ECE 663

P-N Junctions

ECE 663

• So far we learned the basics of semiconductor physics, culminating in the Minority Carrier Diffusion Equation

• We now encounter our simplest electronic device, a diode

• Understanding the principle requires the ability to draw band-diagrams

• Making this quantitative requires ability to solve MCDE (only exponentials!)

• Here we only do the equilibrium analysis

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P-N junction diode

V

I

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P-N junction diode

V

I

I = I0(eqV/kT-1)

pn v

Ip0 = q(ni2/ND) (Lp/p)

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P-N Junctions - Equilibrium

<= P-type, low EF

- = fixed ionized acceptors

+ = mobile holes, p

<= N-type, high EF

+ = fixed ionized donors

- = mobile electrons, n

What happens when these bandstructures collide?

• Fermi energy must be constant at equilibrium, so bands must bend near interface

• Far from the interface, bandstructures must revert

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Time < 0

P-type piece N-type piece

Time < 0: Pieces separated

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At time = 0, slam the two pieces together

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Hole gradient

Jp, diffusion = -qDp dp/dx = current right, holes right

Electron gradient

Jn,diffusion = -qDn dn/dx = current right, electrons right

Gradients drive diffusion

left

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Gradients drive diffusion

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But charges can’t venture toofar from the interface becausetheir Coulomb forces pull them back!

++- -

++++

- -- -

- -- -

-+++++

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Separation of a sea of charge, leavingbehind a charge depleted region

http://scott.club365.net/uploaded_images/Moses-Parts-the-Red-Sea-2-782619.jpg

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V (stopping e - flow)

V (stopping e+ flow)

Resulting in a barrier across a depletion region

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E

Depletion Region

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How much is the Built-in Voltage?

RightiFLeftFibi EEEEqV )()(

P side N side

i

aLeftFi

kTEE

ia

a

nN

kTEE

enN

NpFi

ln)(

)(

i

dRightiF

kTEE

id

d

nN

kTEE

enN

NniF

ln)(

)(

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2ln

lnln

i

dabi

i

d

i

abi

n

NNq

kTV

nN

qkT

nN

qkT

V

Na acceptor level on the p side

Nd donor level on the n side

How much is the Built-in Voltage?

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Special case: One-sided Junctions

• One side very heavily doped so that Fermi level is at band edge.

• e.g. p+-n junction (heavy B implant into lightly doped substrate)

i

dGbi

i

DRightiF

GViLeftFi

nN

qkT

qE

V

nN

kTEE

EEEEE

ln2

ln)(

2/)(

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How wide is the depletion region?

kTEE

i

Fi

enp)(

kTEE

i

iF

enn)(

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Depletion Approximation-step junction

x

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Depletion approximation-step junction

Exponentials replaced with step-functions

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Doping

ChargeDensity

Electric Field

ElectrostaticPotential

NAxp = NDxn

= WD/(NA-1 + ND

-1)

Ks0Em = -qNAxp = -qNDxn

= -qWD/(NA-1 + ND

-1)

Vbi = ½|Em|WD

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Depletion Width

A

DAn

A

Dn

A

Dnnpn N

NNx

NN

xNN

xxxxW 1

21

0

21

0

2

2

biAD

DAS

A

DAbi

DAD

AS

VNN

NNq

KW

NNN

VNNN

Nq

KW

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Maximum Field

Em = 2qVbi/ks0(NA-1+ND

-

1)

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How far does Wd extend into each junction?

Depletion width on the n-side depends on the doping on the p-sideDepletion width on the p-side depends on the doping on the n-side

e.g. if NA>>ND then xn>>xp One-sided junction

DA

An

A

DAn

NNN

Wx

or

NNN

xW

DA

Dp

A

DA

D

Ap

A

DAn

NNN

Wx

or

NNN

NN

xN

NNxW

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"P+ - N" => N a >> Nd => xp << xn

"P - N+" => N a << Nd => xp >> xn

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P-N Junction with applied voltage

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Reverse Bias

• +Voltage to the n side and –Voltage to the p side:

holes (+) out of P-sideVapplied is sucking more:

electrons (-) out of N-side

Depletion region will be larger

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Reverse Bias Band Diagram

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Reverse Bias depletion

2

1

02

revbi

AD

DAS VVNN

NNq

KW

DA

Arevn NN

NVWx

DA

Drevp NN

NVWx

Applied voltage disturbs equilibrium EF no longer constant

Reverse bias adds to the effect of built-in voltage

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Forward Bias

+ -

Negative voltage to n side positive to p side

More electrons supplied to n, more holes to p

Depletion region gets smaller

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Forward Bias Depletion

2

1

02

fwdbi

AD

DAS VVNN

NNq

KW

DA

Afwdn NN

NVWx

DA

Dfwdp NN

NVWx

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General Expression

• Convention = Vappl= + for forward bias

Vappl= - for reverse bias

2

1

02

applbi

AD

DAS VVNN

NNq

KW

DA

Aappln NN

NVWx

DA

Dapplp NN

NVWx

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Bands = plots of electron energy

Voltage = potential energy per (+) charge

Positive voltage pulls bands down- bands are plots of electron energy

Fermi level is not constant Current Flow

Fn

Fpn = nie(Fn-Ei)/kT

p = nie(Ei-Fp)/kT

In summary

A p-n junction at equilibrium sees a depletion widthand a built-in potential barrier. Their values depend on the individual doping concentrations

Forward biasing the junction shrinks the depletion widthand the barrier, allowing thermionic emission and highercurrent. The current is driven by the splitting of the quasi-Fermi levels

Reverse biasing the junction extends the depletion widthand the barrier, cutting off current and creating a strongI-V asymmetry

In the next lecture, we’ll make this analysis quantitative by solving the MCDE with suitable boundary conditions