metallurgical junction

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Dr.-Ing. Erwin SitompulPresident University

Lecture 5Semiconductor Device Physics

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Step junctionidealization

Metallurgical JunctionChapter 5 pn Junction Electrostatics

Doping profile

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Poisson’s EquationChapter 5 pn Junction Electrostatics

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:

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Equilibrium Energy Band DiagramChapter 5 pn Junction Electrostatics

pn-Junction diode

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Band diagram

c ref1 ( )V E Eq

Qualitative ElectrostaticsChapter 5 pn Junction Electrostatics

Equilibrium condition

Electrostatic potential

( )V x dxE

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Electric field

Qualitative ElectrostaticsChapter 5 pn Junction Electrostatics

Equilibrium condition

Charge density

dVdx

E

S 0K

Ex

S 0

( )x xK

E

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Formation of pn Junction and Charge DistributionChapter 5 pn Junction Electrostatics

D A( )q p n N N qNA– qND

+

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Formation of pn Junction and Charge DistributionChapter 5 pn Junction Electrostatics

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DF i n-side

i i

( ) ln ln NnE 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 ln NpE E kT kTn n

A Dbi 2

i

ln N NqV kTn

Built-In Potential Vbi

Chapter 5 pn Junction Electrostatics

For non-degenerately doped material,

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

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A1

S

( ) qNx x c

E

A

S

qNddx

E

Dn

S

( ) ( )qNx x x

E

Ap

S

( ) ( )qNx x x

Ewith boundary E(–xp) 0

with boundary E(xn) 0

The Depletion ApproximationChapter 5 pn Junction Electrostatics

On the p-side, ρ = –qNA

On the n-side, ρ = qND

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A p

D n

, 0 , 0 0,

qN x xqN x x

otherwise

Ap p

S

Dn n

S

( ), 0( )

( ), 0

qN x x x xx qN x x x x

E

2Ap p

S

2Dbi n n

S

( ) , 02( )

( ) , 02

qN x x x xV x qNV x x x x

Step Junction with VA 0Chapter 5 pn Junction Electrostatics

Solution for ρ

Solution for E

Solution for V

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2 2A Dp bi n

S S

( ) ( )2 2qN qNx V x

A p D nN x N x

Step Junction with VA0Chapter 5 pn Junction Electrostatics

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

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Relation between ρ(x), E(x), and V(x)Chapter 5 pn Junction Electrostatics

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 NDxnSolve for xp, xn

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S An bi

D A D

2( )

Nx Vq N N N

S Dp bi

A A D

2( )

Nx Vq N N N

n px x W

Dn

A

N xN

Depletion Layer WidthChapter 5 pn Junction Electrostatics

Eliminating xp,

Eliminating xn,

Summing Sbi

A D

2 1 1 Vq N N

Exact solution, try to derive

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S bi Dn p n

D A

2 , 0V NW x x xq N N

S bi2 VW

q N

S bi Ap n p

A D

2 , 0V NW x x xq N N

One-Sided JunctionsChapter 5 pn Junction Electrostatics

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

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D Abi 2

i

ln N NkTVq n

S bi

D

2 VW

qN

n 0.115 mx W

Dp n

A

Nx xN

Example: Depletion Layer WidthChapter 5 pn Junction Electrostatics

A p+n junction has NA 1020 cm–3 and ND 1017cm–3, at 300 K.

a) What isVbi?

b) What is W?

c) What is xn?

d) What is xp?

17 20

10 2

10 1025.86 mV ln 1.012 V(10 )

1/ 214

19 17

2 11.9 8.854 10 1.012 0.115 m1.602 10 10

30.115 m 10 1.15 Å

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Step Junction with VA 0Chapter 5 pn Junction Electrostatics

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

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Step Junction with VA 0Chapter 5 pn Junction Electrostatics

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.

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A Dbi

i i

ln lnN NkT kTVq n q n

Sp n bi A

A D

2 1 1W x x V Vq N N

S Dp bi A

A A D

2 ,Nx V Vq N N N

S A

n bi AD A D

2 Nx V Vq N N N

Step Junction with VA 0Chapter 5 pn Junction Electrostatics

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

A D2

i

ln N NkTq n

Depletion width W :

,Dp

A D

Nx WN N

WNN

NxDA

An

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Effect of Bias on ElectrostaticsChapter 5 pn Junction Electrostatics

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

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Linearly-Graded JunctionChapter 5 pn Junction Electrostatics

S

1 dx

E V dx E

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1. (6.4)Consider a silicon pn junction at T = 300 K with a p-side doping concentration of NA 1018 cm–3. Determine the n-side doping concentration such that the maximum electric field is |Emax| 3×105 V/cm at a reverse bias voltage of VR 25 V.

Chapter 5 pn Junction Electrostatics

Homework 4

Deadline: Wednesday,16 May 2012, at 14:00.

2. (7.6)Problem 5.4Pierret’s “Semiconductor Device Fundamentals”.