Semiconductor Devices 23

8/8/2019 Semiconductor Devices 23

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Semiconductor Devices - Hour 23 "Ohmic" Contacts / Semiconductor Heterojunctions & Band Diagrams (part I)

Ohmic Contacts - (the downside of metal-semiconductor junctions!)

Last lecture: Occasional role of metal-semi. junctions as useful diodes ("Schottky Barrier Diodes")

PROBLEM: ALWAYS have metal-semiconductor junctionsMOST of the time don't want them to act as diodes - just want them to conduct (!)

Metal-semiconductor junction where I V = "Ohmic Contact"

Must get rid of or circumvent normal energy barriers, how?

Metal P- Semi N-Semi Metal

P

N

Shottky P-N Schottky

Semiconductor_Devices_23.mcd 1 7/30/2010


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How do I get back to simple semiconductor device when don't want Schottky's?

Solution #1) Choose metal w/ work function such that is ~ no barrier for carriers in adjacent semiconductor

q s

q m

For P-type semiconductor:

Want: q m ~ q + (Ec - Ef ) ~ q s + Eg (doping high Ef => Ev, no barriers!)

q sq m

For N-type semiconductor:

Want: q m ~ q + (Ec - Ef ) ~ q s (doping high Ef => Ec, no barriers!)



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Solution #2: Tunneling Semiconductor depletion width is: W2 k o Vbi Vappl( )

q Ndoping=

If semiconductor doping is HIGH, then W is small => Carriers can tunnel THROUGH the barrier

Metal - P+ Semi Metal - N+ Semi

thin

thin

Carrier tunneling

under Reverse Bias:

More Doping => Smaller W => Easier Tunneling



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So "Contact Resistance" (metal to semiconductor) decreases with doping:

Metal 1

Metal 2

Metal 3

Resistance of metal tosemiconductor junction


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Semiconductor Heterojunctions: What if two semiconductors are joined?

Rule # 1: To maintain crystal quality, both semiconductors must have ~ same crystal type and atom spacings

Rule #2: To control doping, both semiconductors must come form same columns of the periodic table

(III-V) (III-V) e.g. GaAs AlAs

(IV) (IV) e.g. Si Ge

(II-VI) (II-VI) e.g. CdTe ZnSe

Otherwise if they mix (usually will during crystal growth) they will dope one another in an uncontrolled manner.

Si atoms can move into GaAs =>

e.g., if Si (IV) is joined with GaAs (III-V):

Ga atoms can move into Si => P-doping

As atoms can move into Si => N-doping

N-doping (on Ga sites)

P-doping (on As sites)

Earliest technological "joining" of semiconductors (late 1960's):

GaAs (Eg = 1.42 eV, atom spacing = 5.65 A)

+ AlAs (Eg = 2.2 eV, atom spacing = 5.66 A)



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However, they weren't restricted to pure GaAs / AlAs

They could also use Alloys = Mixtures of "compatible" semiconductors

BOTH Ga or Al work on the column III site, can use either one or a mixture of both:

GaxAl1-xAs = "Alloy" of GaAs with AlAs "Gallium Aluminum Arsenide"

x = fraction of GaAs

1-x (the remainder) = fraction of AlAs

For example, if x = 0.5 (50% GaAs + 50% AlAs) the alloying process would be:

GaAs AlAs

0.5 x + 0.5 x



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Note:

On the column III sites, theGa and Al atoms can berandomly distributed

=> Ga0.5Al0.5As

Can also get even more complicated with alloys like GaAs with AlP

- III atoms (Ga or and Al) can switch=> GaxAl1-xAsyP1-y "Gallium Aluminum Arsenide Phosphide"

- V atoms (As and P) can switch

III V III V

V III V III

III V III V

III-Site: Al or Ga

x = fraction Ga

V-Site: As or P

y = fraction As



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This works because crystal still retains average of 4 bonding electrons per site

So a III atom (short one electron) must bond to a V atom (w/ extra electron)

Same for II-VI semiconductors, e.g. CdTe, ZnSe...

Doesn't matter for IV-IV alloys because all atoms have 4 electrons

The advantage of a semiconductor alloy? Adjustable bandgap!

2.2 eV

For GaxAl1-xAs alloy: Eg x( )

1.42 eV

01 x (Ga fraction) =>

So now, grow layer(s) of GaxAl1-xAs on top of GaAs crystal substrate:

GaAs substrate GaxAl1-xAs AlAs



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Get bandgap of:2.2 eV

1.42eV

Position =>

= SEMICONDUCTOR HETEROJUNCTION

Main research topic of the Bell Labs department I hired into and later headed

Thumbnail history of heterojunctions (much of it due to Bell Labs):

Late 1960's: GaxAl1-xAs on GaAs heterojunctions (i.e. alloys of GaAs & AlAs grown on GaAs)

Material Developer: Jerry Woodall (then at IBM Labs)

Breakthrough Application: 1st Semiconductor Laser

by Izuo Hiyashi and Mort Panish (my former boss & mentor)



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1970's: InxGa1-xAsyP1-y on InP (i.e. Alloys InAs, GaAs & InP grown on InP substrates)

InxGa1-xAsyP1-y = x * [ InP ] + (1-x) * [ GaAsyP1-y ]________________

Alloy_______________________________

Alloy of Alloys

How on earth did they think up this crazy alloy of alloys? Answer lies in "rules" given above:



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InxGa1-xAsyP1-y: Materials Developers: Many of my Bell Labs colleagues / employees

Breakthrough Application: Long wavelength fiber-optic compatible lasers

Quartz - Fiber Loss (vertical) versus light wavelength (horizontal) - See also the book's figure 8-13:

For fiber-optic long distance communications, want laser colors at the two minima at 1.3 and 1.5 m

- GaxAl1-xAs is too energetic => its bandgap translates into < 1.3 m

- But from diagram above, see that CAN use InGaAsP alloys instead



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1980's: Si / GexSi1-x (alloys of Si and Ge grown on Silicon substrates)

Challenge: These materials don't fit one another

Si: Ge: 4.2 % larger

So, in an alloy of Si / GexSi1-x, the lattice spacing should be x * (4.2%) large than tha of pure Si !

- Difference of (4.2%) x is enough to destroy crystal order

Trick / Breakthrough: Find crystal growth conditions so that GexSi1-x thin layer will squeeze together to fit onto the Si

"Strained - Layer Epitaxy"

(illustrated by figure below)



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Material Developer: J.C. Bean (then at Bell Labs)

14 Patents / ~ 15 years later: "Killer Application" = GeSi Heterojunction Bipolar Transistor

Now in commercial production @ almost all large IC factories



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Band Diagrams: Need these to select heterojunction materials and to design their devices

Going back to GaAs / AlAs structure as an example:

GaAs ("Gallium Arsenide") : AlAs ("Aluminum Arsenide")

Both are III-V semiconductorsBoth have "zincblende" crystal structure

Atomic spacings ~ identical

So can easily grow one on top of the other

But Eg (GaAs) < Eg (AlAs)

And qGaAs does not equal qAlAs

So, line up with respect to Evacuum reference energy:

Evacuum

q 2

q 1

Eg2Eg1



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Now, bring them together:

Ec

Ec

Ev

Ev

Used vacuum level to align, but once together can erase

(electron does not have to go up to vacuum level to cross junction)

"Band discontinuities" Ec, Ev are real and fixed:

All semiconductors don't have to line up this way. Depends on 's.

For different semiconductors, different's could yield:



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"Staggered" "Flat Ec" "Nested" "Flat Ev"

Assume for the moment we have the "Nested" alignment (it is the most common):

Material 1 Material 2

EcEc

Ec = "Conduction Band

Discontinuity"

Eg1 Eg2

Ev = "Valence Band

Discontinuity"EvEv

NOTE: Ec Ev+ Eg2 Eg1= Eg=

If Ec ~ Ev ~ Eg /2 called a "50-50 split" Or

if Ec = 0.4 Egthen must haveEv = 0.6 Eg called a "40-60 split"



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But wait a second!!!

E => Energy gradient => Force

Yes, at interface, can be forces that can drive carriers across the interface

In above drawing, interfacial electrons and holes are BOTH sucked into left material from right material

Could easily happen based on relative arrangements of their conduction and valence bands:

Material 2 Conduction BandMaterial 1 Conduction Band

Material 1 Valence BandMaterial 2 Valence Band

But BOTH carriers going to left??? Why not:

Material 2 holds onto its valence band electrons more strongly than Material 1

Material 2 holds onto its conduction band electrons more weakly than Material 1

Think of it as a tug of war between adjacent different types of atoms at the interface



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Consider how bands join up at a Si to Ge interface:

Both types of atom form covalent bonds

Inside layers Si=Si or Ge=Ge implies pairs of electrons shared equally, midway between the partner atoms

But for Si=Ge, sharing will NOT be exactly equal and expect electrons to be shifted toward one of the partners

That partner has (by definition) lower valence bandedge (where bonding electons ARE)

And, depending on relative bandgaps, either partner could end up having higher conduction bandedge

What is the effect when add these "heterojunctions to diodes? Stay with "Nested alignment" ~ 50-50 split

Look at three cases: N-N, P-P, N-P

Case #1: Both N-N Heterojunction

t = 0: t > 0:



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HOLD IT ! What has happened to the junction ? "Gauss's Law":

= but with V= get

"Poisson's Eqn.": 2V

= but - qV = bandedge +

constant=> Curvature of bands proportional to net charge at that point

EF

Nd(+)x

xn(-)

x



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= (-): Band curvature = (-) = (+): Band curvature = (+)

More on band diagrams (and continuation of cases) next time!



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Semiconductor Devices 23

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