This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 6.012  Electronic Devices and Circuits Lecture 3  Solving The Five Equations  Outline • Announcements
Handouts  1. Lecture; 2. Photoconductivity; 3. Solving the 5 eqs. • Review See website for Items 2 and 3. 5 unknowns: n(x,t), p(x,t), Je(x,t), Jh(x,t), E(x,t)
5 equations: Gauss's law (1), Currents (2), Continuity (2)
What isn't covered: Thermoelectric effects; Peltier cooling • Special cases we can solve (approximately) by hand Carrier concentrations in uniformly doped material
Uniform electric field in uniform material (drift)
Low level uniform optical injection (LLI, τmin)
Photoconductivity
Doping profile problems (depletion approximation)
Nonuniform injection (QNR diffusion/flow) (Lect. 1)
(Lect. 1)
(Lect. 2)
(Lect. 2)
(Lects. 3,4)
(Lect. 5) • Doping profile problems
Electrostatic potential
Poisson's equation
Clif Fonstad, 9/17/09 Lecture 3  Slide 1 Nonuniform doping/excitation: Summary
What we have so far:
Five things we care about (i.e. want to know):
Hole and electron concentrations:
Hole and electron currents:
Electric field: p( x, t ) and n ( x, t )
J hx ( x, t ) and J ex ( x, t )
E x ( x, t ) And, amazingly, we already have five equations relating them: "p( x, t ) 1 "J h ( x, t )
+
= G # R $ Gext ( x, t ) # [ n ( x, t ) p( x, t ) # n i2 ] r( t )
"t ! q "x
"n( x, t ) 1 "J e ( x, t )
Electron continuity: #
= G # R $ Gext ( x, t ) # [ n ( x, t ) p( x, t ) # n i2 ] r( t )
"t
q "x
"p( x, t )
J h ( x, t ) = qµ h p( x, t ) E ( x, t ) # qDh
Hole current density: "x
"n ( x , t )
Electron current density: J e ( x, t ) = qµ e n ( x, t ) E ( x, t ) + qDe
"x
" [&( x ) E x ( x, t )]
Charge conservation: %( x, t ) =
$ q[ p( x, t ) # n ( x, t ) + N d ( x ) # N a ( x )]
"x
Hole continuity: So...we're all set, right? No, and yes..... Clif Fonstad, 9/17/09 We'll see today that it isn't easy to get a general solution, but we can prevail. Lecture 3  Slide 2 ! Thermoelectric effects*  the Seebeck and Peltier effects
(current fluxes caused by temperature gradients, and visa versa) Hole current density, isothermal conditions:
Drift Diffusion $ d [ q# ] '
$ dp '
J h = µ h p &"
) + qDh & " )
% dx (
% dx (
Hole potential
energy gradient Concentration
gradient Hole current density, nonisothermal conditions: ! Drift Diffusion Seebeck Effect $ d [q# ] '
$ dp '
$ dT '
J h = µ h p &"
) + q Dh & " ) + q Sh p&" )
% dx (
% dx (
% dx (
Temperature
gradient Seebeck Effect:
! eltier Effect:
P
Clif Fonstad, 9/17/09 temperature gradient → current Generator current → temperature gradient Cooler/heater * A cultural item; we will only consider isothermal
situations on 6.012 exams and problem sets. Lecture 3  Slide 3 Thermoelectric effects  the Seebeck and Peltier effects
(current fluxes caused by temperature gradients, and visa versa) Two examples:
Right  The hot point probe, an apparatus for determining the carrier type of semiconductor samples. Below  A thermoelectric array like those in thermoelectric generators and solidstate refrigerators. Clif Fonstad, 9/17/09 Ref.: Appendix B in the course text. Lecture 3  Slide 4 Thermoelectric Generators and Coolers  modern examples
Electrical power for a trip to Pluto
Cooling/heating for the
necessities of life Image of thermoelectric wine cooler
removed due to copyright restrictions. Source: NASA. "…electrical power for the New Horizons spacecraft
and science instruments [is] provided by a single
radioisotope thermoelectric generator, or RTG." Thermoelectric Wine Cooler
28 bottles
12˚C  18˚C
Quiet, gas free, vibration free,
environmentally friendly, LED
display, interior light.
Zhongshan Candor Electric Appl. Co. Launched
1/19/2006 http://www.alibaba.com/
Clif Fonstad, 9/17/09 http://pluto.jhuapl.edu/
Source: NASA/Johns Hopkins University Applied Physics
Laboratory/Southwest Research Institute. Lecture 3  Slide 5 Nonuniform doping/excitation: Back to work
(laying the groundwork to model diodes and transistors) What we have:
Five things we care about (i.e. want to know):
Hole and electron concentrations:
Hole and electron currents:
Electric field: p( x, t ) and n ( x, t )
J hx ( x, t ) and J ex ( x, t )
E x ( x, t ) And, five equations relating them: "p( x, t ) 1 "J h ( x, t )
+
= G # R $ Gext ( x, t ) # [ n ( x, t ) p( x, t ) # n i2 ] r( t )
"t !q "x
"n( x, t ) 1 "J e ( x, t )
Electron continuity: #
= G # R $ Gext ( x, t ) # [ n ( x, t ) p( x, t ) # n i2 ] r( t )
"t
q "x
"p( x, t )
J h ( x, t ) = qµ h p( x, t ) E ( x, t ) # qDh
Hole current density: "x
"n ( x , t )
Electron current density: J e ( x, t ) = qµ e n ( x, t ) E ( x, t ) + qDe
"x
" [&( x ) E x ( x, t )]
Charge conservation: %( x, t ) =
$ q[ p( x, t ) # n ( x, t ) + N d ( x ) # N a ( x )]
"x
Hole continuity: We can get approximate analytical solutions in 5 important cases!
Clif Fonstad, 9/17/09 Lecture 3  Slide 6 ! Solving the five equations: special cases we can handle
1. Uniform doping, thermal equilibrium (nopo product, no, po): "
"
= 0,
= 0, gL ( x, t ) = 0, J e = J h = 0
"x
"t Lecture 1 2. Uniform doping and Efield (drift conduction, Ohms law): "
"
= 0,
= 0, gL ( x, t ) = 0, E x constant
"x
"t ! Lecture 1 3. Uniform doping and uniform low level optical injection (τmin):
"
= 0, gL ( t ), n ' << po
"x
! Lecture 2 3'. Uniform doping, optical injection, and Efield (photoconductivity): ! "
= 0, E x constant, gL ( t )
"x Lecture 2 4. Nonuniform doping in thermal equilibrium (junctions, interfaces) ! "
= 0, gL ( x, t ) = 0, J e = J h = 0
"t Lectures 3,4 5. Uniform doping, nonuniform LL injection (QNR diffusion) "N d "N a
"n ' "
=
= 0, n ' # p', n' << p o , J e # qDe
,
# 0 Lecture 5
"
"x
"x "t
!x
Clif Fonstad, 9/17/09
Lecture 3  Slide 7 Nonuniform material with nonuniform excitations (laying the groundwork to model diodes and transistors) Where cases 2, 4, and 5 appear in important semiconductor devices
Junction diodes, LEDs: ptype
Flo 5  Flow
Case w problem Bipolar transistors:
E B F ow r  lem
Clasep5obFlow
Junctiion problem
Case 4 Junction JCnctio4 problem
uase n  Junctions p ntype ntype ntype C F ow r  lems
Clasep5obFlow MOS transistors: G S
n+ n+ ptype Diode
Case 4  Junctionss
Clif Fonstad, 9/17/09 D Drift
Case 2  Drift Depletion approximation
Case 4  Interface (In subthrehsold: Case 5  Flow) Lecture 3  Slide 8 Case 4: Nonuniform doping in thermal equilibrium Doping Profiles and pn Junctions in TE: Na (x), Nd (x)
Any time the doping varies with position, we can no longer assume that
there is charge neutrality everywhere and that ρ(x) = 0. The dopants
are fixed, but the carriers are mobile and diffuse: no(x),
NdNa NdNa Can't say: no ( x) = N D ( x ) ρ(x) < 0
no(x) ! ρ(x) > 0
E(x) x
Electron diffusion
Electron drift
Hole drift
Hole diffusion
Clif Fonstad, 9/17/09 In T.E., the net fluxes must be zero Lecture 3  Slide 9 Nonuniform doping in thermal equilibrium, cont.
To treat nonuniformly doped materials we begin by looking at
them in thermal equilibrium, as we've said.
This is useful because in thermal equilibrium we must have: gL ( x, t ) = 0
n ( x, t ) = n o ( x )
p( x, t ) = po ( x )
J e ( x, t ) = 0
J h ( x, t ) = 0
Consequently, the 2 continuity equations in our 5 equations
reduce to 0 = 0, e.g.: "n( x, t ) 1 "J e! , t )
(x
#
= gL ( x, t ) # [ n ( x, t ) $ p( x, t ) # n o ( x ) $ po ( x )] r(T )
"t
q "x 0
Clif Fonstad, 9/17/09 ! 0 0 0
Lecture 3  Slide 10 Nonuniform doping in thermal equilibrium, cont. The third and fourth equations, the current equations,
give: dn o ( x )
0 = qµ e n o ( x ) E ( x ) + qDe
dx
dpo ( x )
0 = qµ h po ( x ) E ( x ) " qDh
dx
! "
# d# De 1 dn o ( x )
=
dx µ e n o ( x ) dx
d$
Dh 1 dpo ( x )
="
dx
µ h po ( x ) dx And Poisson’s equation becomes: dE ( x )
d 2# ( x )
q
="
= [ po ( x ) " n o ( x ) + N d ( x ) " N a ( x )]
dx
dx 2
$ ! ! In the end, we have three equations in our three remaining
unknowns, no(x), po(x), and φ(x), so all is right with the
world. Clif Fonstad, 9/17/09 Lecture 3  Slide 11 Nonuniform doping in thermal equilibrium, cont. Looking initially at the first of our new set of equations,
we note that both sides can be easily integrated with
respect to position :
x d"
D x 1 dn o ( x )
dx = e # x
dx
# xo dx
µ e o n o ( x ) dx "( x) $ " ( xo ) = De
D
n ( x)
[ln n o ( x ) $ ln n o ( x o )] = e ln o
µe
µe no ( x o ) Next, raising both sides to the e power yields: n o ( x ) = n o ( x o )e ! µe
[ " ( x )#" ( x o )]
De We chose intrinsic material as our zero reference for the
electrostatic potential: !
and arrive at : Clif Fonstad, 9/17/09 ! " ( x ) = 0 where n o ( x ) = n i n o ( x ) = n ie µe
" (x )
De
Lecture 3  Slide 12 Nonuniform doping in thermal equilibrium, cont.
From the corresponding equation for holes we also find : po ( x ) = n ie " Next use the Einstein relation: µh
# (x )
Dh µh µe
q
=
=
Dh De kT Incredibly
Multilingually
rhyming Note: this relationship rhymes as written, as well as when inverted, and also
!
either way in Spanish. It is a very fundamental, and important, relationship! Note : @ R.T. q kT " 40 V #1 and kT q " 25 mV !
Using the Einstein relation we have:
n o ( x ) = n ie q" ( x ) kT ! and po ( x ) = n ie# q" ( x ) kT Finally, putting these in Poisson’s equation, a single
equation for φ(x) in a doped semiconductor in TE
materializes: ! [ d 2" ( x )
q
= # n i (e# q" ( x ) / kT # e q" ( x ) / kT ) + N d ( x ) # N a ( x )
dx 2
$ Clif Fonstad, 9/17/09 Lecture 3  Slide 13 Nonuniform doping in thermal equilibrium, cont. (an aside) What do these equations say? n o ( x ) = n ie q" ( x ) kT and po ( x ) = n ie# q" ( x ) kT To see, consider what they tell us about the ratio of the hole
concentration at x2, where the electrostatic potential is φ2, and
that at x1, φ1:
" q [ # ( x 2 )"# ( x1 )] / kT
! po ( x 2 ) = po ( x1 )e The thermal energy is kT, and the change in potential energy of a
hole moved from x1 to x2 is q(φ2  φ1), so have: ! po ( x 2 ) = po ( x1 )e " #PE x1 $x2 / kT If the potential energy is higher at x2, than at x1, then the population
is lower at x2 by a factor eΔPE/kT.
That is, the population is lower at the top of a potential hill. ! If the potential energy is lower, then the population is higher.
That is, the population is, conversely, higher at the bottom of a potential hill. Clif Fonstad, 9/17/09 Lecture 3  Slide 14 Nonuniform doping in thermal equilibrium, cont.
(continuing the aside) The factor eΔPE/kT is called a Boltzman factor. It is a factor
relating the population densities of particles in many
situations, such as gas molecules in an ideal gas under the
influence of gravity (i.e, the air above the surface of the earth)
and conduction electrons and holes in a semiconductor.*
The potential energy difference for holes is qΔφ, while that for
electrons is qΔφ. Thus when we look at the electron and hole
populations at a point where the electrostatic potential is φ,
relative to those where the potential is zero (and both
populations are ni) we have: n o ( x ) = n ie q" ( x ) kT and po ( x ) = n ie# q" ( x ) kT We will return to this picture of populations on either side of a
potential hill when we examine at the minority carrier
populations on either side of a biased pn junction. ! * Until the doping levels are very high, in which case the
Boltzman factor must be replaced by a Fermi factor.**
Clif Fonstad, 9/17/09 ** Don’t worry about it. Lecture 3  Slide 15 Doing the numbers:
I. D to µ conversions, and visa versa
To convert between D and µ it is convenient to say
in which case q/kT ≈ 40 V1:
Example 1: µe = 1600 cm2/Vs, µh = 600 kT/q ≈ 25 mV,
17˚C/62˚F cm2/Vs De = µe (q kT ) = 1600 / 40 = 40 cm 2 / s
Dh = µh (q kT ) = 600 / 40 = 15 cm 2 / s II. Relating φ to n and p, and visa versa
To calculate φ knowing n or p it is better to say that
!
because then (kT/q)ln10 ≈ 60 mV: kT/q ≈ 26 mV,
28˚C/83˚F Example 1: ntype, ND = Nd  Na = 1016 cm3 kT 1016 kT
kT
"n =
ln 10 =
ln10 6 =
ln10 # log10 6 $ 0.06 ln10 6 = 0.36 eV
q 10
q
q ! Example 2: ptype, NA = Na  Nd = 1017 cm3
kT 1017
kT
" p = # ln 10 = #
ln10 $ log10 7 % #0.06 $ 7 = #0.42 eV
q 10
q ! Example 3: 60 mV rule:
For every order of magnitude the doping is above (below) ni,
the potential increases (decreases) by 60 meV. Clif Fonstad, 9/17/09 Lecture 3  Slide 16 More numbers no[cm3] po[cm3] φ [V] 1019 − 1 03 − 0.42 − 1016 − 1 04 − 0.36 − 1015 − 1 05 − 0.30 − 1 06 − 0.24 − 1013 − 1 07 − 0.18 − 1012 − 1 08 − 0.12 − 1011 − 1 09 − 0.06 − 1010 − 1010 − 0.00 − 109 − 1011 − 0.06 − 108 − 1012 − 0.12 − 107 − 1013 − 0.18 − 106 − 1014 − 0.24 − 105 − 1015 − 0.30 − 104 − 1016 − 0.36 − 103 − 1017 − 0.42 − 102 − 1018 − 0.48 − 101 −
Clif Fonstad, 9/17/09 0.48 − 1014 − ptype 1 02 − 1017 − Intrinsic 0.54 − 1018 − ntype 1 01 − 1019 − 0.54 − Typical range Typical range Lecture 3  Slide 17 Nonuniform doping in thermal equilibrium,cont:
We have reduced our problem to solving one equation for
one unknown, in this case φ(x): [ d 2" ( x )
q
= # n i (e# q" ( x ) / kT # e q" ( x ) / kT ) + N d ( x ) # N a ( x )
dx 2
$ Once we find φ(x) we can find no and po from: n o ( x ) = n ie q" ( x ) kT ! and po ( x ) = n ie# q" ( x ) kT Solving Poisson’s equation for φ(x) is in general nontrivial,
and for precise answers a "Poisson Solver" program must
! be employed. However, in two special cases we can find
very useful, insightful approximate analytical solutions:
Case I: Abrupt changes from p to ntype (i.e., junctions)
also: surfaces (Si to air or other insulator)
interfaces (Si to metal, Si to insulator, or Si to insulator to metal) Case II: Slowly varying doping profiles.
Clif Fonstad, 9/17/09 Lecture 3  Slide 18 Nonuniform doping in thermal equilibrium, cont.: Case I: Abrupt pn junctions
Consider the profile below:
NdNa
NDn x
 NAp
ptype po = N Ap , n o = n i2 N Ap "= # kT
ln( N Ap / n i ) $ " p
q Clif Fonstad, 9/17/09 ntype ? no ( x) = ?
po ( x ) = ?
"( x) = ?
! n o = N Dn , po = n i2 N Dn
kT
"=
ln( N Dn / n i ) # " n
q
Lecture 3  Slide 19 Abrupt pn junctions, cont:
First look why there is a dipole layer in the vicinity of the
junction, and a "builtin" electric field.
no , po
NDn
NDn
NAp
NAp
ni2/NAp
ni2/NDn
x Hole diffusion Electron diffusion ρ(x) Drift balances
diffusion in the
steady state. qNDn
+Q 0 0
x Q
qNAp
Hole drift
Clif Fonstad, 9/17/09 Electron drift
Lecture 3  Slide 20 Abrupt pn junctions, cont:
If the charge density is no longer zero there must be an
electric field: εEx(x) = ∫ρ(x)dx
Ex
0
0
x
Epk
and an electrostatic potential step: φ(x) = ∫Ex(x)dx
φ(x)
φn x
φp
Clif Fonstad, 9/17/09 Ok, but how do we find φ(x)? Lecture 3  Slide 21 Abrupt pn Junctions: the general strategy
We have to solve an nonlinear, second order differential
equation for φ: [ ! d 2" ( x )
$( x )
q
=#
= # n i (e# q" ( x ) / kT # e q" ( x ) / kT ) + N d ( x ) # N a ( x )
dx 2
%
%
$( x )
Or, alternatively
" ( x ) = # &&
dx + Ax 2 + Bx
%
In the case of an abrupt pn junction we have a pretty good
idea of what ρ(x) must look like, and we know the details
will be lost anyway after integrating twice, so we can try
the following iteration strategy: Guess a starting ρ(x).
Integrated once to get E(x), and again to get φ(x).
Use φ(x) to ﬁnd po(x), no(x), and, ultimately, a new ρ(x).
Compare the new ρ(x) to the starting ρ(x).
 If it is not close enough, use the new ρ(x) to iterate again.
 If it is close enough, quit.
Clif Fonstad, 9/17/09 Lecture 3  Slide 22 To figure out a good first guess for ρ(x), look at how quickly
no and po must change by looking first at how φ changes:
φ(x)
φn
60 mV
φn
700 to 900 mV x
60 mV φp A 60 mV change in φ
decreases no and po
10x and ρ increases to
90% of its final value. φp
…and what it
means for ρ(x): ρ(x)
qNDn
90%
+Q 0
90% Clif Fonstad, 9/17/09 The change in ρ must
be much more abrupt!
0
x Q qNAp The observation that ρ changes a lot, when φ changes
a little, is the key to the depletion approximation. Lecture 3  Slide 23 6.012  Electronic Devices and Circuits Lecture 3  Solving The Five Equations  Summary • Nonuniform excitation in nonuniform samples
The 5 unkowns: n(x,t), p(x,t), Je(x,t), Jh(x,t), E(x,t)
The 5 equations: coupled, nonlinear differential equations • Special cases we can solve (approximately)
Carrier concentrations:
Drift: Jdrift = Je,drift + Jh,drift = q (µe no + µh po) E = σo E
Low level optical injection: dn'/dt – n'/tmin ≈ gL(t)
Doping profile problems: junctions and interfaces
Nonuniform injection: QNR flow problems (Lect. 1)
(Lect. 2)
(Lect. 2) • Using the hand solutions to model devices
pn Diodes: two flow problems and a depl. approx.
BJTs: three flow problems and two depl. approx.’s
MOSFETs: three depl. approx.’s and one drift • Nonuniform doping in T.E.
Relating no, po, and electrostatic potential, φ
Poisson's equation: two situations important in devices
Clif Fonstad, 9/17/09 Lecture 3  Slide 24 MIT OpenCourseWare
http://ocw.mit.edu 6.012 Microelectronic Devices and Circuits
Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
View
Full
Document
This note was uploaded on 11/07/2011 for the course COMPUTERSC 6.012 taught by Professor Charlesg.sodini during the Fall '09 term at MIT.
 Fall '09
 CharlesG.Sodini

Click to edit the document details