semiconductor devices - iit bombaysequel/ee207/sd_pn_1.pdf · 2021. 2. 1. · * in integrated...
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Semiconductor Devices
p-n Junctions: Part 1
M.B. [email protected]
www.ee.iitb.ac.in/~sequel
Department of Electrical EngineeringIndian Institute of Technology Bombay
M. B. Patil, IIT Bombay
p-n junction diodes
p n
V
i
schematic diagram
p (anode) n (cathode)
symbol
* A p-n junction is useful as a stand-alone device (the diode).
* It is also an integral part of devices such as transistors, IGBTs, thyristors, etc.
* In integrated circuits, pn junctions are used to provide isolation between devices.
* We will focus on semiconductor p-n junctions first and look at metal-semiconductorjunctions later.
M. B. Patil, IIT Bombay
p-n junction diodes
p n
V
i
schematic diagram
p (anode) n (cathode)
symbol
* A p-n junction is useful as a stand-alone device (the diode).
* It is also an integral part of devices such as transistors, IGBTs, thyristors, etc.
* In integrated circuits, pn junctions are used to provide isolation between devices.
* We will focus on semiconductor p-n junctions first and look at metal-semiconductorjunctions later.
M. B. Patil, IIT Bombay
p-n junction diodes
p n
V
i
schematic diagram
p (anode) n (cathode)
symbol
* A p-n junction is useful as a stand-alone device (the diode).
* It is also an integral part of devices such as transistors, IGBTs, thyristors, etc.
* In integrated circuits, pn junctions are used to provide isolation between devices.
* We will focus on semiconductor p-n junctions first and look at metal-semiconductorjunctions later.
M. B. Patil, IIT Bombay
p-n junction diodes
p n
V
i
schematic diagram
p (anode) n (cathode)
symbol
* A p-n junction is useful as a stand-alone device (the diode).
* It is also an integral part of devices such as transistors, IGBTs, thyristors, etc.
* In integrated circuits, pn junctions are used to provide isolation between devices.
* We will focus on semiconductor p-n junctions first and look at metal-semiconductorjunctions later.
M. B. Patil, IIT Bombay
p-n junction diodes
p n
V
i
schematic diagram
p (anode) n (cathode)
symbol
* A p-n junction is useful as a stand-alone device (the diode).
* It is also an integral part of devices such as transistors, IGBTs, thyristors, etc.
* In integrated circuits, pn junctions are used to provide isolation between devices.
* We will focus on semiconductor p-n junctions first and look at metal-semiconductorjunctions later.
M. B. Patil, IIT Bombay
p-n junction fabrication
Start with n+ substrate, with n epitaxial layer grown on top.
n+ substrate
n epi-layer
M. B. Patil, IIT Bombay
p-n junction fabrication
Deposit SiO2.
n+ substrate
n epi-layer
M. B. Patil, IIT Bombay
p-n junction fabrication
Deposit SiO2.
n+ substrate
n epi-layer
oxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Apply photoresist.
n+ substrate
n epi-layer
oxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Apply photoresist.
n+ substrate
n epi-layer
photoresistoxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Place mask.
n+ substrate
n epi-layer
photoresistoxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Place mask.
n+ substrate
n epi-layer
photoresistmask
oxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Expose to UV light.
n+ substrate
n epi-layer
photoresistmask
oxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Expose to UV light.
n+ substrate
n epi-layer
photoresistmask
oxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Expose to UV light.
n+ substrate
n epi-layer
photoresistmask
oxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Remove mask.
n+ substrate
n epi-layer
photoresistmask
oxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Remove mask.
n+ substrate
n epi-layer
photoresistoxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Develop photoresist.
n+ substrate
n epi-layer
photoresistoxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Develop photoresist.
n+ substrate
n epi-layer
photoresistoxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Etch oxide (in HF).
n+ substrate
n epi-layer
photoresistoxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Etch oxide (in HF).
n+ substrate
n epi-layer
photoresistoxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Remove photoresist.
n+ substrate
n epi-layer
photoresistoxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Remove photoresist.
n+ substrate
n epi-layer
oxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Perform diffusion of Boron.
n+ substrate
n epi-layer
oxide
M. B. Patil, IIT Bombay
p-n junction fabrication
Perform diffusion of Boron.
n+ substrate
n epi-layer
oxide
BH3
M. B. Patil, IIT Bombay
p-n junction fabrication
Perform diffusion of Boron.
n+ substrate
n epi-layer
oxide
BH3
p
M. B. Patil, IIT Bombay
p-n junction fabrication
Add metal contacts (a few steps).
oxide
n+ substrate
n epi-layer
p
M. B. Patil, IIT Bombay
Idealised p-n junction diode structure
n
xj
p
p
n
Doping densitiesFabricated structure
Na
Nd
(Na − Nd) np
xj
x
x
x
Nd
xj
Idealised structure
NdNa
acceptor density
donor densityx
x
Na
* For our analysis, we will consider a simplified structure with p-type doping on one sideand n-type on the other.
* We will assume the doping densities to change abruptly at the junction → “abrupt” pnjunction.
M. B. Patil, IIT Bombay
Idealised p-n junction diode structure
n
xj
p
p
n
Doping densitiesFabricated structure
Na
Nd
(Na − Nd) np
xj
x
x
x
Nd
xj
Idealised structure
NdNa
acceptor density
donor densityx
x
Na
* For our analysis, we will consider a simplified structure with p-type doping on one sideand n-type on the other.
* We will assume the doping densities to change abruptly at the junction → “abrupt” pnjunction.
M. B. Patil, IIT Bombay
Idealised p-n junction diode structure
n
xj
p
p
n
Doping densitiesFabricated structure
Na
Nd
(Na − Nd) np
xj
x
x
x
Nd
xj
Idealised structure
NdNa
acceptor density
donor densityx
x
Na
* For our analysis, we will consider a simplified structure with p-type doping on one sideand n-type on the other.
* We will assume the doping densities to change abruptly at the junction → “abrupt” pnjunction.
M. B. Patil, IIT Bombay
Idealised p-n junction diode structure
n
xj
p
p
n
Doping densitiesFabricated structure
Na
Nd
(Na − Nd) np
xj
x
x
x
Nd
xj
Idealised structure
NdNa
acceptor density
donor densityx
x
Na
* For our analysis, we will consider a simplified structure with p-type doping on one sideand n-type on the other.
* We will assume the doping densities to change abruptly at the junction → “abrupt” pnjunction.
M. B. Patil, IIT Bombay
pn junction in equilibrium Na Nd
xjp n
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Vbi
ψ
0
E
−Em
* There is a “depletion region” in which the potential ψ varies.
* Away from the depletion region, ψ is constant, and theelectric field is zero.
* There is a “built-in” voltage drop between the p and n sides,denoted by Vbi.
* Note that ψ=− 1
q
dEc
dx, and E =− dψ
dx.
* Let us check if this picture is consistent with Poisson’sequation.
M. B. Patil, IIT Bombay
pn junction in equilibrium Na Nd
xjp n
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Vbi
ψ
0
E
−Em
* There is a “depletion region” in which the potential ψ varies.
* Away from the depletion region, ψ is constant, and theelectric field is zero.
* There is a “built-in” voltage drop between the p and n sides,denoted by Vbi.
* Note that ψ=− 1
q
dEc
dx, and E =− dψ
dx.
* Let us check if this picture is consistent with Poisson’sequation.
M. B. Patil, IIT Bombay
pn junction in equilibrium Na Nd
xjp n
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Vbi
ψ
0
E
−Em
* There is a “depletion region” in which the potential ψ varies.
* Away from the depletion region, ψ is constant, and theelectric field is zero.
* There is a “built-in” voltage drop between the p and n sides,denoted by Vbi.
* Note that ψ=− 1
q
dEc
dx, and E =− dψ
dx.
* Let us check if this picture is consistent with Poisson’sequation.
M. B. Patil, IIT Bombay
pn junction in equilibrium Na Nd
xjp n
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Vbi
ψ
0
E
−Em
* There is a “depletion region” in which the potential ψ varies.
* Away from the depletion region, ψ is constant, and theelectric field is zero.
* There is a “built-in” voltage drop between the p and n sides,denoted by Vbi.
* Note that ψ=− 1
q
dEc
dx, and E =− dψ
dx.
* Let us check if this picture is consistent with Poisson’sequation.
M. B. Patil, IIT Bombay
pn junction in equilibrium Na Nd
xjp n
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Vbi
ψ
0
E
−Em
* There is a “depletion region” in which the potential ψ varies.
* Away from the depletion region, ψ is constant, and theelectric field is zero.
* There is a “built-in” voltage drop between the p and n sides,denoted by Vbi.
* Note that ψ=− 1
q
dEc
dx, and E =− dψ
dx.
* Let us check if this picture is consistent with Poisson’sequation.
M. B. Patil, IIT Bombay
pn junction in equilibrium Na Nd
xjp n
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Vbi
ψ
0
E
−Em
* There is a “depletion region” in which the potential ψ varies.
* Away from the depletion region, ψ is constant, and theelectric field is zero.
* There is a “built-in” voltage drop between the p and n sides,denoted by Vbi.
* Note that ψ=− 1
q
dEc
dx, and E =− dψ
dx.
* Let us check if this picture is consistent with Poisson’sequation.
M. B. Patil, IIT Bombay
pn junction in equilibrium Na Nd
xjp n
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Vbi
ψ
0
E
−Em
* There is a “depletion region” in which the potential ψ varies.
* Away from the depletion region, ψ is constant, and theelectric field is zero.
* There is a “built-in” voltage drop between the p and n sides,denoted by Vbi.
* Note that ψ=− 1
q
dEc
dx, and E =− dψ
dx.
* Let us check if this picture is consistent with Poisson’sequation.
M. B. Patil, IIT Bombay
pn junction in equilibrium Na Nd
xjp n
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Vbi
ψ
0
E
−Em
* There is a “depletion region” in which the potential ψ varies.
* Away from the depletion region, ψ is constant, and theelectric field is zero.
* There is a “built-in” voltage drop between the p and n sides,denoted by Vbi.
* Note that ψ=− 1
q
dEc
dx, and E =− dψ
dx.
* Let us check if this picture is consistent with Poisson’sequation.
M. B. Patil, IIT Bombay
pn junction in equilibrium Na Nd
xjp n
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Vbi
ψ
0
E
−Em
* There is a “depletion region” in which the potential ψ varies.
* Away from the depletion region, ψ is constant, and theelectric field is zero.
* There is a “built-in” voltage drop between the p and n sides,denoted by Vbi.
* Note that ψ=− 1
q
dEc
dx, and E =− dψ
dx.
* Let us check if this picture is consistent with Poisson’sequation.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
xnxp xj
102
106
1010
1014
1018
p
n (in cm−3)
p(in 1017 cm−3)
Na
acceptor density
0
2
Nd
donor density
(in 1017 cm−3)
x (µm)
n
0
2
19.7 20 20.3
Charge density:
p(x) = Nv exp−(EF − Ev (x)
kT
),
n(x) = Nc exp−(Ec (x)− EF
kT
).
In the neutral p-region (x < xp), p = N−a ≈ Na.
In the neutral n-region (x > xn), n = N+d ≈ Nd .
Na = Nv exp−(EF − Ev (xp)
kT
),
Nd = Nc exp−(Ec (xn)− EF
kT
).
→ p(x)
Na= exp−
(Ev (xp)− Ev (x)
kT
),
n(x)
Nd= exp−
(Ec (x)− Ec (xn)
kT
).
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
xnxp xj
102
106
1010
1014
1018
p
n (in cm−3)
p(in 1017 cm−3)
Na
acceptor density
0
2
Nd
donor density
(in 1017 cm−3)
x (µm)
n
0
2
19.7 20 20.3
Charge density:
p(x) = Nv exp−(EF − Ev (x)
kT
),
n(x) = Nc exp−(Ec (x)− EF
kT
).
In the neutral p-region (x < xp), p = N−a ≈ Na.
In the neutral n-region (x > xn), n = N+d ≈ Nd .
Na = Nv exp−(EF − Ev (xp)
kT
),
Nd = Nc exp−(Ec (xn)− EF
kT
).
→ p(x)
Na= exp−
(Ev (xp)− Ev (x)
kT
),
n(x)
Nd= exp−
(Ec (x)− Ec (xn)
kT
).
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
xnxp xj
102
106
1010
1014
1018
p
n (in cm−3)
p(in 1017 cm−3)
Na
acceptor density
0
2
Nd
donor density
(in 1017 cm−3)
x (µm)
n
0
2
19.7 20 20.3
Charge density:
p(x) = Nv exp−(EF − Ev (x)
kT
),
n(x) = Nc exp−(Ec (x)− EF
kT
).
In the neutral p-region (x < xp), p = N−a ≈ Na.
In the neutral n-region (x > xn), n = N+d ≈ Nd .
Na = Nv exp−(EF − Ev (xp)
kT
),
Nd = Nc exp−(Ec (xn)− EF
kT
).
→ p(x)
Na= exp−
(Ev (xp)− Ev (x)
kT
),
n(x)
Nd= exp−
(Ec (x)− Ec (xn)
kT
).
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
xnxp xj
102
106
1010
1014
1018
p
n (in cm−3)
p(in 1017 cm−3)
Na
acceptor density
0
2
Nd
donor density
(in 1017 cm−3)
x (µm)
n
0
2
19.7 20 20.3
Charge density:
p(x) = Nv exp−(EF − Ev (x)
kT
),
n(x) = Nc exp−(Ec (x)− EF
kT
).
In the neutral p-region (x < xp), p = N−a ≈ Na.
In the neutral n-region (x > xn), n = N+d ≈ Nd .
Na = Nv exp−(EF − Ev (xp)
kT
),
Nd = Nc exp−(Ec (xn)− EF
kT
).
→ p(x)
Na= exp−
(Ev (xp)− Ev (x)
kT
),
n(x)
Nd= exp−
(Ec (x)− Ec (xn)
kT
).
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
xnxp xj
102
106
1010
1014
1018
p
n (in cm−3)
p(in 1017 cm−3)
Na
acceptor density
0
2
Nd
donor density
(in 1017 cm−3)
x (µm)
n
0
2
19.7 20 20.3
Charge density:
p(x) = Nv exp−(EF − Ev (x)
kT
),
n(x) = Nc exp−(Ec (x)− EF
kT
).
In the neutral p-region (x < xp), p = N−a ≈ Na.
In the neutral n-region (x > xn), n = N+d ≈ Nd .
Na = Nv exp−(EF − Ev (xp)
kT
),
Nd = Nc exp−(Ec (xn)− EF
kT
).
→ p(x)
Na= exp−
(Ev (xp)− Ev (x)
kT
),
n(x)
Nd= exp−
(Ec (x)− Ec (xn)
kT
).
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
xnxp xj
102
106
1010
1014
1018
p
n (in cm−3)
p(in 1017 cm−3)
Na
acceptor density
0
2
Nd
donor density
(in 1017 cm−3)
x (µm)
n
0
2
19.7 20 20.3
Charge density:
p(x) = Nv exp−(EF − Ev (x)
kT
),
n(x) = Nc exp−(Ec (x)− EF
kT
).
In the neutral p-region (x < xp), p = N−a ≈ Na.
In the neutral n-region (x > xn), n = N+d ≈ Nd .
Na = Nv exp−(EF − Ev (xp)
kT
),
Nd = Nc exp−(Ec (xn)− EF
kT
).
→ p(x)
Na= exp−
(Ev (xp)− Ev (x)
kT
),
n(x)
Nd= exp−
(Ec (x)− Ec (xn)
kT
).
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
xnxp xj
102
106
1010
1014
1018
p
n (in cm−3)
p(in 1017 cm−3)
Na
acceptor density
0
2
Nd
donor density
(in 1017 cm−3)
x (µm)
n
0
2
19.7 20 20.3
Charge density:
p(x) = Nv exp−(EF − Ev (x)
kT
),
n(x) = Nc exp−(Ec (x)− EF
kT
).
In the neutral p-region (x < xp), p = N−a ≈ Na.
In the neutral n-region (x > xn), n = N+d ≈ Nd .
Na = Nv exp−(EF − Ev (xp)
kT
),
Nd = Nc exp−(Ec (xn)− EF
kT
).
→ p(x)
Na= exp−
(Ev (xp)− Ev (x)
kT
),
n(x)
Nd= exp−
(Ec (x)− Ec (xn)
kT
).
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
xnxp xj
102
106
1010
1014
1018
p
n (in cm−3)
p(in 1017 cm−3)
Na
acceptor density
0
2
Nd
donor density
(in 1017 cm−3)
x (µm)
n
0
2
19.7 20 20.3
Charge density:
p(x) = Nv exp−(EF − Ev (x)
kT
),
n(x) = Nc exp−(Ec (x)− EF
kT
).
In the neutral p-region (x < xp), p = N−a ≈ Na.
In the neutral n-region (x > xn), n = N+d ≈ Nd .
Na = Nv exp−(EF − Ev (xp)
kT
),
Nd = Nc exp−(Ec (xn)− EF
kT
).
→ p(x)
Na= exp−
(Ev (xp)− Ev (x)
kT
),
n(x)
Nd= exp−
(Ec (x)− Ec (xn)
kT
).
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
p(in 1017 cm−3)
Nd
donor density
(in 1017 cm−3)
Na
acceptor density
0
2
xjxp xn
Ec
EF
Ev
qVbi
depletion region
n
0
2
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
19.7 20 20.3
−Na
Nd
ρ/q (approx)
* Charge density:
ρ(x) = q[N+d (x) + p(x)− N−
a (x)− n(x)]
≈ q [(Nd (x)− n(x))− (Na(x)− p(x))].
* ρ= 0 in the neutral regions.
* Within the depletion region, both n and p are small, i.e., this region isdepleted of carriers → “depletion region”.
* To proceed further analytically, we make the “depletion approximation,”i.e., we assume that the transistions between the neutral regions and thedepletion region are abrupt.
* Since the depletion region has non-zero charge density, it is also called“space charge region.”
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
p(in 1017 cm−3)
Nd
donor density
(in 1017 cm−3)
Na
acceptor density
0
2
xjxp xn
Ec
EF
Ev
qVbi
depletion region
n
0
2
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
19.7 20 20.3
−Na
Nd
ρ/q (approx)
* Charge density:
ρ(x) = q[N+d (x) + p(x)− N−
a (x)− n(x)]
≈ q [(Nd (x)− n(x))− (Na(x)− p(x))].
* ρ= 0 in the neutral regions.
* Within the depletion region, both n and p are small, i.e., this region isdepleted of carriers → “depletion region”.
* To proceed further analytically, we make the “depletion approximation,”i.e., we assume that the transistions between the neutral regions and thedepletion region are abrupt.
* Since the depletion region has non-zero charge density, it is also called“space charge region.”
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
p(in 1017 cm−3)
Nd
donor density
(in 1017 cm−3)
Na
acceptor density
0
2
xjxp xn
Ec
EF
Ev
qVbi
depletion region
n
0
2
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
19.7 20 20.3
−Na
Nd
ρ/q (approx)
* Charge density:
ρ(x) = q[N+d (x) + p(x)− N−
a (x)− n(x)]
≈ q [(Nd (x)− n(x))− (Na(x)− p(x))].
* ρ= 0 in the neutral regions.
* Within the depletion region, both n and p are small, i.e., this region isdepleted of carriers → “depletion region”.
* To proceed further analytically, we make the “depletion approximation,”i.e., we assume that the transistions between the neutral regions and thedepletion region are abrupt.
* Since the depletion region has non-zero charge density, it is also called“space charge region.”
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
p(in 1017 cm−3)
Nd
donor density
(in 1017 cm−3)
Na
acceptor density
0
2
xjxp xn
Ec
EF
Ev
qVbi
depletion region
n
0
2
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
19.7 20 20.3
−Na
Nd
ρ/q (approx)
* Charge density:
ρ(x) = q[N+d (x) + p(x)− N−
a (x)− n(x)]
≈ q [(Nd (x)− n(x))− (Na(x)− p(x))].
* ρ= 0 in the neutral regions.
* Within the depletion region, both n and p are small, i.e., this region isdepleted of carriers → “depletion region”.
* To proceed further analytically, we make the “depletion approximation,”i.e., we assume that the transistions between the neutral regions and thedepletion region are abrupt.
* Since the depletion region has non-zero charge density, it is also called“space charge region.”
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
p(in 1017 cm−3)
Nd
donor density
(in 1017 cm−3)
Na
acceptor density
0
2
xjxp xn
Ec
EF
Ev
qVbi
depletion region
n
0
2
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
19.7 20 20.3
−Na
Nd
ρ/q (approx)
* Charge density:
ρ(x) = q[N+d (x) + p(x)− N−
a (x)− n(x)]
≈ q [(Nd (x)− n(x))− (Na(x)− p(x))].
* ρ= 0 in the neutral regions.
* Within the depletion region, both n and p are small, i.e., this region isdepleted of carriers → “depletion region”.
* To proceed further analytically, we make the “depletion approximation,”i.e., we assume that the transistions between the neutral regions and thedepletion region are abrupt.
* Since the depletion region has non-zero charge density, it is also called“space charge region.”
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
p(in 1017 cm−3)
Nd
donor density
(in 1017 cm−3)
Na
acceptor density
0
2
xjxp xn
Ec
EF
Ev
qVbi
depletion region
n
0
2
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
19.7 20 20.3
−Na
Nd
ρ/q (approx)
* Charge density:
ρ(x) = q[N+d (x) + p(x)− N−
a (x)− n(x)]
≈ q [(Nd (x)− n(x))− (Na(x)− p(x))].
* ρ= 0 in the neutral regions.
* Within the depletion region, both n and p are small, i.e., this region isdepleted of carriers → “depletion region”.
* To proceed further analytically, we make the “depletion approximation,”i.e., we assume that the transistions between the neutral regions and thedepletion region are abrupt.
* Since the depletion region has non-zero charge density, it is also called“space charge region.”
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
p(in 1017 cm−3)
Nd
donor density
(in 1017 cm−3)
Na
acceptor density
0
2
xjxp xn
Ec
EF
Ev
qVbi
depletion region
n
0
2
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
19.7 20 20.3
−Na
Nd
ρ/q (approx)
* Charge density:
ρ(x) = q[N+d (x) + p(x)− N−
a (x)− n(x)]
≈ q [(Nd (x)− n(x))− (Na(x)− p(x))].
* ρ= 0 in the neutral regions.
* Within the depletion region, both n and p are small, i.e., this region isdepleted of carriers → “depletion region”.
* To proceed further analytically, we make the “depletion approximation,”i.e., we assume that the transistions between the neutral regions and thedepletion region are abrupt.
* Since the depletion region has non-zero charge density, it is also called“space charge region.”
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion regionqNd
ρ
0
−qNa
Na Nd
xjnp
0
E
−Em
W
dEdx
=ρ
ǫ
Wp Wn
x
ψ
Vbi
xp xj xn
E =− dψ
dx
* Built-in voltage Vbi:
p(x) = Nv exp
[− EF − Ev (x)
kT
], n(x) = Nc exp
[− Ec (x)− EF
kT
].
→ p(xn)
p(xp)= exp
[− Ev (xp)− Ev (xn)
kT
].
→ p(xn)n(xn)
p(xp)n(xn)=
n2i
NaNd= exp
(− qVbi
kT
).
The built-in voltage Vbi is therefore given by
Vbi =kT
qlog
(NaNd
n2i
)
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion regionqNd
ρ
0
−qNa
Na Nd
xjnp
0
E
−Em
W
dEdx
=ρ
ǫ
Wp Wn
x
ψ
Vbi
xp xj xn
E =− dψ
dx
* Built-in voltage Vbi:
p(x) = Nv exp
[− EF − Ev (x)
kT
], n(x) = Nc exp
[− Ec (x)− EF
kT
].
→ p(xn)
p(xp)= exp
[− Ev (xp)− Ev (xn)
kT
].
→ p(xn)n(xn)
p(xp)n(xn)=
n2i
NaNd= exp
(− qVbi
kT
).
The built-in voltage Vbi is therefore given by
Vbi =kT
qlog
(NaNd
n2i
)
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion regionqNd
ρ
0
−qNa
Na Nd
xjnp
0
E
−Em
W
dEdx
=ρ
ǫ
Wp Wn
x
ψ
Vbi
xp xj xn
E =− dψ
dx
* Built-in voltage Vbi:
p(x) = Nv exp
[− EF − Ev (x)
kT
], n(x) = Nc exp
[− Ec (x)− EF
kT
].
→ p(xn)
p(xp)= exp
[− Ev (xp)− Ev (xn)
kT
].
→ p(xn)n(xn)
p(xp)n(xn)=
n2i
NaNd= exp
(− qVbi
kT
).
The built-in voltage Vbi is therefore given by
Vbi =kT
qlog
(NaNd
n2i
)
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion regionqNd
ρ
0
−qNa
Na Nd
xjnp
0
E
−Em
W
dEdx
=ρ
ǫ
Wp Wn
x
ψ
Vbi
xp xj xn
E =− dψ
dx
* Built-in voltage Vbi:
p(x) = Nv exp
[− EF − Ev (x)
kT
], n(x) = Nc exp
[− Ec (x)− EF
kT
].
→ p(xn)
p(xp)= exp
[− Ev (xp)− Ev (xn)
kT
].
→ p(xn)n(xn)
p(xp)n(xn)=
n2i
NaNd= exp
(− qVbi
kT
).
The built-in voltage Vbi is therefore given by
Vbi =kT
qlog
(NaNd
n2i
)
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion regionqNd
ρ
0
−qNa
Na Nd
xjnp
0
E
−Em
W
dEdx
=ρ
ǫ
Wp Wn
x
ψ
Vbi
xp xj xn
E =− dψ
dx
* Built-in voltage Vbi:
p(x) = Nv exp
[− EF − Ev (x)
kT
], n(x) = Nc exp
[− Ec (x)− EF
kT
].
→ p(xn)
p(xp)= exp
[− Ev (xp)− Ev (xn)
kT
].
→ p(xn)n(xn)
p(xp)n(xn)=
n2i
NaNd= exp
(− qVbi
kT
).
The built-in voltage Vbi is therefore given by
Vbi =kT
qlog
(NaNd
n2i
)
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion regionqNd
ρ
0
−qNa
Na Nd
xjnp
0
E
−Em
W
dEdx
=ρ
ǫ
Wp Wn
x
ψ
Vbi
xp xj xn
E =− dψ
dx
* Built-in voltage Vbi:
p(x) = Nv exp
[− EF − Ev (x)
kT
], n(x) = Nc exp
[− Ec (x)− EF
kT
].
→ p(xn)
p(xp)= exp
[− Ev (xp)− Ev (xn)
kT
].
→ p(xn)n(xn)
p(xp)n(xn)=
n2i
NaNd= exp
(− qVbi
kT
).
The built-in voltage Vbi is therefore given by
Vbi =kT
qlog
(NaNd
n2i
)
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion regionqNd
ρ
0
−qNa
Na Nd
xjnp
0
E
−Em
W
dEdx
=ρ
ǫ
Wp Wn
x
ψ
Vbi
xp xj xn
E =− dψ
dx
* Built-in voltage Vbi:
p(x) = Nv exp
[− EF − Ev (x)
kT
], n(x) = Nc exp
[− Ec (x)− EF
kT
].
→ p(xn)
p(xp)= exp
[− Ev (xp)− Ev (xn)
kT
].
→ p(xn)n(xn)
p(xp)n(xn)=
n2i
NaNd= exp
(− qVbi
kT
).
The built-in voltage Vbi is therefore given by
Vbi =kT
qlog
(NaNd
n2i
)
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Example:
For a silicon pn junction with Na = 5× 1017 cm−3, Nd = 1017 cm−3,compute Vbi at T = 300 K. (ni = 1.5× 1010 cm−3 at 300 K.)
Solution:
Vbi =kT
qlog
NaNd
n2i
= (0.0259 V) log(5× 1017)(1017)
(1.5× 1010)2
= 0.86 V
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Example:
For a silicon pn junction with Na = 5× 1017 cm−3, Nd = 1017 cm−3,compute Vbi at T = 300 K. (ni = 1.5× 1010 cm−3 at 300 K.)
Solution:
Vbi =kT
qlog
NaNd
n2i
= (0.0259 V) log(5× 1017)(1017)
(1.5× 1010)2
= 0.86 V
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Example:
For a silicon pn junction with Na = 5× 1017 cm−3, Nd = 1017 cm−3,compute Vbi at T = 300 K. (ni = 1.5× 1010 cm−3 at 300 K.)
Solution:
Vbi =kT
qlog
NaNd
n2i
= (0.0259 V) log(5× 1017)(1017)
(1.5× 1010)2
= 0.86 V
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Example:
For a silicon pn junction with Na = 5× 1017 cm−3, Nd = 1017 cm−3,compute Vbi at T = 300 K. (ni = 1.5× 1010 cm−3 at 300 K.)
Solution:
Vbi =kT
qlog
NaNd
n2i
= (0.0259 V) log(5× 1017)(1017)
(1.5× 1010)2
= 0.86 V
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Electric field E(x):
* E =dψ
dx= 0 in the neutral regions.
* In the depletion region,
∫ xn
xp
dE =
∫ xn
xp
ρ
εdx .
Since E(xp) = E(xn) = 0, we must have
∫ xn
xp
ρ
εdx = 0,
which means the area under the ρ versus x curve must be zero.
i.e., NaWp = NdWn →Wp
Wn=
Nd
Na.
→ The depletion width is larger on the lightly doped side.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Electric field E(x):
* E =dψ
dx= 0 in the neutral regions.
* In the depletion region,
∫ xn
xp
dE =
∫ xn
xp
ρ
εdx .
Since E(xp) = E(xn) = 0, we must have
∫ xn
xp
ρ
εdx = 0,
which means the area under the ρ versus x curve must be zero.
i.e., NaWp = NdWn →Wp
Wn=
Nd
Na.
→ The depletion width is larger on the lightly doped side.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Electric field E(x):
* E =dψ
dx= 0 in the neutral regions.
* In the depletion region,
∫ xn
xp
dE =
∫ xn
xp
ρ
εdx .
Since E(xp) = E(xn) = 0, we must have
∫ xn
xp
ρ
εdx = 0,
which means the area under the ρ versus x curve must be zero.
i.e., NaWp = NdWn →Wp
Wn=
Nd
Na.
→ The depletion width is larger on the lightly doped side.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Electric field E(x):
* E =dψ
dx= 0 in the neutral regions.
* In the depletion region,
∫ xn
xp
dE =
∫ xn
xp
ρ
εdx .
Since E(xp) = E(xn) = 0, we must have
∫ xn
xp
ρ
εdx = 0,
which means the area under the ρ versus x curve must be zero.
i.e., NaWp = NdWn →Wp
Wn=
Nd
Na.
→ The depletion width is larger on the lightly doped side.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Electric field E(x):
* E =dψ
dx= 0 in the neutral regions.
* In the depletion region,
∫ xn
xp
dE =
∫ xn
xp
ρ
εdx .
Since E(xp) = E(xn) = 0, we must have
∫ xn
xp
ρ
εdx = 0,
which means the area under the ρ versus x curve must be zero.
i.e., NaWp = NdWn →Wp
Wn=
Nd
Na.
→ The depletion width is larger on the lightly doped side.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Electric field E(x):
* Since ρ is piecewise constant, E must be piecewise linear.
* The maximum value (magnitude) of E occurs at x = xj .∫ xj
xp
dE =1
ε
∫ xj
xp
ρdx → −Em − 0 =1
ε(−qNaWp)
→ Em =qNaWp
ε=
qNdWn
ε∵ NaWp = NdWn.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Electric field E(x):
* Since ρ is piecewise constant, E must be piecewise linear.
* The maximum value (magnitude) of E occurs at x = xj .∫ xj
xp
dE =1
ε
∫ xj
xp
ρdx → −Em − 0 =1
ε(−qNaWp)
→ Em =qNaWp
ε=
qNdWn
ε∵ NaWp = NdWn.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Electric field E(x):
* Since ρ is piecewise constant, E must be piecewise linear.
* The maximum value (magnitude) of E occurs at x = xj .∫ xj
xp
dE =1
ε
∫ xj
xp
ρdx → −Em − 0 =1
ε(−qNaWp)
→ Em =qNaWp
ε=
qNdWn
ε∵ NaWp = NdWn.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Potential ψ(x):
* xp < x < xj :dEdx
= − qN−a
ε≈ − qNa
ε→ E(x) = − qNa
εx + k1.
Since E = 0 at x = xp , we get E(x) = − qNa
ε(x − xp).
→ ψ(x) = −∫Edx =
qNa
ε
[x2
2− xpx
]+ k2.
Taking ψ(xp) = 0, we can find k2.
→ ψ(x) =qNa
2ε(x − xp)2.
If xj is taken as 0, i.e., x ← (x − xj ), we get
ψ(x) =qNa
2ε(x + Wp)2.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Potential ψ(x):
* xp < x < xj :dEdx
= − qN−a
ε≈ − qNa
ε→ E(x) = − qNa
εx + k1.
Since E = 0 at x = xp , we get E(x) = − qNa
ε(x − xp).
→ ψ(x) = −∫Edx =
qNa
ε
[x2
2− xpx
]+ k2.
Taking ψ(xp) = 0, we can find k2.
→ ψ(x) =qNa
2ε(x − xp)2.
If xj is taken as 0, i.e., x ← (x − xj ), we get
ψ(x) =qNa
2ε(x + Wp)2.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Potential ψ(x):
* xp < x < xj :dEdx
= − qN−a
ε≈ − qNa
ε→ E(x) = − qNa
εx + k1.
Since E = 0 at x = xp , we get E(x) = − qNa
ε(x − xp).
→ ψ(x) = −∫Edx =
qNa
ε
[x2
2− xpx
]+ k2.
Taking ψ(xp) = 0, we can find k2.
→ ψ(x) =qNa
2ε(x − xp)2.
If xj is taken as 0, i.e., x ← (x − xj ), we get
ψ(x) =qNa
2ε(x + Wp)2.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Potential ψ(x):
* xp < x < xj :dEdx
= − qN−a
ε≈ − qNa
ε→ E(x) = − qNa
εx + k1.
Since E = 0 at x = xp , we get E(x) = − qNa
ε(x − xp).
→ ψ(x) = −∫Edx =
qNa
ε
[x2
2− xpx
]+ k2.
Taking ψ(xp) = 0, we can find k2.
→ ψ(x) =qNa
2ε(x − xp)2.
If xj is taken as 0, i.e., x ← (x − xj ), we get
ψ(x) =qNa
2ε(x + Wp)2.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Potential ψ(x):
* xp < x < xj :dEdx
= − qN−a
ε≈ − qNa
ε→ E(x) = − qNa
εx + k1.
Since E = 0 at x = xp , we get E(x) = − qNa
ε(x − xp).
→ ψ(x) = −∫Edx =
qNa
ε
[x2
2− xpx
]+ k2.
Taking ψ(xp) = 0, we can find k2.
→ ψ(x) =qNa
2ε(x − xp)2.
If xj is taken as 0, i.e., x ← (x − xj ), we get
ψ(x) =qNa
2ε(x + Wp)2.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Potential ψ(x):
* xp < x < xj :dEdx
= − qN−a
ε≈ − qNa
ε→ E(x) = − qNa
εx + k1.
Since E = 0 at x = xp , we get E(x) = − qNa
ε(x − xp).
→ ψ(x) = −∫Edx =
qNa
ε
[x2
2− xpx
]+ k2.
Taking ψ(xp) = 0, we can find k2.
→ ψ(x) =qNa
2ε(x − xp)2.
If xj is taken as 0, i.e., x ← (x − xj ), we get
ψ(x) =qNa
2ε(x + Wp)2.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Potential ψ(x):
* xj < x < xn:
For convenience, let us take xj = 0 → xp =−Wp , xn =Wn.
dEdx
=qN+
d
ε≈ qNd
ε→ E(x) =
qNd
εx + k3.
Since E = 0 at x =Wn, we get E(x) =qNd
ε(x −Wn).
→ ψ(x) = −∫Edx = − qNd
ε
[x2
2−Wnx
]+ k4.
We can find k4 using continuity of ψ at x = 0.
→ ψ(x) =qNd
ε
[Wnx −
x2
2
]+
qNa
2εW 2
p .
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Potential ψ(x):
* xj < x < xn:
For convenience, let us take xj = 0 → xp =−Wp , xn =Wn.
dEdx
=qN+
d
ε≈ qNd
ε→ E(x) =
qNd
εx + k3.
Since E = 0 at x =Wn, we get E(x) =qNd
ε(x −Wn).
→ ψ(x) = −∫Edx = − qNd
ε
[x2
2−Wnx
]+ k4.
We can find k4 using continuity of ψ at x = 0.
→ ψ(x) =qNd
ε
[Wnx −
x2
2
]+
qNa
2εW 2
p .
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Potential ψ(x):
* xj < x < xn:
For convenience, let us take xj = 0 → xp =−Wp , xn =Wn.
dEdx
=qN+
d
ε≈ qNd
ε→ E(x) =
qNd
εx + k3.
Since E = 0 at x =Wn, we get E(x) =qNd
ε(x −Wn).
→ ψ(x) = −∫Edx = − qNd
ε
[x2
2−Wnx
]+ k4.
We can find k4 using continuity of ψ at x = 0.
→ ψ(x) =qNd
ε
[Wnx −
x2
2
]+
qNa
2εW 2
p .
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Potential ψ(x):
* xj < x < xn:
For convenience, let us take xj = 0 → xp =−Wp , xn =Wn.
dEdx
=qN+
d
ε≈ qNd
ε→ E(x) =
qNd
εx + k3.
Since E = 0 at x =Wn, we get E(x) =qNd
ε(x −Wn).
→ ψ(x) = −∫Edx = − qNd
ε
[x2
2−Wnx
]+ k4.
We can find k4 using continuity of ψ at x = 0.
→ ψ(x) =qNd
ε
[Wnx −
x2
2
]+
qNa
2εW 2
p .
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Potential ψ(x):
* xj < x < xn:
For convenience, let us take xj = 0 → xp =−Wp , xn =Wn.
dEdx
=qN+
d
ε≈ qNd
ε→ E(x) =
qNd
εx + k3.
Since E = 0 at x =Wn, we get E(x) =qNd
ε(x −Wn).
→ ψ(x) = −∫Edx = − qNd
ε
[x2
2−Wnx
]+ k4.
We can find k4 using continuity of ψ at x = 0.
→ ψ(x) =qNd
ε
[Wnx −
x2
2
]+
qNa
2εW 2
p .
M. B. Patil, IIT Bombay
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Vbi
ψ
0
E
−EmWp Wn
Numerical Analytical
0
E
−Em
W
x
ψ
Vbi
qNd
ρ
0
−qNa
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
* pn junction in equilibrium: The band diagram is consistent with Poisson’s equation.
M. B. Patil, IIT Bombay
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Vbi
ψ
0
E
−EmWp Wn
Numerical Analytical
0
E
−Em
W
x
ψ
Vbi
qNd
ρ
0
−qNa
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
* pn junction in equilibrium: The band diagram is consistent with Poisson’s equation.
M. B. Patil, IIT Bombay
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Vbi
ψ
0
E
−EmWp Wn
Numerical Analytical
0
E
−Em
W
x
ψ
Vbi
qNd
ρ
0
−qNa
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
* pn junction in equilibrium: Depletion approximation agrees well with numerical results.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Depletion region width W :
The built-in voltage Vbi is given by the area under the E(x) curve.
Vbi =1
2EmWp +
1
2EmWn =
1
2EmW =
1
2
qNaWp
εW .
Since Wn + Wp =W and WnNd =WpNa, we get
Wn =Na
Na + NdW , Wp =
Nd
Na + NdW .
→ Vbi =1
2
q
ε
NaNd
Na + NdW 2, i.e., W =
√2ε
q
(Na + Nd
NaNd
)Vbi.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Depletion region width W :
The built-in voltage Vbi is given by the area under the E(x) curve.
Vbi =1
2EmWp +
1
2EmWn =
1
2EmW =
1
2
qNaWp
εW .
Since Wn + Wp =W and WnNd =WpNa, we get
Wn =Na
Na + NdW , Wp =
Nd
Na + NdW .
→ Vbi =1
2
q
ε
NaNd
Na + NdW 2, i.e., W =
√2ε
q
(Na + Nd
NaNd
)Vbi.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Depletion region width W :
The built-in voltage Vbi is given by the area under the E(x) curve.
Vbi =1
2EmWp +
1
2EmWn =
1
2EmW =
1
2
qNaWp
εW .
Since Wn + Wp =W and WnNd =WpNa, we get
Wn =Na
Na + NdW , Wp =
Nd
Na + NdW .
→ Vbi =1
2
q
ε
NaNd
Na + NdW 2, i.e., W =
√2ε
q
(Na + Nd
NaNd
)Vbi.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Depletion region width W :
The built-in voltage Vbi is given by the area under the E(x) curve.
Vbi =1
2EmWp +
1
2EmWn =
1
2EmW =
1
2
qNaWp
εW .
Since Wn + Wp =W and WnNd =WpNa, we get
Wn =Na
Na + NdW , Wp =
Nd
Na + NdW .
→ Vbi =1
2
q
ε
NaNd
Na + NdW 2, i.e., W =
√2ε
q
(Na + Nd
NaNd
)Vbi.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Depletion region width W :
The built-in voltage Vbi is given by the area under the E(x) curve.
Vbi =1
2EmWp +
1
2EmWn =
1
2EmW =
1
2
qNaWp
εW .
Since Wn + Wp =W and WnNd =WpNa, we get
Wn =Na
Na + NdW , Wp =
Nd
Na + NdW .
→ Vbi =1
2
q
ε
NaNd
Na + NdW 2, i.e., W =
√2ε
q
(Na + Nd
NaNd
)Vbi.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
For an abrupt, uniformly doped silicon pn junction, Na = 5× 1017 cm−3.
Compute Vbi, W , Wn, Wp , and Em for Nd = 1016, 1017, 5× 1017, 1018,
and 5× 1018 cm−3 (T = 300 K).
Solution:
Vbi = VT logNaNd
n2i
= 0.0259 V× log(5× 1017)(1× 1016)
(1.5× 1010)2
= 0.8 V.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
For an abrupt, uniformly doped silicon pn junction, Na = 5× 1017 cm−3.
Compute Vbi, W , Wn, Wp , and Em for Nd = 1016, 1017, 5× 1017, 1018,
and 5× 1018 cm−3 (T = 300 K).
Solution:
Vbi = VT logNaNd
n2i
= 0.0259 V× log(5× 1017)(1× 1016)
(1.5× 1010)2
= 0.8 V.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
W =
√2εr ε0
q
(Na + Nd
NaNd
)Vbi
=
√2× 11.7× 8.85× 10−14
1.6× 10−19
(5× 1017 + 1× 1016
(5× 1017)(1× 1016)
)(0.8)
= 3.24× 10−5 cm
= 0.324µm.
Units:
√F/cm
Coul× cm−3
cm−6× V = cm
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
W =
√2εr ε0
q
(Na + Nd
NaNd
)Vbi
=
√2× 11.7× 8.85× 10−14
1.6× 10−19
(5× 1017 + 1× 1016
(5× 1017)(1× 1016)
)(0.8)
= 3.24× 10−5 cm
= 0.324µm.
Units:
√F/cm
Coul× cm−3
cm−6× V = cm
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
W =
√2εr ε0
q
(Na + Nd
NaNd
)Vbi
=
√2× 11.7× 8.85× 10−14
1.6× 10−19
(5× 1017 + 1× 1016
(5× 1017)(1× 1016)
)(0.8)
= 3.24× 10−5 cm
= 0.324µm.
Units:
√F/cm
Coul× cm−3
cm−6× V = cm
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
W =
√2εr ε0
q
(Na + Nd
NaNd
)Vbi
=
√2× 11.7× 8.85× 10−14
1.6× 10−19
(5× 1017 + 1× 1016
(5× 1017)(1× 1016)
)(0.8)
= 3.24× 10−5 cm
= 0.324µm.
Units:
√F/cm
Coul× cm−3
cm−6× V = cm
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
W =
√2εr ε0
q
(Na + Nd
NaNd
)Vbi
=
√2× 11.7× 8.85× 10−14
1.6× 10−19
(5× 1017 + 1× 1016
(5× 1017)(1× 1016)
)(0.8)
= 3.24× 10−5 cm
= 0.324µm.
Units:
√F/cm
Coul× cm−3
cm−6× V = cm
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Wn =Na
Na + NdW = 0.318µm, Wp =
Nd
Na + NdW = 0.006µm.
Em =qNd
εWn or
qNa
εWp
=1.6× 10−19 Coul× 1016 cm−3
11.7× 8.85× 10−14 F/cm× (3.18× 10−5 cm)
= 4.9× 104 V/cm
= 49 kV/cm.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Wn =Na
Na + NdW = 0.318µm, Wp =
Nd
Na + NdW = 0.006µm.
Em =qNd
εWn or
qNa
εWp
=1.6× 10−19 Coul× 1016 cm−3
11.7× 8.85× 10−14 F/cm× (3.18× 10−5 cm)
= 4.9× 104 V/cm
= 49 kV/cm.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Wn =Na
Na + NdW = 0.318µm, Wp =
Nd
Na + NdW = 0.006µm.
Em =qNd
εWn or
qNa
εWp
=1.6× 10−19 Coul× 1016 cm−3
11.7× 8.85× 10−14 F/cm× (3.18× 10−5 cm)
= 4.9× 104 V/cm
= 49 kV/cm.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Wn =Na
Na + NdW = 0.318µm, Wp =
Nd
Na + NdW = 0.006µm.
Em =qNd
εWn or
qNa
εWp
=1.6× 10−19 Coul× 1016 cm−3
11.7× 8.85× 10−14 F/cm× (3.18× 10−5 cm)
= 4.9× 104 V/cm
= 49 kV/cm.
M. B. Patil, IIT Bombay
pn junction in equilibrium
n regionp regionneutral neutral
xjxp xn
Ec
EF
Ev
qVbi
depletion region
Na Nd
xjnp
Wp Wn
qNd
ρ
0
−qNa
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Wn =Na
Na + NdW = 0.318µm, Wp =
Nd
Na + NdW = 0.006µm.
Em =qNd
εWn or
qNa
εWp
=1.6× 10−19 Coul× 1016 cm−3
11.7× 8.85× 10−14 F/cm× (3.18× 10−5 cm)
= 4.9× 104 V/cm
= 49 kV/cm.
M. B. Patil, IIT Bombay
qNd
ρ
0
−qNa
Na Nd
xjnp
Wp Wn
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Effect of Nd , with Na = 5× 1017 cm−3 held fixed.
(Vbi in Volts, W , Wn, Wp in µm, Em in kV/cm.)
Nd (cm−3) Vbi W Wn Wp Em1.0× 1016 0.80 0.324 0.318 0.006 49
1.0× 1017 0.86 0.115 0.096 0.019 148
5.0× 1017 0.90 0.068 0.034 0.034 263
1.0× 1018 0.92 0.060 0.020 0.040 307
5.0× 1018 0.96 0.052 0.004 0.047 366
* Vbi = VT logNaNd
n2i
, W =
√2ε
q
(Na + Nd
NaNd
)Vbi,
Wn =Na
Na + NdW , Wp =
Nd
Na + NdW .
Nd � Na (p+n junction):
Wn ≈W , and W is determined mainly by Nd .
Na � Nd (n+p junction):
Wp ≈W , and W is determined mainly by Na.
M. B. Patil, IIT Bombay
qNd
ρ
0
−qNa
Na Nd
xjnp
Wp Wn
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Effect of Nd , with Na = 5× 1017 cm−3 held fixed.
(Vbi in Volts, W , Wn, Wp in µm, Em in kV/cm.)
Nd (cm−3) Vbi W Wn Wp Em1.0× 1016 0.80 0.324 0.318 0.006 49
1.0× 1017 0.86 0.115 0.096 0.019 148
5.0× 1017 0.90 0.068 0.034 0.034 263
1.0× 1018 0.92 0.060 0.020 0.040 307
5.0× 1018 0.96 0.052 0.004 0.047 366
* Vbi = VT logNaNd
n2i
, W =
√2ε
q
(Na + Nd
NaNd
)Vbi,
Wn =Na
Na + NdW , Wp =
Nd
Na + NdW .
Nd � Na (p+n junction):
Wn ≈W , and W is determined mainly by Nd .
Na � Nd (n+p junction):
Wp ≈W , and W is determined mainly by Na.
M. B. Patil, IIT Bombay
qNd
ρ
0
−qNa
Na Nd
xjnp
Wp Wn
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Effect of Nd , with Na = 5× 1017 cm−3 held fixed.
(Vbi in Volts, W , Wn, Wp in µm, Em in kV/cm.)
Nd (cm−3) Vbi W Wn Wp Em1.0× 1016 0.80 0.324 0.318 0.006 49
1.0× 1017 0.86 0.115 0.096 0.019 148
5.0× 1017 0.90 0.068 0.034 0.034 263
1.0× 1018 0.92 0.060 0.020 0.040 307
5.0× 1018 0.96 0.052 0.004 0.047 366
* Vbi = VT logNaNd
n2i
, W =
√2ε
q
(Na + Nd
NaNd
)Vbi,
Wn =Na
Na + NdW , Wp =
Nd
Na + NdW .
Nd � Na (p+n junction):
Wn ≈W , and W is determined mainly by Nd .
Na � Nd (n+p junction):
Wp ≈W , and W is determined mainly by Na.
M. B. Patil, IIT Bombay
qNd
ρ
0
−qNa
Na Nd
xjnp
Wp Wn
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Effect of Nd , with Na = 5× 1017 cm−3 held fixed.
(Vbi in Volts, W , Wn, Wp in µm, Em in kV/cm.)
Nd (cm−3) Vbi W Wn Wp Em1.0× 1016 0.80 0.324 0.318 0.006 49
1.0× 1017 0.86 0.115 0.096 0.019 148
5.0× 1017 0.90 0.068 0.034 0.034 263
1.0× 1018 0.92 0.060 0.020 0.040 307
5.0× 1018 0.96 0.052 0.004 0.047 366
* Vbi = VT logNaNd
n2i
, W =
√2ε
q
(Na + Nd
NaNd
)Vbi,
Wn =Na
Na + NdW , Wp =
Nd
Na + NdW .
Nd � Na (p+n junction):
Wn ≈W , and W is determined mainly by Nd .
Na � Nd (n+p junction):
Wp ≈W , and W is determined mainly by Na.
M. B. Patil, IIT Bombay
qNd
ρ
0
−qNa
Na Nd
xjnp
Wp Wn
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Effect of Nd , with Na = 5× 1017 cm−3 held fixed.
(Vbi in Volts, W , Wn, Wp in µm, Em in kV/cm.)
Nd (cm−3) Vbi W Wn Wp Em1.0× 1016 0.80 0.324 0.318 0.006 49
1.0× 1017 0.86 0.115 0.096 0.019 148
5.0× 1017 0.90 0.068 0.034 0.034 263
1.0× 1018 0.92 0.060 0.020 0.040 307
5.0× 1018 0.96 0.052 0.004 0.047 366
* For high doping densities such as 1018 cm−3, degenerate
statistics should be used for higher accuracy, i.e.,
n = Nc2√πF1/2(ηc ), with ηc =
EF − Ec
kT, and
p = Nv2√πF1/2(ηv ), with ηv =
Ev − EF
kT.
M. B. Patil, IIT Bombay
qNd
ρ
0
−qNa
Na Nd
xjnp
Wp Wn
0
E
−Em
W
x
ψ
Vbi
xp xj xn
dEdx
=ρ
ǫ
E =− dψ
dx
Effect of Nd , with Na = 5× 1017 cm−3 held fixed.
(Vbi in Volts, W , Wn, Wp in µm, Em in kV/cm.)
Nd (cm−3) Vbi W Wn Wp Em1.0× 1016 0.80 0.324 0.318 0.006 49
1.0× 1017 0.86 0.115 0.096 0.019 148
5.0× 1017 0.90 0.068 0.034 0.034 263
1.0× 1018 0.92 0.060 0.020 0.040 307
5.0× 1018 0.96 0.052 0.004 0.047 366
* For high doping densities such as 1018 cm−3, degenerate
statistics should be used for higher accuracy, i.e.,
n = Nc2√πF1/2(ηc ), with ηc =
EF − Ec
kT, and
p = Nv2√πF1/2(ηv ), with ηv =
Ev − EF
kT.
M. B. Patil, IIT Bombay
pn junction in equilibrium: current densities
n regionp regionneutral neutral
xj
Na Nd
xjp n
xjxp xn
Ec
EF
Ev
qVbi
depletion region
102
106
1010
1014
1018
x (µm)
p
n (in cm−3)
19.7 20 20.3
* The diffusion currents can be expected to be substantial sincethere is a large change in n or p between the p-side and then-side.
* In equilibrium, the drift and diffusion currents are equal andopposite for eletrons as well as holes, i.e.,
Jdiffn = −Jdrift
n , Jdiffp = −Jdrift
p .
* Qualitatively, we can see that the diffusion and drift currentswill be in opposite directions:
Electrons:
Fdiffn : , E : , Fdrift
n : .
Holes:
Fdiffp : , E : , Fdrift
p : .
M. B. Patil, IIT Bombay
pn junction in equilibrium: current densities
n regionp regionneutral neutral
xj
Na Nd
xjp n
xjxp xn
Ec
EF
Ev
qVbi
depletion region
102
106
1010
1014
1018
x (µm)
p
n (in cm−3)
19.7 20 20.3
* The diffusion currents can be expected to be substantial sincethere is a large change in n or p between the p-side and then-side.
* In equilibrium, the drift and diffusion currents are equal andopposite for eletrons as well as holes, i.e.,
Jdiffn = −Jdrift
n , Jdiffp = −Jdrift
p .
* Qualitatively, we can see that the diffusion and drift currentswill be in opposite directions:
Electrons:
Fdiffn : , E : , Fdrift
n : .
Holes:
Fdiffp : , E : , Fdrift
p : .
M. B. Patil, IIT Bombay
pn junction in equilibrium: current densities
n regionp regionneutral neutral
xj
Na Nd
xjp n
xjxp xn
Ec
EF
Ev
qVbi
depletion region
102
106
1010
1014
1018
x (µm)
p
n (in cm−3)
19.7 20 20.3
* The diffusion currents can be expected to be substantial sincethere is a large change in n or p between the p-side and then-side.
* In equilibrium, the drift and diffusion currents are equal andopposite for eletrons as well as holes, i.e.,
Jdiffn = −Jdrift
n , Jdiffp = −Jdrift
p .
* Qualitatively, we can see that the diffusion and drift currentswill be in opposite directions:
Electrons:
Fdiffn : , E : , Fdrift
n : .
Holes:
Fdiffp : , E : , Fdrift
p : .
M. B. Patil, IIT Bombay
pn junction in equilibrium: current densities
n regionp regionneutral neutral
xj
Na Nd
xjp n
xjxp xn
Ec
EF
Ev
qVbi
depletion region
102
106
1010
1014
1018
x (µm)
p
n (in cm−3)
19.7 20 20.3
* The diffusion currents can be expected to be substantial sincethere is a large change in n or p between the p-side and then-side.
* In equilibrium, the drift and diffusion currents are equal andopposite for eletrons as well as holes, i.e.,
Jdiffn = −Jdrift
n , Jdiffp = −Jdrift
p .
* Qualitatively, we can see that the diffusion and drift currentswill be in opposite directions:
Electrons:
Fdiffn : , E : , Fdrift
n : .
Holes:
Fdiffp : , E : , Fdrift
p : .
M. B. Patil, IIT Bombay
pn junction in equilibrium: current densities
n regionp regionneutral neutral
xj
Na Nd
xjp n
xjxp xn
Ec
EF
Ev
qVbi
depletion region
102
106
1010
1014
1018
x (µm)
p
n (in cm−3)
19.7 20 20.3
* The diffusion currents can be expected to be substantial sincethere is a large change in n or p between the p-side and then-side.
* In equilibrium, the drift and diffusion currents are equal andopposite for eletrons as well as holes, i.e.,
Jdiffn = −Jdrift
n , Jdiffp = −Jdrift
p .
* Qualitatively, we can see that the diffusion and drift currentswill be in opposite directions:
Electrons:
Fdiffn : , E : , Fdrift
n : .
Holes:
Fdiffp : , E : , Fdrift
p : .
M. B. Patil, IIT Bombay
pn junction in equilibrium: current densities
n regionp regionneutral neutral
xj
Na Nd
xjp n
xjxp xn
Ec
EF
Ev
qVbi
depletion region
102
106
1010
1014
1018
x (µm)
p
n (in cm−3)
19.7 20 20.3
* The diffusion currents can be expected to be substantial sincethere is a large change in n or p between the p-side and then-side.
* In equilibrium, the drift and diffusion currents are equal andopposite for eletrons as well as holes, i.e.,
Jdiffn = −Jdrift
n , Jdiffp = −Jdrift
p .
* Qualitatively, we can see that the diffusion and drift currentswill be in opposite directions:
Electrons:
Fdiffn : , E : , Fdrift
n : .
Holes:
Fdiffp : , E : , Fdrift
p : .
M. B. Patil, IIT Bombay
Summary
xn
−Na
xp xj
ρ/q (approx)
(in kV/cm)
(in 105 A/cm2)
(in 1017 cm−3)
(in 104 A/cm2)
(in 1017 cm−3)
102
106
1010
1014
1018
p
n (in cm−3)
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
E
(in kV/cm)
0
−100
E
Jdiffn
Jdriftn
x (µm)
Jdiffp
Jdriftp
0
6
−6
x (µm)
Ec
EF
Ev
qVbi
xjxn
depletionregion
neutraln region
np
xpdepletionregion
neutralp region
0
1
xj xj
0
−100
5
0
−5
2
0
Nd
19.7 20 20.3 20 20.1219.84 20
* There are three regions: p neutral region, n neutral region, and depletion region.
M. B. Patil, IIT Bombay
Summary
xn
−Na
xp xj
ρ/q (approx)
(in kV/cm)
(in 105 A/cm2)
(in 1017 cm−3)
(in 104 A/cm2)
(in 1017 cm−3)
102
106
1010
1014
1018
p
n (in cm−3)
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
E
(in kV/cm)
0
−100
E
Jdiffn
Jdriftn
x (µm)
Jdiffp
Jdriftp
0
6
−6
x (µm)
Ec
EF
Ev
qVbi
xjxn
depletionregion
neutraln region
np
xpdepletionregion
neutralp region
0
1
xj xj
0
−100
5
0
−5
2
0
Nd
19.7 20 20.3 20 20.1219.84 20
* There are three regions: p neutral region, n neutral region, and depletion region.
M. B. Patil, IIT Bombay
Summary
xn
−Na
xp xj
ρ/q (approx)
(in kV/cm)
(in 105 A/cm2)
(in 1017 cm−3)
(in 104 A/cm2)
(in 1017 cm−3)
102
106
1010
1014
1018
p
n (in cm−3)
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
E
(in kV/cm)
0
−100
E
Jdiffn
Jdriftn
x (µm)
Jdiffp
Jdriftp
0
6
−6
x (µm)
Ec
EF
Ev
qVbi
xjxn
depletionregion
neutraln region
np
xpdepletionregion
neutralp region
0
1
xj xj
0
−100
5
0
−5
2
0
Nd
19.7 20 20.3 20 20.1219.84 20
* The electric field is zero in the neutral regions and maximum (in magnitude) at the junction.
M. B. Patil, IIT Bombay
Summary
xn
−Na
xp xj
ρ/q (approx)
(in kV/cm)
(in 105 A/cm2)
(in 1017 cm−3)
(in 104 A/cm2)
(in 1017 cm−3)
102
106
1010
1014
1018
p
n (in cm−3)
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
E
(in kV/cm)
0
−100
E
Jdiffn
Jdriftn
x (µm)
Jdiffp
Jdriftp
0
6
−6
x (µm)
Ec
EF
Ev
qVbi
xjxn
depletionregion
neutraln region
np
xpdepletionregion
neutralp region
0
1
xj xj
0
−100
5
0
−5
2
0
Nd
19.7 20 20.3 20 20.1219.84 20
* The electric field is zero in the neutral regions and maximum (in magnitude) at the junction.
M. B. Patil, IIT Bombay
Summary
xn
−Na
xp xj
ρ/q (approx)
(in kV/cm)
(in 105 A/cm2)
(in 1017 cm−3)
(in 104 A/cm2)
(in 1017 cm−3)
102
106
1010
1014
1018
p
n (in cm−3)
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
E
(in kV/cm)
0
−100
E
Jdiffn
Jdriftn
x (µm)
Jdiffp
Jdriftp
0
6
−6
x (µm)
Ec
EF
Ev
qVbi
xjxn
depletionregion
neutraln region
np
xpdepletionregion
neutralp region
0
1
xj xj
0
−100
5
0
−5
2
0
Nd
19.7 20 20.3 20 20.1219.84 20
* There is a potential difference – the built-in voltage Vbi – between the neutral p and neutral n sides.
M. B. Patil, IIT Bombay
Summary
xn
−Na
xp xj
ρ/q (approx)
(in kV/cm)
(in 105 A/cm2)
(in 1017 cm−3)
(in 104 A/cm2)
(in 1017 cm−3)
102
106
1010
1014
1018
p
n (in cm−3)
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
E
(in kV/cm)
0
−100
E
Jdiffn
Jdriftn
x (µm)
Jdiffp
Jdriftp
0
6
−6
x (µm)
Ec
EF
Ev
qVbi
xjxn
depletionregion
neutraln region
np
xpdepletionregion
neutralp region
0
1
xj xj
0
−100
5
0
−5
2
0
Nd
19.7 20 20.3 20 20.1219.84 20
* There is a potential difference – the built-in voltage Vbi – between the neutral p and neutral n sides.
M. B. Patil, IIT Bombay
Summary
xn
−Na
xp xj
ρ/q (approx)
(in kV/cm)
(in 105 A/cm2)
(in 1017 cm−3)
(in 104 A/cm2)
(in 1017 cm−3)
102
106
1010
1014
1018
p
n (in cm−3)
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
E
(in kV/cm)
0
−100
E
Jdiffn
Jdriftn
x (µm)
Jdiffp
Jdriftp
0
6
−6
x (µm)
Ec
EF
Ev
qVbi
xjxn
depletionregion
neutraln region
np
xpdepletionregion
neutralp region
0
1
xj xj
0
−100
5
0
−5
2
0
Nd
19.7 20 20.3 20 20.1219.84 20
* Jn and Jp are individually zero because the drift and diffusion components cancel out.
M. B. Patil, IIT Bombay
Summary
xn
−Na
xp xj
ρ/q (approx)
(in kV/cm)
(in 105 A/cm2)
(in 1017 cm−3)
(in 104 A/cm2)
(in 1017 cm−3)
102
106
1010
1014
1018
p
n (in cm−3)
x (µm)
−1
2
0
ρ/q (exact)
(in 1017 cm−3)
E
(in kV/cm)
0
−100
E
Jdiffn
Jdriftn
x (µm)
Jdiffp
Jdriftp
0
6
−6
x (µm)
Ec
EF
Ev
qVbi
xjxn
depletionregion
neutraln region
np
xpdepletionregion
neutralp region
0
1
xj xj
0
−100
5
0
−5
2
0
Nd
19.7 20 20.3 20 20.1219.84 20
* Jn and Jp are individually zero because the drift and diffusion components cancel out.
M. B. Patil, IIT Bombay