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P-N Junctions - PowerPoint PPT Presentation


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P-N Junctions. 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

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Presentation Transcript
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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

ECE 663

slide4

P-N junction diode

I

V

I = I0(eqV/hkT-1)

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

pn v

ECE 663

p n junctions equilibrium
P-N Junctions - Equilibrium

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

ECE 663

slide10

-

-

-

+

-

+

-

+

-

+

-

+

-

+

-

-

+

+

-

+

+

+

But charges can’t venture too

far from the interface because

their Coulomb forces pull them back!

ECE 663

slide11

Separation of a sea of charge, leaving

behind a charge depleted region

ECE 663

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

slide13

E

Depletion

Region

E

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how much is the built in voltage15
How much is the Built-in Voltage?

Na acceptor level on the p side

Nd donor level on the n side

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special case one sided junctions
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)

ECE 663

depletion approximation step junction19
Depletion approximation-step junction

Exponentials replaced with step-functions

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slide20

Doping

Charge

Density

Electric

Field

Electrostatic

Potential

NAxp = NDxn

= WD/(NA-1 + ND-1)

Kse0Em = -qNAxp = -qNDxn

= -qWD/(NA-1 + ND-1)

Vbi = ½|Em|WD

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maximum field
Maximum Field

Em = 2qVbi/kse0(NA-1+ND-1)

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how far does w d extend into each junction
How far does Wd extend into each junction?

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

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

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

ECE 663

reverse bias
Reverse Bias
  • +Voltage to the n side and –Voltage to the p side:

Depletion region will be larger

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reverse bias depletion
Reverse Bias depletion

Applied voltage disturbs equilibrium EF no longer constant

Reverse bias adds to the effect of built-in voltage

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forward bias
Forward Bias

+ -

Negative voltage to n side positive to p side

More electrons supplied to n, more holes to p

Depletion region gets smaller

ECE 663

general expression
General Expression
  • Convention = Vappl= + for forward bias

Vappl= - for reverse bias

ECE 663

slide33

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

n = nie(Fn-Ei)/kT

Fp

Fn

p = nie(Ei-Fp)/kT

Fermi level is not constant  Current Flow

ECE 663

slide34

In summary

A p-n junction at equilibrium sees a depletion width

and a built-in potential barrier. Their values depend on

the individual doping concentrations

Forward biasing the junction shrinks the depletion width

and the barrier, allowing thermionic emission and higher

current. The current is driven by the splitting of the

quasi-Fermi levels

Reverse biasing the junction extends the depletion width

and the barrier, cutting off current and creating a strong

I-V asymmetry

In the next lecture, we’ll make this analysis quantitative

by solving the MCDE with suitable boundary conditions