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Lundstrom ECE 305 S15
ECE-305: Spring 2015
Ideal Diode Equation
Professor Mark Lundstrom Electrical and Computer Engineering
Purdue University, West Lafayette, IN USA lundstro@purdue.edu
2/25/15
Pierret, Semiconductor Device Fundamentals (SDF) pp. 235-259
equilibrium e-band diagram
2
EFEF
EC
EV
x
W
E
qVbi
I = 0
VA = 0
−xn xp
e-band diagram under bias
3
EC
EV
x
W
E
−xn xp
V = 0
VA > 0
qVAFn
Fp
q Vbi −VA( )
Fn − Fp = qVA
electrostatics under bias
4
W = 2KSε0q
NA + ND
NDNA
⎛⎝⎜
⎞⎠⎟Vbi −VA( )⎡
⎣⎢
⎤
⎦⎥
1/2
E 0( ) = 2 Vbi −VA( )
W
xn =NA
NA + ND
W
xp =ND
NA + ND
W
N P
ND = 1016
depletion region xp−xn 0
E 0( )
Vbi =kBTqln NDNA
ni2
⎛⎝⎜
⎞⎠⎟
WNA = 1016
Question: n-N junction
5
EF
EC
EVx
E
Ei
x = Lx1x = 0 x2 x3
ND = 1015 ND = 1017Where is the depletion region?
a) b) c) d) e)
0 ≤ x < x1x1 ≤ x < x2x2 ≤ x < x3x1 ≤ x < x3x3 ≤ x < L
outline
6
1) Ideal diode equation (qualitative)
2) Law of the Junction
3) Ideal diode equation (long base)
4) Ideal diode equation (short base)
Lundstrom ECE 305 S15
ECE 255
7
VA
− VA +
I mA( )
0.6 − 0.7 V
I = I0 eqVA kBT −1( )
“ideal diode equation” “Shockley diode equation”
I = I0 eqVA kBT −1( )
questions
8
“ideal diode equation” “Shockley diode equation”
I = I0 eqVA kBT −1( )
1) Why does the current increase exponentially with the applied forward bias?
2) Why is the reverse bias current independent of the reverse bias voltage?
e-band diagram under bias
9
EC
EV
x
E
−xn xp
qVAFn
Fp
q Vbi −VA( )
e-band diagram under bias
10
EC
EV
x
E
−xn xp
qVAFn
Fp
q Vbi −VA( )
P0 = e−ΔE kBT = e−qVbi kBT
NP junction in equilibrium
11
ΔE = qVbiEF
P0
P0 = ?
P0 = e−ΔE kBT = e−qVbi kBT
NP junction in equilibrium
12
ΔE = qVbiEF
P0
ND ×P0 = ni
2 NA( )×1
P0 =
ni2
NDNA
qVbi = kBT lnNDNA
ni2
⎛⎝⎜
⎞⎠⎟
FN→P0
FP→N0
FN→P0 ∝ND ×P0
FP→N0 ∝ ni
2
ND
×1
FN→P0 = FP→N
0
NP junction under bias
13
Fn
q Vbi −VA( )
Fp
P
P = e−ΔE kBT = e−q Vbi−VA( ) kBT =P0eqVA kBT
NP junction under bias
14
Fn
q Vbi −VA( )Fp
P
P = e−ΔE kBT = e−q Vbi−VA( ) kBT =P0eqVA kBT
I = qA FN→P − FP→N( )
FN→P = FN→P0 eqVA kBT
FP→N = FP→N0
FN→P
FP→N
I = qAFN→P FB( )
I = −qAFP→N0 RB( )
questions
15
“ideal diode equation” “Shockley diode equation”
I = I0 eqVA kBT −1( )
1) Why does the current increase exponentially with the applied forward bias?
2) Why is the reverse bias current independent of the reverse bias voltage?
Excess carriers under forward bias
q Vbi −VA( )
p0P ≈ NA
Fp
nP >> n0Pn0P =
ni2
NA
recombination and current
17
minority carriers injected across junction
Fn FPqVA
−VA +ID
Every time a minority electron recombines on the p-side, one electron flows in the external current.
excess electrons on the p-side of junction
18
Fn
q Vbi −VA( )
p0P ≈ NA
nP >> n0Pn0N ≈ ND
Fp
0x
What is Δn(x) on the p-side? Ans. Solve the MCDE.
WP
boundary conditions for the MCDE
19
Fn
q Vbi −VA( )
p0P ≈ NA
Δn WP( ) = 0
Fp
0x
WP
Δn 0( ) = ?
outline
20
1) Ideal diode equation (qualitative)
2) Law of the Junction
3) Ideal diode equation (long base)
4) Ideal diode equation (short base)
Lundstrom ECE 305 S15
NP junction in equilibrium
21
ΔE = qVbi
EF
n0P =ni2
NA
p0P = NA
0x
WP
NP junction under forward bias
22
Fn
q Vbi −VA( )
pP ≈ NA
Fp
0x
WP
np =ni2
NA
eqVA kBT
np 0( ) pp 0( ) = ni2eqVA kBT
“Law of the Junction”
excess carriers
23
Fn
q Vbi −VA( )
pP ≈ NA
Fp
0x
WN
np =ni2
NA
eqVA kBT Δnp 0( ) = ni2
NA
eqVA kBT − np0
“Law of the Junction”
Δnp 0( ) = ni2
NA
eqVA kBT −1( )
Law of the Junction another way
24
Fn
q Vbi −VA( )
pP ≈ NA
Fp
0x
WN
np 0( ) = nie Fn−Ei( ) kBTAssume QFLs are constant across the transition region
pp 0( ) = nie Ei−Fp( ) kBT
np 0( ) pp 0( ) = ni2e Fn−Fp( ) kBT
np 0( ) pp 0( ) = ni2eqVA kBT
outline
25
1) Ideal diode equation (qualitative)
2) Law of the Junction
3) Ideal diode equation (long base)
4) Ideal diode equation (short base)
Lundstrom ECE 305 S15
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