EE 5340 Semiconductor Device Theory Lecture 4 - Fall 2003

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L 04 Sept 04 1 EE 5340 Semiconductor Device Theory Lecture 4 - Fall 2003 Professor Ronald L. Carter [email protected] http://www.uta.edu/ronc

description

EE 5340 Semiconductor Device Theory Lecture 4 - Fall 2003. Professor Ronald L. Carter [email protected] http://www.uta.edu/ronc. Web Pages. If you have not already done so: R. L. Carter’s web page www.uta.edu/ronc/ EE 5340 web page and syllabus www.uta.edu/ronc/5340/syllabus.htm - PowerPoint PPT Presentation

Transcript of EE 5340 Semiconductor Device Theory Lecture 4 - Fall 2003

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EE 5340Semiconductor Device TheoryLecture 4 - Fall 2003

Professor Ronald L. [email protected]

http://www.uta.edu/ronc

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Web Pages* If you have not already done so:• R. L. Carter’s web page

– www.uta.edu/ronc/• EE 5340 web page and syllabus

– www.uta.edu/ronc/5340/syllabus.htm• University and College Ethics Policies

– www2.uta.edu/discipline/– www.uta.edu/ronc/5340/COE_EthicsStatement_Fall02.htm

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Assignment 1: If you have not received response from me ...• Send e-mail to [email protected]

– On the subject line, put “5340 e-mail”– In the body of message include

• email address: ______________________• Your Name*: _______________________• Last four digits of your Student ID: _____

* Your name as it appears in the UTA Record - no more, no less

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Assignment 2: 930AM stu-dents only*• IF you have a class OTHER than 5340

at 8 AM, e-mail to [email protected]– Subject line: “5340 8 AM Class”– In the body of message include

• email address: ______________________• Your Enrollment Name*: ______________• Class you are taking at 8:00 AM: ________

* If you don’t do this, and don’t take Test 1 at 8:00 AM, you will not get credit for Test 1

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Classes ofsemiconductors• Intrinsic: no = po = ni, since Na&Nd << ni

=[NcNvexp(-Eg/kT)]1/2, (not easy to get)• n-type: no > po, since Nd > Na, noNd-Na

• p-type: no < po, since Nd < Na, poNa-Nd

• Compensated: no=po=ni, w/ Na- = Nd

+ > 0• Note: n-type and p-type are usually

partially compensated since there are usually some opposite- type dopants

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Equilibriumconcentrations• Charge neutrality requires q(po + Nd

+) + (-q)(no + Na

-) = 0• Assuming complete ionization, so Nd

+ = Nd and Na

- = Na • Gives two equations to be solved

simultaneously 1. Mass action, no po = ni

2, and 2. Neutrality po + Nd = no + Na

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Equilibriumconc (cont.)• For Nd > Na (taking the + root) no =

(Nd-Na)/2 + {[(Nd-Na)/2]2+ni2}1/2

• For Nd >> Na and Nd >> ni, can use the binomial expansion, giving no = Nd/2 + Nd/2[1 + 2ni

2/Nd2 + … ]

• So no = Nd, and po = ni2/Nd in the

limit of Nd >> Na and Nd >> ni

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Equilibriumconc (cont.)• For Na > Nd (taking the + root) po =

(Na-Nd)/2 + {[(Na-Nd)/2]2+ni2}1/2

• For Na >> Nd and Na >> ni, can use the binomial expansion, giving po = Na/2 + Na/2[1 + 2ni

2/Na2 + … ]

• So po = Na in the limit of Na >> Nd and Na >> ni

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Examplecalculations• For Nd = 3.2E16/cm3, ni = 1.4E10/cm3

no = Nd = 3.2E16/cm3

po = ni2/Nd , (po is always ni

2/no) = (1.4E10/cm3)2/3.2E16/cm3 = 6.125E3/cm3 (comp to ~1E23 Si)

• For po = Na = 4E17/cm3,no = ni

2/Na = (1.4E10/cm3)2/4E17/cm3 = 490/cm3

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Position of theFermi Level• Efi is the Fermi level

when no = po

• Ef shown is a Fermi level for no > po

• Ef < Efi when no < po

• Efi < (Ec + Ev)/2, which is the mid-band

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EF relative to Ec and Ev• Inverting no = Nc exp[-(Ec-EF)/kT] gives

Ec - EF = kT ln(Nc/no) For n-type material: Ec - EF =kTln(Nc/Nd)=kTln[(Ncpo)/ni

2]• Inverting po = Nv exp[-(EF-Ev)/kT]

gives EF - Ev = kT ln(Nv/po) For p-type material: EF - Ev = kT ln(Nv/Na)

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EF relative to Efi• Letting ni = no gives Ef = Efi ni = Nc

exp[-(Ec-Efi)/kT], so Ec - Efi = kT ln(Nc/ni). Thus EF - Efi = kT ln(no/ni) and for n-type EF - Efi = kT ln(Nd/ni)

• Likewise Efi - EF = kT ln(po/ni) and for p-type Efi - EF = kT ln(Na/ni)

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Locating Efi in the bandgap • Since Ec - Efi = kT ln(Nc/ni), and Efi

- Ev = kT ln(Nv/ni) • The 1st equation minus the 2nd gives

Efi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv)• Since Nc = 2.8E19cm-3 > 1.04E19cm-3

= Nv, the intrinsic Fermi level lies below the middle of the band gap

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Samplecalculations• Efi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv), so at

300K, kT = 25.86 meV and Nc/Nv = 2.8/1.04, Efi is 12.8 meV or 1.1% below mid-band

• For Nd = 3E17cm-3, given that Ec - EF = kT ln(Nc/Nd), we have Ec - EF = 25.86 meV ln(280/3), Ec - EF = 0.117 eV =117meV ~3x(Ec - ED) what Nd gives Ec-EF =Ec/3

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Equilibrium electronconc. and energies

ov

2i

vof

io

fiffif

io

cocf

cfco

pNlnkT

nNnlnkTEvE and

;nnlnkTEE or ,kT

EEexpnn

;NnlnkTEE or ,kT

EEexpNn

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Equilibrium hole conc. and energies

oc

2i

cofc

io

ffiffi

io

vo

fvfv

vo

nNlnkT

nNplnkTEE and

;nplnkTEE or ,kT

EEexpnp

;NplnkTEE or ,kT

EEexpNp

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Carrier Mobility• In an electric field, Ex, the velocity

(since ax = Fx/m* = qEx/m*) is vx = axt = (qEx/m*)t, and the displ

x = (qEx/m*)t2/2• If every coll, a collision occurs which

“resets” the velocity to <vx(coll)> = 0, then <vx> = qExcoll/m* = Ex

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Carrier mobility (cont.)• The response function is the

mobility.• The mean time between collisions,

coll, may has several important causal events: Thermal vibrations, donor- or acceptor-like traps and lattice imperfections to name a few.

• Hence thermal = qthermal/m*, etc.

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Carrier mobility (cont.)• If the rate of a single contribution

to the scattering is 1/i, then the total scattering rate, 1/coll is

all

collisions itotal

all

collisions icoll

11by given is mobility total

the and , 11

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Drift Current• The drift current density (amp/cm2) is

given by the point form of Ohm LawJ = (nqn+pqp)(Exi+ Eyj+ Ezk), so

J = (n + p)E = E, where = nqn+pqp defines the conductivity• The net current is

SdJI

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Net silicon extrresistivity (cont.)• Since = (nqn + pqp)-1, and n > p, ( = q/m*) we have

p > n

• Note that since1.6(high conc.) < p/n < 3(low conc.), so1.6(high conc.) < n/p < 3(low conc.)