kasappp3 1617 jf - w3.ualg.ptw3.ualg.pt/~jlongras/aulas/sco_1617_12_160317.pdf · from: s.o. kasap,...

From: S.O. Kasap, Optoelectronics and Photonics: Principles and Practices, Second Edition, © 2013 Pearson Education, USA 1From: S.O. Kasap, Optoelectronics and Photonics: Principles and Practices, Second Edition, © 2013 Pearson Education, USA 1

JF, SCO 1617

Semiconductor Science

S.O. Kasap, Optoelectronics and Photonics: Principles and Practices, Second Edition, © 2013 Pearson Education© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is prote cted by Copyright and written permission should be obtained from the

publisher prior to any prohibited reproduction, sto rage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photo copying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearso n Education, Inc., Upper Saddle River, NJ 07458.

16-Mar-2017


JF, SCO 1617

Intrinsic Semiconductors



16-Mar-2017

From: S.O. Kasap, Optoelectronics and Photonics: Principles and Practices, Second Edition, © 2013 Pearson Education, USA 3

(a) Energy levels in a Li atom are discrete. (b) The energy levels corresponding to outer shells of isolated Li atoms form an energy band inside the crystal, for example the 2s level forms a 2s band.

Energy levels form a quasi continuum of energy within the energy band. Various energy bands overlap to give a single band of energies that is only partially full of electrons. There are states with

energies up to the vacuum level where the electron is free. (c) A simplified energy band diagram and the photoelectric effect.

Energy Bands in Metals


(a) Above 0 K, due to thermal excitation, some of the electrons are at energies above EF. (b) The density of states,g(E) vs. E in the band. (c) The probability of occupancy of a state at an energy E is f(E). (d) The product g(E)f(E) is the number of electrons per unit

energy per unit volume or electron concentration per unit energy. The area under the curve with the energy axis is the concentration of electrons in the band, n.




−+=

Tk

EEEf

B

Fexp1

1)(

2/12/132/3)2(4)( AEEhmE e == −πg

Density of states Fermi-Dirac function

dEEfEnFE

∫Φ+

=0

)()(g

3/22 3

8

=

πn

m

hE

eFO


JF, SCO 1617

(a) A simplified two dimensional view of a region of the Si crystal showing covalent bonds. (b) The energy band diagram of electrons in the Si crystal at absolute zero of

temperature. The bottom of the VB has been assigned a zero of energy.

Energy Bands in Semiconductors


JF, SCO 1617

(a) A photon with an energy hυ greater than Eg can excite an electron from the VB to the CB. (b) Each line between Si-Si atoms is a valence electron in a bond. When a photon breaks a Si-Si bond, a free electron and a hole in the Si-Si bond is created. The result is

the photogeneration of an electron and a hole pair (EHP)

Energy Bands in Semiconductors


JF, SCO 1617

A pictorial illustration of a hole in the valence band (VB) wandering around the crystal due to the tunneling of electrons from neighboring bonds; and its eventual recombination with a wandering electron in the conduction band. A missing electron in a bond represents a hole as in (a). An electron in a neighboring bond can tunnel into this empty state and thereby cause the hole to be displaced as in (a) to (d). The hole is able to wander around in the crystal as if it were free but with a different effective mass than the electron. A wandering electron in the CB meets a hole in the VB in (e), which results in the recombination and the filling of the empty VB state as in (f)

Hole Motion in a Semiconductor


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(a) Energy band diagram.

(b) Density of states (number of states per unit energy per unit volume).

(c) Fermi-Dirac probability function (probability of occupancy of a state).

(d) The product of g(E) and f (E) is the energy density of electrons in the CB (number of electrons per unit energy

per unit volume). The area under nE(E) versus E is the electron concentration.

Semiconductor Statistics


Electron and Hole Drift Velocities

vde = Drift velocity of the electrons, µe = Electron drift mobility, Ex = Applied electric field, vdh = Drift velocity of the holes, µh = Hole drift mobility

vdh = µhEx

Conductivity of a Semiconductor

σ = Conductivity, e= Electronic charge, n = Electron concentration in the CB, µe = Electron drift mobility, p = Hole concentration in the VB, µh = Hole drift mobility

σ = enµe + epµh

vde = µeEx


Electron Concentration in CB

dEEfEnc

c

E

E∫+

=χ

)()(CBg

−−≈Tk

EEEf

B

Fexp)(2/1

2/132/3CB

)(

)()2(4)(

c

ce

EEA

EEhmE

−=

−= −πg

Density of states in the CBFermi-Dirac function ≈Boltzmann function

−−=Tk

EENn

B

Fcc

)(exp


Electron Concentration in CB

n = Electron concentration in the CB Nc = Effective density of states at the CB edge Ec = Conduction band edge, EF = Fermi energy kB = Boltzmann constant, T = Temperature (K)

−−=Tk

EENn

B

Fcc

)(exp

Effective Density of States at CB Edge

Nc = Effective density of states at the CB edge, me* = Effective mass of the electron

in the CB, k = Boltzmann constant, T = Temperature, h = Planck’s constant

2/3

2

*22

=

h

TkmN Be

c

π


Hole Concentration in VB

p = Hole concentration in the VBNv = Effective density of states at the VB edge

Ev = Valence band edge, EF = Fermi energykB = Boltzmann constant, , T = Temperature (K)

−−=Tk

EENp

B

vFv

)(exp

Effective Density of States at VB Edge

Nv = Effective density of states at the VB edge, mh* = Effective mass of a hole in

the VB, k = Boltzmann constant, T = Temperature, h = Planck’s constant

2/3

2

*22

=

h

TkmN Bh

v

π


−==

Tk

ENNnnp

B

gvci exp2

Mass Action Law

Thenp product is a “constant”, ni2, that depends on the material properties Nc, Nv,

Eg, and the temperature. If somehow n is increased (e.g. by doping), p must decrease to keep npconstant.

Mass action law applies

in thermal equilibrium

and

in the dark (no illumination)

ni = intrinsic concentration


Fermi Energy in Intrinsic Semiconductors

EFi = Fermi energy in the intrinsic semiconductorEv = Valence band edge, Eg = Ec − Ev = Bandgap energy

kB = Boltzmann constant, T = temperatureNc = Effective density of states at the CB edgeNv = Effective density of states at the VB edge

−+=

v

cBgvFi N

NTkEEE ln

2

1

2

1

me* = Electron effective mass (CB), mh

* = Hole effective mass (VB)

−+=

*

*

ln4

3

2

1

h

eBgvFi m

mTkEEE


Average Electron Energy in CB

= Average energy of electrons in the CB, Ec = Conduction band edge, kB = Boltzmann constant, T = Temperature

TkEE Bc 2

3CB +=

CBE

(3/2)kBT is also the average kinetic energy per atom in a monatomic gas (kinetic molecular theory) in which the gas atoms move around freely and randomly inside a container.

The electron in the CB behaves as if it were “free” with a mean kinetic energy that is (3/2)kBT and an effective mass me*.

dEEfE

dEEfEEE

c

c

E

E

∫

∫∞

∞

=)()(

)()(

CB

CB

CBg

g

Average is found from


Fermi EnergyFermi energy is a convenient way to represent free carrier concentrations (n in

the CB and p in the VB) on the energy band diagram.

However, the most useful property of EF is in terms of a change in EF.

Any change ∆EF across a material system represents electrical work input or output per electron.

For a semiconductor system in equilibrium, in the dark, and with no applied voltage or no emf generated, ∆EF = 0 and EF must be uniform across the system.

For readers familiar with thermodynamics, its rigorous definition is that EF is the

chemical potential of the electron, that is Gibbs free energy per electron. The

definition of EF anove is in terms of a change in EF.

∆EF = eV


JF, SCO 1617

Extrinsic Semiconductors



16-Mar-2017


Extrinsic Semiconductors: n-Type

(a) The four valence electrons of As allow it to bond just like Si but the fifth electron is left orbiting the As site. The energy required to release to free fifth-electron into the CB is very small. (b) Energy band diagram for an n-type Si doped with 1 ppm As. There are

donor energy levels just below Ec around As+ sites.



edhd

ied eN

N

neeN µµµσ ≈

+=

2

Nd >> ni, then at room temperature, the electron concentration in the CB will nearly be equal to Nd, i.e. n≈ Nd

A small fraction of the large number of electrons in the CB recombine with holes in the VB so as to maintain np= ni

2

n = Nd and p = ni2/Nd

np= ni2


Extrinsic Semiconductors: p-Type

(a) Boron doped Si crystal. B has only three valence electrons. When it substitutes for a Si atom one of its bonds has an electron missing and therefore a hole. (b) Energy band

diagram for a p-type Si doped with 1 ppm B. There are acceptor energy levels just above Ev around B− sites. These acceptor levels accept electrons from the VB and

therefore create holes in the VB.



haea

iha eN

N

neeN µµµσ ≈

+=

2

Na >> ni, then at room temperature, the hole concentration in the VB will nearly be equal to Na, i.e. p≈ Nd

A small fraction of the large number of holes in the VB recombine with electrons in the CB so as to maintain np= ni

2

p = Na and n = ni2/Na

np= ni2


JF, SCO 1617

Intrinsic, i-Sin = p = ni

Semiconductor energy band diagrams

n-typen = Nd

p = ni2/Nd

np = ni2

p-typep = Na

n = ni2/Na

np = ni2


JF, SCO 1617

Semiconductor energy band diagrams

Energy band diagrams for (a) intrinsic (b) n-type and (c) p-type semiconductors. In all cases, np= ni

2. Note that donor and acceptor energy levels are not shown. CB = Conduction band, VB = Valence band, Ec = CB

edge, Ev = VB edge, EFi = Fermi level in intrinsic semiconductor, EFn = Fermi level in n-type semiconductor, EFp = Fermi level in p-type semiconductor. χ is the electron affinity. Φ, Φn and Φp are the work functions for the intrinsic, n-

type and p-type semiconductors


JF, SCO 1617

Semiconductor Properties


JF, SCO 1617 Compensation DopingCompensation doping describes the doping of a semiconductor with both donors and acceptors to control the properties.

Example: A p-type semiconductor doped with Na acceptors can be converted to an n-type semiconductor by simply adding donors until the concentration Nd exceeds Na.

The effect of donors compensates for the effect of acceptors.

The electron concentration n = Nd − Na > ni

When both acceptors and donors are present, electrons from donors recombine with the holes from the acceptors so that the mass action law np = ni

2 is obeyed.

We cannot simultaneously increase the electron and hole concentrations because that leads to an increase in the recombination rate which returns the electron and hole concentrations to values that satisfy np= ni

2.

n = Nd − Na

p = ni2/(Nd − Na)

Nd > Na


Summary of Compensation Doping

ad NNn −=ad

ii

NN

n

n

np

−==

22

iad nNN >>−More donors than acceptors

More acceptors than donorsida nNN >>−

da NNp −=da

ii

NN

n

p

nn

−==

22


JF, SCO 1617

Degenerate Semiconductors



16-Mar-2017


JF, SCO 1617 Degenerate Semiconductors

(a) Degenerate n-type semiconductor. Large number of donors form a band that overlaps the CB. Ec is pushed down and EFn is within the CB. (b) Degenerate p-type semiconductor.


JF, SCO 1617

Energy bands under an applied electric field



16-Mar-2017


JF, SCO 1617

Energy bands bend in an applied fieldEnergy band diagram of an n-typesemiconductor connected to a voltagesupply of V volts.

The whole energy diagram tilts becausethe electron now also has an electrostaticpotential energy.


JF, SCO 1617Example: Fermi levels in semiconductors

An n-type Si wafer has been doped uniformly with 1016 phosphorus (P) atoms cm–3. Calculate the position of the Fermi energy with respect to the Fermi energy EFi in intrinsic Si. The above n-type Si sample is further doped with 2×1017 boron atoms cm–3. Calculate position of the Fermi energy with respect to the Fermi energy EFi in intrinsic Si at room temperature (300 K), and hence with respect to the Fermi energy in the n-type case above.

Solution

P (Group V) gives n-type doping with Nd = 1016 cm–3, and since Nd >> ni ( = 1010

cm–3 from Table 3.1), we haven = Nd = 1016 cm–3. For intrinsic Si,

ni = Ncexp[−(Ec − EFi)/kBT]

whereas for doped Si,

n = Ncexp[−(Ec − EFn)/kBT] = Nd

where EFi and EFn are the Fermi energies in the intrinsic andn-type Si. Dividing the two expressions

Nd /ni = exp[(EFn − EFi)/kBT]

so that EFn − EFi = kBTln(Nd/ni) = (0.0259 eV) ln(1016/1010) = 0.358 eV


JF, SCO 1617Example: Fermi levels in semiconductors

Solution (Continued)

When the wafer is further doped with boron, the acceptor concentration, Na = 2×1017 cm–3 > Nd = 1016 cm–3. The semiconductor is compensation doped and compensation converts the semiconductor to a p-type Si. Thus,

p = Na − Nd = 2×1017−1016 = 1.9×1017 cm–3.

For intrinsic Si,

p = ni = Nvexp[−(EFi − Ev)/kBT],

whereas for doped Si,

p = Nvexp[−(EFp − Ev)/kBT] = Na − Nd

where EFi and EFp are the Fermi energies in the intrinsic and p–type Si respectively Dividing the two expressions,

p/ni = exp[−(EFp − EFi)/kBT]

so that

EFp − EFi = −kBT ln(p/ni)

= −(0.0259 eV)ln(1.9×1017/1.0×1010) = −0.434 eV


JF, SCO 1617Example: Conductivity of n-Si

Consider a pure intrinsic Si crystal. What would be its intrinsic conductivity at 300K? What is the electron and hole concentrations in an n-type Si crystal that has been doped with 1016 cm–3

phosphorus (P) donors. What is the conductivity if the drift mobility of electrons is about 1200 cm2

V-1 s-1 at this concentration of dopants.

Solution

The intrinsic concentration ni = 1×1010 cm-3, so that the intrinsic conductivity is

σ = eni(µe + µh) = (1.6×10-19 C)( 1×1010 cm-3)(1450 + 490 cm2 V-1 s-1)

= 3.1×10-6 Ω-1 cm-1 or 3.1×10-4 Ω-1 m-1

Consider n-type Si. Nd = 1016 cm-3 > ni (= 1010 cm-3), the electron concentration n = Nd = 1016 cm-3 and

p = ni2/Nd = (1010 cm-3)2/(1016 cm-3) = 104 cm-3

and negligible compared to n.

The conductivity is 1111231619 cmΩ 92.1)sVcm 1200)(cm 101)(C 106.1( −−−−−− =××== edeN µσ


JF, SCO 1617

Electrons in Semiconductor Crystals

Direct and Indirect Bandgap Semiconductors



16-Mar-2017


JF, SCO 1617

The electron PE, V(x), inside the crystal is periodic with the same periodicity as that of theCrystal, a. Far away outside the crystal, by choice, V = 0 (the electron is free and PE= 0).



JF, SCO 1617


0)]([2

22

2

=−+ ψψxVE

m

dx

d e

h

V(x) = V(x + ma) ; m = 1, 2, 3…

Periodic Potential Energy

ψk(x) = Uk(x)exp(jkx)Bloch electron wavefunction

Electron wave vectorQuantized


JF, SCO 1617


Periodic Potential Energy

ψk(x) = Uk(x)exp(jkx)

k is quantized. Each k represents a particular ψk

Electron’s crystal momentum = hk

Each k and ψk have an energy Ek

×exp(−jωt)

Ψk(x,t)

External force = d(hk) / dt


JF, SCO 1617


Periodic Potential Energy in 3 D

k is quantized. Each k represents a particular ψk

V(r) = V(r + mT) ; m = 1, 2, 3…

Ek vshk

Crystal structure is very important

ψk(x) = Uk(r)exp(jk⋅r)

Crystal momentum


JF, SCO 1617

The E-k diagram of a direct bandgap semiconductor such as GaAs. The E-k curve consists of many discrete points with each point corresponding to a possible state, wavefunction ψk(x), that is

allowed to exist in the crystal. The points are so close that we normally draw the E-k relationship as a continuous curve. In the energy range Ev to Ec there are no points, i.e. no ψk(x) solutions

E vs. k Diagrams


JF, SCO 1617

Direct and Indirect Bandgap Semiconductors



16-Mar-2017


JF, SCO 1617

(a) In GaAs the minimum of the CB is directly above the maximum of the VB. GaAs is therefore a direct bandgap semiconductor. (b) In Si, the minimum of the CB is displaced

from the maximum of the VB and Si is an indirect bandgap semiconductor. (c) Recombination of an electron and a hole in Si involves a recombination center .

E vs. k Diagrams and Direct and Indirect

Bandgap Semiconductors


JF, SCO 1617

pn Junctions



16-Mar-2017


JF, SCO 1617

Properties of the pn junction

(a) The p- and n- sides of the pn junction before the contact.

(b) The pn junction after contact, in equilibrium and in open circuit.

(c) Carrier concentrations along the whole device, through the pn junction. At all points, npoppo = nnopno = ni

2.

(d) Net space charge density ρnet across the pn junction.

pn Junction


JF, SCO 1617

(e) The electric field across the pn junction is found by integrating ρnet in (d).

(f) The potential V(x) across the device. Contact potentials are not shown at the semiconductor-metal contacts now shown.

(g) Hole and electron potential energy (PE) across the pn junction. Potential energy is charge potential = q V

pn Junction


Ideal pn Junction

Depletion Widths

ndpa WNWN =

ερ )(net x

dx

d =E

Field (E) and net space charge density

Field in depletion region

∫−=x

Wp

dxxx )(1

)( netρε

E

Acceptor concentration Donor concentration

Net space charge density

Permittivity of the medium

Electric Field


Ideal pn JunctionBuilt-in field

εnd

o

WeN−=E

=

2ln

i

dao n

NN

e

kTV

Built-in voltage

Depletion region width

( ) 2/12

+=da

odao NeN

VNNW

ε

ε = εo εr

where Wo = Wn+ Wp is the total width of the depletion region under a zero applied voltage

ni is the intrinsic concentration


Law of the JunctionApply Boltzmann Statistics (can only be used with nondegenerate semiconductors)

N1 = ppo N2 = pno

E1 = PE2 = 0 E2 = PE1 = eVo

−−=Tk

EE

N

N

B

)(exp 12

1

2

−−=Tk

eV

p

p

B

o

po

no )0(exp

−=

Tk

eV

p

p

B

o

po

no exp


Law of the Junction

Apply Boltzmann Statistics

−=

Tk

eV

p

p

B

o

po

no exp

−=

Tk

eV

n

n

B

o

no

po exp

=

=

2lnln

i

daB

no

poBo n

NN

e

Tk

p

p

e

TkV


JF, SCO 1617

Forward biased pn junction and the injection of minority carriers (a) Carrier concentration profiles across the device under forward bias. (b). The hole potential energy with and

without an applied bias. W is the width of the SCL with forward bias

Forward Biased pn Junction


Forward Bias: Apply Boltzmann Statistics

Note: pn(0) is the hole concentration just outside the depletion region on the n-side

N1 = ppo

N2 = pn(0)

−−=Tk

EE

N

N

B

)(exp 12

1

2

−=

−−=Tk

eV

Tk

eV

Tk

VVe

p

p

BB

o

B

o

po

n expexp)(

exp)0(

pno/ppo

=

Tk

eVpp

Bnon exp)0(


Forward BiasLaw of the Junction: Minority Carrier Concentrations and Voltage

( )

( )

=

=

Tk

eVnn

Tk

eVpp

Bpop

Bnon

exp0

exp0

np(0) is the electron concentration just outside the depletion region on the p-side

pn(0) is the hole concentration just outside the depletion region on the n-side


Forward Bias: Minority Carrier Profile

hnn L

xx'px'p

′∆−=∆ δδ )()(

AssumptionsNo field in the neutral region (pure diffusion)Neutral region is very long

1/Lh = Mean probability of recombination per unit distance; or mean distance diffused

Change in excess hole concentration due to recombination over a small distance δx′would be

−∆=∆

hnn L

x'px'p exp)0()(


Forward Bias: Diffusion Current

'

)(hole, dx

x'pdeD

dx'

(x')dpeDJ n

hn

hD

∆−=−=

−∆

=

hn

h

hD L

x'p

L

eDJ exp)0(hole,

−

= 1exp

2

hole, Tk

eV

NL

neDJ

Bdh

ihD



'

)(hole, dx

x'pdeD

dx'

(x')dpeDJ n

hn

hD

∆−=−=

−∆

=

hn

h

hD L

x'p

L

eDJ exp)0(hole,

−

= 1exp

2

hole, Tk

eV

NL

neDJ

Bdh

ihD


JF, SCO 1617

The total current anywhere in the device is constant. Just outside the depletion region it is due to the diffusion of minority carriers.

Forward Bias: Total Current



Hole diffusion current in n-side in the neutral region

−

= 1exp

2

hole, Tk

eV

NL

neDJ

Bdh

ihD

There is a similar expression for the electron diffusion current density JD,elec in the p-region.



Ideal diode (Shockley) equation

−

= 1exp

Tk

eVJJ

Bso

2i

ae

e

dh

hso n

NL

eD

NL

eDJ

+

=

Reverse saturation current



−

+= 1exp2

Tk

eVn

NL

eD

NL

eDJ

Bi

ae

e

dh

h

Intrinsic concentration

−=

Tk

ENNn

B

gvci exp2

Jso depends strongly on the material (e.g.bandgap) and temperature

ni depends strongly on the material (e.g. bandgap) and temperature

Ideal diode (Shockley) equation


JF, SCO 1617

Schematic sketch of the I−V characteristics of Ge, Si, and GaAs pn junctions.

V

I

0.2 0.6 1.0

Ge Si GaAs

Ge, Si and GaAs Diodes: Typical Characteristics

1 mA


Forward Bias: Diffusion CurrentShort Diode

ln and lp represent the lengths of the neutral n- and p-regions outside the depletion region.



Short Diode

−

+= 1exp2

Tk

eVn

Nl

eD

Nl

eDJ

Bi

ap

e

dn

h

ln and lp represent the lengths of the neutral n- and p-regions outside the depletion region.


JF, SCO 1617

Forward Bias: Recombination Current

Forward biased pn junction, the injection of carriers and their recombination in the SCL. Recombination of electrons and holes in the depletion region involves the current

supplying the carriers.


JF, SCO 1617


A symmetricalpn junction for calculating the recombination current.


JF, SCO 1617


he

eBCDeABCJ

ττ+≈recom

−−=Tk

VVe

p

p

B

o

po

M

2

)(exp

=

Tk

eVnp

BiM 2exp

+≈

Tk

eVWWenJ

Bh

n

e

pi

2exp

2recom ττ


Forward Bias: Recombination and Total Current

Recombination Current

( ) ]12/[exprecom −= TkeVJJ Bro

+

=

Tk

eVJ

Tk

eVJJ

Bro

Bso 2

expexp

Total diode current = diffusion + recombination

e

TkV B>

The diode equation

+=

h

n

e

piro

WWenJ

ττ2 where

−

= 1exp

Tk

eVJJ

Bo η

where Jo is a new constant and is an ideality factor

General diode equation


JF, SCO 1617

Typical I-V characteristics of Ge, Si and GaAs pn junctions as log(I) vs. V. The slope indicates e/(kBT)

Typical I-V characteristics of Ge, Si and GaAs pn junctions


JF, SCO 1617

Reverse biased pn junction. (a) Minority carrier profiles and the origin of the reverse

current. (b) Hole PE across the junction under reverse bias

Reverse Bias


Reverse Bias

Total Reverse Current

g

ii

ae

e

dh

h eWnn

NL

eD

NL

eDJ

τ+

+= 2

rev

Mean thermal generation time

Diffusion current in neutral regionsthe Shockley reverse current

Thermal generation in depletion region


JF, SCO 1617

Left: Reverse I-V characteristics of a pn junction (the positive and negative current axes have different scales). Right: Reverse diode current in a Ge pn junction as a function of temperature in a ln(Irev) vs

1/T plot. Above 238 K, Irev is controlled by ni2 and below 238 K it is controlled by ni. The vertical axis

is a logarithmic scale with actual current values. (From D. Scansen and S.O. Kasap, Cnd. J. Physics. 70, 1070-1075, 1992.)

Reverse Current


JF, SCO 1617

(a) Depletion region has negative (−Q) charges in Wp and positive (+Q) charges in Wp, which are separated as in a capacitor. Under a reverse bias Vr, the charge on the n-side is +Q. When the reverse bias is increased by δVr, this charge increases by δQ. (b) Cdepvs voltage across an abrupt pn junction

for 3 different set of dopant concentrations. (Note that the vertical scale is logarithmic.)

Depletion Capacitance: Cdep


JF, SCO 1617

(a) The forward voltage across a pn junction increases by δV, which leads to further minority carrier injection and a larger forward current, which is increases by δI. Additional minority carrier charge δQis injected into the n-side. The increase δQ in charge stored in the n-side with δV appears as a though

there is a capacitance across the diode. (b) The increase δV results in an increase δI in the diode current. The dynamic or incremental resistance rd = δV/δI. (c) A simplified equivalent circuit for a

forward biased pn junction for small signals.

Dynamic Resistance and Capacitance


JF, SCO 1617

Forward Bias Dynamic Resistance and Capacitance

I = Ioexp(eV/kBT)

I

V

dI

dVrd

th== Vth = kBT/e Dynamic resistance

dV

dQC =diff

Diffusion capacitance

I = Q/τh

d

hh

rV

IC

ττ ==th

diff

pn junction current



JF, SCO 1617

Forward Bias Diffusion Capacitance

dV

dQC =diff


d

hh

rV

IC

22 thdiff

ττ ==

d

tt

rV

IC

ττ ==th

diff

Diffusion capacitance, long

diode (ac)

Diffusion capacitance, short diode (ac)

Minority carrier

transit time

τt = ln2 / 2Dh


JF, SCO 1617

Depletion Capacitance: Cdep

dV

dQ

V

QC ==

δδ

dep

2/1

2/1dep )(2

)(

)(

+−==

da

da

o NN

NNe

VV

A

W

AC

εε

Definition

2/1

dep )(2

−=

VV

NeAC

o

dε

Depletion layer capacitance for p+n


EXAMPLE : A direct bandgap pn junction

Consider a symmetrical GaAs pn junction which has the following properties. Na (p-side doping) = Nd (n-side doping) = 1017 cm- 3(or 1023 m- 3), direct recombination coefficientB ≈2×10-16 m3 s-1, cross sectional area A = 1 mm2. Suppose that the forward voltage across the diode V = 0.80 V. What is the diode current due to minority carrier diffusion at 27 °C (300 K)assuming direct recombination. If the mean minority carrier lifetime in the depletion region were to be the same as this lifetime, what would be the recombination component of the diode current? What are the two contributions at V = 1.05 V ? What is your conclusion?Note that at 300 K, GaAs has an intrinsic concentration (ni) of 2.1× 106 cm-3 and a relative permittivity (εr) of 13.0 The hole drift mobility (µh) in the n-side is 250 cm2 V-1 s-1 and electron drift mobility (µe) in the p-side is 5000 cm2 V-1 s-1 (these are at the doping levels given).


EXAMPLE : A direct bandgap pn junction

Solution

Assuming weak injection, we can calculate the recombination times τe and τh for electrons and holes recombining in the neutral p and n-regions respectively. Using S.I. units throughout, and taking kBT/e= 0.02585 V, and since this is a symmetric device,

To find the Shockley current we need the diffusion coefficients and lengths. The Einstein relation gives the diffusion coefficients as

Dh = µhkBT/e= (250×10-4) (0.02585)= 6.46×10-4 m2 s-1,

De = µekBT/e= (5000×10-4) (0.02585) = 1.29×10-2 m2 s-1.

s1000.5)m 10)(1sm 100.2(

11 83231316

−−−− ×=

××=≈=

aeh BN

ττ


Solution (continued)

where kBT /ewas taken as 0.02585 V. The diffusion lengths are easily calculated as Lh = (Dhτh)1/2 = [(6.46×10-4 m2 s-1)(50.0×10-9 s)]1/2 = 5.69×10-6 m,Le = (Deτe) 1/2 = [(1.29×10-2 m2 s-1)(50.0×10-9s)]1/2 = 2.54×10-5 m.Notice that the electrons diffuse much further in the p-side. The reverse

saturation current due to diffusion in the neutral regions is

A 104.4

)101.2)(106.1()10)(1054.2(

1029.1

)10)(1069.5(

1046.6)10(

21

21219235

2

236

46

2

−

−−

−

−

−−

×≈

××

××+

××=

+= i

ae

e

dh

hso en

NL

D

NL

DAI

µA 0.12orA 102.1V 02585.0

V 80.0exp)A 104.4(

exp

721

diff

×=

×=

=

−−

Tk

eVII

Bso

Thus, the forward diffusion current at V = 0.80 V is



Recombination component of the current is quite difficult to calculate because we need to know the mean electron and hole recombination times in the SCL. Suppose that, as a first order, we assume that these recombination times are as above.

The built-in voltage Vo is

Depletion layer width W is

As this is a symmetric diode, Wp = Wn = W /2. The pre-exponential Iro is

23 23

2 12 2

10 10ln (0.02585) ln 1.27 V

(2.1 10 )a dB

oi

N Nk TV

e n

= = = ×

µm. 0.116or m, 101.16

)m )(10m C)(10 10(1.6

V) 80.0)(1.27m 10)(10m F 102(13)(8.85

))((2

7

2/1

32332319

32323112

2/1

−

−−−

−−−

×=

×−+×=

−+=da

oda

NeN

VVNNW

ε

2 2pi n i

roe h e

WAen W Aen WI

τ τ τ

= + =



6 19 12 713

8

(10 )(1.6 10 )(2.1 10 ) 1.16 103.9 10 A

2 5.00 10

− − −−

−

× × ×= ≈ × ×

so that at V = 0.8 V,

The recombination current is more than an order of magnitude greater than the diffusion current. If we repeat the calculation for a voltage of 1.05 V across the device, then we would find Idiff = 1.9 mA and Irecom= 0.18 mA, whereIdiff

dominates the current. Thus, as the voltage increases across a GaAs pn junction, the ideality factor η is initially 2 but then becomes 1 as showm in Figure 3.20.

The EHP recombination that occurs in the SCL and the neutral regions in this GaAs pn junction case would result in photon emission, with a photon energy that is approximately Eg. This direct recombination of injected minority carriers and the resulting emission of photons represent the principle of operation of the Light Emitting Diode (LED).

µA 2.1orA 100.2)V 02585.0(2

V .80exp)A 109.3(

2exp

613

recom

−− ×=

×≈

≈

Tk

eVII

Bro


JF, SCO 1617

LEFT: Considerp- andn-type semiconductor (same material) before the formation of thepn junction,, separated from each and not interacting. RIGHT: After the formation of thepn junction, there is a built-in voltage across the junction.


JF, SCO 1617

Energy band diagrams for a pn junction under (a) open circuit and (b) forward bias


JF, SCO 1617

Energy band diagrams for a pn junction under reverse biasLEFT: Shockley model

RIGHT: Thermal generation in the depletion region


JF, SCO 1617

Light-emitting diodes I



21-Mar-2017

kasappp3 1617 jf - w3.ualg.ptw3.ualg.pt/~jlongras/aulas/sco_1617_12_160317.pdf · from: s.o. kasap,...

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