## Mobility and Drift current:

Let a electric field E (volt/meter) is applied across a metal then electron certainly got acceleration and move towards positive terminal with velocity v (m/sec), so velocity is proportional to electrical field.

v is proportional to E, v=u_{e} x E

u_{e} is called as mobility of electrons, as per definition it defined as movement of majority carriers. Units are square meter/volt-second.

Note: mobility of GaAs is greater than Si and Ge.

Here I don’t want solve entire derivation but important formulas given as follows.

Conductivity = enu_{e} ———1

Where e- charge of electron (c) n-no of electrons/m3

But the conductivity of a semi conductor is different from a metal in the respect that in a semi conductor the charge carriers are electrons as well as holes.

above eq 1 is applied for conductivity due to holes.

Similarly conductivity due to holes as follows

Conductivity (holes) = epu_{h}

Where uh- mobility of holes and p- no of holes/m^{3}

Total conductivity = epu_{e} + epu_{h}

In intrinsic semi conductor: n=p=n_{i} so conductivity = en_{i}[u_{e} + u_{h}]

In extrinsic semi conductor p-type: n= N_{D} and n>>p so conductivity = eN_{D}U_{e}

n-type: p= NP and p>>n so conductivity = eN_{A}U_{h}

Note: other important formulas are current density (J) = charge density x drift velocity (v)

Drift current (I) = envA (where A is area)

**8th: ** if for intrinsic silicon at 27^{o}c, the charge concentration and mobilities of free electrons and holes are 1.5×10^{16} /m^{3}, 0.13 m^{2}/vs and 0.05 m^{2}/vs respectively. Its conductivity will be? ** [IES-2001]**

A. 2.4×10^{-3} B. 3.15×10^{-3} C. 5×10^{-4} D. 4.32×10^{-4}

**Answer: D**

ni = 1.5×1016 /m^{3}

ue = 0.13 m^{2}/vs

uh = 0.05 m^{2}/vs

For intrinsic semi conductor conductivity = en_{i}[u_{e} + u_{h}]

E = 1.6x 10^{-16}

Conductivity = 4.32x 10^{-4}

**9th**: match List-1 and List-2 **[IES-2007]**

List-1 List-2

- Carrier mobility 1.ev ( electron volt)
- Diffusion length 2. m
^{2}/v-sec - Diffusion coefficient 3.m
- Energy gap 4.m
^{2}/sec

Codes: A B C D

a) 4 2 3 1

b) 2 3 4 1

c) 2 3 1 4

d) 4 2 1 3

**Answer: b**

The intrinsic concentration increases with increasing temperature hence conductivity also increases

n_{i}^{2}= A_{o} T^{3} exp.[-E_{G0}/kT]

where A_{0} – constant independent of temperature

k – Boltzmann constant

E_{G0} – energy gap at 0^{0}K

T – Temperature

**10th:** 3 Consider the following statements :

The intrinsic carrier concentration of a semiconductor

1. Depends on doping

2. Increases exponentially with decrease of band gap of the semiconductor

3. Increases non–linearly with increase of temperature

4. Increases linearly with increase of temperature

Which of the above statements are correct? **[IES-2012]**

(A) 1, 2 and 3 (B) 1 and 2 only (C) 2 and 3 only (D) 2 and 4 only

**Answer: C** (from above formula 2 and 3 are correct statements)

**11th**: match list1 and list 2 **[IES-2006]**

List 1

A. Electron mobility around room temperature

B. Energy gap

C. Intrinsic carrier concentration

D. Mole density (gm/mole)

List 2

1. Increase with temperature

2. Decreases with temperature

3. Remains constant as temperature is varied

Codes: A B C D

a) 2 1 1 1

b) 1 2 1 3

c) 2 2 1 3

d) 2 2 1 1

**Answer : c**

Mobility is inversely proportional to temperature, energy gap decreases with increasing temperature, in the above equation indicates intrinsic carrier is raises with temperature raise. Mole density is remains constant as no of molecules remains same for all conditions.

**12th:** If n is the number of electrons per unit volume of the semiconductor and vd is the drift velocity of the electrons, then the current flowing through a semi-conductor is given by **[IES 2012]**

(A) i=n/v_{d} (B) i=nv_{d} (C) i=v_{d}/n (D) i=nv_{d}^{1/2}

**Answer: B**

Reason: current is defined as the charge flowing per second across any normal plane of the conductor so current= (no of electrons/volume)x (velocity of electron)

### The Hall effect:

When a specimen carrying a current I is placed in a transverse magnetic field B, then an electric field E is induced in the direction perpendicular to both I and B, the phenomenon is called Hall effect.

Hall effect is used for (1) whether semi conductor is N-type or P-type

(2) Finding the carrier concentration

(3) Calculating the mobility by measuring the conductivity

Figure shows a semi conductor bar carrying a current I in the positive X direction. Let a magnetic field B is applied in the positive Z direction. Now a force is exerted on the charge carriers (whether electrons or holes) in the negative Y direction. Due to this force, charge carriers are pressed downwards towards face 1. For example, in N-type specimen the charge carriers are electrons which are accumulated on face 1. So face 1 become negatively charged with respect to face 2.therefore a potential difference V_{H} is developed between 1 and 2 which is called as hall effect. The polarity of hall voltage enables us to determine whether the semi conductor specimen is N-type or P-type.

If voltage between face 2 and 1 are positive then it is N-type

If voltage between face 2 and 1 are negative then it is P-type.

Important formulas in this topic:

Hall voltage V_{H} = (flux density x current)/ ( charge density x width of specimen )= BI/pw

Hall coefficient R_{H} = 1/p = V_{H}w/BI

Mobility U = conductivity x Hall co-efficient

From the above formulas you can find any unknown parameter with other known parameters.

Importance of hall effect: it gives following information

- It gives the information regarding semi conductor type
- To measure semi conductor parameters like electron or hole concentration and mobility
- It gives whether the given metal is insulator, conductor or semi conductor
- It can be used to measure magnetic flux density
- It can used to measure power in em wave

**13th:** in a hall effect experiment, a p-type semi conductor sample with hole concentration p1 is used. The measured value of the hall voltage is V_{H1}. If the p-type sample is now replaced by another p-type sample with hole concentration p_{2} where p_{2}= 2p_{1}, what is the new hall voltage? **[IES-2006]**

(A) 2V_{H1} (B) 4V_{H1 } (C) 0.5 V_{H1 } (D) 0.25 V_{H1}

**Answer: C**

Reason: hall voltage is inversely proportional to charge density from above formula. If hole concentration increase by twice then its density also increase by twice so hall voltage will decrease by half.

I covered most of the important topics still there are some topics but not so important in the point of view of competitive exams. i sort out all important formulas please mug up formulas.for the next class i will come up with concept of P-N junction diode and related objectives in gate and ies. thank you for reading.