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Lecture 5

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P-N Junctions Part 1: Concepts, Charges and Fields

The most basic device of microelectronics, the P-N junction

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Vbi has nothing to do with voltages that are applied via external batteries or power

supplies

Vbi is spontaneous internal voltage developed by the rearrangement of holes &

electrons!

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Now know the energy / voltage step formed - But have no handle on how wide the "depletion layer"

is

The NET charge in the depletion layer (between the p and n regions) is now due acceptors & donorsalone

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But from equation 6

using this and equation 6 to solve for xp:

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All of the Above: No external voltages applied!! Vbi is an internal, naturally generated,

voltage

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NOW FOR THE FIRST TIME: Apply an external voltage (with a battery or DC

power supply) end to end:

1) Apply "Reverse Voltage" Reversed in the sense: + Applied to N-side

- Applied to P-side

Remember that Voltage = Potential Energy per positive charge = opposite of ourelectron bands

So positive voltage pulls DOWN electron energy

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What has changed from previous case with our application of Vreverse?

Just have wider regions of exposed Na- and Nd+

So parabolic profiles simply extend longer and steps become higher!

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So with that justification get revisions:

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2) Apply Forward Voltage: Forward in the sense: + Applied to P-side- Applied to N-side

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Calculate Capacitance of this junction:

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P-N Junction Current / Equation for Diode with Thick Layers

Calculate current density, J, across the diode as a function of the applied voltage

1) Assume "Steady-State

Vapplied constant in time (or has been at present value a "long" time)

Then total current, Jtotal = Jp + Jn, must be constant throughout the diode

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This is assumed OUTSIDE the depletion region (W)

Inside the depletion layer has been massive change in majority carrier concentration-they are all gone

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Assumption 3) All carriers that start across the junction make it to the other side

Electric field pushes minority carriers across

Will move so fast have no time to recombine!

MAJORITY Carriers: Because minority carriers (above) whip across junction, thereare ~ 0 there at any time~ Nothing for majority carriers to recombine WITH! Majority also likely to makeacross

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NOTATION

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If ~ all carriers make it across => same density of electrons on both sides /

same density of holes on both sides

Restate arguments (for electrons)

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Restate arguments (for electrons)

Form of both of these equations:

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Form of both of these equations:

If Vapplied were turned off (=> 0), go back to overall equilibrium

But in strict equilibrium, we KNOW what the minority carrier concentrations (left sides) are

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Plug equations 3 and 4 into equations 1 and 2, respectively

Or graphically

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Remember the "Minority Carrier Continuity Equations" Time to put them to truly

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Remember the Minority Carrier Continuity Equations Time to put them to trulyuseful work

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"Thick Diode" or more precisely, "Thick Layer" Case

If layers are thick compared to diffusion lengths:

One term falls to zero Makes sense: carriers diffuse deeper and recombine

One terms goes to Makes NO sense: minority carriers increasing awayfrom junction

Solution is to set coefficient of growing terms equal to zero (B=0 and D=0 above)

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Diode with THIN layers

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y

At least one side of a diode is much THINNER than diffusion length of itsminority carriers!!

Comes from the way real diodes are made:

1) Start with layer (either thick or thin) of one type

of semiconductor

2) Implant other dopant short distance into surface:

3) P-dopant penetrates very short distance

(1 micron or less)

but in high concentration that overwhelmsN-type dopant:

In an integrated circuit, layer below may have been created same way

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In an integrated circuit, layer below may have been created same way=> BOTH layers "thin"

How do we deal with this? Go back to last lecture at point we had derivedgeneral minority equations

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Now assume P-layer on left is much thinner than Ln:

Then where ever we are in the left layer ( x < xp), we know that in the exponentxp + x is

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-"The Law of the Junction":

Would then know the slope if we could figure out value at LEFT face of P-layer

(the surface)

Front surface of crystal = excellent place for minority carriers to recombine(like continuous "traps"

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(like continuous traps

Can thus assume excess electron concentration => 0 at surface. If P-layerthickness is xp_layer:

where np(-xp) is theConstant

Use this minority carrier profile to calculate the diffusion (gradient driven)current of minority carrier electrons

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current of minority carrier electrons

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plugging in boundary condition

Evaluate this equation at x = -xp to get electron current flowing across junction

Then need hole current flowing across junction into the N-layer to get total current indiode:

if N-layer is thick (as have assumed to this point and already derived

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or a ayer, use sma er o wo: s m nor y carr er us on eng s un ep e e

thickness

REAL DIODES

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REAL DIODES: Must also take into account

1) FORWARD BIAS (lowered step => high current in junction)

1a) Recombination of carriers within the depletion region

- Are so MANY crossing at any moment that there ARE plenty to

recombine with- So expect some recombination, and thus loss of current within thejunction!

1b) Resistive loss in thick layers

We apply voltage at both ends of the diode

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pp y g

But as with right N-layer (thick), that face is a LONG wayaway from junction

Current flowing thorough thick layers => resistive voltage

drop

Voltage at junction is thus reduced below full applied voltage!

2) REVERSE BIAS (increased step and Wdepletion, very large electric fields)

2a) Thermal generation in thickened depletion layer

Depletion region can become quite thick (1.6 microns at -20 Volts in

example above)

There is still heat!Heat => Generation of new carriers - electrons boiling out of valence

band to conduction band

Can thus get INCREASE in current as it crosses the depletion region!

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2b) Tunneling "Zener Breakdown"

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Abrupt increase in reverse current, depends on junction width - which in turndepends on doping

2c) Breakdown due to avalanche multiplication of carriers crossing junction

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Electrons and holes are "falling" over energy step

Where is their lost energy going? Energy => crystal lattice

THIS energy can be used to create new pairs of electrons and holes

=> More electron / hole flow

=> More dissipated energy

=> More electrons and holes AVALANCHE!

Once it starts, multiplies VERY rapidly (as in real avalanche) => Similar abrupt

increase in reverse current!