ECE3080-L-11-P-N Junction 1users.ece.gatech.edu/alan/ECE3080/Lectures/ECE3080-L-11-P-NJunc… ·...

ECE 3080 - Dr. Alan DoolittleGeorgia Tech

Lecture 11

P-N Junction Diodes: Part 1

How do they work? (postponing the math)

Reading:

(Cont’d) Notes and Anderson2 sections Part 2 preface and sections 5.0-5.3


Our First Device: p-n Junction Diode

N-type material P-type material

N-type material P-type material

A p-n junction diode is made by forming a p-type region of material directly next to a n-type region.



ND-NA

x-NA

ND

Ec

Ei

Ef

Ev

Ec

Ei

Ef

Ev

In regions far away from the “junction” the band diagram looks like:

Junction



Ec

Ei

Ef

Ev

Ec

Ei

Ef

Ev

But when the device has no external applied forces, no current can flow. Thus, the fermi-level must be flat!

We can then fill in the junction region of the band diagram as:

or…



Ec

Ei

Ef

Ev


Ec

Ei

Ef

Ev


Electrostatic Potential, V = -(1/q)(Ec-Eref)

X

VBI or the “built in potential”

-qVBI



Electrostatic Potential, V=-(1/q)(Ec-Eref)

X

VBI or the “built in potential”

X

Electric Field

E=-dV/dx



Poisson’s Equation:

osos KdxdEDinor

KE

ερ

ερ

==•∇ ,1

Electric Field Charge Density (NOT resistivity)

Permittivity of free spaceRelative Permittivity of Semiconductor

(previously referred to as εR)

ρ=q( p – n + ND – NA )



X

Electric Field, E=-dV/dx

X

dxdEK osερ =


Energy

Potential

Electric Field

Charge Density

-1/q

-dV/dx

dxdEK osε




How do they work? (A little bit of math)


Fig Pierret 5.5

Space Charge or Depletion Region

Movement of electrons and holes when forming the junctionCircles are charges free to move (electrons and holes)

Squares are charges NOT free to move (ionized donor or acceptor atoms)

High electron Concentration

High hole Concentration

Electron diffusion

Hole diffusion

Local region of positive charge

due to imbalance in

electron-donor concentrations

Local region of negative

charge due to imbalance in

hole-acceptor concentrations


Movement of electrons and holes when forming the junction

E= - dV/dx

-Edx=dV

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−==−=

−=−=

=+=

=−−==−

∫∫

∫∫

−−

−−

)()(ln

...

0

...

)()(

)(

)(

)(

)(

p

nxn

xn

x

xbi

n

N

NnN

bipn

xV

xV

x

x

xnxn

qkTdx

ndxdn

qkTEdxV

thusndxdn

qkT

ndxdn

DE

dxdnqDnEqJ

but

VxVxVdVEdx

n

p

n

p

n

p

n

p

µ

µNo net current flow in equilibrium


⎥⎦

⎤⎢⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

−=

2

2

ln

ln)(

)(ln

i

DAbi

A

i

D

p

nbi

nNN

qkTV

Nn

Nq

kTxnxn

qkTV

Movement of electrons and holes when forming the junction

For NA=ND=1015/cm-3 in silicon at room temperature, Vbi~0.6 V*

For a non-degenerate semiconductor, |-qVbi|<|Eg|

*Note to those familiar with a diode turn on voltage: This is not the diode turn on voltage! This is the voltage required to reach a flat band diagram and sets an upper limit (typically an overestimate) for the voltage that can be applied to a diode before it burns itself up.


Movement of electrons and holes when forming the junctionDepletion Region Approximation

++++++++++++++++++++++

- - - - - - - - - - - - - - - - - - - -

Depletion Region Approximation states that approximately no free carriers exist in the space charge region and no net charge exists outside of the depletion region (known as the quasi-neutral region). Thus,

regionechspacethewithinNNK

qdxdEbecomes

regionneutralquasithewithinNNnpK

qKdx

dE

ADoS

ADoSoS

arg)(

...

0)(

−=

−=−+−==

ε

εερ


Movement of electrons and holes when forming the junctionDepletion Region Approximation: Step Junction Solution

Fig. 5.9 a and b

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

≥−≤

≤≤

≤≤−−

=

⎪⎩

⎪⎨

⎧

≥−≤≤≤

≤≤−−=

np

noS

D

poS

A

np

nD

pA

xxandxxfor

xxforKqN

xxforKqN

dxdE

thus

xxandxxforxxforqN

xxforqN

0

0

0

,

00

0

ε

ε

ρ

oSKdxdE

ερ

=

Where we have used:


( )

( ) nnoS

D

n

x

xoS

DxE

ppoS

A

p

x

xoS

AxE

xxforxxKqNxE

xxfordxKqNdE

and

xxforxxKqNxE

xxfordxKqNdE

n

p

≤≤−−

=

≤≤=

≤≤−+−

=

≤≤−−

=

∫∫

∫∫ −

0)(

0''

0)(

0''

0

)(

)(

0

ε

ε

ε

ε


Since E(x=0-)=E(x=0+)

NAxp=NDxn


( )

( )

( )

( )

( )

( )⎪⎪⎩

⎪⎪⎨

⎧

≤≤−−

≤≤−+=

≤≤−=

≤≤−+=

⎪⎪⎩

⎪⎪⎨

⎧

≤≤−

≤≤−+=

−=

∫∫

∫∫ −

nnoS

Dbi

ppoS

A

n

x

x noS

DV

xV

p

x

x poS

AxV

nnoS

D

ppoS

A

xxforxxKqN

V

xxforxxKqN

xV

xxfordxxxKqNdV

xxfordxxxKqN

dV

or

xxforxxKqN

xxforxxKqN

dxdV

dxdVE

nBi

p

02

02)(

0'''

0'''

,

0

0

2

2

)(

)(

0

ε

ε

ε

ε

ε

ε


V=VBi

V=0


( ) ( )

( )

( ) ( )( )

biDA

DAoSnp

biDAA

DoSpbi

DAD

AoSn

A

Dnp

noS

Dbip

oS

A

VNNNN

qK

xxW

VNNN

Nq

KxandV

NNNN

qK

x

NNxxgU

xKqN

VxKqN

+=+=

+=

+=

=

−=

ε

εε

εε

2

22

,sin

2222


At x=0,



Vbi

Vbi

|VA|Vbi

|VA|

Vbi

VA=0 : No Bias VA<0 : Reverse Bias VA>0 : Forward Bias



( ) ( ) ( ) ( )

( ) ( )AbiDA

DAoSnp

AbiDAA

DoSpAbi

DAD

AoSn

VVNNNN

qK

xxW

VVNNN

Nq

KxandVV

NNNN

qK

x

−+

=+=

−+

=−+

=

ε

εε

2

22

Thus, only the boundary conditions change resulting in direct replacement of Vbi with (Vbi-VA)


Movement of electrons and holes when forming the junctionStep Junction Solution: What does it mean?

Consider a p+ -n junction (heavily doped p-side, normal or lightly doped n side).


Movement of electrons and holes when forming the junctionStep Junction Solution: What does it mean?

Efp-Efn=-qVA

Fermi-level only applies to equilibrium (no current flowing)

Majority carrier Quasi-fermi levels



Current Flowing through a Diode


P-n Junction I-V Characteristics

Electron Diffusion Current

Electron Drift Current

Hole Diffusion Current

Hole Drift Current

In Equilibrium, the Total current balances due to the sum of the individual components


Electron Diffusion CurrentElectron Drift

Current

Hole Diffusion Current

Hole Drift Current

Current flow is proportional to e(Va/Vref) due to the exponential decay of carriers into the majority carrier bands


Current flow is dominated by majority carriers flowing across the junction and becoming minority carriers

QuickTime Movie



Electron Diffusion Current negligible due to large energy barrier

Electron Drift Current

Hole Diffusion Current negligible due to large energy barrier

Hole Drift Current

Current flow is constant due to thermally generated carriers swept out by E-fields in the depletion region

Current flow is dominated by minority carriers flowing across the junction and becoming majority carriers

QuickTime Movie



Where does the reverse bias current come from? Generation near the depletion region edges “replenishes” the current source.


P-n Junction I-V CharacteristicsPutting it all together

Reverse Bias: Current flow is constant due to thermally generated carriers swept out by E-fields in the depletion region

Forward Bias: Current flow is proportional to e(Va/Vref) due to the exponential decay of carriers into the majority carrier bands

Current flow is zero at no applied voltage

I=Io(eVa/Vref - 1)



Quantitative Analysis (Math, math and more math)


Quantitative p-n Diode Solution

Assumptions:1) steady state conditions2) non- degenerate doping3) one- dimensional analysis4) low- level injection5) no light (GL = 0)

Current equations:J=Jp(x)+Jn(x)

Jn =q µnnE +qDn(dn/dx)

Jp = q µppE - qDp (dp/dx) VA

Quasi-Neutral Regions

Depletion Region

p-type n-type


p-type n-type


Depletion Region

-xp xn ∞

0)()(0

)()()(

2

2

2

2

+∆

−∂∆∂

=

+∆

−∂∆∂

=∂∆∂

p

nnP

Lp

nnP

n

pxpD

GpxpD

tp

τ

τ

n

ppN

Ln

ppN

p

nxn

D

Gn

xn

Dtn

τ

τ

)()(0

)()()(

2

2

2

2

∆−

∂

∆∂=

+∆

−∂

∆∂=

∂

∆∂ Since electric fields exist in the depletion region, the minority

carrier diffusion equation does not

apply here.

Application of the Minority Carrier Diffusion Equation

∞−

0)(:

=−∞→∆ xnConditionBoundary

p ?)(:

=−=∆ pp xxnConditionBoundary

?)(:

==∆ nn xxpConditionBoundary

0)(:

=∞→∆ xpConditionBoundary

n

0≠E0=E 0=E


p-type n-type


Depletion Region

-xp xn ∞Application of the Minority Carrier Diffusion Equation

∞−

?)(:

=−=∆ pp xxnConditionBoundary

?)(:

==∆ nn xxpConditionBoundary

0≠E0=E 0=E

( ) ( )

( )

⎟⎠⎞

⎜⎝⎛ −==∆=⎟

⎠⎞

⎜⎝⎛ −=−=∆

−=−=∆

=−=

=−==−=−=

=−=−=

==−

−−

1)(1)(

)(

)(

)()()(

)()(

22

2

2

2

2

kTqV

D

innn

kTqV

A

ipp

okT

qV

A

ipp

kTqV

A

ipp

kTqV

iApppppp

kTFF

ipppp

kTFE

ikT

EF

i

AA

A

A

A

PN

PiiN

eNn

xxpxxatsimilarlyandeNn

xxn

neNn

xxn

eNn

xxn

enNxxnxxpxxn

enxxpxxn

enpandenn


p-type n-type


Depletion Region

-xp xn∞− ∞

Application of the Current Continuity Equation

( )

dx

dx

dx

pn

pon

nnn

ndqD

nndqD

dnDnq J

∆=

∆+=

⎟⎠⎞

⎜⎝⎛ +Ε= µ

( )

dxp

dx

dx

np

nop

ppp

dqD

ppdqD

dpDpq J

∆−=

∆+−=

⎟⎠⎞

⎜⎝⎛ −Ε= µ

0≠E0=E 0=E

?


p-type n-type


Depletion Region

-xp xn∞− ∞

No thermal recombination and generation implies Jn and Jp are constant throughout the depletion region. Thus, the total current can be define in terms of only the current at the

depletion region edges.

Application of the Current Continuity Equation: Depletion Region

)()( nppn xJxJJ +−=

xJ

q

Jq

tn

tnJ

qtn

N

N

etclightassuchprocessesotherAllGenerationncombinatioN

∂∂

=

⋅∇=

∂∂

+∂∂

+⋅∇=∂∂

−

10

10

1...,

Re

0≠E0=E 0=E

xJ

q

Jq

tp

tpJ

qtp

P

P

etclightassuchprocessesotherAllGenerationncombinatioP

∂∂

−=

⋅∇−=

∂∂

+∂∂

+⋅∇−=∂∂

−

10

10

1...,

Re


p-type n-type


Depletion Region

x’=0∞ ∞

Approach:

•Solve minority carrier diffusion equation in quasi-neutral regions

•Determine minority carrier currents from continuity equation

•Evaluate currents at the depletion region edges

•Add these together and multiply by area to determine the total current through the device.

•Use translated axes, x x’ and -x x’’ in our solution.

0≠E0=E 0=E

x’’=0


p-type n-type


Depletion Region

x’=0∞ ∞

0≠E0=E 0=E

x’’=0

p

nnP

px

pDτ

)('

)(0 2

2 ∆−

∂∆∂

=

⎟⎠⎞

⎜⎝⎛ −==∆==

⎟⎠⎞

⎜⎝⎛ −==∆

=∞→∆

1)0'(0

1)0'(

0)'(:

2

2

kTqV

D

in

kTqV

D

in

n

A

A

eNnxpAandB

eNnxp

xpConditionsBoundary

( ) ( )ppP

LxLxn DLwhereBeAexp PP τ≡+=∆ +− /'/')'(

( ) 0'1)'( /'2

≥⎟⎠⎞

⎜⎝⎛ −=∆ − xforee

Nnxp P

A LxkTqV

D

in


( ) 0'1)'( /'2

≥⎟⎠⎞

⎜⎝⎛ −=∆ − xforee

Nnxp P

A LxkTqV

D

in

p-type n-type


Depletion Region

x’=0∞ ∞

0≠E0=E 0=E

x’’=0

( ) 0'1 J

dqD J

/'2

p

npp

≥⎟⎠⎞

⎜⎝⎛ −=

∆−=

− xforeeNLnD

q

dxp

PA LxkT

qV

Dp

ip


( ) 0''1)''( /''2

≥⎟⎠⎞

⎜⎝⎛ −=∆ − xforee

Nnxn n

A LxkTqV

A

ip

p-type n-type


Depletion Region

x’=0∞ ∞

0≠E0=E 0=E

x’’=0

( ) 0''1 J

dqD J

/''2

n

pnn

≥⎟⎠⎞

⎜⎝⎛ −=

∆−=

− xforeeNLnDq

dxn

nA LxkT

qV

An

in

Similarly for electrons on the p-side…


p-type n-type


Depletion Region0≠E0=E 0=E

pn JJJ +=

( )nLxn eJ /''−∝

( )pLxp eJ /'−∝

np JJJ −= pn JJJ −=

Total on current is constant throughout the device. Thus, we can characterize the current flow components as…

x’=0∞ ∞x’’=0

Recombination Recombination


Thus, evaluating the current components at the depletion region edges, we have…


""

1 I

1 J

0)(x'J0)(x'J0)'(x'J0)'(x'J0)(x'J0)'(x'JJ

22

22

pnpnpn

currentsaturationreversetheisI

NLnD

NLnDqAIwhereeI

or

xallforeNLnD

NLnDq

o

Dp

ip

An

ino

kTqV

o

kTqV

Dp

ip

An

in

A

A

⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟

⎠⎞

⎜⎝⎛ −=

⎟⎠⎞

⎜⎝⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛+=

=+===+===+==

Note: Vref from our previous qualitative analysis equation is the thermal voltage, kT/q


Quantitative p-n Diode SolutionExamples: Diode in a circuit

R=1000 ohms

V=9V, 5V, 2V, -9V

VA

I

pAIwhereeI okT

qV

o

A

11 I =⎟⎠⎞

⎜⎝⎛ −=

V1=IR

A0259.0

A0259.0

0259.0

A

V191 9V

V(1000)1121 9V

1121 I

VI(1000) 9V

+⎟⎠⎞

⎜⎝⎛ −−=

+⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −−=

⎟⎠⎞

⎜⎝⎛ −−=

+=

VV

VV

VV

A

A

A

ee

ee

or

ee

-1 pA-9.0V-9V

1.5 mA0.55V2V

4.4 mA0.58V5V

8.4 mA0.59V9V

IVAV

SolutionsIn forward bias (VA>0) the VA is ~constant for large differences in current

In reverse bias (VA<0) the current is ~constant (=saturation current)

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