semiconductor devices - iit bombaysequel/ee207/sd_pn_1.pdf · 2021. 2. 1. · * in integrated...

Semiconductor Devices

p-n Junctions: Part 1

M.B. Patilmbpatil@ee.iitb.ac.in

www.ee.iitb.ac.in/~sequel

Department of Electrical EngineeringIndian Institute of Technology Bombay

M. B. Patil, IIT Bombay

p-n junction diodes

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.

p-n junction diodes

schematic diagram

symbol

p-n junction diodes

schematic diagram

symbol

p-n junction diodes

schematic diagram

symbol

p-n junction diodes

schematic diagram

symbol

p-n junction fabrication

Start with n+ substrate, with n epitaxial layer grown on top.

n+ substrate

n epi-layer

Deposit SiO2.

n+ substrate

n epi-layer

Deposit SiO2.

n+ substrate

n epi-layer

Apply photoresist.

n+ substrate

n epi-layer

Apply photoresist.

n+ substrate

n epi-layer

photoresistoxide

Place mask.

n+ substrate

n epi-layer

photoresistoxide

Place mask.

n+ substrate

n epi-layer

photoresistmask

Expose to UV light.

n+ substrate

n epi-layer

photoresistmask

Expose to UV light.

n+ substrate

n epi-layer

photoresistmask

Expose to UV light.

n+ substrate

n epi-layer

photoresistmask

Remove mask.

n+ substrate

n epi-layer

photoresistmask

Remove mask.

n+ substrate

n epi-layer

photoresistoxide

Develop photoresist.

n+ substrate

n epi-layer

photoresistoxide

Develop photoresist.

n+ substrate

n epi-layer

photoresistoxide

Etch oxide (in HF).

n+ substrate

n epi-layer

photoresistoxide

Etch oxide (in HF).

n+ substrate

n epi-layer

photoresistoxide

Remove photoresist.

n+ substrate

n epi-layer

photoresistoxide

Remove photoresist.

n+ substrate

n epi-layer

Perform diffusion of Boron.

n+ substrate

n epi-layer

n+ substrate

n epi-layer

n+ substrate

n epi-layer

Add metal contacts (a few steps).

n+ substrate

n epi-layer

Idealised p-n junction diode structure

Doping densitiesFabricated structure

(Na − Nd) np

Idealised structure

acceptor density

donor densityx

* 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.

(Na − Nd) np

Idealised structure

acceptor density

donor densityx

(Na − Nd) np

Idealised structure

acceptor density

donor densityx

(Na − Nd) np

Idealised structure

acceptor density

donor densityx

pn junction in equilibrium Na Nd

n regionp regionneutral neutral

xjxp xn

depletion region

* 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

dx, and E =− dψ

* Let us check if this picture is consistent with Poisson’sequation.

xjxp xn

depletion region

dx, and E =− dψ

xjxp xn

depletion region

dx, and E =− dψ

xjxp xn

depletion region

dx, and E =− dψ

xjxp xn

depletion region

dx, and E =− dψ

xjxp xn

depletion region

dx, and E =− dψ

xjxp xn

depletion region

dx, and E =− dψ

xjxp xn

depletion region

dx, and E =− dψ

xjxp xn

depletion region

dx, and E =− dψ

pn junction in equilibrium

xjxp xn

depletion region

xnxp xj

n (in cm−3)

p(in 1017 cm−3)

acceptor density

donor density

(in 1017 cm−3)

x (µm)

19.7 20 20.3

Charge density:

p(x) = Nv exp−(EF − Ev (x)

n(x) = Nc exp−(Ec (x)− EF

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)

Nd = Nc exp−(Ec (xn)− EF

→ p(x)

Na= exp−

(Ev (xp)− Ev (x)

Nd= exp−

(Ec (x)− Ec (xn)

xjxp xn

depletion region

xnxp xj

n (in cm−3)

p(in 1017 cm−3)

acceptor density

donor density

(in 1017 cm−3)

x (µm)

19.7 20 20.3

Charge density:

→ p(x)

Na= exp−

(Ev (xp)− Ev (x)

Nd= exp−

(Ec (x)− Ec (xn)

xjxp xn

depletion region

xnxp xj

n (in cm−3)

p(in 1017 cm−3)

acceptor density

donor density

(in 1017 cm−3)

x (µm)

19.7 20 20.3

Charge density:

→ p(x)

Na= exp−

(Ev (xp)− Ev (x)

Nd= exp−

(Ec (x)− Ec (xn)

xjxp xn

depletion region

xnxp xj

n (in cm−3)

p(in 1017 cm−3)

acceptor density

donor density

(in 1017 cm−3)

x (µm)

19.7 20 20.3

Charge density:

→ p(x)

Na= exp−

(Ev (xp)− Ev (x)

Nd= exp−

(Ec (x)− Ec (xn)

xjxp xn

depletion region

xnxp xj

n (in cm−3)

p(in 1017 cm−3)

acceptor density

donor density

(in 1017 cm−3)

x (µm)

19.7 20 20.3

Charge density:

→ p(x)

Na= exp−

(Ev (xp)− Ev (x)

Nd= exp−

(Ec (x)− Ec (xn)

xjxp xn

depletion region

xnxp xj

n (in cm−3)

p(in 1017 cm−3)

acceptor density

donor density

(in 1017 cm−3)

x (µm)

19.7 20 20.3

Charge density:

→ p(x)

Na= exp−

(Ev (xp)− Ev (x)

Nd= exp−

(Ec (x)− Ec (xn)

xjxp xn

depletion region

xnxp xj

n (in cm−3)

p(in 1017 cm−3)

acceptor density

donor density

(in 1017 cm−3)

x (µm)

19.7 20 20.3

Charge density:

→ p(x)

Na= exp−

(Ev (xp)− Ev (x)

Nd= exp−

(Ec (x)− Ec (xn)

xjxp xn

depletion region

xnxp xj

n (in cm−3)

p(in 1017 cm−3)

acceptor density

donor density

(in 1017 cm−3)

x (µm)

19.7 20 20.3

Charge density:

→ p(x)

Na= exp−

(Ev (xp)− Ev (x)

Nd= exp−

(Ec (x)− Ec (xn)

p(in 1017 cm−3)

donor density

(in 1017 cm−3)

acceptor density

xjxp xn

depletion region

x (µm)

ρ/q (exact)

(in 1017 cm−3)

19.7 20 20.3

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.”

p(in 1017 cm−3)

donor density

(in 1017 cm−3)

acceptor density

xjxp xn

depletion region

x (µm)

ρ/q (exact)

(in 1017 cm−3)

19.7 20 20.3

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))].

p(in 1017 cm−3)

donor density

(in 1017 cm−3)

acceptor density

xjxp xn

depletion region

x (µm)

ρ/q (exact)

(in 1017 cm−3)

19.7 20 20.3

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))].

p(in 1017 cm−3)

donor density

(in 1017 cm−3)

acceptor density

xjxp xn

depletion region

x (µm)

ρ/q (exact)

(in 1017 cm−3)

19.7 20 20.3

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))].

p(in 1017 cm−3)

donor density

(in 1017 cm−3)

acceptor density

xjxp xn

depletion region

x (µm)

ρ/q (exact)

(in 1017 cm−3)

19.7 20 20.3

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))].

p(in 1017 cm−3)

donor density

(in 1017 cm−3)

acceptor density

xjxp xn

depletion region

x (µm)

ρ/q (exact)

(in 1017 cm−3)

19.7 20 20.3

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))].

p(in 1017 cm−3)

donor density

(in 1017 cm−3)

acceptor density

xjxp xn

depletion region

x (µm)

ρ/q (exact)

(in 1017 cm−3)

19.7 20 20.3

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))].

xjxp xn

depletion regionqNd

−qNa

xp xj xn

E =− dψ

* Built-in voltage Vbi:

p(x) = Nv exp

[− EF − Ev (x)

], n(x) = Nc exp

[− Ec (x)− EF

→ p(xn)

p(xp)= exp

[− Ev (xp)− Ev (xn)

→ p(xn)n(xn)

p(xp)n(xn)=

NaNd= exp

(− qVbi

The built-in voltage Vbi is therefore given by

Vbi =kT

xjxp xn

depletion regionqNd

−qNa

xp xj xn

E =− dψ

p(x) = Nv exp

[− EF − Ev (x)

], n(x) = Nc exp

[− Ec (x)− EF

→ p(xn)

p(xp)= exp

→ p(xn)n(xn)

p(xp)n(xn)=

NaNd= exp

(− qVbi

Vbi =kT

xjxp xn

depletion regionqNd

−qNa

xp xj xn

E =− dψ

p(x) = Nv exp

[− EF − Ev (x)

], n(x) = Nc exp

[− Ec (x)− EF

→ p(xn)

p(xp)= exp

→ p(xn)n(xn)

p(xp)n(xn)=

NaNd= exp

(− qVbi

Vbi =kT

xjxp xn

depletion regionqNd

−qNa

xp xj xn

E =− dψ

p(x) = Nv exp

[− EF − Ev (x)

], n(x) = Nc exp

[− Ec (x)− EF

→ p(xn)

p(xp)= exp

→ p(xn)n(xn)

p(xp)n(xn)=

NaNd= exp

(− qVbi

Vbi =kT

xjxp xn

depletion regionqNd

−qNa

xp xj xn

E =− dψ

p(x) = Nv exp

[− EF − Ev (x)

], n(x) = Nc exp

[− Ec (x)− EF

→ p(xn)

p(xp)= exp

→ p(xn)n(xn)

p(xp)n(xn)=

NaNd= exp

(− qVbi

Vbi =kT

xjxp xn

depletion regionqNd

−qNa

xp xj xn

E =− dψ

p(x) = Nv exp

[− EF − Ev (x)

], n(x) = Nc exp

[− Ec (x)− EF

→ p(xn)

p(xp)= exp

→ p(xn)n(xn)

p(xp)n(xn)=

NaNd= exp

(− qVbi

Vbi =kT

xjxp xn

depletion regionqNd

−qNa

xp xj xn

E =− dψ

p(x) = Nv exp

[− EF − Ev (x)

], n(x) = Nc exp

[− Ec (x)− EF

→ p(xn)

p(xp)= exp

→ p(xn)n(xn)

p(xp)n(xn)=

NaNd= exp

(− qVbi

Vbi =kT

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

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

= (0.0259 V) log(5× 1017)(1017)

(1.5× 1010)2

= 0.86 V

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Example:

Solution:

Vbi =kT

= (0.0259 V) log(5× 1017)(1017)

(1.5× 1010)2

= 0.86 V

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Example:

Solution:

Vbi =kT

= (0.0259 V) log(5× 1017)(1017)

(1.5× 1010)2

= 0.86 V

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Example:

Solution:

Vbi =kT

= (0.0259 V) log(5× 1017)(1017)

(1.5× 1010)2

= 0.86 V

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Electric field E(x):

* E =dψ

dx= 0 in the neutral regions.

* In the depletion region,

∫ xn

εdx .

Since E(xp) = E(xn) = 0, we must have

∫ xn

εdx = 0,

which means the area under the ρ versus x curve must be zero.

i.e., NaWp = NdWn →Wp

→ The depletion width is larger on the lightly doped side.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

* E =dψ

∫ xn

εdx .

∫ xn

εdx = 0,

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

* E =dψ

∫ xn

εdx .

∫ xn

εdx = 0,

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

* E =dψ

∫ xn

εdx .

∫ xn

εdx = 0,

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

* E =dψ

∫ xn

εdx .

∫ xn

εdx = 0,

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

* Since ρ is piecewise constant, E must be piecewise linear.

* The maximum value (magnitude) of E occurs at x = xj .∫ xj

∫ xj

ρdx → −Em − 0 =1

ε(−qNaWp)

→ Em =qNaWp

ε∵ NaWp = NdWn.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

∫ xj

ρdx → −Em − 0 =1

ε(−qNaWp)

→ Em =qNaWp

ε∵ NaWp = NdWn.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

∫ xj

ρdx → −Em − 0 =1

ε(−qNaWp)

→ Em =qNaWp

ε∵ NaWp = NdWn.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

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 =

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.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Potential ψ(x):

* xp < x < xj :dEdx

= − qN−a

ε≈ − qNa

εx + k1.

ε(x − xp).

→ ψ(x) = −∫Edx =

2− xpx

]+ k2.

→ ψ(x) =qNa

2ε(x − xp)2.

ψ(x) =qNa

2ε(x + Wp)2.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Potential ψ(x):

* xp < x < xj :dEdx

= − qN−a

ε≈ − qNa

εx + k1.

ε(x − xp).

→ ψ(x) = −∫Edx =

2− xpx

]+ k2.

→ ψ(x) =qNa

2ε(x − xp)2.

ψ(x) =qNa

2ε(x + Wp)2.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Potential ψ(x):

* xp < x < xj :dEdx

= − qN−a

ε≈ − qNa

εx + k1.

ε(x − xp).

→ ψ(x) = −∫Edx =

2− xpx

]+ k2.

→ ψ(x) =qNa

2ε(x − xp)2.

ψ(x) =qNa

2ε(x + Wp)2.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Potential ψ(x):

* xp < x < xj :dEdx

= − qN−a

ε≈ − qNa

εx + k1.

ε(x − xp).

→ ψ(x) = −∫Edx =

2− xpx

]+ k2.

→ ψ(x) =qNa

2ε(x − xp)2.

ψ(x) =qNa

2ε(x + Wp)2.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Potential ψ(x):

* xp < x < xj :dEdx

= − qN−a

ε≈ − qNa

εx + k1.

ε(x − xp).

→ ψ(x) = −∫Edx =

2− xpx

]+ k2.

→ ψ(x) =qNa

2ε(x − xp)2.

ψ(x) =qNa

2ε(x + Wp)2.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Potential ψ(x):

* xj < x < xn:

For convenience, let us take xj = 0 → xp =−Wp , xn =Wn.

ε≈ qNd

ε→ E(x) =

εx + k3.

Since E = 0 at x =Wn, we get E(x) =qNd

ε(x −Wn).

→ ψ(x) = −∫Edx = − qNd

2−Wnx

]+ k4.

We can find k4 using continuity of ψ at x = 0.

→ ψ(x) =qNd

[Wnx −

2εW 2

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Potential ψ(x):

* xj < x < xn:

ε≈ qNd

ε→ E(x) =

εx + k3.

ε(x −Wn).

→ ψ(x) = −∫Edx = − qNd

2−Wnx

]+ k4.

→ ψ(x) =qNd

[Wnx −

2εW 2

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Potential ψ(x):

* xj < x < xn:

ε≈ qNd

ε→ E(x) =

εx + k3.

ε(x −Wn).

→ ψ(x) = −∫Edx = − qNd

2−Wnx

]+ k4.

→ ψ(x) =qNd

[Wnx −

2εW 2

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Potential ψ(x):

* xj < x < xn:

ε≈ qNd

ε→ E(x) =

εx + k3.

ε(x −Wn).

→ ψ(x) = −∫Edx = − qNd

2−Wnx

]+ k4.

→ ψ(x) =qNd

[Wnx −

2εW 2

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Potential ψ(x):

* xj < x < xn:

ε≈ qNd

ε→ E(x) =

εx + k3.

ε(x −Wn).

→ ψ(x) = −∫Edx = − qNd

2−Wnx

]+ k4.

→ ψ(x) =qNd

[Wnx −

2εW 2

xjxp xn

depletion region

−EmWp Wn

Numerical Analytical

−qNa

xp xj xn

E =− dψ

* pn junction in equilibrium: The band diagram is consistent with Poisson’s equation.

xjxp xn

depletion region

−EmWp Wn

−qNa

xp xj xn

E =− dψ

* pn junction in equilibrium: The band diagram is consistent with Poisson’s equation.

xjxp xn

depletion region

−EmWp Wn

−qNa

xp xj xn

E =− dψ

* pn junction in equilibrium: Depletion approximation agrees well with numerical results.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Depletion region width W :

The built-in voltage Vbi is given by the area under the E(x) curve.

Vbi =1

2EmWp +

2EmWn =

2EmW =

Since Wn + Wp =W and WnNd =WpNa, we get

Wn =Na

Na + NdW , Wp =

Na + NdW .

→ Vbi =1

Na + NdW 2, i.e., W =

√2ε

(Na + Nd

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Vbi =1

2EmWp +

2EmWn =

2EmW =

Wn =Na

Na + NdW , Wp =

Na + NdW .

→ Vbi =1

Na + NdW 2, i.e., W =

√2ε

(Na + Nd

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Vbi =1

2EmWp +

2EmWn =

2EmW =

Wn =Na

Na + NdW , Wp =

Na + NdW .

→ Vbi =1

Na + NdW 2, i.e., W =

√2ε

(Na + Nd

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Vbi =1

2EmWp +

2EmWn =

2EmW =

Wn =Na

Na + NdW , Wp =

Na + NdW .

→ Vbi =1

Na + NdW 2, i.e., W =

√2ε

(Na + Nd

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Vbi =1

2EmWp +

2EmWn =

2EmW =

Wn =Na

Na + NdW , Wp =

Na + NdW .

→ Vbi =1

Na + NdW 2, i.e., W =

√2ε

(Na + Nd

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

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

= 0.0259 V× log(5× 1017)(1× 1016)

(1.5× 1010)2

= 0.8 V.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

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

= 0.0259 V× log(5× 1017)(1× 1016)

(1.5× 1010)2

= 0.8 V.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

√2εr ε0

(Na + Nd

√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

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

√2εr ε0

(Na + Nd

√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

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

√2εr ε0

(Na + Nd

√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

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

√2εr ε0

(Na + Nd

√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

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

√2εr ε0

(Na + Nd

√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

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Wn =Na

Na + NdW = 0.318µm, Wp =

Na + NdW = 0.006µm.

Em =qNd

εWn or

=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.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Wn =Na

Na + NdW = 0.318µm, Wp =

Na + NdW = 0.006µm.

Em =qNd

εWn or

=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.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Wn =Na

Na + NdW = 0.318µm, Wp =

Na + NdW = 0.006µm.

Em =qNd

εWn or

=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.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Wn =Na

Na + NdW = 0.318µm, Wp =

Na + NdW = 0.006µm.

Em =qNd

εWn or

=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.

xjxp xn

depletion region

−qNa

xp xj xn

E =− dψ

Wn =Na

Na + NdW = 0.318µm, Wp =

Na + NdW = 0.006µm.

Em =qNd

εWn or

=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.

−qNa

xp xj xn

E =− dψ

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

√2ε

(Na + Nd

Wn =Na

Na + NdW , Wp =

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.

−qNa

xp xj xn

E =− dψ

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

√2ε

(Na + Nd

Wn =Na

Na + NdW , Wp =

Na + NdW .

−qNa

xp xj xn

E =− dψ

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

√2ε

(Na + Nd

Wn =Na

Na + NdW , Wp =

Na + NdW .

−qNa

xp xj xn

E =− dψ

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

√2ε

(Na + Nd

Wn =Na

Na + NdW , Wp =

Na + NdW .

−qNa

xp xj xn

E =− dψ

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

−qNa

xp xj xn

E =− dψ

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

pn junction in equilibrium: current densities

xjxp xn

depletion region

x (µm)

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

* Qualitatively, we can see that the diffusion and drift currentswill be in opposite directions:

Electrons:

Fdiffn : , E : , Fdrift

Holes:

Fdiffp : , E : , Fdrift

xjxp xn

depletion region

x (µm)

n (in cm−3)

19.7 20 20.3

Jdiffn = −Jdrift

Electrons:

Holes:

xjxp xn

depletion region

x (µm)

n (in cm−3)

19.7 20 20.3

Jdiffn = −Jdrift

Electrons:

Holes:

xjxp xn

depletion region

x (µm)

n (in cm−3)

19.7 20 20.3

Jdiffn = −Jdrift

Electrons:

Holes:

xjxp xn

depletion region

x (µm)

n (in cm−3)

19.7 20 20.3

Jdiffn = −Jdrift

Electrons:

Holes:

xjxp xn

depletion region

x (µm)

n (in cm−3)

19.7 20 20.3

Jdiffn = −Jdrift

Electrons:

Holes:

Summary

ρ/q (approx)

(in kV/cm)

(in 105 A/cm2)

(in 1017 cm−3)

(in 104 A/cm2)

(in 1017 cm−3)

n (in cm−3)

x (µm)

ρ/q (exact)

(in 1017 cm−3)

(in kV/cm)

−100

Jdiffn

Jdriftn

x (µm)

Jdiffp

Jdriftp

x (µm)

depletionregion

neutraln region

xpdepletionregion

neutralp region

−100

19.7 20 20.3 20 20.1219.84 20

* There are three regions: p neutral region, n neutral region, and depletion region.

Summary

ρ/q (approx)

(in kV/cm)

(in 105 A/cm2)

(in 1017 cm−3)

(in 104 A/cm2)

(in 1017 cm−3)

n (in cm−3)

x (µm)

ρ/q (exact)

(in 1017 cm−3)

(in kV/cm)

−100

Jdiffn

Jdriftn

x (µm)

Jdiffp

Jdriftp

x (µm)

depletionregion

neutraln region

xpdepletionregion

neutralp region

−100

19.7 20 20.3 20 20.1219.84 20

* There are three regions: p neutral region, n neutral region, and depletion region.

Summary

ρ/q (approx)

(in kV/cm)

(in 105 A/cm2)

(in 1017 cm−3)

(in 104 A/cm2)

(in 1017 cm−3)

n (in cm−3)

x (µm)

ρ/q (exact)

(in 1017 cm−3)

(in kV/cm)

−100

Jdiffn

Jdriftn

x (µm)

Jdiffp

Jdriftp

x (µm)

depletionregion

neutraln region

xpdepletionregion

neutralp region

−100

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.

Summary

ρ/q (approx)

(in kV/cm)

(in 105 A/cm2)

(in 1017 cm−3)

(in 104 A/cm2)

(in 1017 cm−3)

n (in cm−3)

x (µm)

ρ/q (exact)

(in 1017 cm−3)

(in kV/cm)

−100

Jdiffn

Jdriftn

x (µm)

Jdiffp

Jdriftp

x (µm)

depletionregion

neutraln region

xpdepletionregion

neutralp region

−100

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.

Summary

ρ/q (approx)

(in kV/cm)

(in 105 A/cm2)

(in 1017 cm−3)

(in 104 A/cm2)

(in 1017 cm−3)

n (in cm−3)

x (µm)

ρ/q (exact)

(in 1017 cm−3)

(in kV/cm)

−100

Jdiffn

Jdriftn

x (µm)

Jdiffp

Jdriftp

x (µm)

depletionregion

neutraln region

xpdepletionregion

neutralp region

−100

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.

Summary

ρ/q (approx)

(in kV/cm)

(in 105 A/cm2)

(in 1017 cm−3)

(in 104 A/cm2)

(in 1017 cm−3)

n (in cm−3)

x (µm)

ρ/q (exact)

(in 1017 cm−3)

(in kV/cm)

−100

Jdiffn

Jdriftn

x (µm)

Jdiffp

Jdriftp

x (µm)

depletionregion

neutraln region

xpdepletionregion

neutralp region

−100

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.

Summary

ρ/q (approx)

(in kV/cm)

(in 105 A/cm2)

(in 1017 cm−3)

(in 104 A/cm2)

(in 1017 cm−3)

n (in cm−3)

x (µm)

ρ/q (exact)

(in 1017 cm−3)

(in kV/cm)

−100

Jdiffn

Jdriftn

x (µm)

Jdiffp

Jdriftp

x (µm)

depletionregion

neutraln region

xpdepletionregion

neutralp region

−100

19.7 20 20.3 20 20.1219.84 20

* Jn and Jp are individually zero because the drift and diffusion components cancel out.

Summary

ρ/q (approx)

(in kV/cm)

(in 105 A/cm2)

(in 1017 cm−3)

(in 104 A/cm2)

(in 1017 cm−3)

n (in cm−3)

x (µm)

ρ/q (exact)

(in 1017 cm−3)

(in kV/cm)

−100

Jdiffn

Jdriftn

x (µm)

Jdiffp

Jdriftp

x (µm)

depletionregion

neutraln region

xpdepletionregion

neutralp region

−100

19.7 20 20.3 20 20.1219.84 20

* Jn and Jp are individually zero because the drift and diffusion components cancel out.

semiconductor devices - iit bombaysequel/ee207/sd_pn_1.pdf · 2021. 2. 1. · * in integrated...

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