6. Metal-Semiconductor
L5
BAND BENDING!
metal SC (n-type)
qφm
qφn
qχ
EC
EF
EV
metal SC (n-type)
EG
qφn
qχ
EC
EF
EG
EV
qφm
qV qφ bi B
“Built-in Voltage”
Vbi = φm - φs
φB = φm - χ
Barrier Height
8. Metal-Semiconductor
L5
metal SC (p-type)
qφm
qφp
qχ
EC
EG
EF
EV
qφp
metal SC (p-type)
qχ
EC
EV
EF
EG
qφm
qVbi
qφB
Vbi = φm - φp
φB = EG/q - (φm - χ)
9. Metal-Semiconductor
L5
φB = φm - χ Why doesn't this rule hold?
Metal – SC interface effects
● Lattice Mis-match
● High density of defects
● Interface states in gap
15. Semiconductor-Semiconductor
L5
qφp
qχp
EG
EF
p n
qφn
qχn
EF
EG
EV
p-n homojunction
p n
qVbi
EF EF
Ei
Ei
Defining Electrostatic potential, Ψ, for a SC
● Difference between Ei and Ef
qΨn
qΨp
Vbi = Ψn - Ψp
Electrostatic Potential, Ψ
Electron energy, E
KNOW HOW TO DRAW THIS!
Coming back here shortly to
get a more physical picture
16. Semiconductor-Semiconductor
L5
Everybody's favourite homojunction - Silicon
p-type n-type
http://electronics.howstuffworks.com/diode1.htm
Every bit of electronics you own is jam
packed with silicon homojunctions
Extrinsically doped
iPhone 5:
1 billion transistors!
~ 45 nm wide
Mono-C solar cells
~750 μm thick
cell size = 1 cm2
17. Semiconductor-Semiconductor
L5
p-n heterojunction: Different band gaps
qφp
qχp
EFp
p n
qφn
qχn
EFn
ΔEC
ΔEV
wide gap narrow gap
EF
1. Align the Fermi level. Leave
some space for transition region.
18. Semiconductor-Semiconductor
L5
p-n heterojunction: Different band gaps
qφp
qχp
EFp
p n
qφn
qχn
EFn
ΔEC
ΔEV
wide gap narrow gap
EF
2. Mark out ΔEC and ΔEC at mid-way
points
ΔEC
ΔEC
19. Semiconductor-Semiconductor
L5
p-n heterojunction: Different band gaps
qφp
qχp
EFp
p n
qφn
qχn
EFn
ΔEC
ΔEV
wide gap narrow gap
EF
3. Connect the C.B. and V.B.
keeping the band gap constant in
each material
Difference in band gaps give rise to
discontinuities in band diagrams. This
limits the carrier transport by introducing
potential barriers at the junction
Barrier to hole transport
20. Semiconductor-Semiconductor
L5
Examples of p-n heterojections
Most thin-film PV technologies
http://pubs.rsc.org/en/content/articlehtml/2013/ee/c3ee41981a#cit111
21. Semiconductor-Semiconductor
L5
Examples of p-n heterojections
III-V multi-junction devices
Being able to control the band offsets of a material can help eliminate barriers
22. Physical Picture
L5
Drawing band schematics for junctions all day is fun, but what is actually going on here?
http://www.youtube.com/watch?v=JBtEckh3L9Q
Excellent qualitative description
6:27 – Didactic Model of Junction formation
23. Physical Picture
L5
Equilibrium Fermi Levels – Explained – our one rule for drawing band schematics!
In thermal equilibrium, i.e. steady state: not net current flows
J=J (drift )+J (diffusion)=0
http://en.wikipedia.org/wiki/File:Pn-junction-equilibrium.png
notice position
of “metallurgical”
junction. Usually
call this x=0
24. Physical Picture
L5
Lets look at hole current first:
J p = J p(drift)+J p(diffusion)
= qμp p ξ−qDp
dp
dx
25. Physical Picture
L5
Lets look at hole current first:
J p = J p(drift)+J p(diffusion)
= qμp p ξ−qDp
dp
dx
mobility: hole distribution: e-field: diffusion coefficient
μp=
e τ
mh
p=ni e(Ei−EF)/ kB T
OK to use Boltzmann as
approx to Fermi-Dirac
dp
dx
p
(dE= i
k T dx
B−
dEF
dx )
ξ=1e
dEi
dx
Dp=(k BT /e )μp
Einstein relation
will use this shortly
26. Physical Picture
L5
J p = J p(drift)+J p(diffusion)
= qμp p ξ−qDp
dp
dx
= μp p
dEi
dx
−μp p(dEi
dx
−
dEF
dx )
0 = μp p
dEF
dx
Condition for steady state
27. Physical Picture
L5
dEF
dx
=0
TA DA!
The same is true from consideration of net electron current, Jn
28. Abrupt Junction
L5
charge density
Hook & Hall, “Soild State Physics”, 2nd ed. Chap 6, p. 171
outside depletion region:
p = NA and n = ND
Space-charge condition
NAwp = Ndwn
i.e. areas of rectangles must be the same
electric field
electrostatic
potential
Max field (at x=0):
ξm = qNDxn/εs = qNAxp/εs
I've already called this Vbi
ξm
Area = Vbi
Vbi=ϕn−ϕp=
k BT
q
ln(NA ND
2 )
ni
29. Abrupt Junction
L5
charge density
2 ) Lets derive this!
Hook & Hall, “Soild State Physics”, 2nd ed. Chap 6, p. 171
outside depletion region:
p = NA and n = ND
Space-charge condition
NAwp = Ndwn
i.e. areas of rectangles must be the same
electric field
electrostatic
potential
Max field (at x=0):
ξm = qNDxn/εs = qNAxp/εs
I've already called this Vbi
ξm
Area = Vbi
Vbi=ϕn−ϕp=
k T
ln(NNBA D
q
ni
(but skip if I'm being slow)
30. Abrupt Junction
L5
From Earlier:
ϕ =−1e
(Ei−EF )
What is in ϕ p and n regions away from junction?
Here
p=ni e(Ei−EF)/ kB T n=ni e(Ei−EF )/ kB T and
ϕp≡−1
e
(Ei−EF)|x⩽−x p
=−
k BT
e (NA
ni )
ϕn≡−1e
(Ei−EF)|x⩾−xn
=−
k BT
e (ND
ni )
S. M. Sze, “Semiconductor
Devices: Physics and
Technology”, Chap 4, p.90
31. Abrupt Junction
L5
Hence:
Vbi=ϕn−ϕp=
k BT
q
ln(NA ND
2 )
ni
want to know what this is for Si?
Go to: http://pveducation.org/pvcdrom/pn-junction/intrinsic-carrier-concentration
Next Question:
How do I calculate ϕ
in the depletion region? (And why would I want to?)
Must solve Poisson's Equation
∇2ϕ=−ρε
Go read some electrostatics if you're interested
32. Abrupt Junction
L5
∇2ϕ=−ρε
x)={−Ne
A ρ(+Ne
D 0
−wp< x<0
0< x<wn
elsewhere
ξ=−
d ϕ
dx
={−
N A e
ε ( x+wp)
ND e
ε ( x+wn)
−wp< x<0
0< x<wn
ϕ( x)={−
NA e
2ε
(x+wp)2
ND e
2ε
(x+wn)2
INTEGRATE ONCE
INTEGRATE TWICE
−wp< x<0
0< x<wn
33. Abrupt Junction
L5
Vbi=ϕ(x=wn)−ϕ( x=−wp)
Vbi=
e
2 ε
2+ND wn 2
(NA wp
)
Vbi
using result:
wn=√ 2 ε NAV bi
eND (NA+ND)
wp=√ 2 εNDV bi
eN A(N A+ND)
and with some jiggery pokery:
W=wp+wn=√2 ε
e (N A+ND
N A ND )
NA wp=NDwn
34. Under Bias
L5
What happens to our ideal p-n
junction if we put a forward or reverse
bias across it?
36. Under Bias
L5
p n
W
0 wn -wp
EF
FORWARD BIAS
Vbi - V
EF
+ -
● Electron energy levels in p side
lowered relative to those in n side
● Energy barrier qVbi reduced
● Flow of electrons from n to p
increases
● Flow of holes from p to n increases
● Depletion width decreases
37. Under Bias
L5
p n
W
0 wn -wp
EF
REVERSE BIAS
Vbi + V
EF
- +
● Electron energy levels in p side
raised relative to those in n side
● Energy barrier qVbi increase by V
● Flow of electrons from n to p
reduced
● Junction width increases
38. L5
Under Bias
Of course, you already know all about this:
39. Ideal dark JV curve
L2
J0 = 1 x 10-10 A/cm2
J=J 0 [exp( eV
k BT )−1 ]
http://pveducation.org/pvcdrom/pn-junction/diode-equation
40. Ideal dark JV curve
L2
J0 = 1 x 10-10 A/cm2
J=J 0 [exp( eV
k BT )−1 ]
Lets derive this by considering
our band schematic. Fun fun
fun!
http://pveducation.org/pvcdrom/pn-junction/diode-equation
41. L5
Deriving Shockley Equation
Steady State again (i.e. no bias)
Consider electron drift current
– Ieo from p to n
42. L5
Deriving Shockley Equation
Steady State again (i.e. no bias)
Consider electron drift current
– Ieo from p to n
I e0=e×(generationrate /volume)
×(volume within Le of depletion zone)
W
Le
J e 0=e (
np
τ p
) Le
43. L5
Deriving Shockley Equation
Steady State again (i.e. no bias)
Consider electron drift current
– Ieo from p to n
I e0=e×(generationrate /volume)
×(volumewithin Le of depletion zone)
J e 0=e (
np
τ p
) Le
W
Le
number of electrons on p side
electron lifetime
Le=√De τ p
diffusion length
44. L5
Deriving Shockley Equation
Steady State again (i.e. no bias)
Consider electron drift current
– Ieo from p to n
W
Le
By assuming that all acceptors on
p side are ionized:
np=
n2
i
p p
=
n2
i
NA
J e 0=
2
eDe ni
Le NA
can re-write Je0 as:
45. L5
Deriving Shockley Equation
What happens under forward bias?
● Drift (p to n)? - Nothing. No
potential barrier in this
direction
● Diffusion (n to p)? - Increases
by exponential factor:
k BT ) If we assume occupation
exp ( eV
of states in CB is given
by Boltzmann distribution
46. L5
Deriving Shockley Equation
What happens under forward bias?
● Drift (p to n)? - Nothing. No
potential barrier in this
direction
● Diffusion (n to p)? - Increases
by exponential factor:
k BT ) If we assume occupation
exp ( eV
of states in CB is given
by Boltzmann distribution
Net electron current:
J e =−J e 0+J e 0 exp(eV
k BT )
= J e 0[exp(eV
k BT )−1]
47. L5
Deriving Shockley Equation
Can do the same for the net hole current:
J h=J h 0 [exp( eV
k BT )−1] J h 0=
2
eDH ni
Lh ND
where
48. L5
Deriving Shockley Equation
Can do the same for the net hole current:
J h=J h 0 [exp( eV
k BT )−1] J h 0=
2
eDH ni
Lh ND
where
Therefore: Net Current!
J=J e+Jh=J0 [exp( eV
kBT )−1] J 0=J e 0+J h0=eni
2( De
Ln NA
+
Dh
LhN A )
ideal diode equation reverse saturation current,
a.k.a “dark current”
49. L5
Deriving Shockley Equation
Previous derivation ignores the recombination and generation of
carriers within the depletion layer itself!
J=J 0[exp( eV
n k BT )−1]
50. L5
Deriving Shockley Equation
Previous derivation ignores the recombination and generation of
carriers within the depletion layer itself!
J=J 0[exp( eV
n k BT )−1]
ideality factor – a complete FUDGE
Worth reading up about – but don't stress out about it.
● S. M. Sze, “Semiconductor Devices”, 2nd ed. Chap 4, pp 109 – 113
● Hook & Hall “Solid State Physics”, 2nd ed. Chap 6, pp 178 – 179
I have these books if you want them.