Semiconductor Physics
INF 5440 - CMOS billedsensorer
CHARGE CARRIERS
GENERATION
AND
RECOMBINATION
AO
10V
2.1
Semiconductor Physics
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INJECTION LEVEL
ND=1016
1018
nn0
Equilibrium: nn0pno = ni2 = 104 x 1016 = 10 20
1016
Low injection level (e.g. due to light): ∆n = ∆p << ND
Carrier concentration (cm-3)
pn = 10 + 10 ≈ 10
nn = 1016 + 1012 ≈ 1016
12
pn
1014
Adding 1012 carriers /cm3
4
nn
nn
12
High injection level:∆n = ∆p ≈ ND (not relevant)
Adding 1016 carriers /cm3
pn = 104 + 1016 ≈ 1016
nn = 1016 + 1016 = 2x1016
pn
1012
ni, Si @ 300 ºK
1010
108
106
pn0
104
102
Equilibrium
Low
injection level
High
injection level
Ref: Grove
AO
10V
2.2
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GENERATION / RECOMBINATION
Definitions:
Gth: Dark (thermal) generation rate
R: Recombination rate
U = R - Gth: Net recombination rate
GL: Generation rate due to absorption of light
Uniformly illuminated semiconductor.
Net change in carrier concentration pn:
dp n ( t )
---------------- = G L + G th – R = G L – U
dt
Turning light off:
GL = 0
p n ( t ) – p n0
dp n ( t )
---------------- = – -------------------------dt
τp
Solving the differential equation. Initial condition pn(0)=pL:
p n ( t ) = p n0 + ( p L – p n0 )e
(2.1)
Assuming U is proportional to excess carrier
concentration (concentration beyond equilibrium).
Equilibrium concentration: pn0. Lifetime: τp
1
(2.2)
U = ----- ( p n ( t ) – p n0 )
τp
Steady state (GL=U):
dp n ( t )
--------------- = 0
dt
pn ( t )
AO
= p n0 + τ p G L ≡ p L
ss
(2.5)
1
0.9
0.8
0.7
pn(t)-pn0
pL-pn0
Combining these gives the differential equation:
dp n ( t )
p n ( t ) – p n0
(2.3)
--------------- = G L – -------------------------dt
τp
–t ⁄ τp
τp=10
0.6
0.5
0.4
0.3
τp=1
0.2
τp=0.1
0.1
0
(2.4)
0
1
2
3
4
5
6
7
8
9
10
t
Ref: Grove
10V
2.3
Semiconductor Physics
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Surface recombination:
Charge carriers diffuse towards the surface.
Diffusion Flux:
∂p n ( x, t )
F p = – D p ---------------------∂x
Surface
recombination
(2.6)
e
h
x
∂F
∂p n ( x, t )
--------------------- = – --------p- + G L – U
∂t
∂x
2
∂p n ( x, t )
∂ p n ( x, t )
p n – p n0
+ G L – --------------------------------------- = D p -----------------------2
∂t
τ
p
∂x
(2.7)
Minority carrier
Concentration
Concentration is given by the ‘transport equation’:
pn(x)
τpGL
x
Ref: Grove
Transport equation:
F(x)
K(x,t)
∂K
∆x ------- = F ( x ) – F ( x + ∆x )
∂t
F(x+∆x)
∂K
F ( x ) – F ( x + ∆x )
------- = -----------------------------------------∂t
∆x
∆x
AO
10V
∆x → 0
∂F
= -----∂x
2.4
Semiconductor Physics
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Steady state and boundary conditions.
∂p n ( x, t )
--------------------- = 0
∂t
p ( x = ∞) = p L = p n0 + τ p G L
∂p n
D p --------= s p [ p n ( 0 ) – p n0 ]
∂x
(2.8)
1.2
1
x=0
sp is a proportionality constant, surface
recombination rate. Diffusion is proportional to
the excess carrier concentration.
0.1
1
0.8
pn(x)-pno
pL-pno
Carriers which reach the surface (x=0)
recombine there.
spτp/Lp = 0
10
0.6
spτp/Lp = OO
0.4
Differential equation has the solution:
0.2
sp τ p ⁄ Lp –x ⁄ Lp
p n ( x ) = p L – ( p L – p n0 ) ------------------------------ e
1 + sp τp ⁄ Lp
(2.9)
0
Lp ≡ Dp τp
0
1
2
3
4
5
x/Lp
Lp: Diffusion length
Ref: Grove
AO
10V
2.5
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LATTICE DISLOCATIONS
Dislocations deviates from the perfect periodicity. The surface is an obvious example. Energy states in the band
gap becomes recombination centres, “stepping stones”. These increases the probability of recombination, i.e.
reduce the lifetime of free charge carriers.
Probability of occupied centre at energy level ET:
1
f ( E T ) = ----------------------------------------( E T – E F ) ⁄ kT
1+e
(2.10)
k is Boltzmanns constant and T is absolute temperature.
Probability of an unoccupied centre: 1-f
v th =
5m
T
3k ---- ≈ 10 ---s
m
σ n ≈ 10
– 17
Transition rate a:
m
EC
Termal velosity
ET
capture cross section
r a = v th σ n nN t ( 1 – f ( E T ) )
(2.11)
Transition rate b:
rb = en Nt f ( ET )
(2.12)
Transition rate c:
r c = v th σ p pN t f ( E T )
(2.13)
Transition rate d:
rd = ep Nt ( 1 – f ( ET ) )
(2.14)
EV
a
b
c
d
k = 1.3805 x 10 -23 J / ºK
Ref: Grove
en and ep are emission probability (depend on the distance to conduction band and valence band respectively)
AO
10V
2.6
Semiconductor Physics
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Emission probability:
At equilibrium, process a and b have equal rate and en and ep can be found by setting ra = rb:
v th σ n nN t ( 1 – f ( E T ) ) = e n N t f ( E T )
1 – f ( ET )
( E T – E F ) ⁄ kT
– ( E C – E F ) ⁄ kT
e n = v th σ n ----------------------- n = v th σ n e
Nc e
f ( ET )
e n = v th σ n N c e
– ( E C – E T ) ⁄ kT
(2.15)
Corresponding result for ep:
e p = v th σ p N v e
– ( E T – E V ) ⁄ kT
(2.16)
Ref: Grove
AO
10V
2.7
Semiconductor Physics
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ILLUMINATED SEMICONDUCTOR
Uniform light and steady state *
Electrons enter and leave the conduction band at the same rate:
dn n
--------- = G L – ( r a – r b ) = 0
dt
ra
rb
GL
Holes enter and leave the valence band at the same rate.:
dp n
--------- = G L – ( r c – r d ) = 0
dt
rc
rd
No accumulation of electrons in the recombination centres:
ra – rb = rc – rd
Solve for the occupation factor f, using the expressions for the processes a, b, c, d [(2.11), (2.12), (2.13), (2.14)]
It follows the Fermi-model **
– ( E T – E V ) ⁄ kT
nσ n + σ p N v e
f ( E T ) = -------------------------------------------------------------------------------------------------------------------------------------–
(
E
–
E
)
⁄
kT
–
(
E
–
E
)
⁄
kT
C
T
 + σ p + N e T V

σn  n + Nc e
p
v



AO
(2.17)
Ref: Grove
*
Equilibrium = steady state with out external influence
**
Cannot use the expression for Fermi energy because this is based on equilibrium But we can use the model.
10V
2.8
Semiconductor Physics
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We have from chapter 1:
n = Nc e
p = Nv e
– ( E C – E F ) ⁄ kT
– ( E F – E V ) ⁄ kT
2
ni = Nv Nc e
n = ni e
p = ni e
( E F – E i ) ⁄ kT
( E i – E F ) ⁄ kT
– E G ⁄ ( kT )
EF is valid at equilibrium only, but the model can be used for ET and can be written as:
( E i – E T ) ⁄ kT
nσ n + σ p n i e
f ( E T ) = ------------------------------------------------------------------------------------------------------------------------(
E
–
E
)
⁄
kT
(
E
–
E
)
⁄
kT
T
i
 + σ p + n e i T

σn  n + ni e
p
i



(2.18)
Replacing f(E) in the individual processes, get the net recombination rate: U = ra - r b = rc - r d
2
σ n σ p v th N t [ pn – n i ]
U = -------------------------------------------------------------------------------------------------------------------------------------–
(
E
–
E
)
⁄
kT
–
(
E
–
E
)
⁄
kT
C
T
 + σ p + N e T V

σn  n + Nc e
p
v



(2.19)
Alternatively:
2
σ n σ p v th N t [ pn – n i ]
U = ------------------------------------------------------------------------------------------------------------------------(
E
–
E
)
⁄
kT
(
E
–
E
)
⁄
kT
T
i
 + σ p + n e i T

σn  n + ni e
p
i



(Recall that pn=ni2 in equilibrium only)
AO
10V
(2.20)
Ref: Grove
2.9
Semiconductor Physics
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Special case: σp = σn = σ
2
pn – n i
-------------------------------------------------------------U = v th σN t
ET – Ei
n + p + 2n i cosh  ------------------
 kT 
(2.21)
1.5
pn-ni2 is the deviation from equilibrium and the driving
force for recombination
Maximum recombination rate when ET = Ei,
i.e. ET has the value in the middle of the energy gap.
1
-1
-0.5
0
0.5
x
1
–x
e +e
cosh ( x ) = -------------------2
Ref: Grove
AO
10V
2.10
Semiconductor Physics
INF 5440 - CMOS billedsensorer
We have the expression for net recombination: (2.20)
Using (2.2):
2
σ n σ p v th N t [ pn – n i ]
U = ------------------------------------------------------------------------------------------------------------------------(
E
–
E
)
⁄
kT
(
E
–
E
)
⁄
kT
T
i
 + σ p + n e i T

σn  n + ni e
p
i



p n – p n0
U = -------------------,
τp
Expression for minority carriers in N-type (holes):
Knowing that at low level injection in N-type
semiconductor
nn » pn
nn » ni e
1
τ p = -------------------σ p v th N t
(2.23)
Similar for P-type:
E T – E i ⁄ ( kT )
U ≈ σ n v th N t [ n p – n p0 ]
we can simplify (2.20) for N-type such that:
and lifetime for minority carriers (electrons):
1
τ n = -------------------σ n v th N t
2
σ n σ p v th N t [ p n n n – n i ]
U ≈ -------------------------------------------------------- = σ p v th N t [ p n – p n0 ] (2.22)
σn nn
(2.24)
(2.25)
Given these conditions, the recombination rate
independent on the majority carrier concentration.
That is, the minority carrier concentration
determines the recombination rate.
(Recall that both ni and pn0 represent equilibrium).
Ref: Grove
AO
10V
2.11
Semiconductor Physics
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EXAMPLES ON THE ORIGIN OF RECOMBINATION CENTRES
• Impurities from other groups than III and V in the periodic table gives energy states in the band gap.
• Surface states due to the lattice non-regularity.
Approximately atoms/area (~1015 cm-2). Lower density on oxidized surface (~1011 cm-2)
Ref: Grove
AO
10V
2.12
Photo Diode
INF 5440 - CMOS Image Sensors
THE DIODE
AS
PHOTO SENSOR
AO
10V
2.13
Photo Diode
INF 5440 - CMOS Image Sensors
CURRENT - VOLTAGE CHARACTERISTICS
There are two sources of reverse biased current:
• Generation current
Carriers generated in the depletion region.
The field sweeps carriers out of the depletion
region, electrons to the n-region and holes to
the p-region.
• Diffusion current
Carriers generated outside the depletion
region, but within a diffusion length from the
depletion region. Minority carriers diffuse to
the edge of the depletion region and swept
across by the field.
Lp
V+dV
n
W
Ln
p
Diffusion current
Carriers that are swept across becomes majority
carriers.
There are (almost) zero free carriers in the depletion
region and therefore low probability for
recombination there.
Generation current
AO
10V
2.14
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GENERATION CURRENT:
For σp = σn = σ:
Carriers are quickly driven out of the depletion region.
• pn << ni
σv th N t n i
U = – -------------------------------------Ei – ET
2 cosh  ------------------
 kT 
• No recombination
• VR >> kT/q
Using (2.20), recombination = generation.
(2.27)
Generation current (dark current):
I gen = q U WA j
Generation rate is therefore (p,n<<ni):
(2.28)
2
– σ n σ p v th N t n i
n
U = -------------------------------------------------------------------------------------------- ≡ – -------i- (2.26)
( E T – E i ) ⁄ kT
( E i – E T ) ⁄ kT
2τ 0
σn ni e
+ σp ni e
( E T – E i ) ⁄ kT
1 ni
I gen = --- q ----- WA j
2 τ0
(2.29)
Aj is the cross section of the depletion region
( E i – E T ) ⁄ kT
σn e
+ σp e
τ 0 ≡ ---------------------------------------------------------------------------------2σ n σ p v th N t
We see that “step stones” close to Ei gives the largest
contribution to the generation current.
Ref: Grove
AO
10V
2.15
Semiconductor Physics
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DIFFUSION CURRENT:
Diffusion current (differentiate):
We apply the differential equation (2.7) for surface
recombination on minority carriers.
• The concentration at the edge is 0 due to the
field which sweeps the carriers across the
junction.
( n p0 + τ n G L )
dn p


-------------------------------- Aj
-------I diff, n = – q  – D n
 A j = qD n
dx
L

n
x = 0
(2.31)
• Steady state, no variation with time.
Corresponding for holes in n-region:
( p n0 + τ p G L )
I diff, p = qD p --------------------------------- A j
Lp
2
∂ n
n p – n p0
- = 0
D p -----------p + G L – ------------------2
τ
n
∂x
Far from the depletion region (2.4):
n p ( ∞ ) = n p0 + τ n G L
(2.32)
No injection (no light):
2
n p0
ni
I diff, n = qD n -------- A j = qD n --------------- A j
Ln
NA L
n
At the edge of the depletion region:
np ( 0 ) = 0
2
ni
Solution:
n p ( x ) = ( n p0 + τ n G L ) ( 1 – e
–x ⁄ Ln
)
p n0
I diff, p = qD p -------- A j = qD p --------------- A j
Lp
ND L
p
(2.30)
Ln= diffusion length for the electron in the p-region.
AO
(2.33)
Ref: Grove
10V
2.16
Photo Diode
INF 5440 - CMOS Image Sensors
REVERSE CURRENT VS. TEMPERATURE
VR= 1V
100µA
Circles: Generation current
Squares:Diffusion current
Solid line:Temperature dependence of ni.
Dotted line:Temperature dependence of ni2
10µA
1µA
Generation current is proportional to ni,
Diffusion current is proportional to ni2
• Low temperature:
The generation current dominates
100nA
IR
• High temperature:
The diffusion current dominates.
10nA
1nA
100pA
10pA
1pA
0
1
2
3
4
5
6
1000/T(ºK)
Ref: Grove
AO
10V
2.17
Photo Diode
INF 5440 - CMOS Image Sensors
Light intensity
Illuminance (light flux) has the unit [Lux]=[Lumen/m2]
vs.
Irradiance (power) has the unit [W/m2
Wave length
λ (nm)
380
390
400
410
420
430
440
450
460
470
480
490
500
507
510
520
530
540
550
555
560
AO
Photopic conversion
lm/W
W/lm
0.027
37.03704
0.082
12.19512
0.270
3.70370
0.826
1.21065
2.732
0.36603
7.923
0.12621
15.709
0.06366
25.954
0.03853
40.980
0.02440
62.139
0.01609
94.951
0.01053
142.078
0.00704
220.609
0.00453
303.464
0.00330
343.549
0.00291
484.930
0.00206
588.746
0.00170
651.582
0.00153
679.551
0.00147
683.000
0.00146
679.585
0.00147
10V
Wave length
λ (nm)
570
580
590
600
610
620
630
640
650
660
670
680
690
700
710
720
730
740
750
760
770
Photopic conversion
lm/W
W/lm
650.216
0.00154
594.210
0.00168
517.031
0.00193
430.973
0.00232
343.549
0.00291
260.223
0.00384
180.995
0.00553
119.525
0.00837
73.081
0.01368
41.663
0.02400
21.856
0.04575
11.611
0.08613
5.607
0.17835
2.802
0.35689
1.428
0.70028
0.715
1.39860
0.355
2.81690
0.170
5.88235
0.082 12.19512
0.041 24.39024
0.020 50.00000
2.18
Photo Diode
INF 5440 - CMOS Image Sensors
Sensor efficiency
• Photons with energy larger than the bandgap can generate e-h pair.
• Some of the photons are reflected at the surface and do not contribute.
• Some of the photons are reflected at the silicon oxide surface or Passivation surface and do not contribute.
• Some of the generated charge carriers that are collected recombine fast, and are therefore lost.
AO
10V
2.19
Photo Diode
INF 5440 - CMOS Image Sensors
Reflection factor:
 n 1 – n 2 2
Reflexed
intensity
R = --------------------------------------------- =  ------------------
Incoming intensity
 n 1 + n 2
(2.34)
where n1 and n2 is the refractive index to the interfacing
materials.
Example: SiO2 (n=1.45) and Si (n=4) result in R=22%
Efficient light = incoming light x (1-R)
Passivation
SiO2
Si
AO
10V
2.20
Photo Diode
INF 5440 - CMOS Image Sensors
Fill Factor
Ratio photosensitive area to total pixel area
FF =(APD/Apixel)100%
AO
10V
2.21
Photo Diode
INF 5440 - CMOS Image Sensors
Photon energy
Plank and black body radiation:
x 10
Atoms in a heated body behaves like
harmonic oscillators,
each oscillator can absorb or emit energy, in an
amount proportional to its frequency:
E = hν
(2.35)
where ν is the frequency,
h is Plancks constant: 6.626 x 10-34 Js
5
4
3.5
4500 °C
3
E(λ)
[J/m4]
2.5
4000 °C
2
1.5
The energy is quantized:
3500 °C
1
E n = nhν
3000 °C
0.5
where n is a positive integer: Number of photons.
2
4
6
8
10
λ [m]
Photon wavelength λ = c/ν
12
14
x 10
-7
Visible light: 450nm-700nm
1
8πhc
E ( ν ) = ------------- ----------------------------5 hc ⁄ λkT
λ e
–1
Ref. Alonso-Finn
AO
10V
2.22
Photo Diode
INF 5440 - CMOS Image Sensors
Absorption
Silicon
dΦ = αΦdx
Φ ph ( x ) = Φ 0 e
– αx
(2.36)
Absorption coefficient α depends on the energy, i.e. is
larger for shorter wave lengths.
Photons in the blue range of the spectrum
• Short wave length - high energy
• High probability of e-h pair generation.
• Easily absorbed
10-2
107
10-1
[µm]
106
Photons in the red range of the spectrum
300 ºK
77 ºK
105
104
103
• Short penetration depth.
1
101
102
0.4
0.6 0.8 1.0 1.2
Wave length [µm]
Penetration depth 1/α
Optical absorption coefficient a [m-1]
The flux of photons, with energy higher than the band
gap, decreases as the photons are absorbed and e-h
pair are generated. Thus, the photon flux, Φ(x), decreases with the penetration depth.
108
103
• Long wave length - low energy
• Low probability of e-h pair generation.
• Passes more material before absorption take place.
• Long penetration depth.
AO
Ref. Sze
10V
2.23
Photo Diode
INF 5440 - CMOS Image Sensors
Foveon principle
www.foveon.com
AO
10V
2.24
Photo Diode
INF 5440 - CMOS Image Sensors
Foveon (cont.)
AO
10V
2.25
Photo Diode
INF 5440 - CMOS Image Sensors
Response limits
The responsivity has a lower limit.
The band gap must be exceeded (excitation of electrons).
Lower limit is given by the wave length. Apply (2.35):
hν ≥ E g
E g, Si = 1,11 eV
hc
------ ≥ E g
λ
1 eV = 1,602 ⋅ 10
– 34
– 19
J
8
hc
6,626 ⋅ 10 [ Js ]3 ⋅ 10 [ m ⁄ s ]
λ max = ------ = --------------------------------------------------------------------------- = 1,06 µm
– 19
Eg
1,11 ⋅ 1,602 ⋅ 10 [ J ]
(2.37)
The Responsivity has an upper limit given by the penetration depth.
Recombination rate is high at the surface,
where short wave photons are absorbed,
due to high density of energy levels in the band gap at the surface.
(typically Nts=1011 so ~ 150 results in ts = 1/so ~ 0.7 µs/µm)
AO
10V
2.26
Photo Diode
INF 5440 - CMOS Image Sensors
Relative response
Sensitivity vs. wave length
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
400
Human Eye
Silicon
500
600
700
800
900
1000
1100
Wavelength (nm)
AO
10V
2.27
Photo Diode
INF 5440 - CMOS Image Sensors
Quantum Efficiency (QE)
One definition [ref. Sze]:
Number of generated electron-hole pair per photon hitting the sensor (pixel).
I ph ⁄ q
η = -------------------P opt ⁄ hν
elektrons per time unit
-----------------------------------------------------photons per time unit
(2.38)
Common definition of QE: Number of collected electrons per photon.
Responsivity: The ratio photo current to optical input power [ref. Sze]
I ph
A
ηλ 6
ηq
----R = --------- = ------ = ---------- 10
W
1,24
P opt
hν
(2.39)
– 19
q

λ
λ
qλ
1,602 ⋅ 10
 ------ = ------ = --------------------------------------------------- = --------------------------
 hν hc 6,626 ⋅ 10 – 34 3 ⋅ 10 8 1,24 ⋅ 10 – 6
Responsivity is proportional to the wave length.
AO
10V
2.28
Photo Diode
INF 5440 - CMOS Image Sensors
Conversion gain
C=dQ/dV
q
V = ---- ⋅ number of electrons
C
Voltage per collected electrons is defined as
“Conversion Gain”:
µV
q
-------CG = ---e
C
(2.40)
From optical power to voltage:
P opt
V = CG ⋅ QE ⋅ t ⋅ ---------hν
(2.41)
Full Well
Maximum number of electrons that can be stored in the pixel:
V max
1
N sat = --q
∫
C PD ( V ) dV
(2.42)
V Reset
AO
10V
2.29
Photo Diode
INF 5440 - CMOS Image Sensors
Pinned Photo diode
0
n-region buried in a p-substrate.
Reduces the effect of surface recombination
• Improved response in blue range
• Reduced dark current.
Vp
Potential
p
Tom “well”
n
Full “well”
p
Pinned voltage = the voltage that results in
a complete depletion of the n-region.
Pinning voltage depends on doping
concentration and implantation range.
Nd3 > Nd2
Pinning Voltage, Vp
Added process step to
standard CMOS process.
Patented: Eastman-Kodak/Motorola
(ImageMOSTM).
Nd2 > Nd1
Nd1
Implant thickness (nd)
AO
10V
2.30
Photo Diode
INF 5440 - CMOS Image Sensors
Example 1 - photo diode
D=1 µm
NA
ND
ND
EPI=10 µm
Ø=5 µm
2V
NA
W
p++
Incoming light 550nm (green). Exposure time: 10 ms. External reverse bias VR = 2V.
• Dark current and signal from a non illuminated sensor?
• Responsivity?
• Required intensity to achieve a signal of 1 V?
AO
10V
2.31
Photo Diode
INF 5440 - CMOS Image Sensors
Example 1 cont.
Physical data
NA
1016 cm-3
σp , σn
10-15 cm2 = 10-19 m2
k
1.38 x 10-23 J / ºK
ND
1018 cm-3
µe
0.135 m2/Vs
q
1.602 x 10-19 C
Nt
3 x 1011 cm-3
µp
0.048 m2/Vs
ε0
8.85 x 10-12 C2 / Nm2
Et
Ei
εr,Si
11.7
η
0.75
R
0.3 (reflection coefficient)
AO
10V
2.32
References
INF 5440 - CMOS billedsensorer
Grove
Physics and Technology of Semiconductor Devices
A.S. Grove
John Wiley & sons
Alonso-Finn
Fundamental University Physics
III Quantum and statistical physics
Marcelo Alonso and Edward J. Finn
Addison-Wesley Publishing company
Sze
Physics of Semiconductor Devices
S. M. Sze
John Wiley & Sons
AO
10V
2.33