pn Junction Diodes - KOCWcontents.kocw.net/KOCW/document/2014/konkuk/minyosep2/10.pdf ·...

Prof. Yo-Sep Min Electronic Materials: Semiconductor Physics & Devices Chapt. 5 - Lec 10-1

pn Junction Diodes

• A systematic analysis of pn junction diodes is typically divided

into four segments:

electrostatics (charge density, E-field, and potential): Chapt. 5

steady-state (d.c.) response: Chapt. 6

small-signal (a.c.) response: Chapt. 7 skipped

transient (pulsed) response: Chapt. 8 skipped


Chapter 5. pn Junction Electrostatics

• You will also learn about:

Poisson’s Equation

Built-in Potential

Depletion Approximation

Step-Junction Solution

In this chapter, you will learn about pn junction electrostatics:

charge density, electric field and electrostatic potential existing

inside the diode under equilibrium and steady state conditions.


pn-junction fabrication

pn-Junctions are created by

several processes including:

1. Diffusion

2. Ion-implantation

3. Epitaxial deposition

Each process results in different

doping profiles.


Ideal step-junction doping profile

Step junction Linearly-graded junction

Approximation of an ion-implantation

or shallow diffusion into a lightly

doped wafer

Approximation of deep diffusion into

a heavily doped wafer


Poisson’s Equation

ε

ρ

dx

d

E = charge density

= Ks o

Ks = dielectric constant of semiconductor

• According to the Poisson’s equation, in one dimensional

problem, the charge density (ρ) is proportional to dE/dx.

• Therefore, a diagram of ρ versus x can be deduced from an

E-x plot by simply noting the slope.

• Assuming the dopants to be totally ionized, the charge

density inside a semiconductor is given by

AD NNnpqρ

For a uniformly-doped semiconductor in equilibrium,

neutrality charge the todue 0ρ


Drawing energy band diagram of pn junction in equilibrium

kT

EE

enpFi

i

EF = same everywhere

under equilibrium

Join the two sides of

the bands by a smooth

curve.

kT

EE

enniF

i

p-type n-type

Band bending ???


Electrostatic variables of pn junction in equilibrium

x

E

qx

E

q d

d1

d

d1 iC E

• Potential energy of electrons,

P.E.(electron) = (EC–Eref)

xd

dE

Electrostatic

potential

E-field

Charge

density

Where this charge come from?

• Potential energy difference between

the two sides is EC = –qVbi (< 0).

• There is a built-in potential (Vbi) in

the junction which is equal to

(1/q)(EC).

p-type n-type


Conceptual pn-junction formation

The ionized dopant atoms will build

up an electric field which will prevent

further movement of carriers.

p and n type regions

before junction formation

Majority carriers diffuse to the

opposite side leaving ionized

dopant atoms (immobile charges)

Where this charge come from?

Immobile dopant ions

Space charge region

Depletion region


The built-in potential, Vbi

When the junction is formed, electrons from the n-side and

holes from the p-side will diffuse leaving behind charged

dopant atoms. Remember that the dopant atoms cannot

move! Electrons will leave behind positively-charged donor

atoms and holes will leave behind negatively-charged

acceptor atoms.

The net result is the build up of an electric field from the

positively-charged atoms to the negatively-charged atoms,

i.e., from the n-side to p-side. When steady state condition is reached after the formation of junction, the built-in E-field (or

the built-in potential) will prevent further diffusion of electrons

and holes.

Is there net current across the pn junction?


Equilibrium conditions

Under equilibrium conditions, the net electron current and

hole current will be zero in the pn junction.

electron drift current

opposite to electron flux

electron diffusion current

opposite to electron flux

hole diffusion current

hole drift current net = 0

net = 0

p-type n-type



• The voltage drop across the depletion region under

equilibrium is called built-in potential.

• By the definition of electric field, dx

dVE

Integrating across the depletion region gives

bipn

xV

xV

x

xVxVxVdVdx

n

p

n

p

)()(

)(

)(E

Under equilibrium condition,

0dx

dnqDnqJ nnn E

Solving for E,

n

dxdn

q

kT

n

dxdnD

n

n //

E

p-type n-type



n

dxdn

q

kT

n

dxdnD

n

n //

E

bipn

xV

xV

x

xVxVxVdV

n

p

n

p

)()(

)(

)(E

Combining two equations,

n

p

n

p

x

x

x

xbi dx

n

dxdn

q

kTdxV E

/

)(

)(ln

p

nx

x xn

xn

q

kT

n

dn

q

kTn

p

For a non-degenerately doped step junction (total ionization),

Dn Nxn )(A

ip

N

nxn

2

)( and

2ln

i

DAbi

n

NN

q

kTVTherefore



2ln

i

DAbi

n

NN

q

kTV

Ex) For a pn junction of Si (NA = ND = 1015/cm3) at 300 K

Vq

eVVbi

6.0

10

1010ln

026.020

1515

• In pn junctions of non-degenerately doped semiconductors,

q

EV G

bi generally

• The electrostatic potential (V) in Volts is numerically equal to

the energy (E) in eV.



iFi ln

n

pkTEE

iiF ln

n

nkTEE

q Vbi = (Ei EF)p-side + (EF Ei)n-side

kT

EE

i

iF

enn

kT

EE

i

Fi

enp

p-type n-type

Alternative derivation of Vbi by using the energy band diagram



Vbi = (1/q) {(Ei EF)p-side + (EF Ei)n-side }

22lnln

lnln

i

DA

i

np

ii

n

NN

q

kT

n

np

q

kT

n

n

q

kT

n

p

q

kT

kT

biVq

n

n

p

pexp

p

n

n

pProve this relation!!

q Vbi = (Ei EF)p-side + (EF Ei)n-side

sidenonionconcentratelectron nand

sideponionconcentrat holepwhere

n

p


Built-in potential in a p+n junction

Ex) Derive an equation for the Vbi for a p+n junction,

Because EF = EV in the p+ side,

Vbi = (1/q) {(Ei EF)p-side + (EF Ei)n-side }

(Ei EF)p-side = EG/2q

i

DGbi

n

N

q

kT

q

EV ln

2

Similarly, for n+p junction,

i

AGbi

n

N

q

kT

q

EV ln

2



To obtain a quantitative solution for the electrostatic variables (potential, E-field and charge density), the Poisson’s equation

should be solved.

ε

ρ

dx

d

E

dx

dVE

AD

ss

NNnpεK

q

εK

ρ

ε

ρ

dx

d

00

E

In order to write down ρ as a function of x, we must know

explicit expression for n and p as a function of x.

Unfortunately, the n and p in E ≠ 0 regions, like the depletion

region, are not specified prior to solving the Poisson’s equation.

We need an approximation method.



The carrier depletion tends to be greatest in the immediate

vicinity of the metallurgical junction and then tails off as one

proceeds away from the junction.

However, in the depletion approximation, we assume as

below:

(1) In the depletion region (–xp ≤ x ≤ xn), the

n and p are negligible compared to the net

doping concentration (ND – NA).

(2) The charge density outside the depletion

region (x ≤ –xp and x ≥ xn) is identically zero.

AD

s

NNnpεK

q

dx

d

0

E


Depletion approximation

Poisson equation

np xxx

dx

dE AD

s

NNεK

q

0

0 else everywhere

Note that the charge density has

precisely the same functional

form as ND-NA within the

depletion region.


Announcements

• Next lecture: p. 209 ~ 226

• Mid-term EXAM

Chapters 1, 2, 3, and 5.1

13:00 ~ 15:00; October 20,

별232

Supplementary sheet and calculator

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