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Lundstrom ECE 305 S16
ECE-606: Fall 2016
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
Lundstrom ECE 305 S16
e-band diagram under bias
3
EC
EV
x
W
E
−xn xp
V = 0
VA > 0
qVAFn
Fp
q Vbi −VA( )
Lundstrom ECE 305 S16
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
Lundstrom ECE 305 S16
outline
5
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 S16
ECE 255
6
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
7
“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
8
EC
EV
x
E
−xn xp
qVAFn
Fp
q Vbi −VA( )
Lundstrom ECE 305 S16
e-band diagram under bias
9
EC
EV
x
E
−xn xp
qVAFn
Fp
q Vbi −VA( )
Lundstrom ECE 305 S16
P0 = e−ΔE kBT = e−qVbi kBT
NP junction in equilibrium
10
ΔE = qVbiEF
P0
P0 = ?
Lundstrom ECE 305 S16
P0 = e−ΔE kBT = e−qVbi kBT
NP junction in equilibrium
11
ΔE = qVbiEF
P0
P0 =
n0Pn0N
=ni2 NA( )ND
= ni2
NAND
qVbi = kBT lnNDNA
ni2
⎛⎝⎜
⎞⎠⎟
FN→P0
FP→N0
n0P =ni2
NA
Lundstrom ECE 305 S16
NP junction under bias
12
Fn
q Vbi −VA( )
Fp
P
P = e−ΔE kBT = e−q Vbi−VA( ) kBT =P0eqVA kBT
Lundstrom ECE 305 S16
NP junction under bias
13
Fn
q Vbi −VA( )Fp
P
P =P0eqVA kBT
I = qA FN→P − FP→N( )
FN→P = FN→P0 eqVA kBT
FP→N = FP→N0
FN→P
FP→N
Lundstrom ECE 305 S16
questions
14
“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
Lundstrom ECE 305 S16 15
recombination and current
16
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
17
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
18
Fn
q Vbi −VA( )
p0P ≈ NA
Δn WP( ) = 0
Fp
0x
WP
Δn 0( ) = ?
Lundstrom ECE 305 S16
outline
19
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 S16
NP junction in equilibrium
20
ΔE = qVbi
EF
n0P =ni2
NA
p0P = NA
0x
WP
Lundstrom ECE 305 S16
NP junction under forward bias
21
Fn
q Vbi −VA( )
pP ≈ NA
Fp
0x
WP
np =ni2
NA
eqVA kBT
np 0( ) pp 0( ) = ni2eqVA kBT
“Law of the Junction”
Lundstrom ECE 305 S16
excess carriers
22
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( )
Lundstrom ECE 305 S16
Law of the Junction another way
23
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
Lundstrom ECE 305 S16
outline
24
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 S16
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