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ECE-305: Fall 2016 PN Junctions Professor Peter Bermel Electrical and Computer Engineering Purdue University, West Lafayette, IN USA [email protected] Pierret, Semiconductor Device Fundamentals (SDF) Chapter 4 (pp. 195-213) 9/26/2016 1
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Bermel ECE 305 F16
ECE-305: Fall 2016
PN Junctions
Professor Peter BermelElectrical and Computer Engineering
Purdue University, West Lafayette, IN [email protected]
Pierret, Semiconductor Device Fundamentals (SDF)Chapter 4 (pp. 195-213)
9/26/2016 1
outline
1) PN Junctions qualitative
2) PN Junctions quantitative
Bermel ECE 305 F169/26/2016 2
NP junction (equilibrium)
N P
xp-xn 0
“transition region”
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NP junction
N P
transition region
electric field, electrostatic potential, n(x), p(x), rho(x)
xp-xn 0
+
-
EVL >VR
r < 0
NA-
r > 0
ND+
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energy band approach
EC
EV
EF
Ei V = 0
EC
EV
Ei
1) Fermi-level must be constant in equilibrium.
2) Positive electrostatic potentials lower the electron energy
3) Left side must develop a positive potential, Vbi.
EF
qVbiV = Vbi
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eq. energy band diagram
EFEF
1) Begin with EF
2) Draw the E-bands where you know the carrier density3) Electrostatic potential by flipping E-band upside down. 4) E-field from slope5) n(x), p(x) from the E-band diagram6) rho(x) from n(x) and p(x)7) diffusion current from (5) or from (6)
EC x( ) = EC- ref - qV x( )
E x( ) =
1
qdEC x( ) dx
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energy band diagram
7
EF
EC
EV
x
E
Ei
x = xpx = 0x = -xn
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“read” the e-band diagram
1) Electrostatic potential vs. position
2) Electric field vs. position
3) Electron and hole densities vs. position
4) Space-charge density vs. position
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electrostatics: V(x)
V
x
N P
xp-xn
qVbi
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electrostatics: E (x)
E
xN P
r = q p0 x( ) - n0 x( ) + ND+ x( ) - NA
- x( )éë ùû
xp-xn
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carrier densities vs. x
log10 n x( ), log10 p x( )
xN P
xp-xn
p0P = NA
p0N = ni2 ND
n0N = ND
n0 p = ni2 NA
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electrostatics: rho(x)
r
x
N P
r = q p0 x( ) - n0 x( ) + ND+ x( ) - NA
- x( )éë ùû
xp
-xn
qND
-qNA
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the built-in potential
EC
EV
EFP
Ei V = 0
EC
EV
Ei
EFN
qVbiV = Vbi
n0 = nieEFN -Ei( ) kBT p0 = nie
Ei-EFP( ) kBT
n0p0= e EFN -EFP( ) kBT = eqVbi kBT Vbi =
kBT
qlnn0p0ni2
æ
èçö
ø÷=kBT
qlnNDNAni2
æ
èçö
ø÷
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outline
1) PN Junctions qualitative
2) PN Junctions quantitative
Bermel ECE 305 F16
✓
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Using “the depletion approximation”
15
“the semiconductor equations”
Three equations in three unknowns:
In steady state equilibrium, we only need to solve the Poisson equation
How do we calculate rho(x), E(x), and V(x)?
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NP junction (equilibrium)
N P
xp-xn 0
“transition (depletion) region”
Bermel ECE 305 F16
Vbi =kBT
qlnNDNAni2
æ
èçö
ø÷
V = 0V =Vbi
1) What is the width of the depletion region?2) What is the maximum electric field?
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the Poisson equation
d
dxeSE( ) = r x( )
dE
dx=r x( )eS
=r x( )KSe0
dE
dx=r x( )KSe0
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the “depletion approximation”
dE
dx=r x( )KSe0
r
x
N P
-xn
r = +qND
xp
r = -qNA
qNDxn = qNAxp
NDxn = NAxp
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the electric field
dE
dx=r x( )KSe0
r
xN P
xp
-xn
r = -qNA
dE
dx=qNDKSe0
dE
dx= -
qNAKSe0
E x( ) > 0
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the electrostatic potential
dE
dx=r x( )KSe0
xN P
xp-xn
E
E x( ) = -
dV
dx
W = xn + xP
E 0( ) =qNDKSe0
xn
NDxn = NAxPxn =NA
NA + NDW
dE
dx=qNDKSe0
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calculating �(�) from ℰ(�)
xP
xp-xn
E
V x( ) = - E x( )x
xp
ò dx
V x( )
V = 0
V =Vbi
See Pierret, SDF, pp. 212-213
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Vbi = E x( )- xn
xp
ò dx
Vbi =
1
2E 0( )W
ℰ(0)
�
E x( ) = -
dV
dx
summary
xN P
xp-xn
E
W = xn + xP
NDxn = NAxP
xn =NA
NA + NDW
xp =ND
NA + NDW
W =2KSe0q
NA + NDNDNA
æ
èçö
ø÷Vbi
é
ëê
ù
ûú
1/2
E 0( ) =2qVbiKse0
NDNANA + ND
æ
èçö
ø÷é
ëê
ù
ûú
1/2
Vbi =kBT
qln
NDNAni2
æ
èçö
ø÷
E 0( ) =
2VbiW
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conclusions
Developed a qualitative procedure to sketch ℰ, �, �, and � for PN junctions
Used Poisson’s equation and ‘depletion approximation’ to quantify these values
This approach also gives us the width of the ‘depletion region’ on both sides of the junction (�� and ��), plus the ‘built-
in’ voltage ���
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