semiconductor device modeling and characterization ee5342, lecture 7-spring 2004

L7 February 10 1

Semiconductor Device Modeling and CharacterizationEE5342, Lecture 7-Spring 2004

Professor Ronald L. Carterronc@uta.edu

http://www.uta.edu/ronc/

L7 February 10 2

MidTerm andProject Tests• MidTerm on Thursday 2/12

– Cover sheet to be posted at http://www.uta.edu/ronc/5342/tests/

• Project 1 draft assignment will be posted 2/13.– Project report to be used in doing:– Project 1 Test on Thursday 3/11– Cover sheet will be posted as above

L7 February 10 3

Ideal diodeequation• Assumptions:

– low-level injection– Maxwell Boltzman statistics– Depletion approximation– Neglect gen/rec effects in DR– Steady-state solution only

• Current dens, Jx = Js expd(Va/Vt)

– where expd(x) = [exp(x) -1]

L7 February 10 4

Ideal diodeequation (cont.)• Js = Js,p + Js,n = hole curr + ele curr

Js,p = qni2Dp coth(Wn/Lp)/(NdLp) =

qni2Dp/(NdWn), Wn << Lp, “short” =

qni2Dp/(NdLp), Wn >> Lp, “long”

Js,n = qni2Dn coth(Wp/Ln)/(NaLn) =

qni2Dn/(NaWp), Wp << Ln, “short” =

qni2Dn/(NaLn), Wp >> Ln, “long”

Js,n << Js,p when Na >> Nd

L7 February 10 5

Diffnt’l, one-sided diode cond. (cont.)

DQ

t

dQd

QDDQt

DQQd

tat

tQs

Va

DQd

tastasD

IV

g1

Vr ,resistance diode The

. VII where ,V

IVg then

, VV If . V

VVexpI

dV

dIVg

VVdexpIVVdexpAJJAI

Q

L7 February 10 6

Cap. of a (1-sided) short diode (cont.)

p

x

x p

ntransitQQ

transitt

DQ

pt

DQQ

taaa

a

Ddx

Jp

qVV

V

I

DV

IV

VVddVdV

dVA

nc

n2W

Cr So,

. 2W

C ,V V When

exp2

WqApd2

)W(xpqAd

dQC Define area. diode A ,Q'Q

2n

dd

2n

dta

nn0nnn

pdpp

L7 February 10 7

General time-constant

np

a

nnnn

a

pppp

pnVa

pn

Va

DQd

CCC ecapacitanc diode total

the and ,dVdQ

Cg and ,dV

dQCg

that so time sticcharacteri a always is There

ggdV

JJdA

dVdI

Vg

econductanc the short, or long diodes, all For

QQ

L7 February 10 8

General time-constant (cont.)

times.-life carr. min. respective the

, and side, diode long

the For times. transit charge physical

the ,D2

W and ,

D2W

side, diode short the For

n0np0p

n

2p

transn,np

2n

transp,p

L7 February 10 9

General time-constant (cont.)

Fdd

transitminF

gC

and 111

by given average

the is time transition effective The

sided-one usually are diodes Practical

L7 February 10 10

Effect of non-zero E in the CNR• This is usually not a factor in a short

diode, but when E is finite -> resistor• In a long diode, there is an additional

ohmic resistance (usually called the parasitic diode series resistance, Rs)

• Rs = L/(nqnA) for a p+n long diode.

• L=Wn-Lp (so the current is diode-like for Lp and the resistive otherwise).

L7 February 10 11

)pn( ,ppp and ,nnn where

kTEfiE

coshn2np

npnU

dtpd

dtnd

GRU

oo

oT

i

2i

Effect of carrierrecombination in DR• The S-R-H rate (no = po = o) is

L7 February 10 12

Effect of carrierrec. in DR (cont.)• For low Va ~ 10 Vt

• In DR, n and p are still > ni

• The net recombination rate, U, is still finite so there is net carrier recomb.– reduces the carriers available for the

ideal diode current– adds an additional current component

L7 February 10 13

eff,o

taieffavgrec

o

taimaxfpfna

fnfii

fifni

x

xeffavgrec

2V2/Vexpn

qWxqUJ

2V2/Vexpn

U ,EEqV w/

,kT/EEexpnp

and ,kT/EEexpnn cesin

xqUqUdxJ curr, ecRn

p

Effect of carrierrec. in DR (cont.)

L7 February 10 14

High level injection effects• Law of the junction remains in the same

form, [pnnn]xn=ni

2exp(Va/Vt), etc.

• However, now pn = nn become >> nno = Nd, etc.

• Consequently, the l.o.t.j. reaches the limiting form pnnn = ni

2exp(Va/Vt)

• Giving, pn(xn) = niexp(Va/(2Vt)), or np(-xp) = niexp(Va/(2Vt)),

L7 February 10 15

High level injeffects (cont.)

KFKFKFsinj lh,s

i

at

i

dtKFa

appdnn

a

tainj lh,sinj lh

VJJ ,JJJ :Note

nN

lnV2 or ,n

NlnV2VV Thus

Nx-n or ,Nxp giving

V of range the for important is This

V2/VexpJJ

:is density current injection level-High

L7 February 10 16

Summary of Va > 0 current density eqns.• Ideal diode, Jsexpd(Va/(Vt))

– ideality factor,

• Recombination, Js,recexp(Va/(2Vt))– appears in parallel with ideal term

• High-level injection, (Js*JKF)

1/2exp(Va/(2Vt))

– SPICE model by modulating ideal Js term

• Va = Vext - J*A*Rs = Vext - Idiode*Rs

L7 February 10 17

1N ,

V2NV

t

aexp~

1N ,

VNV

t

aexp~

Vext

ln(J)

data Effect of Rs

2NR ,

VNRV

t

aexp~

VKF

Plot of typical Va > 0 current density equations

Sexta RAJ-VV

KFS JJln

recsJln ,

SJln

KFJln

L7 February 10 18

Reverse bias (Va<0)=> carrier gen in DR• Va < 0 gives the net rec rate,

U = -ni/, = mean min carr g/r l.t.

NNN/NNN and

qN

VV2W where ,

2Wqn

J

(const.) U- G where ,qGdxJ

dadaeff

eff

abi

0

igen

x

xgen

n

p

L7 February 10 19

Reverse bias (Va< 0),carr gen in DR (cont.)

gens

gen

gengensrev

JJJ

JSPICE

JJJJJ

or of largest the set then ,0

V when 0 since :note model

VV where ,

current generation the plus bias negative

for current diode ideal the of value The

current the to components two are there

bias, reverse ,)0V(V for lyConsequent

a

abi

ra

L7 February 10 20

Reverse biasjunction breakdown• Avalanche breakdown

– Electric field accelerates electrons to sufficient energy to initiate multiplication of impact ionization of valence bonding electrons

– field dependence shown on next slide

• Heavily doped narrow junction will allow tunneling - see Neamen*, p. 274– Zener breakdown

L7 February 10 21

Reverse biasjunction breakdown• Assume -Va = VR >> Vbi, so Vbi-Va-->VR

• Since Emax~ 2VR/W = (2qN-VR/())1/2, and

VR = BV when Emax = Ecrit (N- is doping of

lightly doped side ~ Neff)

BV = (Ecrit )2/(2qN-)

• Remember, this is a 1-dim calculation

L7 February 10 22

Reverse biasjunction breakdown

8/3

4/3

0

4/3

2/3

20

161/

1.1/ 120 so

,161/

1.1/ 60 gives *,***

usually , 2

D.A. theand diode sided-one a Assuming

EN

EqNVE

EN

EVBVCasey

BVqN

EBV

g

Sicrit

B

g

icritSi

i

L7 February 10 23

Ecrit for reverse breakdown (M&K**)

Taken from p. 198, M&K**

Casey Model for Ecrit

L7 February 10 24

Junction curvatureeffect on breakdown• The field due to a sphere, R, with

charge, Q is Er = Q/(4r2) for (r > R)

• V(R) = Q/(4R), (V at the surface)• So, for constant potential, V, the field,

Er(R) = V/R (E field at surface increases for smaller spheres)

Note: corners of a jctn of depth xj are like 1/8 spheres of radius ~ xj

L7 February 10 25

BV for reverse breakdown (M&K**)

Taken from Figure 4.13, p. 198, M&K**

Breakdown voltage of a one-sided, plan, silicon step junction showing the effect of junction curvature.4,5

L7 February 10 26

rpc

rprj

rnrnc

Gauss’ Law

Surface r

rErdSE0

Surfacein Enclosed2 Q)(4

2

3

amax

33a2

3

qN so

,3

4qN 4

j

pjr

Surface

pr

r

rrEE

rrErdSE

L7 February 10 27

Spherical DiodeFields calculations

2

3

d2

2

max 3

qN

r

rr

r

rEE jj

r Setting Er = 0 at r = rn, we get

3

d

max

qN

31

jjn r

Err

Note that the equivalent of the lever law for this spherical diode is

33d

33a NN jnpj rrrr

For rj < ro ≤ rn,

L7 February 10 28

Spherical DiodeFields calculations

Assume Na >> Nd, so rn – rj d >> rj – rp. Further, setting the usual definition for the potential difference, and evaluating the potential difference at breakdown, we havePHIi – Va = BV and Emax = Em = Ecrit = Ec. We also define = 3eEm/qNd[cm].

njj

njjnj rr

rrr

rrr11

E11E

2

E BV 2

c3c22c

L7 February 10 29

Showing therj ∞ limit

C1. Solve for rn – rj = as a function of Emax and solve

for the value of in the limit of rj . The solution for

rn is given below.

theorem.binomial apply the limit, thegwhen takin

11 so

,qN

E3 , 1

1/3

,0d

crit

1/3

jjjn

Sirj

jn

rrrr

rrr

.

L7 February 10 30

Solving for theBreakdown (BV)

Solve for BV = [i – Va]Emax = Ecrit,

and solve for the value of BV in the limit of rj . The solution for BV is given

below.

L7 February 10 31

Spherical diodeBreakdown Voltage

1.0

10.0

100.0

1.00E+14 1.00E+15 1.00E+16 1.00E+17

Substrate Concentration (cm^-3)

Bre

ak

do

wn

Vo

lta

ge

(V

olt

)

rj = 0.1 micron

rj = 0.2 micron

rj = 0.5 micron

rj = 1.0 micron

L7 February 10 32

Example calculations• Assume throughout that p+n jctn with Na

= 3e19cm-3 and Nd = 1e17cm-3

• From graph of Pierret mobility model, p

= 331 cm2/V-sec and Dp = Vtp = ? • Why p and Dp?

• Neff = ?

• Vbi = ?

L7 February 10 33

0

500

1000

1500

1.E+13 1.E+14 1.E+15 1.E+16 1.E+17 1.E+18 1.E+19 1.E+20

Doping Concentration (cm̂ - 3)

Mob

ility

(cm̂

2/V

-se

c)P As B n(Pierret) p(Pierret)

L7 February 10 34

Parameters forexamples• Get min from the model used in Project

2 min = (45 sec) 1+(7.7E-18cm3Ni+(4.5E-36cm6Ni

2

• For Nd = 1E17cm3, p = 25 sec

– Why Nd and p ?

• Lp = ?

L7 February 10 35

Hole lifetimes, taken from Shur***, p. 101.

L7 February 10 36

Example

• Js,long, = ?

• If xnc, = 2 micron, Js,short, = ?

L7 February 10 37

Example(cont.)• Estimate VKF

• Estimate IKF

L7 February 10 38

Example(cont.)• Estimate Js,rec

• Estimate Rs if xnc is 100 micron

L7 February 10 39

Example(cont.)• Estimate Jgen for 10 V reverse bias

• Estimate BV

L7 February 10 40

Diode equivalentcircuit (small sig)

ID

VDVQ

IQ

t

Q

dd

VD

D

V

I

r1

gdVdI

Q

is the practical

“ideality factor”

Q

tdiff

t

Qdiffusion

mintrdd

IV

r , V

IC

long) for short, for ( , Cr

L7 February 10 41

Small-signal eqcircuit

CdiffCdep

l

rdiff

Cdiff and

Cdepl are both charged by

Va = VQQa

2/1

bi

ajojdepl VV ,

VV

1CCC

Va

L7 February 10 42

Diode Switching

• Consider the charging and discharging of a Pn diode – (Na > Nd)

– Wd << Lp

– For t < 0, apply the Thevenin pair VF and RF, so that in steady state • IF = (VF - Va)/RF, VF >> Va , so current source

– For t > 0, apply VR and RR

• IR = (VR + Va)/RR, VR >> Va, so current source

L7 February 10 43

Diode switching(cont.)

+

+ VF

VR

DRR

RF

Sw

R: t > 0

F: t < 0

ItI s

F

FF R

VI0tI

VF,VR >>

Va

F

F

F

aFQ R

VR

VVI

0,t for

L7 February 10 44

Diode chargefor t < 0

xn xncx

pn

pno

Dp2W

,IWV,xqp'Q

2N

TR

TRFnFnndiff,p

D

2i

noV/V

noFn Nn

p ,epV,xp tF

dxdp

qDJ since ,qAD

Idxdp

ppp

F

L7 February 10 45

Diode charge fort >>> 0 (long times)

xn xncx

pn

pno

tF V/Vnon ep0t,xp

t,xp

sppp

S Jdxdp

qDJ since ,qADI

dxdp

L7 February 10 46

Equationsummary

Q discharge to flows

R/VI current, a 0, but small, t For

RV

I ,qAD

Idxdp

AJI ,AqD

I

JqD1

dxdp

RRR

F

FF

p

F

0t,F

ssp

s

,ppt,R

L7 February 10 47

Snapshot for tbarely > 0

xn xncx

pn

pno

p

F

qADI

dxdp

p

RqAD

Idxdp

tF V/Vnon ep0t,xp

0t,xp Total charge removed, Qdis=IRt

st,xp

L7 February 10 48

I(t) for diodeswitching

ID

t

IF

-IR

ts ts+trr

- 0.1 IR

sRdischarge

p

Rs

tIQ

constant, a is qAD

Idxdp

,tt 0 For

pnp

p2is L/WtanhL

DqnI

L7 February 10 49

References

* Semiconductor Physics and Devices, 2nd ed., by Neamen, Irwin, Boston, 1997.

**Device Electronics for Integrated Circuits, 2nd ed., by Muller and Kamins, John Wiley, New York, 1986.

***Physics of Semiconductor Devices, Shur, Prentice-Hall, 1990.