Eece 352notes

Only available on StudyMode
  • Topic: Diode, Semiconductor, Semiconductors
  • Pages : 20 (2179 words )
  • Download(s) : 65
  • Published : March 6, 2013
Open Document
Text Preview
EECE 352: 352 PN Junctions

Peyman Servati y

Outline
• • • • Diffusion and drift currents Current continuity PN Junctions Diodes, photodetectors, solar cells

Diffusion of Electrons
Movement of electrons and holes (charge carriers) results in ( g ) conduction in materials. Electrons and holes in solids move based on Brownian motion (random walk). In this random movement, electrons effectively move from a movement location where the concentration is higher to where the concentration is lower. 1D random walk

2D random walk

Δx

x − Δx x + Δx

x

x

Diffusion Current
The current carried due to the diffusion of electrons is proportional to the gradient of electron density:

dn( x ) dn( x ) J n ( x ) = qDn dx dx The coefficient Dn is the diffusion constant for electrons with a unit of cm2/Vs.

J n ( x) ∝

Since electron charge is negative the sign for current is positive.

The current carried due to the diffusion of holes is proportional to the gradient of hole density:

n(x)
p(x) ( )

J p ( x ) = − qD p

The coefficient D p is the diffusion constant for holes with a unit of cm2/Vs.

x

dp ( x ) dx

Total Current
The total current carried in a semiconductor is the sum of diffusion and drift currents for both electrons and holes:

J ( x) = J n ( x) + J p ( x)
dn ( x ) dx drift diffusion dp ( x ) J p ( x ) = qμ p p ( x )E ( x ) − q p qD dx d J n ( x ) = q μ n n ( x ) E ( x ) + qD n D The diffusion constant and drift mobility are related.

n(x)
p(x) ( )

Δx 2 Dn = 2τ
The kinetic energy of electron is given by:

K .E . =

* mn vavg g

2

2

= kT
Dn Dp 2/s) (cm2/s) (cm
Ge Si GaAs 100 35 220 50 12.5 10

Einstein relation

μn
(cm2/Vs)
3900 1350 8500

(cm2/Vs)
1900 480 400

Dn

kT = = Vth q μn

Dp

kT = μp q

x

μp

Examples
1. 1 An intrinsic Si is doped with donor atoms with a density of (a) In equilibrium J(x) = 0 what is electric field E(x)? (b) Sketch a band diagram.

N D (x) = N 0e−ax

E (x)

EC
Ei EV
x

Drift Mobility
Mobility should be constant ideally. In reality it is not. Note that if mobility is constant the conductivity will change only with carrier density density.

Dependence of mobility on lattice scattering and impurity scattering. Ionized impurity scattering ( ) Lattice scattering ( )

Electric field dependence of mobility. • Scattering limited drift velocity • Hot electrons • GaAs negative differential conductivity

t coll

−1

= tiion + tllatt

−1

−1

Streetman & Banerjee, Solid State Electronic Devices

Sze & Ng, Physics of Semiconductor Devices

PN Junctions
The PN junction is formed by putting together p-type and n-type semiconductor pieces. Let’s assume that we have an acceptor dopant density NA and a donor dopant density ND in these pieces.

EC

p
Ei
E Fp EV

n

EC E Fn Ei

NA

ND

EV

p p0 = N A Majority carriers: holes Minority carriers: electrons n p 0 = ni 2 / N A Fermi energy: EV − E Fp = kT ln

Majority carriers: electrons nn 0 = N D Minority carriers: holes pn 0 = ni 2 / N D

NA N or Ei − E Fp = kT ln A NV ni

E Fn − EC = kT ln

ND ND or E Fn − Ei = kT ln NC ni

As soon as the p and n pieces are put together, electrons and holes rush and recombine in the middle, leaving a depletion region. Depletion region The depletion region is filled with fixed ions left after depletion of free electrons and holes. Acceptor atoms (p-type) leave negative ions and p n donor atoms (n-type) leave positive ions. As a result the charge density in the n-type depletion region is +qND and in the p-type depletion region is -qNA. In the remaining regions of the PN junction, the material is neutral and charge density is zero. E0 V When no voltage is applied and no current is flowing, the Fermi energy is the same in all parts of the device. This Fermi energy alignment can EC only be achieved by a band bending that causes a potential barrier V0 qV 0 at the junction.

0

qV0...
tracking img