, the polarity of V0 reinforces this drift current.

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• Because these holes and free electrons are produced by thermal energy, IS is heavily dependent on temperature

• Any depletion-layer voltage, regardless of how small, will cause the transition across junction. Therefore IS is independent of V0.

• drift current (IS) – is the movement of these minority carriers.

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0 barrier voltage thermal voltage

acceptor doping concentration donor doping concentration

concentration of free electrons... ...in intrinsic

2

sem

0

ic

(eq3.22)

TA

Di

VV

NN

n

A DT

i

N NV V

n

==

==

=

=

ln

onductor

• pn-junction built-in voltage (V0) – is the equilibrium value of barrier voltage

• Generally, it takes on a value between 0.6 and 0.9V for silicon at room temperature

• This voltage is applied across depletion region, not terminals of pn junction• Power cannot be drawn from V0

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Note that the magnitude of drift current (IS) is unaffected by level of diffusion and / or V0. It will be, however, affected by temperature.

Figure: The pn junction with no applied voltage (open-circuited terminals).

diffusion current (ID)

drift current (IS)

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• Q: Is the depletion region always symmetrical? As shown on previous slides?

• A: The short answer is no.• Q: Why?

• A: Because, typically NA > ND.• When the concentration of doping agents (NA, ND) is

unequal, the width of depletion region will differ from side to side.

• The depletion region will extend deeper in to the “less doped” material, a requirement to uncover the same amount of charge.

• xp = width of depletion p-region• xn = width of depletion n-region

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The depletion region will extend further in to region with “less” doping. However, the “number” of uncovered charges is the same.

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3.4.2: Operation withOpen-Circuit Terminals

• Width of and Charge Stored in the Depletion Region• the question we ask here is, what happens once the open-circuit pn junction

reaches equilibrium???• typically NA > ND

• minority carrier concentrations at equilibrium (no voltage applied) are denoted by np0 and pn0

• because concentration of doping agents (NA, ND) is unequal, the width of depletion region will differfrom side to side

• the depletion region will extend deeper in to the “less doped” material, a requirement to uncover the same amount of charge

• xp = width of depletion p-region

• xn = width of depletion n-region

Oxford University PublishingMicroelectronic Circuits by Adel S. Sedra and Kenneth C.

dv/dx is dependent of Q/W

charge is equal, but width is different

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pp

p

pp

p

pp

magnitude of charghe on -side of junction magnitude of electric charge

PP

P

cross-sectional area of junction penetration of depletion region into -side

co

nP

(eq3.23)

n

D

Q nq

x

n

An

D

N

Q qAx N

+ ===

==

+ =

pp

-pp

pp

pp

centration of donor atoms

magnitude of charghe on -side of junction magnitude of electric charge

cross-sectional area of junction penetration

P

P

P

fP

o

-(eq3.24)

p

Q nq

A

p A

x

Q qAx N=

==

=

=

pp

pp

depletion region into -side concentration of acceptor a

PsPtomA

pN =

• Q: How is the charge stored in both sides of the depletion region defined?

• A: Refer to equations to right. Note that these values should equal one another.

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(eq3.25) n Ap A n D

p D

x NqAx N qAx N

x N= → =

• Q: What information can be derived from this equality?• A: In reality, the depletion region exists almost entirely

on one side of the pn-junction – due to great disparity between NA > ND.

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ppp

pp

0

p width of depletion region electrical permiability of silicon (11.7 1.04 12 )

magnitude of electron charge concentration of acceptor ato

PP/

Pm

0

s

(eq3.22 1

6) 1

S

A

W

qF cm

Sn p

A D

N

W x x Vq N N

ε ε

ε

== = −

==

= + = +

E

pp

pp

pp0

concentration of donor atoms barrier / junction built-in volta Pge

PP

(eq3.27)

(eq3.28)

D

An

A D

Dp

NV

A D

Nx W

N N

Nx W

N N

==

=+

=+

• Note that both xp and xn

may be defined in terms of the depletion region width (W).

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0

(eq3.29)

(eq3.30) 2

A DJ

A D

A DJ S

A D

N NQ Q Aq W

N N

N NQ A q V

N Nε

±

= = +

= +

• Note, also, the charge on either side of the depletion region may be calculated via (1.47) and (1.51).

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• Q: What has been learned about the pn-junction? • A: composition

• The pn junction is composed of two silicon-based semiconductors, one doped to be p-type and the other n-type.

• A: majority carriers• Are generated by doping.• Holes are present on p-side, free electrons are

present on n-side.

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• Q: What has been learned about the pn-junction? • A: bound charges

• Charge of majority carriers are neutralized electrically by bound charges.

• A: diffusion current ID• Those majority carriers close to the junction will

diffuse across, resulting in their elimination.

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• Q: What has been learned about the pn-junction? • A: depletion region

• As these carriers disappear, they release bound charges and effect a voltage differential V0.

• A: depletion-layer voltage• As diffusion continues, the depletion layer voltage (V0)

grows, making diffusion more difficult and eventually bringing it to halt.

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• Q: What has been learned about the pn-junction? • A: minority carriers

• Are generated thermally.• Free electrons are present on p-side, holes are

present on n-side.• A: drift current IS

• The depletion-layer voltage (V0) facilitates the flow of minority carriers to opposite side.

• A: open circuit equilibrium ID = IS

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1.11.1. Qualitative Description of Junction Operation

• Figure to right shows pn-junction under three conditions:• (a) open-circuit – where a barrier voltage V0 exists.• (b) reverse bias – where a dc voltage VR is applied.• (c) forward bias – where a dc voltage VF is applied.

Figure 1.38: The pn junction in: (a) equilibrium; (b) reverse bias; (c) forward bias.

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• Figure to right shows pn-junction under three conditions:• (a) open-circuit – where a barrier voltage V0 exists.• (b) reverse bias – where a dc voltage VR is applied.• (c) forward bias – where a dc voltage VF is applied.

Figure 3.11: The pn junction in: (a) equilibrium; (b) reverse bias;

(c) forward bias.

1) no voltage applied

2) voltage differential across depletion zone is V0

3) ID = IS

1) negative voltage applied

2) voltage differential across depletion zone is V0 + VR

3) ID < IS

1) positive voltage applied

2) voltage differential across depletion zone is V0 - VF

3) ID > IS

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• reverse bias case• the externally applied voltage VR

adds to (aka. reinforces) the barrier voltage V0

• …increase effective barrier• this reduces rate of diffusion,

reducing ID• if VR > 1V, ID will fall to 0A

• the drift current IS is unaffected, but dependent on temperature

• result is that pn junction will conduct small drift current IS

• forward bias case• the externally applied voltage VF

subtracts from the barriervoltage V0

• …decrease effective barrier• this increases rate of diffusion,

increasing ID• k

• the drift current IS is unaffected, but dependent on temperature

• result is that pn junction will conduct significant current ID - IS

minimal current flows in reverse-bias case

significant current flows in forward-bias case

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Reverse-Bias Casepp

p0p

00

width of depletion region electrical permiability of silicon (11.7 1.04 12 )

magn

replace with

itude of electron ch/

0

arge

PP

(eq3.31)2 1 1

( )

S

R

F cm

Sn p R

VV V

W

q

A D

W x x V Vq N N

ε ε

ε

== = −

+

=

= + = + +

action:

E

pp

pp

pp

p0 p

pp

concentration of acceptor atoms concentration of donor atoms

barrier / junction built-in voltage externally applied reverse-bias volta

P

PPg Pe

P

(eq3.3 22)

A

D

R

NN

V

J

V

Q A

==

==

=

0

pp

0

magnitude of charge store

0

d on either side of depletion re

replace with

gi Pon

( )

J

R

VV V

A DS R

A

Q

D

N Nq V V

N Nε

+

=

+ + action:

• Observe that increased barrier voltage will be accompanied by…

• (1) increase in stored uncovered charge on both sides of junction

• (2) wider depletion region

• Width of depletion region shown to right.

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Forward-Bias Case

00

pp

pp

pp

0

width of depletion region electrical permiability of silicon (11.7 1.04 12 )

magnitude of electron charge con

replac

PP

P/

e with

02 1 1

( )

A

F

S F c

V

W

qm

Sn p F

A D

N

V V

W x x V Vq N N

ε ε

ε

== = −

=

=

= + = + −

action:

E

pp

pp

pp0

pp

centration of acceptor atoms concentration of donor atoms

barrier / junction built-in voltage externally applied forward-bias voltage

PP

P

0

P

2 (

D

F

A DJ S F

A D

NV

V

N NQ A q V V

N Nε

==

=

= − +

0

pp

0

magnitude of charge stored on either side of

rep

dep

lace wit

letion region

P

h

)

J

FV V

Q

V

=

action:

• Observe that decreased barriervoltage will be accompanied by…

• (1) decrease in stored uncovered charge on both sides of junction

• (2) smaller depletion region

• Width of depletion region shown to right.

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1.11.2. The Current-Voltage Relationship of the Junction

• Q: What happens, exactly, when a forward-bias voltage (VF) is applied to the pn-junction?

• step #1: Initially, a small forward-bias voltage (VF) is applied. It, because of its polarity, pushes majority carriers (holes in p-region and electrons in n-region) toward the junction and reduces width of the depletion zone.

• Note, however, that this force is opposed by the built-in voltage built in voltage V0.

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step #1: Initially, a small forward-bias voltage (VF) is applied. It, because of its polarity, pushes majority (holes in p-region and electrons in n-region) toward the junction and reduces width of the depletion zone.

Figure: The pn junction with applied voltage.

VF

Note that, in this figure, the smaller circles represent minority carriers and not bound charges – which are not considered here.

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step #2: As the magnitude of VF increases, the depletion zone becomes thin enough such that the barrier voltage (V0 – VF) cannot stop diffusion current – as described in previous slides.

Figure: The pn junction with applied voltage.

VF

Note that removing barrier voltage does not facilitate diffusion, it only removes the electromotive force which opposes it.

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step #3: Majority carriers (free electrons in n-region and holes in p-region) cross the junction and become minority charge carriers in the near-neutral region.

Figure: The pn junction with applied voltage.

diffusion current (ID)

drift current (IS)

VF

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step #4: The concentration of minority charge carriers increases on either side of the junction. A steady-state gradient is reached as rate of majority carriers crossing the junction equals that of recombination.

Figure: The pn junction with applied voltage.

min

ority

car

rier

conc

entr

atio

n

location (x)

VF

For the open-circuit condition, minority carriers are evenly distributed throughout the non-depletion regions. This

concentration is defined as either np0 or pn0.

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step #4: The concentration of minority charge carriers increases on either side of the junction. A steady-state gradient is reached as rate of majority carriers crossing the junction equals that of recombination.

Figure: The pn junction with applied voltage

VF

min

ority

car

rier

conc

entr

atio

n

location (x)

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step #5+: Diffusion current is maintained – in spite low diffusion lengths (e.g. microns) and recombination – by constant flow of both free electrons and holes towards the junction.

Figure: The pn junction with applied voltage

VF

flow of holes flow of electrons

flow of diffusion current (ID)

recombination

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40

0

( ) = concentration of holes in -region as function of = thermal equilibrium concentration

2

0

= applied foward-bias voltage = thermal volt

/0

(eq3.7)

(eq3.33

( ))

n n n

n

p

pp

T

T

pp

p

PP

x n xpV

in

A

V Vn n n

VP

np

N

x p e

=

=

p

p

a

/0 0

e

/

g

0

excessconcentra

excess(eq3.34)

concentratio

tion

( 1)

(eq )

n

3.34

pp

V VTn n

P

V VTn

p e p

p e

= −

= −

• Q: How is the relationship between forward-bias voltage applied (V.) and minority-carrier holes and electrons defined?

• step #1: Employ (1.57).• This function describes

maximum minority carrier concentration at junction.

• step #2: Subtract pn0 from pn(x) to calculate the excess minority charge carriers.

The key aspect of (1.57) is that it relates the minority-charge carrier concentration at the junction boundary in terms of majority-charge

carrier on the opposite side.

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41

0

/0

( ) /0

( ) //0 0

( ) = concentration of holes in -region as function of , = thermal eq

( 1)

(eq3.35

(eq3.35) (

( ) ( )

( 1)

)

)

n p

n n n n

Vn

n

V

p

T

x x Ln n n

x x LV VT

x

n

x p

p e

n n n

n

x p excess concentrati

px

on e

p e e

− −

− −

= +

= + −

p

p

p

uilibrium concentration = point of interest, edge of depletion region, = diffusion lengthn Px x L=

• Q: How is the relationship between forward-bias voltage applied (V.) and minority-carrier holes and electrons defined?

• step #3: Refer to (1.59).• This function describes the minority carrier concentration as a

function of location (x), boundary of depletion region (xn), and diffusion length (Lp).

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42

3.5.2: The Current-VoltageRelationship of the Junction

Oxford University PublishingMicroelectronic Circuits by Adel S. Sedra and Kenneth C. Smith (0195323033)

steady-state minority carrier concentration on both sides of a pn-junction for which NA >> ND

excess concentration

“base” concentration

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43

3.5.2: The Current-VoltageRelationship of the Junction

Oxford University PublishingMicroelectronic Circuits by Adel S. Sedra and Kenneth C. Smith (0195323033)

These excess concentrations effect steady-state diffusion current. However, how is this diffusion current defined?

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44

[ ]

( ) //0

take derivative of

0

( ) //0

( ) /

0

( 1)

0

/

( )

( )

( 1)

(eq3.36) ( 1)

n pT

n p

x x LV VTn n p

n

p

T

nn

x x LV Vn

x x LV Vn

x

pe e

p pp

L

d x dp

dx dx

dp e e

dx

pJ qD e e

L

− −

− −

− −

− −

= +

+ −

= − − −

action: p

p

substitute in value from above

calculate ma

(

ximum

)

/0( ) ( 1)

n

Tp V V

d xdx

p np

DJ q p e

L= −

action:

actio

p

n:

max

• Q: For forward-biased case, how is diffusion current (ID) defined?

• step #1: Take derivative of (1.59) to define component of diffusion current attributed to flow of holes.

• step #2: Note that this value is maximum at x= xn. Oxford University Publishing

Microelectronic Circuits by Adel S. Sedra and Kenneth C. Smith (0195323033)

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45

Q: For forward-biased case, how is diffusion current defined?

/0

/0

(eq3.37) maximum hole - diffusion concentration:

(eq3.38) maximum electron - diffusion concentration:

( ) ( 1)

( ) ( 1 )

T

T

p V Vp n n

p

V Vnn p p

n

Dx q p e

L

Dx q n e

L

+ = −

− = −

J

J

• step #3: Define the component of maximum diffusion current attributed to minority-carrier electrons – in method similar above.

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46

Q: For forward-biased case, how is diffusion current defined?

+

= =

= + + −

= + −

= + −

2 2

0 0

subtitute in valuesfor ( ) and (- )

subtitut

/0

e/ and

/

/

0

2

( ) ( )

( 1)

( 1)

p n n p

n i D p i

T

T

A

p n n p

p V Vnn p

p n

p V Vni

p D n

x x

p n N n n N

A

I A x x

D DI A q p q n e

L L

D DI Aqn e

L N L N

action: J J

action:

J J

= +

= −

2

total current ( ) through junction is equal to area ( ) timesmaximum hole ( )

subt

and electron-diffusion ( ) current densities

/

itute

( 1)

p ns i

p

n

D

T

n A

p

D DI A

I

qnL N L

V VS

N

AJ J

I I eaction:

• step #4: Define total diffusion current as sum of components attributed to free electrons and holes.

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2 / /( 1)(eq3. ( 140) )T T

S

p V V V Vni

n

I

Sp D A

D DI Aqn e I e

L N L N

= + − = −

• Q: For forward-biased case, how is diffusion current (ID) defined?

• A: This is an important equation which will be employed in future chapters.

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• Q: Why is diffusion current (ID) dependent on the concentration gradient of minority (as opposed to majority) charge carriers?

• A: Essentially, it isn’t.• Equation (1.57) defines the minority-charge carrier

concentration in terms of the majority-charge carrier concentrations in “other” region.

• As such, the diffusion current (ID) is most dependent on two factors: applied forward-bias voltage (VF) and doping.

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49

/(eq3.40) ( 1)TV VSI I e= −

• saturation current (IS) – is the maximum reverse current which will flow through pn-junction.

• It is proportional to cross-section of junction (A).

• Typical value is 10-18A.Figure 1.40: The pn junction I–V

characteristic.