Semiconductor Devices 16

8/10/2019 Semiconductor Devices 16

1/18

Semiconductor Devices - Hour 16 P-N Junctions Part 1: Concepts, Charges and Fields

Today: Apply our bag of tools to the most basic device of microelectronics, the P-N junction

- Basic element of all transistors -Device onto itself in form of diode / detector / solar cell

Consider separate pieces of P-type and N-type semiconductor

Ec

EiP-type: =>EF

Ev

- Neutral Si atoms (not shown)

- Fixed negative acceptor ions

- Mobile positive holes

By catching electrons, acceptors pull

down Fermi Energy (electron filling level)

EcEF

N-type: => Ei

Ev- Neutral Si atoms (not shown)

- Fixed positive donorions

- Mobile negative electrons

By giving up electrons, donors push

up Fermi Energy (electron filling level)

Semiconductor_Devices_16.xmcd 1 2/2/2009


2/18

Or in terms of equations for P-type material's lowered Fermi Level:

Using E offsets at left: Ec Using E offset at right:

Ec EF

no Nc e

Ec EF( )k T= no ni e

EF Ei

k T=EiE

iE

F

EFEF Ev

po Nv e

EF Ev( )k T= po ni e

Ei EF

k T=Ev

Same equations work for N-typematerial's raised Fermi Level

Using E offset at right:Using E offsets at left: Ec

Ec EF

EF no ni e

EF Ei

k T=no Nc e

Ec EF( )k T= Ei EFEi

EF Evpo ni e

Ei EF

k T=po Nv e

EF Ev( )k T= Ev



3/18

Position pieces of P-type and N-type material side by side (and draw EVERYTHING!)

Ec

Ev

p+

n-

Nd+

Na-

Total charge = 0!! total Total charge = 0!!



4/18

Bring the pieces together and (magically) look so quickly that there is no time for rearrangement

Ec Abrupt step in EF

Ev

p+


5/18

Donors and Acceptors are out of luck: They are atoms that simply can't move at room temperature!

But conduction band electrons & valence band holes can and WILL move!

Hole gradient:

Jdiffusion_p q Dpx

pd

d= => huge hole diffusion to right!P-side N-side

=> Holes entering N-type side find MANY electrons to recombine with

Electron gradient:

Jdiffusion_n q Dn

x

nd

d

= => huge electron diffusion to left!!


6/18

"Steady-state" configuration Region at junction totally depleted of carriers!

xp xn "Depletion Layer"

Total width = W

Width on p-side = xp

Width on n-side = xn

Ec W

Ev

p+

n-

Nd+

Na-

Dipole of ionstotal

Blocking Electric Field!



7/18

1) Steps in carrier concentration (also manifested by step in EF) => massive carrier flow across the junction

2) Electronsentering P-side recombine with holes - Holesentering N-side recombine with electrons

3) Outward flow of carriers + loss to recombination => regions depleted of carriers leaving Na- / Nd

+ exposed

4) Na-


8/18

On P-side: Ec

p ni

e

Ei EF

k T=Ei

Ei EF If acceptor concentration moderate (< Nv)

p = Na-~ Na_total combine with above:

EF

Ev

Na_total ni e

Ei EF

k T=

Inverting this to get energy offset: Offsetp Ei EF= k T lnNa_total

ni

=

On N-side: Ecn ni e

EF Ei

k T=EFEF Ei

Ei But if donor concentration is moderate (< Nc)

n = Nd+

~ Nd_total combine with above:

Ev

Nd_total ni e

EF Ei

k T=

Inverting to get energy offset: Offsetn EF Ei= k T ln

Nd_total

ni

=



9/18

Add these two offsets to get total offset / step in energy going across junction

Etotal Offsetp Offsetn+= k T lnNa_total

ni

k T ln

Nd_total

ni

+=

Using ln(a) + ln(b) = ln(ab)

Etotal k T lnNa_total Nd_total

ni2

= Etotal k T lnNa Nd

ni2

=or in more common notation

Divide by q to convert to the voltage spontaneously formed voltage step across the junction

Important:

Na= Acceptor concentration on P-side

Nd= Donor concentration on N-side

Vbik T

qln

Na Nd

ni2

= "Built-in Junction Voltage"

TEST ALERT: Vbihas nothing to do with voltages that are applied via external batteries or power supplies

Vbiis spontaneous internal voltage developed by the rearrangement of holes & electrons!

Misunderstanding of this = One of John's Most Popular Test Errors!



10/18

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

The "Depletion Approximation"

From carrier formulas, know that: Carrier concentration e(displacement Fermi Energy)/kT

W Carriers

pEi =>

EF

n

Because carrier concentration changes as EXPONENTIAL of EF-Ei, carrier changes are more abrupt

DEPLETION APPROXIMATION = Assume edges of carrier profiles are in VERTICAL:

p

p

=>

n n



11/18

Then the NET charge in the depletion layer (between the p and n regions) is now due acceptors & donors alone

Net charge: Or:

Charge xnWNd

xp xn

x

NaxpP-side N-side

Can now solve for (x), W, xpand xn

For simplicity, reset origin to the "Metallurgical Junction"between P-N (boundary between Naand Nd)

Nd

xpx

And again haul out Gauss's Law:

x( ) x( )

=xn

Na

Integrate once:



12/18

x( )xp

x

x'

x'( )

d= For xp x 0 , know charge density is -qNa

x( )

xp

x

x'q Na

d=q Na x

q Na xp( )

+=q Na x xp+( )

= for xp x 0

x( )q Na x xp+( )

= for xp x 0 which gives: 0( )

q Na xp

= (equations 1 & 2)

Integrating from 0 to xn:

x( ) x' x'( )

d= But for 0 x xn , know that charge density is +q Nd

x( ) x'q Nd

d=q Nd x

C+= Solving for (0) and setting equal to (0) found in equation 2

Cq Na xp

= which then gives



13/18

x( )q Nd x

q Na xp

= for 0 x xn Giving global results for

x( )q Na x xp+( )

= for

xp x 0

0( )q Na xp

=

x( )q Nd x

q Na xp

= for 0 x xn

(equations 3, 4 & 5)

Know that must be constant outside these ranges - and that constant must be zero (no charge!)

From first equation, at left end of depletion layer: xp( ) 0= fine!

From third equation, at right end of depletion layer: xn( )q Nd xn

q Na xp

= must also = 0!

Implies: Nd xn Na xp= (equation 6)

Total charge left of junction: Qleft q Na xp= Total charge right of junction: Qright q Nd xn=



14/18

So equation 6 is equivalent to saying the magnitude of the total charge on each side of the junction is equal

Makes perfect sense: We built up uncovered ionic charges as we lost electrons and holes

But when we lost a hole we also lost an electron (they recombined!)

Summarizing and Q:

Nd

Qright xp xnxp

xnQleft

Na

Qleft Qright= max 0( )=q Na xn

=q Nd xn

=

Now recalling that = - Voltage, can integrate to get V(x)

xp x 0 V x( )xp

x

x' x'( )

d= Substituting in from equation 3 above



15/18

V x( )xp

x

x'

q Na x' xp+( )

d=

q Na

xp

x

x'x' xp+( )

d=

q Na

0

x xp+

x'x'

d=

q Na

x xp+( )2

2=

xp x< 0V x( )q Na

x xp+( )2

2= (equation 7)

And similarly integrating to right of x = 0

0 x xn V x( ) x' x'( )d= x'

q Nd

x'

q Na

xp

d= use equation 6: Na xp Nd xn=

V x( ) x'q Nd x'

q Nd xn

d=q Nd

x'x' xn

d=q Nd

x2

2

xn x

C'+=

V x( )q Nd

x2

2xn x

C'+= 0 x xn< (equation 8)



16/18

But Voltage must be continuous at x = 0 (or would have infinite ) so equations 7 & 8 must be equal there

q Na

0 xp+( )2

2

q Nd

02

2xn x

C'+= =>

q Na xp2

2 C'=

Putting this into equation 8, and summarizing

x < xp V = 0

xp x 0 V x( )q Na

2x xp+( )

2=

0 x xn V x( )q Nd

xn x

x2

2

q Na

2 xp

2+=

V x( )

parabolic curvature down

parabolic curvature up

x

xp

xn0



17/18

But Voltage = Potential Energy / Positive Charge

So negative of this must be: Potential Energy / Negative Charge = Our Band Diagrams!

So net Voltage step = Net junction energy step / q = Vbifrom early in lecture!

Solve for voltage at xn:

V xn( )q Nd

xn

2xn

2

2

q Na

2 xp

2+=q Nd

2xn

2q Na

2xp

2+=

Vbi

q Nd

2 xn

2q Na

2 xp

2+= (two unknowns xn& xp)

Compared to earlier derivation of Vbi

:



18/18

Vbik T

qln

Na Nd

ni2

=(no unknowns)

Combined with equation 6: Na xp Nd xn= comparison will yield explicit values of xp, xnand W !

Next Time!


Semiconductor Devices 16

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