MIT6_012F09_lec05

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6.012 - Electronic Devices and Circuits

Lecture 5 - p-n Junction Injection and Flow - Outline

• ReviewDepletion approximation for an abrupt p-n junctionDepletion charge storage and depletion capacitance (Rec. Fri.)

qDP(v AB) = – AqN Apxp = – A[2!q("b-v AB){N ApNDn/(N Ap+NDn)}]1/2

Cdp(V AB) #!qDP/!v AB|V AB = A[!q{N ApNDn/(N Ap+NDn)}/2("b-V AB)]1/2

• Biased p-n DiodesDepletion regions change (Lecture 4)Currents flow: two components

– flow issues in quasi-neutral regions (Today) – boundary conditions on p' and n' at -xp and xn (Lecture 6)

• Minority carrier flow in quasi-neutral regions

The importance of minor ity carrier diffusionBoundary conditionsMinority carrier profiles and currents in QNRs

– Short base situations – Long base situations – Intermediate situations

Clif Fonstad, 9/24/09 Lecture 5 - Slide 1

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x-x p

xn

-x p

-qNAp

qNDn

The Depletion Approximation: an informed first estimate of ! (x)

Assume full depletion for -xp < x < xn, where xp and xn aretwo unknowns yet to be determined. This leads to:

" ( x) =

0

#qN ApqN Dn

0

for

for

for

for

x < # x p# x p < x < 0

0 < x < xn

xn < x

$

%

& &

'

& &

%(x)

Integrating the charge once gives the electric field

E ( x) =

0 for x < " x p

"

qN Ap

# Si x+

x p( ) for " x p<

x<

0

qN Dn

# Si

x " xn( ) for 0

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The Depletion Approximation, cont.:

Integrating again gives the electrostatic potential:

"(x)

"n

-x px

xn

" p

equation relating our unknowns, xn and xp: -x p$(x)

x

N Ap x p = N Dn xn

xn

1E(0) = -qNApx p/!Si

= -qNDnxn/!Si

Requir ing that the potential be continuous at x = 0 givesus our second relationship between xn and xp:

" ( x) =

" p for x < # x p

" p +qN Ap

2$ Si x + x p( )

2

for - x p

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Comparing the depletion approximation

with a full solution:

Example: An unbiased abrupt p-n junction with N Ap= 10

17 cm-3, NDn= 5 x 1016 cm-3

Charge

E-field

Potential

nie ±q "(x)/kT po(x), no(x)

Clif Fonstad, 9/24 /09 Lecture 5 - Slide 4

Courtesy of Prof. Peter Hagelstein. Used with permission.

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Depletion approximation: Appl ied bias

Forward bias, v AB > 0:

vAB-w p

wn-x p 0 xn

( b-vAB) x

No dropin wire

No dropat contact No drop

in QNR

No dropin QNR

No dropat contact

No dropin wire

In a well built diode, all the appliedvoltage appears as a change in thethe voltage step crossing the SCL

-w p xwn-x p 0 xn

vAB

( b-vAB)

Reverse bias, v AB < 0:

Note: With applied bias we are no longer in thermal equilibrium so

it is no longer true that n(x) = ni eq (x)/kT and p(x) = n i e

-q (x)/kT.


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The Depletion Approximation: Appl ied bias, cont.

Adding v AB to our earlier sketches: assume reverse bias, v AB < 0

%(x)

xn

-x p

-qNAp

qNDn

w

xnx p

$(x)

x

x

w =2" Si # b $ v AB( )

q

N Ap + N Dn( ) N Ap N Dn

x p = N Dnw

N Ap + N Dn( ), xn =

N Apw

N Ap + N Dn( )

xn

-x p

|E pk |

"# = # b $ v AB

and # b =kT

qln N Dn N Ap

ni2

E pk =2q " b # v AB( )

$ Si

N Ap N

Dn

N Ap + N Dn( )"(x)

xn

-x p

(" b-vAB)

x


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The Depletion Approximation: comparison cont.

Example: Same sample, reverse

biased -2.4 V

Charge

E-field

Potential

nie ±q "(x)/kT p (x)

n (x)


Courtesy of Prof. Peter Hagelstein. Used with permission.

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The Depletion Approximation: comparison cont.

Example: Same sample, forward

biased 0.6 V

Charge

E-field

Potential

nie ±q "(x)/kT p (x)

n (x)

Clif Fonstad, 9/24/09 Lecture 5 - Slide 8Courtesy of Prof. Peter Hagelstein. Used with permission.

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The value of the depletion approximation

The plots look good, but equally important is that1.

It gives an excellent model for making hand calculationsand gives us good values for quantities we care about:

• Depletion region width

• Peak electric field

•

Potential step

2.

It gives us the proper dependences of these quantities onthe doping levels (relative and absolute) and the bias voltage.

Apply bias; what happens?

Two things happen

1. The depletion width changes•

( b - v AB) replaces b in the Depletion Approximation Eqs.

2. Currents flow

• This is the main topic of today’s lecture


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Bias applied, cont.: With v AB 0, it is not true that n(x) = n i eq (x)/kT

and p(x) = ni e-q (x)/kT because we are no longer in TE. However,

outside of the depletion region things are in quasi-equil ibrium, and

we can define local electrostatic potentials for which the equil ibriumrelationships hold for the majori ty carriers, assuming LLI.

Forward bias, v AB > 0:

In this region n(x) !ni eq QNRn /kT

Reverse bias, v AB < 0:

-w p x

wn-x p 0 xn

vAB

QNRn

( b-vAB)

QNRp

vAB

vAB

vAB

x

-w p

wn0 xn

vAB

QNRn

( b-vAB)

QNRp

vAB

vAB

-x p vAB

In this region p(x) !ni e-q QNRp /kT


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Current Flow q

Unbiased

junctionPopulation in

equilibrium withbarrier

q

Forward bias

on junction

Barrier lowered so carriers to left can

cross over it.

q

Reverse bias

on junction Barrier raised so the

few carriers on top spill back down it.


x

x

x

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Current flow: finding the relationship between iD and v ABThere are two pieces to the problem:

•

Minority carrier flow in the QNRs is what limits the current.•

Carrier equilibrium across the SCR determines n'(-xp) and p'(xn),the boundary conditions of the QNR minority carrier flow problems.

Ohmic Uniform p-type Uniform n-type

-wp

-xp 0 xn wnx

p n

Ohmiccontact

contact

A BiD

+ -v AB

Quasineutral Space charge Quasineutralregion I region region II

Minority carrier flowhere determines the

electron current- Today's Lecture -

The values of n' at-xp and p' at xn areestablished here.


hole current- Today's Lecture -

Clif Fonstad, 9/24/09 - Lecture 6 - Lecture 5 - Slide 13

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Solving the five equations: special cases we can handle

, po):

Lecture 1

1. Uniform doping, thermal equilibr ium (nopo product, no

"

" x= 0,

"

" t = 0, g

L ( x, t ) = 0, J

e= J

h= 0

2. Uniform doping and E-field (drift conduction, Ohms law):

Lecture 1"

" x= 0,

"

" t = 0, g

L( x,t ) = 0, E

x constant

3. Uniform doping and uniform low level optical injection (

"

" x= 0, g L (t ), n'

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QNR Flow: Uniform doping, non-uniform LL injection

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:

Hole continuity:

Electron continuity:

Hole current density:

Electron current density:

Charge conservation:

" p( x, t )

" t +

1

q

" J h ( x,t )

" x= G # R $ Gext ( x,t ) # n( x,t ) p( x, t )# ni

2[ ]r (t )

" n( x, t )

" t # 1

q

" J e ( x,t )

" x= G # R $ Gext ( x, t )# n( x,t ) p( x,t ) # ni

2[ ]r (t )

J h ( x,t ) = qµh p( x,t ) E ( x,t )# qDh" p( x,t )

" x

J e ( x,t ) = qµe n( x, t ) E ( x,t ) + qDe" n( x,t )

" x

% ( x, t ) =" & ( x) E x ( x,t )[ ]

" x$ q p( x,t ) # n( x,t )+ N d ( x)# N a ( x)[ ]

We can get approximate analytical solutions if 5 conditions are met!Clif Fonstad, 9/24/09 Lecture 5 - Slide 15

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QNR Flow, cont.: Uniform doping, non-uniform LL injection

Five unknowns, five equations, five flow problem conditions:1. Uniform doping dno

dx=

dpo

dx= 0 "

# n

# x=

# n'

# x,

# p

# x=

# p'

# x

po " no + N d " N a = 0 # $ = q p " n + N d " N a( ) = q p'"n'( )

2. Low level injection(in p-type, for example)

3. Quasineutrality holds

n'" p', # n'

# x" # p'

# x

n'

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With these first four conditions our five equations become:

(assuming for purposes of discussion that we have a p-type sample)

In preparation for continuing to our fifth condi tion, we notethat combining Equations 1 and 3 yields one equation inn'(x,t):

1,2:" p'( x, t )

" t +

1

q

" J h ( x, t )

" x=

" n'( x,t )

" t #

1

q

" J e ( x, t )

" x= g L ( x, t )#

n'( x, t )

$ e

3 : J e( x, t ) % +qDe" n'( x,t )

" x

4 : J h ( x, t ) = qµh p( x, t ) E ( x, t ) + qDh" p'( x,t )

" x

5 :" E ( x, t )

" x=q

& p'( x, t ) # n'( x, t )[ ]

" n'( x, t )

" t # De

" 2n'( x, t )

" x2

= g L ( x, t ) # n'( x, t )

$ e

! The time dependent diffusion equation


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The time dependent d iffusion equation, which is repeatedbelow, is in general still very diff icul t to solve

" n'( x, t )

" t # De

" 2n'( x,t )

" x2

= g L ( x,t ) # n'( x, t )

$ e

but things get much easier if we impose a fifth constraint:

5. Quasi-static excitation

g L ( x, t ) such that all

"

" t # 0

With this constraint the above equation becomes a second

" Ded

2n'( x)

dx 2= g L ( x)"

n'( x)

#

e

order linear di fferential equation:

which in turn becomes, after rearranging the terms :

! The steady state diffusion equation

d 2n'( x)

dx2

" n'( x)

De# e= "

1

Deg L ( x)


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QNR Flow, cont.: Solving the steady state diffusion equation

The steady state diffusion equation in p-type material is:

d 2n'( x)

dx2

" n'( x)

Le2

= " 1

Deg L ( x)

and for n-type material it is:

d 2 p'( x)

dx2

" p'( x)

Lh2

= " 1

Dhg L ( x)

In wri ting these expressions we have introduced Le and Lh,

the minority carrier diffusion lengths for holes and

electrons, as: L

e " D

e#

e

Lh " D

h# h

We'll see that the minority carrier diffusion length tells us howfar the average minority carrier diffuses before it recombines.

In a basic p-n diode, we have gL = 0 which means we only need

the homogenous solutions, i.e. expressions that satisfy:

d 2 p'( x)

dx2

" p'( x)

Lh2

= 0 d

2n'( x)

dx2

" n'( x)

Le

2= 0

n-side: p-side:


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Sketching and comparing the limit ing cases: wn>>Lh, wn Ln (the situation in LEDs)

p' (x) [cm-3] Jh(x) [A/cm2]

x [cm-3]wn

qDhp'n(xn) Lhe-x/Lh

0 xn xn+Lhwn0 xn xn+Lh

n

x [cm-3]

p'n(x

n)

e-x/Lh

Case II - Short base: w

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The four other unknowns

)

Solving the steady state diffusion equation gives n’ .) Knowing n'..... we can easily get p’, Je, Jh, and Ex:

First find Je:

Then find Jh:

Next find Ex:

Then find p’ :

E x ( x) " 1

qµh po J h ( x) #

Dh

De J e ( x)

$

% &

'

( )

p'( x) " n '( x) +#

qdE x

( x

)

dx

J e ( x) " qDedn'(t )

dx

J h ( x)=

J Tot " J e ( x)

Finally, go back and check that all of the five condit ions are

met by the solution.

! Once we solve the diffusion equation to get

the minority excess, n', we know everything.Clif Fonstad, 9/24/09 Lecture 5 - Slide 24

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Current flow: finding the relationship between iD and v ABThere are two pieces to the problem:

•

Minority carrier flow in the QNRs is what limits the current.•

Carrier equilibrium across the SCR determines n'(-xp) and p'(xn),the boundary conditions of the QNR minority carrier flow problems.

Ohmic Uniform p-type Uniform n-type

-wp

-xp 0 xn wnx

p n

Ohmiccontact

contact

A BiD

+ -v AB

Quasineutral Space charge Quasineutralregion I region region II


electron currentThe values of n' at-xp and p' at xn areestablished here.


hole current

Clif Fonstad, 9/24/09 - Lecture 6 next Tuesday - Lecture 5 - Slide 25

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The p-n Junct ion Diode: the game plan for getting iD(v AB)

We have two QNR's and a flow problem in each:

Quasineutral QuasineutralOhmic

n

Ohmiccontact

B-

region II

p

contact

AiD

+vAB

region I

xx-w p -x p 0 0 xn wn

p'(xn) = ?

p'(wn) = 0

n'(-x p) = ? n' p p'n

n'(-w p) = 0

x

-w p -x p 0 0 xn wn

If we knew n'(-xp) and p'(xn), we could solve the flow problems

and we could get n'(x) for -wp

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6.012 - Electronic Devices and Circuits

Lecture 5 - p-n Junction Injection and Flow - Summary

• Biased p-n DiodesDepletion regions change (Lecture 4)Currents flow: two components

– flow issues in quasi-neutral regions – boundary conditions on p' and n' at -xp and xn

• Minority carrier flow in quasi-neutral regionsThe importance of minor ity carrier diffusion

– minority carrier drift is negligible

Boundary conditions Minority carrier profiles and currents in QNRs

– Short base situations – Long base situations

• Carrier populations across the depletion region (Lecture 6)Potential barriers and carrier populationsRelating minority populations at -xp and xn to v ABExcess minori ty carriers at -xp and xn


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6.012 Microelectronic Devices and Circuits

Fall 2009

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MIT6_012F09_lec05