Ch1 3 PN Electrostatic

8/11/2019 Ch1 3 PN Electrostatic

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PN Junction

PN Junction Electrostatics

Built-in potential, depletion approximation

Current-Voltage characteristics

Ideal I-V characteristics

Deviation from the ideal

Breakdown, high-level injection



PN Junction Basics

Physical view of a PN junction

Step junction

metal

n+ p

oxide

n+ p

Cross-section

ND-NA

x

Step junction

idealization

Actual profile




Qualitative Analysis: Thermal Equil ibrium

Energy Band Diagram at Thermal

Equilibrium

Fermi level should be aligned Vacuum level (and hence EC, EV,

and Ei) should be continuous





Electrostatic Variables at Thermal Equilibrium

The V vs x should have the same functional

form of “upside down” of EC (EV, or Ei)

The E vs x can be drawn from the derivation

of V over x (E=-dV/dx)

The ρ vs x can be drawn from the slope of E

)()( iFnFpibi E E E E qV

)ln()()(

2

i

A DiFnFpi

bi

n

N N

q

kT

q

E E E E V

“Built-In Potential”

Since , we have





Depletion region -- how charge

redistributed?

Electrons diffusion from n-region to p-

region; while holes diffusion from p-region to

n-region Unbalanced dopants are left and form built-

in field which directs from n-region to p-

region

Equilibrium condition is met when the

electrons and holes diffusions are balancedby their drifts due to built-in field.



Depletion approximation

Bulk p and n regions assumed as

charge neutral

Transition region assumed to be

depleted of mobile carriers (depletionregion)

Depletion region can be treated as

insulator to withstand voltage, but ischarged due to unbalanced dopants

Quite accurate



Depletion

approximation




Quantitative Analysis: Thermal Equil ibrium

Charge density

In n region near the joint, as electrons

are depleted (depletionapproximation), the region should be

positive charged and equal to qND

Similarly, the p-region near the joint

should be negative charged andequal to –qN A

So, we have charge density:

)(0

)0(

)0()(0

n

n D

p A

p

x x

x xqN

x xqN x x





Electric Field0

sdx

dE

By integration, we have

)()(0

x xqN

x E p

S

A

)()(0

x xqN x E n

S

D

for –x p x 0

for 0 x xn

0 E for x>xn or x -x p

A note: at x=0, the E-fields equal fromabove two equations, thus

n D p A x N x N

(can also obtained from charge neutrality)

n D p A xqN xqN





Electric Potential

By integration, we have

A note: From electrical potential continuity atx=0, one obtains

dx

dV E

With boundary conditions

nbi

p

x xV

x x

V ,

,0

2

0

)(2

)( x xqN

xV p

S

A

2

0

)(2

)( x xqN

V xV n

S

Dbi

for –x p x 0

for 0 x xn

2

0

2

0 22 n

S

D

p

S

A

bi xqN

xqN

V





Depletion width

Recall: from continues of E-field and E-

potential at x=0, we have

xn and xp can be calculated as below

2

0

2

0 22 n

S

D

p

S

A

bi xqN

xqN

V

n D p A x N x N

2/1

0

)(

2

bi D A D

AS

n

V N N N

N

q x

2/1

0

)(

2

bi

D A A

DS p V

N N N

N

q x





Depletion width

The total depletion width, W=xn+xp, can be

given as below

2/1

0 )(2

bi

D A

D AS

pn V N N

N N

q x xW




Quantitative Analysis: V A Applied









Using (Vbi-V A) to replace Vbi





Depletion capacitance

W

AC S 0




Example: Asymmetric Junction

Very often we have N+/P or P+/N junction

For example:ni/pi ~ 1010cm-3

ND+/N A+ ~ 1018cm-3

ND/N A ~ 1015cm-3

For N+/P, only x p has to be considered

N A x p=

2SiV bi

q1( )

1/2

xn~ 0

x p

N D

N A

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Ch1 3 PN Electrostatic