University of Tripoli - Engineering Faculty
Computer Engineering Department
EC 310: Electronic Materials & Devices
Dr. Amna Elhawil
A.elhawil@uot.edu.ly
CHAPTER (3): PN JUNCTION
Textbook: Modern Semiconductor Devices for Integrated Circuits, Chenming Calvin Hu
Chapter (4.1) to (4.11) of textbook
pn Junction
Part (1)
Chapter (4.1) to (4.4) and (4.5) of textbook
CH(3): PN JUNCTION
2
Outline
▪Idealized p-n junction
1.
2.
3.
Built-in potential barrier in thermal equilibrium
Electric field at equilibrium
Space charge width: depletion layer widths for n-side and p-side
▪Reverse bias
1.
2.
Space charge width of reverse bias
Electric field of reverse bias
▪Forward bias
1.
2.
Space charge width of reverse bias
Electric field of reverse bias
CH(3): PN JUNCTION
3
The pn junction
▪The p-n junction is the basic element of all bipolar devices.
▪Its main electrical property is that it rectifies (allow current to flow easily in one direction only).
▪A p-n junction cannot be produced by simply pushing two pieces together or by welding etc….
Because it gives rise to discontinuities across the crystal structure.
▪Special fabrication techniques are adopted to form a p-n junction.
CH(3): PN JUNCTION
4
▪ Diodes, BJT and FET transistors are made of p-n junctions
Diodes
BJT transistors
CH(3): PN JUNCTION
FET transistors
5
p-type
n-type
EC
Eİ
EC
Ef
Eİ
Ef
EV
EC
pn junction
EV
n-type
p-type
Eİ
EC
Ef
Eİ
Ef
EV
EV
Fig. 3.1
CH(3): PN JUNCTION
6
Fig. 3.2
CH(3): PN JUNCTION
7
▪ Lots of electrons majority carriers in n-side want to diffuse to the p-side.
▪ Lots of holes (majority carriers) in the p-side of the junction want to move to the n-side.
▪ Both are called diffusion currents.
High p
concertation
High n
concertation
Fig. 3.3
CH(3): PN JUNCTION
8
pn junction under thermal equilibrium
1.
2.
3.
4.
5.
6.
Majority carrier electrons in the n region will begin diffusing into the p region, and majority
carrier holes in the p region will begin diffusing into the n region.
this diffusion process cannot continue indefinitely. As electrons diffuse from the n region,
positively charged donor atoms are left behind.
Similarly, as holes diffuse from the p region, they uncover negatively charged acceptor
atoms.
The net positive and negative charges in the n and p regions induce an electric field in the
region near the metallurgical junction, in the direction from the positive to the negative
charge, or from the n to the p region.
These two regions are referred to as the space charge region. Essentially all electrons and
holes are swept out of the space charge region by the electric field.
Since the space charge region is depleted of any mobile charge, this region is also referred to
as the depletion region.
CH(3): PN JUNCTION
9
Electric field & drift current
Diffusion current
Hole
Movement
Hole
Movement
n-type
+++++
+++++
+++++
+++++
+++++
+++++
+++++
+++++
---------------------------------
p-type
Electron
Movement
Electron
Movement
Fig. 3.5
++++
++++
++++
Fixed positive
space-charge
----------
Fixed negative
space-charge
Ohmic
end-contact
Fig. 3.4
CH(3): PN JUNCTION
10
What stops the diffusion currents?
▪ With the formation of the depletion region, an electric field emerges.
▪ The field tends to force positive charge flow from n to p region whereas the concentration
gradients necessitate the flow of holes from p to n (and electrons from n to p).
▪ We therefore surmise that the junction reaches equilibrium:
➢ Once the electric field is strong enough to completely stop the diffusion currents.
➢ Alternatively, we can say, in equilibrium, the drift currents resulting from the electric field
exactly cancel the diffusion currents due to the gradients.
𝐽𝑑𝑟𝑖𝑓𝑡 = 𝐽𝑑𝑖𝑓𝑓
.
CH(3): PN JUNCTION
11
Fig. 3.6
CH(3): PN JUNCTION
12
CH(3): PN JUNCTION
13
Applying bias to p-n junction
When a battery is connected to pn junction it basically works in two modes:
1. Reverse bias mode ( negative terminal connected to p-region and positive
terminal connected to n region)
2. Forward bias mode ( positive terminal connected to p-region and negative
terminal connected to n region)
Fig. 3.7
CH(3): PN JUNCTION
14
Potential energy of electron
▪ The electrons receive energy from their position .
▪ The energy received from the position is called potential energy (P.E).
𝑃. 𝐸 = −𝑞𝑉
(J)
(3.1)
CH(3): PN JUNCTION
15
The eV unit of energy
P.𝐸 = 𝑞𝑉
▪ The potential energy of electron gained from 1 Volt is:
P.E = qV = (1.6 x 10-19 C) (1 V) = 1.6 x 10-19 J
So the electron possesses a tiny amount of energy. However, the eV (electron vole) unit is
introduced to define small amounts of power.
1 eV = 1.6 x 10-19 J
CH(3): PN JUNCTION
16
1) Zero applied bias (thermal equilibrium)
Energy-band diagram of a pn junction in thermal equilibrium.
From (3.1)
𝑃. 𝐸 = 𝑞𝑉
q𝐴 = 𝐸𝑐 − 𝐸𝐹𝑛
(eV)
q𝐵 = 𝐸𝑐 − 𝐸𝐹𝑝
(eV)
=➔ 𝑞∅𝑏𝑖 = 𝑞𝐵 − 𝑞𝐴
(eV)
(3.2)
(V)
(3.3)
The potential voltage :
∅𝑏𝑖 = 𝐵 − 𝐴
∅𝑏𝑖 is called built-in potential voltage
Fig. 3.8
CH(3): PN JUNCTION
17
1.1) Built-in potential barrier in thermal equilibrium
➢ The electrons in the conduction band of the n region see a potential barrier in trying to move into the conduction
band of the p region.
Holes
Electrons
𝑛 = 𝑁𝐷 =
−
𝑛𝑖2
[
𝑛=
= 𝑁𝑐 𝑒
𝑁𝐴
− 𝐸𝑐 −𝐸𝐹
[
]
𝑘𝑇
𝑁𝑐 𝑒
𝐴 = 𝐸𝑐 − 𝐸𝐹𝑛
𝑘𝑇
𝑁𝑐
=
× 𝑙𝑛( )
𝑞
𝑁𝐷
∅𝑏𝑖
𝐵 = 𝐸𝑐 − 𝐸𝐹𝑝
𝐸𝑐 −𝐸𝑓𝑝
]
𝑘𝑇
𝑘𝑇
𝑁𝑐𝑁𝐴
=
× 𝑙𝑛( 2 )
𝑞
𝑛𝑖
𝑘𝑇
𝑁𝐷 𝑁𝐴
=𝐵−𝐴=
× ln
𝑞
𝑛𝑖2
∅𝑏𝑖
𝑁𝐷 𝑁𝐴
= 𝑉𝑡 ln( 2 )
𝑛𝑖
CH(3): PN JUNCTION
(3.4)
18
𝑉𝑏𝑖 = ∅𝑏𝑖
𝑁𝐷 𝑁𝐴
= 𝑉𝑡 ln
𝑛𝑖2
(𝑉)
(3.4)
Note:
The thermal voltage Vt = 𝒌𝑻/q = 0.0259 ≈ 0.026 V at 300 K
Proof:
𝑘𝑇
8.62 × 10−5 𝑒𝑉 Τ𝐾 × 300 𝐾
0.0259 𝑒𝑉
𝑉𝑡 =
=
=
= 0.0259 𝑉
−19
−19
𝑞
1.6 × 10 𝐶
1.6 × 10 𝐶
∵ 1 𝑒𝑉 = (1.6 × 10−19 𝐶)(1 𝑉)
CH(3): PN JUNCTION
19
Example (1)
Consider a silicon pn junction at T = 300 K with doping concentration of NA = 2 x 1017 cm-3 and ND = 1015 cm-3.
Calculate the built-in potential energy in a pn junction
Solution
From (3.4)
𝑁𝐷 𝑁𝐴
1015 × 2 x 1017
∅bi = 𝑉𝑡 ln
= 0.0259 ln
= 0.713 𝑉
1.5 × 1010 2
𝑛𝑖2
If we change the doping concentration to NA = 1016 cm-3 and ND = 1015 cm-3, then the built-in potential
barrier becomes ∅bi = 0.635 V.
Comment: the built-in potential barrier changes only slightly as the doping concentration change
by order of magnitude because of the logarithmic dependence.
CH(3): PN JUNCTION
20
p-type
-
Charge density
n-type +
+
+
+
-
xn
Depletion
Region
xp
w
xn is the width of the depletion region in n-side
xp is the width of the depletion region in p-side
W = total depletion region
CH(3): PN JUNCTION
21
Poisson’s Equation
The charge density (𝜌) is proportional to dE /dx:
𝑑𝐸(𝑥) 𝜌(𝑥)
=
𝑑𝑥
𝜀𝑠
(3.5)
E(x) is the electric field,
𝜌(x) is the volume charge density (C/cm3),
𝜀𝑠 the permittivity of the semiconductor
Ionized acceptors
have negative sign
𝜌 = −𝑞𝑁𝐴
➔ In p side at 0 ≤ 𝑥 ≤ 𝑥𝑝
𝜌 = 𝑞𝑁𝐷
➔ In n side at 𝑥𝑛 ≤ 𝑥 ≤ 0
Fig. 3.9
𝜀𝑠 = 𝜀0 𝜀𝑟
𝜀0 =permittivity of vacuum = 8.85 x10-14 F/cm
𝜀𝑟 = relative permittivity of Silicon = 11.7
CH(3): PN JUNCTION
22
1.2) Electric field at equilibrium
The electric field is determined from Poisson’s equation,
𝜌(𝑥)
E x =න
𝑑𝑥
𝜀𝑠
𝜌(𝑥) 𝑑𝐸(𝑥)
=
𝜀𝑠
𝑑𝑥
• An electric field is created in the depletion region by the
separation of positive and negative space charge densities.
E x =න
𝑞𝑁𝐷
𝑞𝑁𝐷
𝑑𝑥 =
𝑥 + 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑥𝑛 ≤ 𝑥 ≤ 0
𝜀𝑠
𝜀𝑠
To find 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡, set E = 0 at x = xn.
𝑞𝑁𝐷
𝐸=
(𝑥 − 𝑥𝑛 )
𝜀𝑠
(3.6)
−𝑞𝑁𝐴
−𝑞𝑁𝐴
0≤ 𝑥 ≤ 𝑥𝑝
E x =න
𝑑𝑥 =
𝑥+𝐶
𝜀𝑠
𝜀𝑠
𝑞𝑁𝐴
(3.7)
To find C, E = 0 at x = xp.
𝐸=
(𝑥𝑝 − 𝑥)
𝜀𝑠
CH(3): PN JUNCTION
Space-charge profile
Fig. 3.10
23
On the N-side of the depletion layer, the field is
𝑞𝑁𝐷
(𝑥 − 𝑥𝑛 )
𝜀𝑠
On the P-side of the depletion layer, the field is
𝑞𝑁𝐴
𝐸=
(𝑥𝑝 − 𝑥)
𝜀𝑠
𝐸=
𝑥𝑛 ≤ 𝑥 ≤ 0
(3.6)
0≤ 𝑥 ≤ 𝑥𝑝
(3.7)
Fig. 3.11 Electric field distribution
• The E-field is a linear function of distance through the junction.
• An electric field exists in the depletion region even when no voltage is applied between the p and n regions.
Note that by setting x=0 in (3.6) and (3.7) we get
𝑁𝐷 𝑥𝑛 = 𝑁𝐴 𝑥𝑝
(3.8)
This equation states that the number of negative charges per unit area in the p region is equal to the number of
positive charges per unit area in the n region.
CH(3): PN JUNCTION
24
The abrupt(step) junction & Linearly graded junction
Ideal case
Acceptors and
donors
concentrations
are constant up
to the junction
Acceptors and
donors
concentrations varies
linearly with the
distance from the
junction
Abrupt junction
Fig. 3.12
CH(3): PN JUNCTION
Linearly graded junction
25
Abrupt junction:
𝑵𝑨 = 𝑵 𝑫
p-type
-
xn = x p
Charge density
n-type +
+
+
+
-
xn
xp
Depletion
Region
w
Space-charge profile diagram
Fig. 3.13
CH(3): PN JUNCTION
26
Abrupt junction: 𝑵𝑨 ≠ 𝑵𝑫
When ND >> NA ➔ xp >> xn ➔ W ≈ xp
When NA >> ND ➔ xn >> xp ➔ W ≈ xn
heavily doped p-type
p-type
-
+
+
n-type
+++
+++
heavily doped n-type
N A  N D
n-type
+
+
p-type
-
+
+
x
x
- - - -
-
xn
N D  N A
xp
xn
xp
Fig. 3.14
CH(3): PN JUNCTION
27
1.3) Voltage distribution at equilibrium
𝐸=−
𝑑𝑉
𝑑𝑥
𝑉 = − න 𝐸 𝑑𝑥
at 𝒙𝒏 ≤ 𝒙 ≤ 𝟎
at 𝟎 ≤ 𝒙 ≤ 𝒙𝒑
𝑞𝑁𝐴
𝐸=
(𝑥𝑝 − 𝑥)
𝜀𝑠
𝑞𝑁𝐴
𝑉 = −න
𝑥𝑝 − 𝑥 𝑑𝑥
𝜀𝑠
𝑉=
−𝑞𝑁𝐴
(𝑥𝑝 − 𝑥)2 +𝐶𝑜𝑛𝑠𝑡
2𝜀𝑠
𝐸=
𝑞𝑁𝐷
(𝑥 − 𝑥𝑛 )
𝜀𝑠
𝑉=
𝑉 = −න
−𝑞𝑁𝐷
(𝑥 − 𝑥𝑛 )2 +𝐶𝑜𝑛𝑠𝑡
2𝜀𝑠
At 𝑥 = 𝑥𝑝 , the voltage V = 0
At 𝑥 = 𝑥𝑛 , the voltage V(𝑥𝑛 ) = ∅𝑏𝑖
−𝑞𝑁𝐴
𝑉=
(𝑥𝑝 − 𝑥)2
2𝜀𝑠
𝑞𝑁𝐷
𝑉 = ∅𝑏𝑖 −
(𝑥 − 𝑥𝑛 )2
2𝜀𝑠
(3.9)
𝑞𝑁𝐷
𝑥 − 𝑥𝑛 𝑑𝑥
𝜀𝑠
(3.10)
at x = 0, eq (3.9 ) and (3.10) are equal
𝑉=
−𝑞𝑁𝐴
𝑞𝑁𝐷
(𝑥𝑝 − 𝑥)2 = ∅𝑏𝑖 −
(𝑥 − 𝑥𝑛 )2
2𝜀𝑠
2𝜀𝑠
CH(3): PN JUNCTION
(3.11)
28
Fig. 3.15
Electric potential diagram
CH(3): PN JUNCTION
29
1.4) Space charge width: depletion layer widths for n-side and p-side
And from (3.8) 𝑁𝐷 𝑥𝑛 = 𝑁𝐴 𝑥𝑝 we get the total depletion or space charge width W
𝑊 = 𝑥𝑝 −𝑥𝑛 =
Using the form:
2𝜀𝑠 ∅𝑏𝑖 1
1
+
𝑞
𝑁𝐴 𝑁𝐷
cm
(3.12)
𝑁𝐷 𝑥𝑛 = 𝑁𝐴 𝑥𝑝
𝑥𝑛 =
𝑥𝑝 =
2𝜀𝑠 ∅𝑏𝑖 𝑁𝐴
1
𝑞
𝑁𝐷 𝑁𝐴 + 𝑁𝐷
2𝜀𝑠 ∅𝑏𝑖 𝑁𝐷
1
𝑞
𝑁𝐴 𝑁𝐴 + 𝑁𝐷
(3.13)
(3.14)
CH(3): PN JUNCTION
W
𝜀𝑠 = 𝜀0 𝜀𝑟
𝜀0 =permittivity of vacuum = 8.85 x10-14 F/cm
𝜀𝑟 = relative permittivity of Silicon = 11.7
30
n-type
p-type
-
+
+
n-type
N A  N D
+
+
p-type
-
+
+
+++
+++
xn
xp
If 𝑁𝐴 >> 𝑁𝐷 , as in a
P+N
x
- - - -
x
-
xn
N D  N A
xp
If 𝑁𝐷 >> 𝑁𝐴 , as in a N+P junction,
junction,
CH(3): PN JUNCTION
31
1.5) Maximum Electric Field
The maximum (magnitude) electric field occurs at the metallurgical junction (at x = 0)
is, from (3.6), (3.7) and (3.8)
𝐸𝑚𝑎𝑥 =
2𝑞𝑉𝑏𝑖 𝑁𝐴 𝑁𝐷
𝜀𝑠 𝑁𝐴 + 𝑁𝐷
𝐸𝑚𝑎𝑥
2𝑉𝑏𝑖
=
𝑊
CH(3): PN JUNCTION
(3.15)
(3.16)
Electric field distribution
32
Example (2)
Calculate ∅𝑏𝑖 , Emax , W, xn and xp for a silicon pn junction at zero bias and T = 300 K for doping concentrations
of Na = 2 x 1017 cm-3, Nd = 1016 cm-3 .
Solution
From (3.8)
∅𝑏𝑖
𝑘𝑇
𝑁𝐷 𝑁𝐴
1016 × 2 × 1017
=
× ln
= 0.0259 ln
= 0.772 𝑉
𝑞
1.0 × 1010 2
𝑛𝑖2
From (3.15),
|𝐸𝑚𝑎𝑥 | =
2𝑞∅𝑏𝑖 𝑁𝐴 𝑁𝐷
=
𝜀𝑠 𝑁𝐴 + 𝑁𝐷
2 × 1.6 × 10−19 × 0.772 × (1016 × 2 × 1017 )
= 4.77 × 104 𝑉/𝑐𝑚
−14
16
17
11.9 × 8.85 × 10 (10 + 2 × 10 )
From (3.12), The space charge width
𝑊=
2𝜀𝑠 ∅𝑏𝑖 1
1
+
=
𝑞
𝑁𝐴 𝑁𝐷
2 × 12 × 8.85 × 10−14 × 0.772
1
1
+
= 0.324 𝜇𝑚
1.6 × 10−19
2 × 1017 1016
CH(3): PN JUNCTION
33
In a P+N junction, nearly the entire depletion layer exists on the N-side
𝑥𝑛 ≈ 𝑊 = 0.324 𝜇𝑚
From (3.13),
𝑥𝑝 = 𝑥𝑛
𝑁𝐷
1016
= 0.324 ×
= 0.016 𝜇𝑚
𝑁𝐴
2 × 1017
Note: The peak electric field in the space charge region of a pn junction is quite
large. We must keep in mind, however, that there is no mobile charge in this
region; hence there will be no drift current.
CH(3): PN JUNCTION
34
1.6) Capacitance-voltage Characteristics
The depletion layer and the neutral N and P regions may be viewed as an insulator
and two conductors ➔ PN junction may be modelled as a parallel-plate capacitor
𝜀𝑠
𝐶=𝐴
𝑊
(3.16)
Cdep is the depletion-layer capacitance
A is the area.
From (3.13):
𝑊 = 𝑥𝑝 −𝑥𝑛 =
2𝜀𝑠 𝑉𝑏𝑖 1
1
+
𝑞
𝑁𝐴 𝑁𝐷
1
𝑊2
2𝑉𝑏𝑖
=
=
𝐶 2 𝐴2 𝜀𝑠2 𝑞𝑁𝜀𝑠 𝐴2
1
1
where N = 𝑁 + 𝑁
𝐴
𝜀𝑠 = 𝜀0 𝜀𝑟
𝐷
𝜀0 =permittivity of vacuum = 8.85 x10-12 F/m
CH(3): PN JUNCTION
35
Relative permittivity
𝜺𝒓
Si
11.7
GaAs
13.1
Ge
16.0
𝜀𝑠 = 𝜀0 𝜀𝑟
𝜀0 =permittivity of vacuum = 8.85 x10-12 F/m
CH(3): PN JUNCTION
36
2) REVERSE-BIASED PN JUNCTION
CH(3): PN JUNCTION
37
Fig. 3.17
CH(3): PN JUNCTION
38
(a)
(b)
Fig. 3.18 Depletion region with a) reversed and b) forward -biased diode
CH(3): PN JUNCTION
39
The pn junction under reverse-biased condition
▪ When a negative terminal voltage is applied (the diode is
reversed-biased), the negative potential attracts the holes
away from the edge of the junction on the p side, while the
positive potential attracts the electrons away from the edge of
the junction on the n side.
▪ This action increases both the depletion region width and the
barrier voltage. This decreases the diffusion current (current
flow across the junction by majority carriers).
▪ The potential of the n region is positive with respect to the p
region so the Fermi energy in the n region is lower than that in
the p region.
Fig. 3.19
▪ The total potential barrier is now larger than that for the zerobias case.
CH(3): PN JUNCTION
40
2.1) Depletion layer width of reverse bias
From (3.13), the depletion or space charge width W
𝑊=
2𝜀𝑠 𝑉𝑏𝑖 1
1
+
𝑞
𝑁𝐴 𝑁𝐷
In the reverse bias, the ∅𝑏𝑖 term is replaced with 𝑉𝑏𝑖 + 𝑉𝑟
𝑊=
2𝜀𝑠 (𝑉𝑏𝑖 +𝑉𝑟 ) 1
1
+
𝑞
𝑁𝐴 𝑁𝐷
(3.17)
CH(3): PN JUNCTION
Fig. 3.20
41
2.2) Capacitance-voltage Characteristics
1
𝑊2
2(𝑉𝑏𝑖 + 𝑉𝑟 )
(3.18)
=
=
2
2
2
2
𝐶
𝑞𝑁𝜀𝑠 𝐴
𝐴 𝜀𝑠
1
1
where N = 𝑁 + 𝑁
𝐴
𝐷
CH(3): PN JUNCTION
42
3) Forward bias
1.
2.
3.
4.
Charge density
Minority carrier distribution
Diffusion current
Generation & recombination current
CH(3): PN JUNCTION
43
Fig. 3.21
CH(3): PN JUNCTION
44
Forward Bias
⚫
⚫
⚫
Forward bias causes an exponential increase in the number of carriers with
sufficient energy to penetrate barrier
Diffusion current increases exponentially.
Drift current decreases.
➔ Net result: Large forward current
CH(3): PN JUNCTION
45
The pn junction under forward-biased condition
• When a positive voltage is applied to the p region with
respect to the n region.
• The total potential barrier is now reduced.
• The Fermi level in the p region is now lower than that in
the n region.
• The smaller potential barrier means that the electric
field in the depletion region is also reduced.
• The smaller electric field means that the electrons and
holes are no longer held back in the n and p regions,
respectively.
• There will be a diffusion of holes from the p region
across the space charge region where they will flow into
the n region.
• Similarly, there will be a diffusion of electrons from the
n region across the space charge region where they will
flow into the p region.
CH(3): PN JUNCTION
Fig. 3.22
46
3.1) Carrier density at equilibrium
∅𝑏𝑖
𝑘𝑇
𝑁𝐷 𝑁𝐴
=
× ln
𝑞
𝑛𝑖2
∅𝑏𝑖 =
𝑘𝑇
𝑁𝐷 𝑁𝐴
× ln
𝑞
𝑛𝑖2
or
or
𝑛𝑖2
𝑁𝐷 𝑁𝐴
𝑛𝑖2
= 𝑒 −(𝑞∅𝑏𝑖 /𝑘𝑇)
𝑁𝐷 𝑁𝐴
= 𝑒 −(𝑞∅𝑏𝑖 /𝑘𝑇)
In n-region, the majority carriers :
𝑛𝑛0 = 𝑁𝐷
In p-region, the minority carriers: 𝑛𝑝0 =
𝑛𝑝0 = 𝑛𝑛0 𝑒 −𝑞𝑉𝑏𝑖 /𝑘𝑇
𝑛𝑖2
𝑁𝐴
In p-region, the majority carriers :
𝑝𝑝0 = 𝑁𝐴
In n-region, the minority carriers: 𝑝𝑛0 =
(3.19)
𝑛𝑛0 : Majority concentration in n-type at equilibrium
𝑛𝑝0 : Minority concentration in p-type at equilibrium
CH(3): PN JUNCTION
𝑝𝑛0 = 𝑝𝑝0 𝑒 −𝑞𝑉𝑏𝑖 /𝑘𝑇
𝑛𝑖2
𝑁𝐷
(3.20)
𝑝𝑝0 : Majority concentration in p-type at equilibrium
𝑝𝑛0 : Minority concentration in n-type at equilibrium
47
3.1) Charge density at forward
• The potential barrier Vbi is replaced by (Vbi - Va) when the junction is forward biased.
At forward bias:
At forward bias:
𝑛𝑝 = 𝑛𝑛0 𝑒 −𝑞(𝑉𝑏𝑖 −𝑉𝑎 )/𝑘𝑇
𝑝𝑛 = 𝑝𝑝0 𝑒 −𝑞(𝑉𝑏𝑖 −𝑉𝑎 )/𝑘𝑇
𝑛𝑝 = 𝑛𝑛0 𝑒 −𝑞𝑉𝑏𝑖 /𝑘𝑇 𝑒 𝑞𝑉𝑎 /𝑘𝑇
𝑝𝑛 = 𝑝𝑝0 𝑒 −𝑞𝑉𝑏𝑖 /𝑘𝑇 𝑒 𝑞𝑉𝑎 /𝑘𝑇
From (3.23)
From (3.22)
𝑝𝑛 = 𝑝𝑛0 𝑒 𝑞𝑉𝑎 /𝑘𝑇
𝑛𝑝 = 𝑛𝑝0 𝑒 𝑞𝑉𝑎/𝑘𝑇 (3.21)
𝑛𝑛0 : Majority concentration in n-type at equilibrium
𝑛𝑝0 : Minority concentration in p-type at equilibrium
𝑛𝑝 ∶ Minority concentration in p−type at forward bias
CH(3): PN JUNCTION
(3.22)
𝑝𝑝0 : Majority concentration in p-type at equilibrium
𝑝𝑛0 : Minority concentration in n-type at equilibrium
𝑝𝑛 ∶ Minority concentration in n−type at forward bias
48
Total minority carrier at the edges of the depletion region
Minority electrons
at forward bias
Minority holes at
forward bias
Minority electrons at
equilibrium
𝑛𝑝 = 𝑛𝑝0 𝑒 𝑞𝑉𝑎/𝑘𝑇
At x=xp (p region)
𝑝𝑛 = 𝑝𝑛0 𝑒 𝑞𝑉𝑎/𝑘𝑇
At x=-xn (n region)
Minority holes at
equilibrium
Fig. 3.23
➢ the total minority carrier concentration in the n and p regions are now greater
than the thermal equilibrium value.
➢ The forward-bias voltage lowers the potential barrier so that majority carrier
electrons from the n region are injected across the junction into the p region,
and vise versa, thereby increasing the minority carrier electron concentration.
CH(3): PN JUNCTION
49
3.2) Minority carrier distribution
• To obtain I-V characteristics of pn junction, we use the ambipolar transport equation:
𝑑2Δ𝑝𝑛 𝑥
∆𝑝𝑛 𝑥
−
=0
2
𝑑𝑥
𝐷𝑝 𝜏𝑝
in the n-region, where
∆𝑝𝑛 = 𝑝𝑛 − 𝑝𝑛0
• The solution of this equation
∆𝑝𝑛 𝑥 = 𝐴𝑒
𝑤ℎ𝑒𝑟𝑒 𝐿𝑝 =
•
For p-region:
+ 𝐵𝑒
−𝑥
𝐿𝑝
(3.23)
𝐷𝑝 𝜏𝑝 is the diffusion length of holes (minority carriers) in the n-region.
Fig. 3.23
𝑑2Δ𝑛𝑝 𝑥
∆𝑛𝑝 𝑥
−
=0
𝑑𝑥2
𝐷𝑛 𝜏𝑛
∆𝑛𝑝 𝑥 =
𝑤ℎ𝑒𝑟𝑒 𝐿𝑛 =
𝑥
𝐿𝑝
𝑥
𝐶𝑒 𝐿𝑛
+
−𝑥
𝐷𝑒 𝐿𝑛
(3.24)
𝐷𝑛 𝜏𝑛 is the diffusion length of electron in the p-region.
CH(3): PN JUNCTION
Dn: electron diffusion coefficient (cm2/s)
Dp: hole diffusion coefficient (cm2/s)
τp: minority carrier hole lifetime.
τn: minority carrier electron lifetime.
50
• Diffusion length: the length of path that an electron travels before recombining:
𝐿𝑛 =
𝐷𝑛 𝜏𝑛
• Dn: electron diffusion coefficient (cm2/s)
• 𝜏𝑛 : electron lifetime.
(𝑐𝑚)
𝐷𝑛 =
(3.25)
𝐾𝑇
𝜇𝑛
𝑞
• A similar diffusion length can be determined for holes as:
𝐿𝑝 =
• Dp: hole diffusion coefficient
• 𝜏𝑝 : hole lifetime.
𝐷𝑝 𝜏𝑝
(cm2/s)
(𝑐𝑚) (3.26)
𝐾𝑇
𝐷𝑝 =
𝜇𝑝
𝑞
CH(3): PN JUNCTION
51
∆𝑝𝑛 𝑥
𝑥
−𝑥
𝑝
= 𝐴𝑒 𝐿𝑥𝑝 + 𝐵𝑒 𝐿−𝑥
= 𝐶𝑒 𝐿𝑛 + 𝐷𝑒 𝐿𝑛
∆𝑛𝑝 𝑥
• To find A, B, C and D we use the following boundary
conditions for the total minority carrier concentrations:
(1)
𝑝𝑛 (𝑥 → ∞) = 𝑝𝑛0
(2)
∆𝑝𝑛 =0
𝑥 ≤ 𝑥𝑛
𝑝𝑛 (𝑥 = 𝑥𝑛 ) = 𝑝𝑛0 𝑒 𝑞𝑉𝑎 /𝑘𝑇
(3)
𝑛𝑝 (𝑥 → −∞) = 𝑛𝑝0
(4)
𝑛𝑝 (𝑥 = −𝑥𝑝 ) = 𝑛𝑝0 𝑒 𝑞𝑉𝑎 /𝑘𝑇 𝑥 ≥ 𝑥𝑝
∆𝑛𝑝 =0
• the boundary condition (1) gives A = 0
• the boundary condition (3) gives D = 0
• To find A, use the boundary condition (2):
∆𝑝𝑛 𝑥 = 𝐵𝑒
−𝑥
𝐿𝑝
Fig. 3.23
∆𝑝𝑛 𝑥 = 𝑝𝑛 (𝑥) − 𝑝𝑛0 (𝑥) = 𝐵𝑒
−𝑥
𝐿𝑝
∆𝑝𝑛 𝑥𝑛 = 𝑝𝑛0 𝑒 𝑞𝑉𝑎 /𝑘𝑇 − 𝑝𝑛0 = 𝐵𝑒
• To find C, use the boundary condition (3)
∆𝑛𝑝 𝑥 =
𝑥
𝐶𝑒 𝐿𝑛
∆𝑛𝑝 𝑥 = 𝑛𝑝 (𝑥) − 𝑛𝑝0 (𝑥) =
𝐵 = 𝑝𝑛0 𝑒 𝑞𝑉𝑎 /𝑘𝑇 − 1 𝑒
𝑥
𝐶𝑒 𝐿𝑛
∆𝑛𝑝 𝑥𝑝 = 𝑛𝑝0 𝑒 𝑞𝑉𝑎/𝑘𝑇 − 𝑛𝑝0 =
CH(3): PN JUNCTION
−𝑥𝑛
𝐿𝑝
𝑥𝑝
𝐶𝑒 𝐿𝑛
C= 𝑛𝑝0
𝑒 𝑞𝑉𝑎/𝑘𝑇
−1 𝑒
𝑥𝑛
𝐿𝑝
−𝑥𝑝
𝐿𝑛
52
p-side
n-side
∆𝑝𝑛 𝑥𝑛
∆𝑛𝑝 𝑥𝑝
𝑛𝑝 𝑥 = ∆𝑛𝑝 𝑥 + 𝑛𝑝0
𝑝𝑛 𝑥 = ∆𝑝𝑛 (𝑥) + 𝑝𝑛0
𝑥𝑝 𝑥𝑛
CH(3): PN JUNCTION
53
Thus, equation (3.23)
∆𝑝𝑛 𝑥 =
𝑥
𝐿
𝐴𝑒 𝑛
∆𝑝𝑛 𝑥 =
𝑞𝑉𝑎
𝑝𝑛0 𝑒 𝑘𝑇
becomes
+
−𝑥
𝐵𝑒 𝐿𝑛
𝐵 = 𝑝𝑛0 𝑒 𝑞𝑉𝑎 /𝑘𝑇 − 1 𝑒
𝐴=0,
− 1 𝑒 −(𝑥−𝑥𝑛)Τ𝐿𝑝
𝑥 < 𝑥𝑛
(3.27)
𝑒 𝑞𝑉𝑎 /𝑘𝑇
−𝑥𝑝
𝐿𝑛
𝑥𝑛
𝐿𝑝
Also, equation (3.24)
∆𝑛𝑝 𝑥 = 𝐶𝑒
becomes
∆𝑛𝑝 𝑥 =
𝑥
𝐿𝑝
+ 𝐷𝑒
𝑞𝑉𝑎
𝑛𝑝0 𝑒 𝑘𝑇
−𝑥
𝐿𝑝
𝐶 = 𝑛𝑝0
− 1 𝑒 (𝑥−𝑥𝑝 )Τ𝐿𝑛
𝑥 > 𝑥𝑝
CH(3): PN JUNCTION
−1 𝑒
𝐷 =0
(3.28)
54
3.3) Current density
The total junction current :
𝐽𝑡𝑜𝑡𝑎𝑙 = 𝐽
𝑑𝑟𝑖𝑓𝑡
+ 𝐽(𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛)
• For the minority carriers the drift current can be neglected compared with the
diffusion current, thus
𝐽𝑡𝑜𝑡𝑎𝑙 = 𝐽(𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛)
𝑑𝑛𝑝
𝐽𝑛𝑝 = 𝑞𝐷𝑛
𝑑𝑥
𝑑𝑝𝑛
𝐽𝑝𝑛 = −𝑞𝐷𝑝
𝑑𝑥
• Dp and Dn are the minorty carrier diffusion coefficients.
CH(3): PN JUNCTION
55
1. At equilibrium (no bias)
▪
The drift and diffusion currents are flowing all the time. But, in thermal equilibrium, the net
current flow is zero since the currents oppose each other.
𝐽
𝑑𝑟𝑖𝑓𝑡
= 𝐽(𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛)
2. At non-equilibrium (with bias)
▪
Under non-equilibrium condition, one of the current flow mechanism is going to dominate
over the other, resulting a net current flow.
CH(3): PN JUNCTION
56
3.4) Ideal pn junction current
∆𝑝𝑛 𝑥 =
• Diffusion current density of holes:
∆𝑛𝑝 𝑥 =
We have
𝐽𝑝(𝑑𝑖𝑓𝑓)
𝑞𝑉𝑎
𝑝𝑛0 𝑒 𝑘𝑇
𝑞𝑉𝑎
𝑛𝑝0 𝑒 𝑘𝑇
− 1 𝑒 (𝑥−𝑥𝑛)Τ𝐿𝑝
− 1 𝑒 (𝑥𝑝 −𝑥)Τ𝐿𝑛
𝑑𝑝𝑛 (𝑥)
= −𝑞𝐷𝑝
ቤ
𝑑𝑥 𝑥
𝑛
𝐽𝑝(𝑑𝑖𝑓𝑓)
𝐷𝑝 𝑃𝑛0 𝑞𝑉𝑎
𝑑𝑝𝑛 𝑥
𝑑
= −𝑞𝐷𝑝
= −𝑞𝐷𝑝
𝛿𝑝𝑛 𝑥𝑛 = +𝑞
𝑒 𝑘𝑇 − 1 𝑒 (𝑥−𝑥𝑛Τ)𝐿𝑝
𝑑𝑥
𝑑𝑥
𝐿𝑝
Diffusion current density at 𝑥 = 𝑥𝑛
𝐷𝑝 𝑃𝑛0 𝑞𝑉𝑎
𝐽𝑝 (𝑥𝑛 ) = 𝑞
𝑒 𝑘𝑇 − 1
𝐿𝑝
(3.29)
• Diffusion current density of electrons :
𝐽𝑛(𝑑𝑖𝑓𝑓)
𝐷𝑛 𝑛𝑝0 𝑞𝑉𝑎
𝑑𝑛
𝑑
= 𝑞𝐷𝑛 ቤ = 𝑞𝐷𝑛
𝛿𝑛 𝑥𝑝 = 𝑞
𝑒 𝑘𝑇 − 1 𝑒 (𝑥𝑝−𝑥)Τ𝐿𝑛
𝑑𝑥 𝑥
𝑑𝑥
𝐿𝑛
𝑝
Diffusion current density at 𝑥 = 𝑥𝑝
𝐷𝑛 𝑛𝑝0 𝑞𝑉𝑎
𝐽𝑛 (𝑥𝑝 ) = 𝑞
𝑒 𝑘𝑇 − 1
𝐿𝑛
Fig. 3.26
(3.30)
CH(3): PN JUNCTION
57
𝐷𝑝 𝑃𝑛0 𝑞𝑉𝑎
𝐽𝑝 (𝑥𝑛 ) = 𝑞
𝑒 𝑘𝑇 − 1
𝐿𝑝
(3.31)
𝐷𝑛 𝑛𝑝0 𝑞𝑉𝑎
𝐽𝑛 (𝑥𝑝 ) = 𝑞
𝑒 𝑘𝑇 − 1
𝐿𝑛
(3.32)
• In forward biasing, one of the current flow mechanism is going to
dominate over the other, resulting a net current flow.
• The total pn junction current will be the minority carrier hole
diffusion current plus the minority carrier electron diffusion current.
𝐽𝑑𝑖𝑓𝑓
Let
𝐷𝑛 𝑛𝑝0 𝐷𝑝 𝑃𝑛0
= 𝐽𝑝 (𝑥𝑛 ) + 𝐽𝑛 (𝑥𝑝 ) = 𝑞
+
𝐿𝑛
𝐿𝑝
𝑞𝑉𝑎
𝑒 𝑘𝑇
−1
𝐷𝑛 𝑛𝑝0 𝐷𝑝 𝑃𝑛0
𝐽𝑠 = 𝑞
+
𝐿𝑛
𝐿𝑝
Js is then referred to as the reverse-saturation current density.
𝐽𝑡𝑜𝑡𝑎𝑙 = 𝐽𝑑𝑖𝑓𝑓 = 𝐽𝑠
𝑞𝑉𝑎
𝑒 𝑘𝑇
−1
CH(3): PN JUNCTION
(3.33)
Fig. 3.27
58
Note #1
Note #2
The saturation current density Js
The saturation current density Js
𝐷𝑛 𝑛𝑝0 𝐷𝑝 𝑃𝑛0
𝐽𝑠 = 𝑞
+
𝐿𝑛
𝐿𝑝
where
𝑛𝑝0 = 𝑁𝐷
𝐽𝑠 = 𝑞
𝑛𝑝0
𝑒 −𝑞𝑉𝑏𝑖/𝑘𝑇
𝐷𝑛 𝑛𝑝0 𝐷𝑝 𝑃𝑛0
+
𝐿𝑛
𝐿𝑝
𝑛𝑖2
=
𝑁𝐴
𝑝𝑛0 = 𝑁𝐴 𝑒 −𝑞𝑉𝑏𝑖 /𝑘𝑇
𝐽𝑠 = 𝑞𝑛𝑖2
➔ Js is temperature dependence.
𝑝𝑛0
𝑛𝑖2
=
𝑁𝐷
𝐷𝑝
𝐷𝑛
+
𝐿𝑛 𝑁𝐴 𝐿𝑝 𝑁𝐷
➔ Js is function of 𝑛𝑖2 , NA, and ND
CH(3): PN JUNCTION
59
𝐽𝑑𝑖𝑓𝑓 =
𝐼=
𝑞𝑉𝑎
𝐽𝑠 𝑒 𝑘𝑇
𝑞𝑉𝑎
𝐼𝑠 𝑒 𝑘𝑇
−1
−1
(3.34)
(3.35)
• Equation (3.35), known as the ideal-diode equation.
• It gives the current–voltage characteristics of the pn junction
over a wide range of currents and voltages.
CH(3): PN JUNCTION
Fig. 3.28
60
General diode equation
• In general, the diode current–voltage relationship is written as
𝐼=
𝑞𝑉𝑎
𝐼𝑠 𝑒 𝑛𝑘𝑇
−1
(3.36)
where the parameter n is called the diode quality factor, n is between 1 and 2.
CH(3): PN JUNCTION
61
For the reverse-biased pn junction
• Although Equation (3.36) was derived assuming a forward-bias voltage (Va > 0),
there is nothing to prevent Va from being negative (reverse bias).
• If the voltage Va becomes negative (reverse bias) by a few kT/q, then the reversebiased current density becomes independent of the reverse-biased voltage.
CH(3): PN JUNCTION
62
Example (3)
A PN junction has Na = 1019 cm–3 and Nd = 1016 cm–3.
a. With V = 0, what are the minority carrier densities at the depletion region edges?
b.
c.
d.
e.
Assume V = 0.6 V for (b)–(d).
What are the minority carrier densities at the depletion region edges?
What are the excess minority carrier densities?
What are the majority carrier densities?
Under the reverse bias of 1.8 V, what are the minority carrier concentrations at the
depletion region edges?
CH(3): PN JUNCTION
63
Solution
p-side
𝑛𝑝0
𝑛𝑖2 1020
=
= 19 = 10 𝑐𝑚−3
𝑁𝐴 10
𝑝𝑛0
𝑛𝑖2 1020
=
= 16 = 104 𝑐𝑚−3
𝑁𝐷 10
Zero biasing
n-side
b. The minority carrier densities at the depletion region edges:
V= 0.6 V
p-side
𝑛𝑝 (𝑥𝑝 ) = 𝑛𝑝0 𝑒 𝑞𝑉𝑎/𝑘𝑇 = 10 × 𝑒 0.6Τ0.0259 = 1011 𝑐𝑚−3
n-side
𝑝𝑛 (𝑥𝑛 ) = 𝑝𝑛0 𝑒 𝑞𝑉𝑎 /𝑘𝑇 = 104 × 𝑒 0.6Τ0.0259 = 1014 𝑐𝑚−3
c. The excess minority carrier densities:
p-side
∆𝑛𝑝 𝑥𝑝 = 𝑛𝑝 𝑥𝑝 − 𝑛𝑝0 = 1011 − 10 = 1011 𝑐𝑚−3
n-side
∆𝑝𝑛 𝑥𝑛 = 𝑝𝑛 𝑥𝑛 − 𝑝𝑛0 = 1014 −104 = 1014 𝑐𝑚−3
CH(3): PN JUNCTION
64
d. The majority carrier densities:
V= 0.6 V
p-side
𝑝 𝑥𝑝 = 𝑁𝑎 = 1019 𝑐𝑚−3
n-side
n 𝑥𝑛 = 𝑁𝑑 = 1016 𝑐𝑚−3
e. the reverse bias of 1.8 V, the minority carrier concentrations at the depletion
region edges:
V= -1.8 V
p-side
𝑛𝑝 (𝑥𝑝 ) = 𝑛𝑝0 𝑒 𝑞𝑉𝑅 /𝑘𝑇 = 10 × 𝑒 −1.8Τ0.0259 = 10−29 𝑐𝑚−3
n-side
𝑝𝑛 (𝑥𝑛 ) = 𝑝𝑛0 𝑒 𝑞𝑉𝑅 /𝑘𝑇 = 104 × 𝑒 −1.8Τ0.0259 = 10−26 𝑐𝑚−3
➔We conclude that n = p = 0 at the junction edge under reverse bias.
CH(3): PN JUNCTION
65
Part (2)
OPTOELECTRONIC DEVICES
Chapter (4.14) & (4.15) of textbook
CH(3): PN JUNCTION
66
OPTOELECTRONIC DEVICES
1. Solar Cell
2. Photodiodes
3. Light-Emitting Diodes (LED)
4. Laser Diode
CH(3): PN JUNCTION
67
OPTOELECTRONIC DEVICES
▪ Photon energy (eV)
ℎ𝑐
𝐸=
𝜆
(3.36)
where
➢ h is the Planck’s constant, 6.62618 × 10−34 J/s = 4.13 × 10 -15 eV·s.
➢ c speed of light = 2.99×108 m/s
➢ 𝜆 is the wave's wavelength.
CH(3): PN JUNCTION
68
Ec
Eph = Eg
Ev
Eg
• If Eph = Eg, photon can promote an electron across gap from
valence to conduction band. i.e. absorption of a photon
a) Photon absorption
Ec
• Recombination of electron in conduction band with hole in
valence band; energy released emitted as photon with energy
Eph = Eg.
Eg
Eph = Eg
Ev
b) Photon emission
CH(3): PN JUNCTION
69
1) Solar Cell Basics
• It is a pn junction coverts sunlight to electricity.
• As photons hit the n-junction, some of these photons create electron-hole pairs.
• The electrons in the valence band are energized to jump to the conduction band.
• The electrons can be swept across the junction by the built-in field, and cause a current
to flow out of the P terminal through the external short circuit..
CH(3): PN JUNCTION
70
• The total diode (solar cell) current
𝐼 = 𝐼0 𝑒 𝑞𝑉Τ𝑘𝑇 − 1 − 𝐼𝑠𝑐
Current without
light
(3.37)
Current with
light
CH(3): PN JUNCTION
71
• Solar cells are connected in series (to increase the total voltage) and parallel (to increase
the current)
CH(3): PN JUNCTION
72
2) Photodiodes
▪A photodiode is a semiconductor device that converts light into current.
▪It’s a reverse-biased pn junction in which reverse current increases when the junction is
exposed to light.
▪The current is generated when photons are absorbed in the photodiode.
▪A small amount of current is also produced when no light is present.
CH(3): PN JUNCTION
73
CH(3): PN JUNCTION
74
• Only photons with sufficient energy to excite an electron across the bandgap of the
material will produce significant energy to develop the current from the photodiode.
Material
Wavelength
sensitivity (nm)
Germanium
800 - 1700
Indium gallium arsenide
800 - 2600
Lead sulphide
~1000 - 3500
Silicon
190 - 1100
CH(3): PN JUNCTION
75
CH(3): PN JUNCTION
76
3) Light-Emitting Diodes (LED)
• LED emits light when current flows through it.
• It consists of pn junction.
• When LED is forward biased electrons from n-region diffuse across depletion layer into narrow
region just inside p-side. Here, they recombine with holes, emitting photons.
• Wavelength λ of light determined by energy gap Eg.
Eg =
hc

12.4 × 10−7
λ=
𝐸𝑔
CH(3): PN JUNCTION
77
3) Light-Emitting Diodes (LED)
• In germanium and silicon diodes (called indirect semiconductors), almost the entire
energy is given up in the form of heat and emitted light is insignificant.
• However, in materials like gallium arsenide (called direct semiconductors), the
number of photons of light energy is sufficient to produce quite intense visible light.
✓ For GaAs: Eg = 1.42 eV → λ = 870 nm (near infrared)
✓ For GaP: Eg = 2.24 eV → λ = 550 nm (red region)
CH(3): PN JUNCTION
78
Direct-Gap and Indirect-Gap Semiconductors
• In a direct semiconductor such as GaAs, an electron in the
conduction band can fall to an empty state in the valence band,
giving off the energy difference Eg as a photon of light.
• In an indirect semiconductor such as Si an electron in the
conduction band of cannot fall directly to the valence band but
must undergo a momentum change as well as changing its energy.
Part of the energy is generally given up as heat to the lattice rather
than as an emitted photon.
CH(3): PN JUNCTION
79
Examples of direct & indirect semiconductor
Material
Formula
Band gap (eV)
Gap type
Silicon
Si
1.12
Indirect
Germanium
Ge
0.67
Indirect
Aluminum nitride
AlN
6.28
Direct
Gallium arsenide
GaAs
1.34
Direct
CH(3): PN JUNCTION
80
LED materials and colors
Color
Wavelength
(nm)
Voltage (V)
Semiconductor Material
Infrared
λ > 760
ΔV < 1.9
Gallium arsenide (GaAs) Aluminium gallium arsenide (AlGaAs)
Red
610 < λ < 760
1.63 < ΔV < 2.03
Aluminium gallium arsenide (AlGaAs) Gallium arsenide phosphide
(GaAsP) Aluminium gallium indium phosphide (AlGaInP) Gallium(III)
phosphide (GaP)
Orange
590 < λ < 610
2.03 < ΔV < 2.10
Gallium arsenide phosphide (GaAsP) Aluminium gallium indium
phosphide (AlGaInP)Gallium(III) phosphide (GaP)
Yellow
570 < λ < 590
2.10 < ΔV < 2.18
Gallium arsenide phosphide (GaAsP) Aluminium gallium indium
phosphide (AlGaInP) Gallium(III) phosphide (GaP)
Green
500 < λ < 570
1.9 < ΔV < 4.0
Indium gallium nitride (InGaN) / Gallium(III) nitride (GaN) Gallium(III)
phosphide (GaP)Aluminium gallium indium phosphide (AlGaInP)
Aluminium gallium phosphide (AlGaP)
Blue
450 < λ < 500
2.48 < ΔV < 3.7
Zinc selenide (ZnSe), Indium gallium nitride (InGaN), Silicon carbide
(SiC) as substrate, Silicon (Si)
Violet
400 < λ < 450
2.76 < ΔV < 4.0
Indium gallium nitride (InGaN)
Purple
multiple types
2.48 < ΔV < 3.7
Dual blue/red LEDs,blue with red phosphor,or white with purple plastic
Ultraviolet
λ < 400
3.1 < ΔV < 4.4
diamond (235 nm), Boron nitride (215 nm) , Aluminum nitride (AlN)
(210 nm) Aluminium gallium nitride (AlGaN) (AlGaInN) — (to 210 nm)
White
Broad
spectrum
ΔV = 3.5
Blue/UV diode with yellow phosphor
CH(3): PN JUNCTION
81
CH(3): PN JUNCTION
82
Advantages of LED
LEDs have replaced incandescent lamps in many applications because they have the following
advantages :
▪
▪
▪
Low voltage
Longer life (more than 20 years)
Fast on-off switching
CH(3): PN JUNCTION
83
4) Laser Diode
• Laser stands for Light Amplification by Stimulated Emission of Radiation
• Laser diode is a semiconductor device that produces coherent radiation (in which the
waves are all at the same frequency and phase)
• It is a pn-junction devices used under a forward bias.
• The light emission process in laser diodes is more complicated than that in LEDs, where
light produced in a spontaneous emission process.
CH(3): PN JUNCTION
84
The LASER
▪ Light Amplification by ‘Stimulated Emission’ and
Radiation (L A S E R).
▪ Its wavelengths are from 180 nanometers to 1
millimeter.
▪ The output of a laser is
1. Coherent
2. Monochromatic
3. Collimated
CH(3): PN JUNCTION
85
Coherent light means that all the waves
have the same frequency and phase
CH(3): PN JUNCTION
86
Monochromatic light means that , the light from a laser contains exactly
one color or wavelength rather than a lot of different wavelengths.
CH(3): PN JUNCTION
87
Collimated light while light waves spread out in all directions, laser
light waves all travel in the same direction, exactly parallel to one
another.
CH(3): PN JUNCTION
88
Stimulated Emission
1.
When an incident photon is absorbed an electron moves from an energy state E1 to an
energy state E2 (Fig. a). This process is known as absorption.
2.
If the electron makes the transition back to the lower energy level with a photon being
emitted, we have a spontaneous emission process . (Fig. b).
3.
If there is an incident photon at a time when an electron is in the higher energy state, the
incident photon can interact with the electron, causing the electron to make a transition
downward. The downward transition produces a photon (Fig. c).
4.
Since this process was initiated by the incident photon, the process is called stimulated
emission.
5.
Note that this stimulated emission process has produced two photons; thus, we can have
optical gain or amplification. The two emitted photons are in phase so that the spectral
output will be coherent.
CH(3): PN JUNCTION
89
𝐸 =ℎ𝑓
CH(3): PN JUNCTION
90
Structure of Laser diode
▪ A p-n junction formed by two gallium arsenide layers.
▪ Both layers of the junction are degenerately doped
▪ The two ends of the structure need to be optically flat and parallel with one end
mirrored and one partially reflective.
▪ The length of the junction must be precisely related to the wavelength of the light to
be emitted.
▪ The junction is forward biased and the recombination process produces light as in the
LED.
CH(3): PN JUNCTION
91
CH(3): PN JUNCTION
92
Common materials for semiconductor lasers are
•
•
•
•
•
•
•
•
•
GaAs (gallium arsenide)
AlGaAs (aluminum gallium arsenide)
GaP (gallium phosphide)
InGaP (indium gallium phosphide)
GaN (gallium nitride)
InGaAs (indium gallium arsenide)
GaInNAs (indium gallium arsenide nitride)
InP (indium phosphide)
GaInP (gallium indium phosphide)
Table. 7.2
Type
GaAs
AlGaAs
GaInAsP
Peak Power Wavelength
Application
5 mW
840 nm
CD Players
50 mW
760 nm
Laser printers
20 mW
1300 nm Fiber communications
CH(3): PN JUNCTION
93
http://uk.rs-online.com/web/p/laser-diodes/7587810/
CH(3): PN JUNCTION
94
Applications of laser diode
• The main applications of laser Diodes is medical applications.
• Laser printers
• Fiber communication
CH(3): PN JUNCTION
95
What is the difference between Lasers and LED's diode?
• LEDs are cheap and easy to reproduce .
• Peak power output of laser diodes is measured in watts, while that of LED's, is
measured in milliwatts.
• LED's are small in size, longer life, reliable and require little power.
• A wide range of wavelengths LEDs is available.
• Laser diode’s response is faster than LED.
CH(3): PN JUNCTION
96
Summary
Emitting diodes
Name
Symbol
Difference between them
• It emits incoherent narrow-spectrum light
LED
Laser diodes
• Laser diodes emit coherent light.
Photodiodes
A photodiode is made to detect light.
CH(3): PN JUNCTION
97