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Lecture 5 
Junctions! 
R. Treharne 
Nov 6th 2014
Lecture Outline 
L5 
● Drawing Junctions 
● Junction Formation – A Physical Picture 
● Abrupt Junction 
● Deriving Important Junction parameters 
● Deriving diode equation 
● Ideality Factor - Recombination
Metal-Semiconductor 
L5 
metal SC (n-type) 
qφm 
qφn 
qχ 
EC 
EF 
EG 
EV 
Schottky Junction
Metal-Semiconductor 
L5 
metal SC (n-type) 
qφm 
qφn 
qχ 
EC 
EF EF 
EG 
EV 
One Rule! 
● Fermi level must be in equilibrium
Metal-Semiconductor 
L5 
“Intimate Contact” 
metal SC (n-type) 
qφm 
qφn 
qχ 
EC 
EF EF 
EV 
qφm 
qφn 
qχ 
EC 
EV 
metal SC (n-type) 
EG EG
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
Metal-Semiconductor 
L5 
metal SC (p-type) 
qφm 
qφp 
qχ 
EC 
EG 
EF 
EV
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)
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
Semiconductor-Semiconductor 
L5 
qφp 
qχp 
EG 
EF 
p n 
qφn 
qχn 
EF 
EG 
EV 
p-n homojunction 
EC
Semiconductor-Semiconductor 
L5 
qφp 
qχp 
EG 
EF 
p n 
qφn 
qχn 
EF 
EG 
EV 
p-n homojunction 
EF EF 
EC
Semiconductor-Semiconductor 
L5 
qφp 
qχp 
EG 
EF 
p n 
qφn 
qχn 
EF 
EG 
EV 
p-n homojunction 
qφp 
qχp 
EF 
qφn 
qχn 
EG 
EF 
p n 
EG 
EV 
EC 
EC 
EC 
EV
Semiconductor-Semiconductor 
L5 
qφp 
qχp 
EG 
EF 
p n 
qφn 
qχn 
EF 
EG 
EV 
p-n homojunction 
p n 
qVbi 
EF EF
Semiconductor-Semiconductor 
L5 
qφp 
qχp 
EG 
EF 
p n 
qφn 
qχn 
EF 
EG 
EV 
p-n homojunction 
p n 
qVbi 
EF EF 
Ei 
Ei
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
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
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.
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
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
Semiconductor-Semiconductor 
L5 
Examples of p-n heterojections 
Most thin-film PV technologies 
http://pubs.rsc.org/en/content/articlehtml/2013/ee/c3ee41981a#cit111
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
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
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
Physical Picture 
L5 
Lets look at hole current first: 
J p = J p(drift)+J p(diffusion) 
= qμp p ξ−qDp 
dp 
dx
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
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
Physical Picture 
L5 
dEF 
dx 
=0 
TA DA! 
The same is true from consideration of net electron current, Jn
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
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)
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
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
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
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
Under Bias 
L5 
What happens to our ideal p-n 
junction if we put a forward or reverse 
bias across it?
Under Bias 
L5 
p n 
W 
0 wn -wp 
EF 
ZERO BIAS 
Vbi
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
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
L5 
Under Bias 
Of course, you already know all about this:
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
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
L5 
Deriving Shockley Equation 
Steady State again (i.e. no bias) 
Consider electron drift current 
– Ieo from p to n
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
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
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:
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
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]
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
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”
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]
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.
Lecture Summary 
L5 
● Drawing Junctions 
● Junction Formation – A Physical Picture 
● Abrupt Junction 
● Deriving Important Junction parameters 
● Deriving diode equation 
● Ideality Factor - Recombination

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Lecture 5: Junctions

  • 1. Lecture 5 Junctions! R. Treharne Nov 6th 2014
  • 2. Lecture Outline L5 ● Drawing Junctions ● Junction Formation – A Physical Picture ● Abrupt Junction ● Deriving Important Junction parameters ● Deriving diode equation ● Ideality Factor - Recombination
  • 3. Metal-Semiconductor L5 metal SC (n-type) qφm qφn qχ EC EF EG EV Schottky Junction
  • 4. Metal-Semiconductor L5 metal SC (n-type) qφm qφn qχ EC EF EF EG EV One Rule! ● Fermi level must be in equilibrium
  • 5. Metal-Semiconductor L5 “Intimate Contact” metal SC (n-type) qφm qφn qχ EC EF EF EV qφm qφn qχ EC EV metal SC (n-type) EG EG
  • 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
  • 7. Metal-Semiconductor L5 metal SC (p-type) qφm qφp qχ EC EG EF EV
  • 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
  • 10. Semiconductor-Semiconductor L5 qφp qχp EG EF p n qφn qχn EF EG EV p-n homojunction EC
  • 11. Semiconductor-Semiconductor L5 qφp qχp EG EF p n qφn qχn EF EG EV p-n homojunction EF EF EC
  • 12. Semiconductor-Semiconductor L5 qφp qχp EG EF p n qφn qχn EF EG EV p-n homojunction qφp qχp EF qφn qχn EG EF p n EG EV EC EC EC EV
  • 13. Semiconductor-Semiconductor L5 qφp qχp EG EF p n qφn qχn EF EG EV p-n homojunction p n qVbi EF EF
  • 14. Semiconductor-Semiconductor L5 qφp qχp EG EF p n qφn qχn EF EG EV p-n homojunction p n qVbi EF EF Ei Ei
  • 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?
  • 35. Under Bias L5 p n W 0 wn -wp EF ZERO BIAS Vbi
  • 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.
  • 51. Lecture Summary L5 ● Drawing Junctions ● Junction Formation – A Physical Picture ● Abrupt Junction ● Deriving Important Junction parameters ● Deriving diode equation ● Ideality Factor - Recombination