Chap-2

Basic Electronics (GTU)

2-1

Transport Phenomena in Semiconductors

Chapter 2 : Transport Phenomena in Semiconductors Section 2.7 : Ex. 2.7.4 :

The resist resistivi ivity ty of intrin intrinsic sic silico silicon n is 3 × 105 Ω -cm -cm at 30°C. Calcul Calculate ate the intrin intrinsic sic ° concentration at 100 C. Assume the following data : 1.

µ

n

= 0.13 m2 / V-sec. at 30°C

2.

µ

p

= 0.05 m2 / V-sec at 30°C.

.Page No. 2-23. 2-23.

Soln. : Step Ste p 1 : Intrin Intrinsic sic conce concentr ntrati ation on ni at 30 C or 303 K :

To calculate n i at 303°K we are going to use the following equation :

σ

i

= ni ( µ

n

+µ

p

)q

...(1)

But to calculate n i we need to find the value of σ i.

∴

σ

i

= = = 3.33 × 10 – 6 (Ω – cm) – 1

σ

i

= 3.33 × 10 – 4 (Ω – m) – 1

...(2)

Hence intrinsic concentration, ni = = 1.16 × 1016 per m3 at 303°K

...(3)

Step Ste p 2 : Intrin Intrinsic sic conc concent entrat ration ion at 100 100 C or 373 K :

To calculate n i at 373°K let us use the following equation (refer to Equation (2.11.9)) : = Ao T3 e – EGO / kT where,

...(4)

EGO = 1.21 eV k = Bolt Boltzm zman ann' n'ss con const stan antt = 8.6 8.62 2 × 10 – 5 eV / °K

and

T1 = 303°K

∴ First of all let us calculate “A o” by substituting the temperature to be 303 °K and the value of n i at 303°K from Equation (3). ∴ Ao = = 6.37 × 1044 Ao is a constant the value of which does not depend on temperature. So let us calculate “n i” at 373°K.

∴ ( ni )2 = Ao e – EGO / kT2

...(5)


2-2


Substituting the values we get, ( ni )2 = 6.37 × 1044 × (373)3 e – ( 1.21 / 8.62

× 10– 5 × 373 )

= 1.497 × 1036

∴

ni = 1.223 × 1018 per m3

...Ans.

This is the intrinsic concentration at 100 °C or 373 °K. Ex. 2.7.5 :

A bar of intrinsic silicon having a cross-sectional area of 2.5 × 10 – 4 m2 has an electron density of 1.5 × 1016 electrons/m3. If the electron mobility is 0.14 m2 / V-sec and hole mobility is 0.05 m2 / V-sec, what is the length of the bar in order to have a current of 1.2 mA when 9 Volts are applied across its ends ? .Page No. 2-23.

Soln. : Given :

1.

A = 2.5 × 10 – 4 m2

3.

µ

n

2.

= 0.14 m 2 / V-s and µ

p

= 0.05 m 2 / V-s

n = 1.5 × 1016 / m3 4.

I = 1.2 mA, V = 9

Volts Step 1 : Conductivity of the intrinsic silicon :

As we know, the conductivity of intrinsic silicon is given by :

σ

= ni ( µ

+µ

)q

...(1)

ni = 1.5 × 1016 / m3

...(2)

i

n

p

For an intrinsic semiconductor n = p = ni

∴

Substituting the values into Equation (1) we get,

∴

σ

i

= 1.5 × 1016 (0.14 + 0.05) × 1.6 × 10 – 19

σ

i

= 4.56 × 10 – 4 (Ω – m) – 1

...(3)

=

...(4)

Step 2 : Resistivity of the intrinsic silicon :

Resistivity, ρ

i

Substituting the value of σ i from Equation (3) we get,

∴

ρ

i

=

ρ

i

= 2193 Ω – m

...(5)

Step 3 : Length “l ” of the semiconductor bar :

Resistance of the semiconductor bar = = = 7500 Ω

...(6)

And expression for resistance is R =

...(7)

Substituting the values we get 7500 =

∴ Ex. 2.7.6 :

l

= 8.55 × 10 – 4 m. = 0.855 mm

...Ans.

Calculate the intrinsic carrier concentration of Silicon and Germanium at 400°K. .Page No. 2-23.

Soln. : The value of A o and EGO for the Silicon and Germanium are as follows : Material

Value of Ao

EGO


2-3

Silicon


×

2.735

Germanium

10

31

10

30

×

2.755

1.21 eV 0.785 eV

k = 8.62 × 10 – 5 eV / °K and T = 400 °K

• 1.

We know that = Ao T3 e – EGO / kT For Silicon :

= 2.735 × 1031 × (400)3 e – 1.1 / 8.62

× 10– 5 × 400

= 2.44 × 1025 ni = 4.94 × 1012 per m3

∴ 2.

... Ans.

For Germanium :

= 2.755 × 1030 × (400)3 e – 0.785 / 8.62

∴ ∴

× 10– 5 × 400

= 2.2845 × 1028 ni = 1.51 × 1014 per m3

... Ans.

Comment : At the same temperature, the intrinsic carrier concentration of Germanium is higher than that of silicon. Ex. 2.7.7 :

The intrinsic concentration of Silicon at 400°K is 4.94 × 1012 / m3. Calculate its carrier concentration at 500°K if EGO for silicon is 1.21 eV. .Page No. 2-23.

Soln. : Given :

( ni )1 = 4.94 × 1012 / m3, T2 = 500°K

To Find :

EGO = 1.21 eV,

T1 = 400°K

and K = 8.62 × 10 – 5 eV / °K

( ni )2 at 500°K

We know that, = Ao T3 e – EGO / kT

...(1)

So and

( ni1 )2 = Ao e – EGO / kT1

...(2)

( ni2 )2 = Ao e – EGO / kT2

...(3)

Divide Equation (3) by Equation (2) to get, = = = 2181.989

∴

( ni2 )2 = 2181.989 × ( ni1 )2 = 2181.989 × ( 4.94 × 1012 )2 = 5.3248 × 1028

∴ Comment :

ni2 = 2.3075 × 1014 / m3

...Ans.


•

2-4


The value of n i2 shows that the intrinsic concentration increases with increase in temperature.

Section 2.8 : Ex. 2.8.2 :

A sample of germanium is doped with 1014 / cm3 of donor impurity and 7 × 103 / cm3 of acceptor impurity. Resistivity of pure Ge is 60 ohms-cm at room temperature. If the total current density of the sample is 52 mA/cm2 determine the applied electric field. (Assume : µ

n

= 3800 cm2 / V-s and µ

p

= 1800 cm2 / V-sec).

.Page No. 2-27.

Soln. : Given that, Donor impurities, ND

1014 / cm3

=

Acceptor impurities, N A =

7 × 1013 / cm3

Resistivity ρ of pure, Ge =

60 Ω -cm

Total current density, J

52 mA/cm 2

=

What is E ? The total current density J is given by : J = (nµ

n

+pµ

p

) qE

...(1)

In this equation we can substitute, J = 52 mA / cm 2, q = 1.6 × 10 – 19 C, µ

n

= 3800 cm 2 / V-sec and µ

p

= 1800 cm2 / V-sec.

But to obtain the value of “E” we must find the values of n and p. Step 1 : To obtain the values of n and p :

For an intrinsic (pure) material, the conductivity is given by, Also

σ

i

= ni q ( µ

σ

i

= = = 0.0167 (Ω -m) – 1

n

+µ

p

)

...(2) ...(3)

Substitute this and other values in Equation (2) to get, ni = = 1.86 × 1013 / cm3

...(4)

ni is the intrinsic concentration. Now use the mass action law to write, n× p =

...(5)

Referring to Equation (2.6.3) we can write that, p + N D = n + NA

∴ ∴ Substituting the values we get,

n – p = N D – NA (n – p) = 1014 – 7 × 103 = 9.99 × 1013 or

n = ( p + 9.99 × 1013 )

Substitute this into Equation (5) we get, ( p + 9.99 × 1013 ) p = = ( 1.86 × 1013 )2

∴ ∴

p2 + 9.99 × 1013 p = 3.46 × 1026

p2 + 9.99 × 1013 p – 3.46 × 1026 = 0

Solving this quadratic equation we get,

...(6)


2-5


p = 3.35 × 1012 / cm3

...(7)

n = p + 9.99 × 1013

and

= 10.325 × 1013 / cm3

...(8)

Thus we have obtained the values of n and p. Step 2 : To obtain the value of E :

Now substitute these values of n and p along with other values into Equation (1) to write,

Ex. 2.8.3 :

∴

E =

∴

E = 0.8185 Volt/cm

...Ans.

A sample of silicon is doped to the extent of 1014 donor atoms per cm3 and 7 × 1013 acceptor atoms per cm3. At the room temperature the resistivity of pure silicon is 60 Ω -cm. If the applied electrostatic field is 2 V/cm., find the total current density. Given : µ

n

= 1300 cm2 / V-sec, µ

p

= 500 cm2 / V-sec.

.Page No. 2-27.

Soln. : Given :

Donor impurity ND = 1014 / cm3,Acceptor impurity NA = 7 × 1013 / cm3 Resistivity of pure silicon, ρ = 60 Ω -cm, Electrostatic field E = 2 V/cm What is J = ?

•

The total current density J is given by, J = (nµ

•

n

+pµ

p

)qE

...(1)

In this equation we know the values of µ n, µ p, q and E. But we do not know the values of n and p.

Step 1 : To obtain the values of n and p :

•

For an intrinsic (pure) material, the conductivity is given by, Also

σ

i

= ni q ( µ

σ

i

= = = 0.0167 (Ω -m) – 1

n

+µ

p

)

...(2) ...(3)


•

2-6


Substituting this and other values in Equation (2) we get, ni = = 5.798 × 1013 / cm3

•

...(4)

Now use the law of mass action to write, n× p =

...(5)

Referring to Equation (2.6.3) we can write that, p + N D = n + NA

∴ •

n – p = N D – NA

Substituting the values we get, n – p = 1014 – 7 × 1013 = 3 × 1013 n = ( p + 3 × 1013 )

or

•

...(6)

Substitute this value of n into Equation (5) we get, ( p + 3 × 1013 ) · p = ( 5.798 × 1013 )2

∴ p2 + 3 × 1013 p = 3.36 × 1027 ∴ p2 + 3 × 1013 p – 3.36 × 1027 = 0

...(7)

Let us solve it as, p =

•

Neglecting the negative sign. p = 4.48 × 1013 per cm3

∴ •

...(8)

Substitute this value in Equation (6) to get, n = ( 4.48 × 1013 + 3 × 1013 ) = 7.48 × 1013

...(9)

Thus we have obtained the values of n and p. Step 2 : To calculate the total current density (J) :

Substituting the values into Equation (1) we get, J =

Ex. 2.8.4 :

∴

J = 0.0382848 Amp/cm 2 = Amp/m2

∴

J = 382.84 Amp/m2

...Ans.

A bar of silicon 0.1 cm long has a cross-sectional area of 8 × 10 – 8 m2, heavily doped with phosphorus. What will be the majority carrier density resulting from doping if the bar is to have a resistance of 1.5 kΩ ? Given for silicon at room temperature : µ n = 0.14 m2 / V-sec, µ p = 0.05 m2 / V-sec, ni = 1.5 × 1010 / cm3. .Page No. 2-28.

Soln. : Given :

l =

0.1 cm = 1 × 10 – 3 m,

A = 8 × 10 – 8 m2

= 0.14 m 2 / V-sec.

R = 1.5 k Ω ,

µ

µ p = 0.05 m2 / V-sec.

ni = 1.5 × 1010 / cm3

n

The majority carrier density ( N D ) as the phosphorous is a donor impurity can be calculated from the following equation :

σ OR

n

= nn µ

n

q


2-7

σ We do not know σ To obtain σ

n

n

n

= ND µ

Transport Phenomena in Semiconductors n

q

...(1)

so let us calculate it.

we will have to calculate resistivity ρ n.

Step 1 : To obtain the value of resistivity (

Resistance,

n

):

R =

∴

ρ

n

=

∴

ρ

n

= = 0.12 Ω -m

Substituting the values we get,

Step 2 : To obtain the value of conductivity (

σ

n

n

...(2)

):

= = = 8.33 (Ω -m) – 1

...(3)

Step 3 : To obtain the value of ND :

From Equation (1) we can write that, ND = Substituting the values we get, ND = = 3.719 × 1020 / m3 Ex. 2.8.5 :

...Ans.

Find the resistivity of intrinsic silicon. What will be the resistivity when the silicon is doped with a pentavalent impurity to the extent of 1 impurity atom for each 108 atoms of silicon ? Given for silicon : 1.

Number of silicon atoms per cm3 = 5 × 1022

2.

Intrinsic carrier concentration = 1.52 × 1010 / cm3 = 1.52 × 1016 / m3

3.

Electron charge = 1.6 × 10 – 19 Coulomb.

4.

Electron mobility = 0.135 m2 / Volt-sec. Hole mobility = 0.048 m2 / V-s.

Resistivity of intrinsic silicon and after doping is to be obtained.

.Page No. 2-28.

Soln. : Step 1 : Resistivity of intrinsic silicon :

The conductivity of intrinsic silicon is given by,

σ

i

= ni ( µ

n

+µ

p

)q

...(1)


2-8



∴

σ

i

= 1.52 × 1016 (0.135 + 0.048) × 1.6 × 10 – 19

σ

i

= 4.45 × 10 – 4 (Ω -m) – 1

∴ Resistivity of the intrinsic silicon = ρ i = = = 2247.2 Ω -m

...(2) ...Ans.

Step 2 : Resistivity of doped silicon :

•

Referring to Equation (2.8.14) we can write that the conductivity of an “n”-type semiconductor is given by,

σ

n

= ND µ

n

q

...(3)

•

This equation has been written with an assumption that the concentration of electrons in an n-type semiconductor is much higher than that of the holes.

•

The number of impurity atoms is 1 in 10 8. Therefore the number of impurity (donor) atoms in 5 × 1022 silicon atoms is given by, ND = = 5 × 1014 / cm3

•

Convert it into atoms/m 3 as, ND = 5 × 1014 × 106 / m3

∴ •

ND = 5 × 1020 / m3

Substituting this value in Equation (3) we get,

σ •

Ex. 2.8.6 :

Soln. :

= 5 × 1020 × 0.135 × 1.6 × 10 – 19 = 10.8 ( Ω -m) – 1

ρ

n

= = = 0.09259 = 92.59 × 10 – 3 Ω -m.

ρ

n

= 92.59 × 10 – 3 Ω -m

...Ans.

A donor impurity is added to intrinsic silicon and the resistivity at room temperature is observed to be 9.6 Ω -cm. Calculate the ratio of donor atoms to silicon atoms per unit volume. Assume µ n as 1300 cm2 / V-s. .Page No. 2-28.

The given data is as follows : Resistivity,

ρ

Mobility,

µ

= 9.6 Ω -cm = 9.6 × 10 – 2 Ω -m. n

= 1300 cm2 / V-s = 0.13 m 2 / V-s

In order to calculate the ratio of donor atoms to silicon atoms per unit volume, we must obtain the value of “ND” i.e. the concentration of donor atoms first. Let us use the equation for the conductivity of “n” type material which states that :

σ •

n

Therefore resistivity of doped silicon is given by,

∴

•

...(4)

n

= ND µ

n

q

(Refer to Equation (2.8.14))

...(1)

But we do not know “ σ n”, so let us find it as follows :

σ Therefore,

∴

n

= = = 10.42 (Ω -m) – 1

ND = = ND = 5 × 1020 atoms/m3

•

The atom density of silicon is 5 × 1022 atoms/cm3 that means 5 × 1028 atoms/m3.

•

Therefore the ratio of donor atoms to silicon atoms per unit volume = = 10 – 8

Ex. 2.8.7 :

...(2)

...(3) ...Ans.

Derive an expression for conductivity of extrinsic semiconductor. Prove that the resistivity of intrinsic germanium is 45 Ω -cm.


2-9


Given for Ge at room temperature : Intrinsic concentration = 2.5 × 1013 / cm3 Mobility for hole = 1800 cm2 / V-s Mobility for electrons = 3800 cm2 / V-s Charge on an electron = 1.6 × 10 – 19 C

Soln. :

.Page No. 2-28.

The conductivity of an intrinsic semiconductor is given by,

σ

i

= ni ( µ

n

+µ

p

)q

...(1)

Solution to the problem : To obtain the resistivity of the intrinsic germanium, we need to calculate its conductivity. Substitute the values into Equation (1) to get,

σ

i

= 2.5 × 1013 (3800 + 1800) × 1.6 × 10 – 19 = 0.0224 (Ω -cm) – 1

The resistivity is given by,

ρ

i

=

= = 44.64 (Ω -cm ) ≈ 45 Ω -cm.

...Ans.

Hence proved. Ex. 2.8.8 :

A bar of silicon has intrinsic concentration of 1.2 × 1016 per m3. The silicon is so doped that the hole concentration becomes 1022 per m3. Determine : 1.

Electron concentration of doped silicon

2.

Conductivity of doped silicon

3.

Type of resulting silicon (p or n type)

Given : µ

n

= 1350 cm2 / V-sec, µ

p

= 500 cm2 / V-sec.

.Page No. 2-28.

Soln : Step 1 : To calculate the electron concentration (n) :

We know that intrinsic concentration, n i is given by, ( ni )2 = n × p

...(1)

∴ The electron concentration, n = = n = 1.44 × 1010 per m3

∴

...Ans.

Step 2 : Type of resulting silicon :

The hole concentration p = 10 22 per m3 and the electron concentration n = 1.44 × 1010 per m3. As the concentration of holes is higher than that of electrons, the resulting silicon is “p type” semiconductor. Step 3 : Conductivity of doped silicon :

Conductivity

σ µ

and

= (nµ n

n

+pµ

p

)q

= 1350 cm2 / V-sec. = 0.135 m 2 / V-sec

µ p = 500 cm2 / V-sec = 0.05 m 2 / V-sec. q = 1.6 × 10 – 19 C.

Substituting all these values into Equation (2) we get,

...(2)


2-10


σ

= [ ( 1.44 × 1010 × 0.135 ) + ( 1 × 1022 × 0.05 ) ] 1.6 × 10 – 19

σ

= 80 (Ω -m) – 1

...Ans. Ex. 2.8.9 :

A 1 mm long bar of silicon is doped with donor impurity of 5 × 1010 / cm 3. Find the voltage drop across the bar if the current density within the bar is 1 mA / cm2. Given : For Si at 300°K,

µ

n

ni = 1.5 × 1010 / cm3.

= 1300 cm2 / V-sec, µ

p

= 500 cm2 / V-sec

.Page No. 2-28.

Soln. : Given :

l =

ND = 5 × 1010 / cm3

1 mm,

J = 1 mA/cm 2 ,

µ n = 1300 cm 2 / V-sec,

ni = 1.5 × 1010 / cm3

µ p = 500 cm2 / V-sec

Steps to be followed : Step 1 : Calculate the hole concentration (p) :

p = /N

∴

where ND = n = 5 × 1010

D

p = = 4.5 × 109

...(1)

Step 2 : Calculate E :

We know that the total current density J is given by J = Jn + J p = nq µ

∴

J = qE [ n µ

Substitute n = N D

∴

J = 1.6 × 10 – 19 × E

But

J = 1 mA / cm 2

∴

E = = 92.93 V/cm

n

n

E + pq µ

+pµ

p

]

p

E


2-11


Step 3 : Calculate the voltage drop :

•

The electric field E is defined as Volts/cm. Its value is 92 V/cm. That means the voltage across a 1 cm piece of semiconductor would be 92 V.

•

But the given piece of semiconductor is 1 mm long.

∴ Voltage across it = = 9.293 Volts

...Ans.

Ex. 2.8.10 : Show that the resistivity of intrinsic Germanium at 300° K is 44.64 Ω cm. If a donor impurity is added to an extent of 1 atom per 108 Germanium atoms, prove that the resistivity drops to 3.7 Ω cm. Assume :

µ

n

= 3800 cm2 / V-sec

µ

ni = 2.5 × 1013 / cm3

p

= 1800 cm2 / V-sec

Atoms per cm3 for Ge = 4.4 × 1022.

.Page No. 2-28.

Soln. : For the first part of the problem refer to the solution of Ex. 2.8.7. Part II Given :

µ n = 3800 cm 2 / V-sec,

µ p = 1800 cm 2 / V-sec

ni = 2.5 × 1013 / cm3 ,

Atoms/cm3 for Ge = 4.4 × 1022

Step 1 : Calculate the donor concentration ND :

ND = n = 1 atom per 108

∴

ND = n =

∴

ND = n = 4.4 × 1014 / cm3

Step 2 : Conductivity of doped germanium :

•

The given semiconductor is an n - type semiconductor. Its conductivity is given by,

σ

n

= nq µ

n

= ND q µ

n

= ( 4.4 × 1014 × 106 / m3 ) × 1.6 × 10 – 19 × 3800 × 10 – 4 m2 / V-sec = 26.752 (Ω -m) – 1 Step 3 : Resistivity of doped germanium :

ρ

n

= = = 0.03738 Ω -m = 0.03738 × 102 Ω -cm = 3.738 Ω -cm

...Ans.

Thus we have proved that the resistivity drops to 3.738 Ω -cm. Ex. 2.8.11 : The intrinsic carrier concentration of silicon sample of 300° K is 1.5 × 1016 / m3. If after doping, the number of majority carriers is 5 × 1020 / m3, calculate the minority carrier density. .Page No. 2-29.

Soln. : Given :

ni = 1.5 × 1016 / m3

n = 5 × 1020 / m3

p=?

•

Let the semiconductor after doping be n type. So the majority carriers will be electrons and minority carriers will be holes.

•

Using the law of mass-action we get, = n× p


2-12

∴


p = = = 4.5 × 1011 / m3

•

...Ans.

This is the minority carrier density.

Ex. 2.8.12 : A sample of silicon of 1 cm length and 1 mm2 cross-sectional area is doped with Nd = 8 × 1015 cm – 3 at 27°C. 1.

Calculate electron and hole concentrations.

2.

If excess holes and electrons are generated with concentrations of δ p = δ n = 1014 cm – 3 , determine the total concentrations of holes and electrons.

3.

For the same semiconductor sample with excess charge carriers, calculate current flowing through it if the voltage applied across its length is 2 V.

For silicon : ni = 1.5 × 1010 cm – 3

µ

n

= 1400 cm2 / V-sec

µ

p

= 500 cm2 / V-sec

.Page No. 2-29.

Soln. : Given :

ni = 1.5 × 1010 per cm3,

µ

µ p = 500 cm2 / V-sec,

l =

n

= 1400 cm 2 / V-sec 1 cm, a = 1 mm 2

Part I : Electron and hole concentration : Given :

•

Nd = 8 × 1015 cm – 3 at T = 27°C or 300 ° K Since Nd >> ni we assume that the electron concentration n ≈ Nd

∴ •

n = Nd = 8 × 1015 electrons per cm 3

...Ans.

Using the law of mass action we can write, = n× p

∴

Hole concentration p = =

∴

p = 2.8125 × 104 holes per cm 3

...Ans.

Part II : Total concentration of electrons and holes :

•

Due to the excess holes and electrons the electron and hole concentrations are, nT = N d + δ and

pT = p + δ

n

= ( 8 × 1015 ) + 1014 = 8.1 × 1015 cm – 3

p

= ( 2.8125 × 104 ) + 1014 = 1 × 1014 cm – 3


2-13


Part III : Calculate current : Given : Voltage across the semiconductor = 2 V. Step 1 : Calculate conductivity :

σ

Conductivity

= ( nT µ

n

+ pT µ p ) q

= [ (8.1 × 1015 × 1400) + (1 × 1014 × 500) ] × 1.6 × 10 – 19 = 1.8224 (Ω -cm) – 1 Step 2 : Calculate resistivity and resistance :

ρ

Resistivity

= = 0.5487 (Ω -cm)

R = = 0.5487 ( Ω -cm) ×

Resistance

∴

R = 54.87 Ω

Step 3 : Calculate current I :

I = = = 0.03644 Amp = 36.44 mA

...Ans.

Ex. 2.8.13 : A silicon sample is non-uniformly doped with donor impurity of 1014 m – 3. A current density of 10 mA / cm2 is generated when electric field of 3V/cm is applied across it. Find the concentration gradient at 27°C. Given : µ

n

= 1500 cm2 / V-sec.

.Page No. 2-29.

Soln. : Given :

ND = 1014 per m3, Jn = 10 mA / cm 2 , E = 3 V / cm T = 27° C = 300 ° K, µ

To find :

n

= 1500 cm 2 / V-sec

Concentration gradient dn/dx

Step 1 : Calculate the diffusion constant Dn :

As per Einstein's relationship

µ ∴

n

= 39 Dn ...at room temperature

Dn = = = 38.46

Step 2 : Calculate the concentration gradient :

•

The total current density due to electrons is given by, Jn = nq µ

• ∴

n

E + q Dn

Here n = N D = 1014 m – 3 = [ 1014 × 1.6 × 10 – 19 × 1500 × 10 – 4 × 3 × 10+ 2 ] + [ 1.6 × 10 – 19 × 38.46 × ]

∴

100 = 7.2 × 10 – 4 + ( 6.1536 × 10 – 18 )

∴

= 1.625 × 1019

...Ans.

Section 2.9 : Ex. 2.9.1 :

An n-type silicon bar is used in Hall experiment has ND = 1013 / cm3, BZ = 0.2 Wb / m2 , d = 5 mm and E = 5 V/cm. What is the magnitude of Hall voltage VH ? .Page No. 2-31.

Soln. : We know that the Hall voltage V H is given by,


2-14


VH = Step 1 : To obtain the charge density “

...(1)

”:

Charge density “ ρ ” = n × q = ND × q

∴

= 1013 × 1.6 × 10 – 19 = 1.6 × 10 – 6 C/ cm3

ρ

...(2)

Step 2 : To obtain the value of current density “J” :

Referring to Equation (2.9.4) we can write that, J = ρ v v = Drift velocity = µ

Substituting

J = ρ µ Assuming

...(3)

µ

n

n

n

E we get,

E

= 1300 cm2 / V-sec we get,

J = 1.6 × 10 – 6 × 1300 × 5 = 0.0104 Amp / cm 2

...(4)

Step 3 : To obtain the hall voltage VH :

Substitute Equations (2) and (4) into Equation (1) and substitute the values of “B” and “d” into Equation (1). B = 0.2 Wb / m2 = 0.2 × 10 – 4 Wb/cm2 and

Ex. 2.9.2 :

d = 5 mm = 5 × 10 – 1 cm

∴

VH =

∴

VH = 0.065 volts or 65 mV

...Ans.

The Hall experiment is used for a silicon bar known to be p-type. The resistivity of the bar is 220 × 103 Ω -cm . Width of the bar is 2 mm and distance between the two surfaces of the bar is 2.2 mm. The magnetic field used has intensity of 0.1 Wb/m2. If measured value of current and Hall voltage are 5 micro-amp and 28 mV respectively calculate the mobility of holes. .Page No. 2-31.

Soln. : Given :

1.

ρ = 220 × 103 Ω -cm = 220 × 10 Ω -m . 2.

d = 2.2 mm = 2.2 × 10 – 3 m.

3.

w = width = 2 mm = 2 × 10 – 3 m.

B = 0.1 Wb/m2,

5.

I = 5 µ amp = 5 × 10 – 6 Amp

4.

6

VH = 28 mV = 28 × 10 –

3

Volts Step 1 : To calculate the mobility of holes :

Mobility Where,

σ

µ

= σ R H

= Conductivity and

...(1)

R H = Hall coefficient.

We do not know both these values so let us find them. Step 2 : Conductivity (

):

Conductivity, σ

= = = 4.54 × 10 – 4 (Ω -m) – 1

...(2)

Step 3 : To find the Hall coefficient ( R H ) :

R H =

...(3)


2-15



R H = = 112

...(4)

Section 2.13 : Ex. 2.13.1 : The hole concentration in a semiconductor specimen is shown in Fig. P. 2.13.1(a). Find the expression and sketch the hole current density Jp (x) for the case in which there is no externally applied electric field. .Page No. 2-42.

Fig. P. 2.13.1(a)

Soln. : As there is no externally applied field, we are expected to obtain the expression for the diffusion hole current density J p (x). Refer to section 2.12.2 for the derivation. We get the following expression for J p (x). J p (x) = – q D p

...(1)

The negative sign indicates that the concentration “p” decreases with increase in “x”. We can obtain the value of as follows : =

for 0 ≤ x ≤ W

= 0

for x > W

…(2)

Substitute Equation (2) in Equation (1) to get, J p (x) = – q D p = 0

for 0 ≤ x ≤ W for x > W

This is the required expression for the hole current density.

Fig. P. 2.13.1(b) : Sketch for J p (x) 

Chap-2

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