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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)
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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
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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!!
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Bring the pieces together and (magically) look so quickly that there is no time for rearrangement
Ec Abrupt step in EF
Ev
p+
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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!!
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"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!
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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-
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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
=
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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!
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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
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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:
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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
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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=
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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
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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)
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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
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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
:
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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!
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