Semiconductor Device Physics

Lecture 6Dr. Gaurav Trivedi,EEE Department,

IIT Guwahati

Doping profile

Metallurgical Junction

Poisson’s Equation

S 0K

E v DD E

S 0K

S 0x K

E

Poisson’s equation is a well-known relationship in electricity and magnetism.

It is now used because it often contains the starting point in obtaining quantitative solutions for the electrostatic variables.

In one-dimensional problems, Poisson’s equation simplifies to:

Equilibrium Energy Band Diagrampn-Junction diode

Qualitative Electrostatics

Band diagram

c ref

1( )V E E

q

Equilibrium condition

Electrostatic potential

( )V x dx E

Qualitative Electrostatics

Electric field

Equilibrium condition

Charge density

dV

dxE

S 0K

Ex

S 0

( )x xK

E

Formation of pn Junction and Charge DistributionFormation of pn Junction and Charge Distribution

D A( )q p n N N qNA– qND

+

Formation of pn Junction and Charge Distribution

Built-In Potential Vbi

DF i n-side

i i

( ) ln lnNn

E E kT kTn n

bi F i n side i F p side( ) ( )qV E E E E

Ai F p-side

i i

( ) ln lnNp

E E kT kTn n

A Dbi 2

i

lnN N

qV kTn

For non-degenerately doped material,

• Vbi for several materials:Ge ≤ 0.66 VSi ≤ 1.12 VGeAs ≤ 1.42 V

The Depletion Approximation

A1

S

( )qN

x x c

E

A

S

qNd

dx

E

Dn

S

( ) ( )qN

x x x

E

Ap

S

( ) ( )qN

x x x

E

with boundary E(–xp) = 0

with boundary E(xn) = 0

The Depletion ApproximationOn the p-side, ρ = –qNA

On the n-side, ρ = qND

Step Junction with VA = 0

A p

D n

, 0 , 0 0,

qN x xqN x x

otherwise

Ap p

S

Dn n

S

( ), 0

( )( ), 0

qNx x x x

xqN

x x x x

E

2Ap p

S

2Dbi n n

S

( ) , 02

( )( ) , 0

2

qNx x x x

V xqN

V x x x x

Solution for ρ

Solution for E

Solution for V

Step Junction with VA = 0

2 2A Dp bi n

S S

( ) ( )2 2

qN qNx V x

A p D nN x N x

At x = 0, expressions for p-side and n-side for the solutions of E and V must be equal:

Relation between ρ(x), E(x), and V(x)

1.Find the profile of the built-in potential Vbi

2.Use the depletion approximation ρ(x) With depletion-layer widths xp, xn unknown

3.Integrate ρ(x) to find E(x) Boundary conditions E(–xp) 0, E(xn)0

4.Integrate E(x) to obtain V(x) Boundary conditions V(–xp) 0, V(xn) Vbi

5.For E(x) to be continuous at x 0, NAxp NDxn

Solve for xp, xn

Depletion Layer Width

S An bi

D A D

2

( )

Nx V

q N N N

S Dp bi

A A D

2

( )

Nx V

q N N N

n px x W

Dn

A

Nx

N

Eliminating xp,

Eliminating xn,

Summing Sbi

A D

2 1 1V

q N N

Exact solution, try to derive

One-Sided Junctions

S bi Dn p n

D A

2 , 0

V NW x x x

q N N

S bi2 VW

q N

S bi Ap n p

A D

2 , 0

V NW x x x

q N N

If NA >> ND as in a p+n junction,

If ND >> NA as in a n+p junction,

Simplifying,

where N denotes the lighter dopant density

Step Junction with VA 0

• To ensure low-level injection conditions, reasonable current levels must be maintained VA should be small

Step Junction with VA 0

In the quasineutral, regions extending from the contacts to the edges of the depletion region, minority carrier diffusion equations can be applied since E ≈ 0.

In the depletion region, the continuity equations are applied.

Step Junction with VA 0

A Dbi

i i

ln lnN NkT kT

Vq n q n

Sp n bi A

A D

2 1 1W x x V V

q N N

S Dp bi A

A A D

2,

Nx V V

q N N N

S An bi A

D A D

2 Nx V V

q N N N

Built-in potential Vbi (non-degenerate doping):

A D2

i

lnN NkT

q n

Depletion width W :

,Dp

A D

Nx W

N N

W

NN

Nx

DA

An

Effect of Bias on Electrostatics

• If voltage drop â, then depletion width â• If voltage drop á, then depletion width á

Quasi-Fermi Levels

F i i F( ) ( ) i i, E E kT E E kTn n e p n e

N i( )i

F E kTn n e i P( )i

E F kTp n e

P ii

lnp

F E kTn

N i

i

lnn

F E kTn

Whenever Δn = Δp ≠ 0 then np ≠ ni2 and we are at non-

equilibrium conditions. In this situation, now we would like to preserve and use the

relations:

On the other hand, both equations imply np = ni2, which does

not apply anymore.The solution is to introduce to quasi-Fermi levels FN and FP such that:

• The quasi-Fermi levels is useful to describe the carrier concentrations under non-equilibrium conditions

Example: Quasi-Fermi Levels

17 30 D 10 cm ,n N

23 3i

00

10 cmn

pn

17 14 17 30 10 +10 10 cmn n n

3 14 14 30 10 +10 10 cmp p p

17 14 31 310 10 =10 cmnp

a) What are p and n?

b) What is the np product?

Example: Quasi-Fermi LevelsConsider a Si sample at 300 K with ND = 1017 cm–3 and

Δn = Δp = 1014 cm–3.

c) Find FN and FP?

N i ilnF E kT n n

5 17 10N i 8.62 10 300 ln 10 10F E

0.417 eV

P i ilnF E kT p n

5 14 10i P 8.62 10 300 ln 10 10E F

0.238 eV

N i i P

i iF E kT E F kTnp n e n e

0.417 0.23810 100.02586 0.0258610 10e e

311.000257 10 31 310 cm

Ec

Ev

Ei

FP

0.238 eV

FN

0.417 eV

Linearly-Graded Junction

S

1dx

E V dx E

Linearly-Graded Junction Homework Assignment

Derive the relations of electrostatic variables, charge density and builtin voltage Vbi for the linearly graded junction.