Friday, 13 October 2017

VIVA QUESTIONS : ENGG PHYSICS

VIVA QUESTIONS
Experiment name : pn junction diode
1) What are semiconductors?
     A semiconductor material has an electrical conductivity value falling between conductor and an insulator 
Ex: Conductor such as copper and insulator such as glass. 
Their resistance decreases as their temperature increases, this behavior is opposite to that of a metal.
Their conducting properties may be altered in useful ways.
In semiconductors, two differently-doped regions can exist in the same crystal, forming a semiconductor junction. 
The behavior of  charge carriers which include electrons, ions and electron holes at these junctions is the basis of diodes, transistors and all modern electronics.
2) What are the properties of semiconductors?
Following are the properties of semiconductor

  1.    Preparation of semiconductor materials.
  2.  Energy bands and electrical conduction.
  3.  Charge carriers (electrons and holes)
  4.  Doping.

3) What are intrinsic and extrinsic semiconductors?
Intrinsic semiconductor are pure semiconductor.their conductivity is poor and as such they do not find much application. 
Extrinsic semiconductor are semiconductors when a trivalent or pentavalent impurity is added to a pure semiconductor ,extrinsic semiconductor is obtained.
4) How p and n types of semiconductors are formed?
P-type semiconductors
 p-type semiconductors have a larger hole concentration than electron concentration. The term p-type refers to the positive charge of the hole. In p-type semiconductors, holes are the majority carriers and electrons are the minority carriers. P-type semiconductors are created by doping an intrinsic semiconductor with acceptor impurities. A common p-type dopant for silicon is boron. For p-type semiconductors the Fermi level is below the intrinsic Fermi level and lies closer to the valence band than the conduction band.
N-type semiconductors
N-type semiconductors have a larger electron concentration than hole concentration. The term n-type comes from the negative charge of the electron. In n-type semiconductors, electrons are the majority carriers and holes are the minority carriers. N-type semiconductors are created by doping an intrinsic semiconductor with donor impurities. A common dopant for n-type silicon is phosphorus. In an n-type semiconductor, the Fermi level is greater than that of the intrinsic semiconductor and lies closer to the conduction band than the valence band.
5) What is band gap?
The band gap generally refers to the energy difference (in eV) between the top of the valence band and the bottom of the conduction band in insulators and semiconductors.
6) What do you mean by energy gap in semiconductors?
The difference of energy between the bottom of the conduction band and the top of the valence band of the electrons in a crystalline solid.

For values below about 2eV the substance is considered to be a semiconductor where as the higher values it is considered to be an insulator.
7) What does the V-I characteristics indicate?
By supplying positive voltage, the electrons get enough energy to overcome the potential barrier (depletion layer) and cross the junction and the same thing happens with the holes as well. The amount of energy required by the electrons and holes for crossing the junction is equal to the barrier potential.

 0.3 V for Ge and 0.7 V for Si, 1.2V for GaAs. 

This is also known as Voltage drop. The voltage drop across the diode occurs due to internal resistance. This can be observed in the below graph. 


8) What is the role of depletion region?
The depletion region, also called depletion layer or junction region,  is an insulating region within a conductive, doped semiconductor material where the mobile charge carriers have been diffused away, or have been forced away by an electric field. The only elements left in the depletion region are ionized donor or acceptor impurities.
The depletion region is so named because it is formed from a conducting region by removal of all free charge carriers, leaving none to carry a current. 

9) What is biasing of the diode?
Biasing of diode means by applying a positive voltage in order to cross the Potential Barrier. Diode can be biased in two types first is forward biasing and Reverse biasing.
10) Applications of diode.
Diode Applications. Diodes have a huge range of uses. You'll find a diode of some type in just about every circuit.
·       The most common and important application of a diode is the rectification of AC power to DC power.
·       Diode as a Rectifier. 
·       Diodes in Clamping Circuits
·       Diodes in Logic Gates
·       A single or combination of four diodes is used in most of the power conversion applications.
Top list of various types of diodes and their applications - Zener, Light Emitting Diode(LED), Schottky, Shockley, Tunnel, Varactor, Laser etc.
11) What is effect of temperature on diode in both biased condition?
The effect of temperature can be very much seen in the reverse biased condition because the reverse saturation current of a silicon diode doubles for every 10°C rise in temperature. It can be written as
I2=I1 * 2 ( T2-T1/10)
In the forward bias , it's shifts to 2.5 mV per °C.
12) What are "majority" and "minority" charge carriers?
The holes (majority charge carriers) carry most of the electric charge or electric current in the p-type semiconductor. In p-type semiconductor, very small number of free electrons is present. Hence, free electrons are the minority charge carriers in the p-type semiconductor.

Thursday, 12 October 2017

PRACTICAL EXAMINATION : FORMAT FOR EXAMINATION : (PAPER FORMAT)

PRACTICAL EXAMINATION FORMAT (PAPER FORMAT)

1. AIM(Given by Invigilator/Teacher)
2. APPARATUS
3. FORMULA/ FORMULAE (with description of variables)
4. CIRCUIT DIAGRAM / RAY DIAGRAM
5. OBSERVATIONS
6. OBSERVATION TABLE
7. GRAPH (if any)
7. CALCULATIONS (from Graph if any)
8. RESULT (with SI unit)
9. PRECAUTIONS (At least Two)   

INSTRUCTIONS TO CANDIDATES:(Follow carefully)
1) USE OF SCIENTIFIC CALCULATORS ARE ALLOWED.
2) PERFORMANCE OF EXPERIMENT ALLOTTED IS COMPULSORY.
3) YOU MUST BRING YOUR OWN GEOMETRY BOX DURING THE EXAMINATION
4) ONLY EXAMINATION MATERIALS APPROVED BY THE FACULTY ARE PERMITTED. 
5) BOOK CONTROL TAKES PLACE IN THE EXAM VENUE BEFORE THE START OF THE EXAMINATION.
6) IF SUPPORT MATERIAL, OTHER THAN THAT WHICH IS SPECIFICALLY PERMITTED, IS FOUND AT OR BY THE DESK, IT MAY BE TREATED AS AN ATTEMPT TO CHEAT AND RELEVANT PROCEDURES FOR CHEATING WILL BE FOLLOWED. 
7) ELECTRONIC DEVICES MUST BE TURNED OFF AND PUT AWAY, AND CANNOT BE STORED IN COATS OR POCKETS.
8) ARRIVE AT THE EXAMINATION VENUE (LABORATORY) AT LEAST 15 MINUTES BEFORE THE START OF THE EXAMINATION.



 

Friday, 6 October 2017

Class 12 : MATHS : SOLUTIONS:Chapter #3 : Trigonometric Functions


SOLUTIONS : CHAPTER 3 # TRIGONOMETRIC FUNCTIONS
Subject : Maths                                                         Class XII
Total Marks: 15                                                Time : 40 Min
Note: Following are the solution for The Question Bank   https://spectrum6.blogspot.in/2017/08/trigonometry-problems-simple-level-v.html
ANS : Q 1 Answer the following questions                       (5M)
a) The solutions of a trigonometric equation of an unknown angle x, between 0 to 2π are called (ii)Principle Solution.
b) Which of the following is not a trigonometric equation.
    iii) 3x = 17.
c) 1 + Cot2Ө = i) Sec2Ө. 
d) If Cos Ө = 0 then Ө = ii) (2n+1)π/2.
e) For Sin x = ½ then the general solution of x is iii) π/6,5π/6,13π/6
ANS : Q2 Prove the following equations (any three)       (6M)
a) If tan 2x =  tan 2β then x = nπ  β
           As tan 2x =  tan 2β             
           Cos 2x  = Cos 2β
    using result Cos x  = Cos β , x=2nπ β
                  2x  =  2nπ
                 x  =   nπ β   
     tan 2x =  tan 2β  implies  x  =   nπ β  where n       
b) Find the Principle solution of Sin x = 1/2
       Sin x = 1/2 = Sin π /4
       also ,Sin (π-x) = Sin x
    Sin π /4   = Sin (x- π /4) = Sin 3π /4      
    Sin π /4  =  Sin 3π /4 = 1/2
       such that 0 < π /4  < 2π  and 0 < 3π /4 < 2π
    The required solutions are 
         x = π /4 and x = 3π /4.
c) Find the general Solution of CosӨ = 0
        here using CosӨ = 0 = Cos π /2 = Cos 3π /2 
        also using the periodicity, we get
        Cos π /2 = Cos (2π + π /2) =  Cos (4 π + π /2) = ......
     Cos (2nπ + π /2) = 0   for  nZ

d) Find the Principle solution of Cot x = 1/3
                                          tan x  = 3
                                          tan x  = 3 = tan π /3 by using allied    
                                                                                        angles
                                 tan (π + x) =  tan x
                                tan (2π + x)=  tan x
                                 tan (π/3) =  tan (π + π/3) = 3            
                               tan (π/3) =  tan (2π + π/3) = 3
                                   tan (4π/3) =  tan (/3) = 3
Hence x = 4π/3 and 7π/3 are the required Principle solutions of the given equation Cot x = 1/3.

ANS : Q3 Solve the following  (any three)                         (9M)
a) If   3 Cosec x = 2 then find the general Solution of equation. 
                 Cosec x = 2/3
                             sin x = 3/2 = sin π/3
                      sin x = sin π/3
         ∴Using the result that  sin x = sin α implies x= nπ + (-1)nα
        We get,                        x = nπ + (-1)n π/3

∴ The required General Solution is
                    x = {nπ + (-1)n π/3} , where n ϵ Z

b) Find the Principle solution of tan x = -1/3   
     tan π /6 = -1/3   and by using allied angles
     tan (π - x) = - tan x  and
     tan (2π - x) = - tan x
     - tan π /6 = tan (π - π/6) = -1/3
   - tan π /6 = tan (2π - π/6) = -1/3
     tan 5π/6 π /6π /6 π /6= -1/3 =   tan 11π/6
  Hence x = 5π/6 and 11π/6 are the required principal solutions   of given equation.

c) Convert 40020’10’’ into radian measurement.
   Angle in Radian 
         = (π / 180) X Angle in Degree
         = (π / 180) X 40020’10’’
         = (π / 180) X [400 + (20’/ 60)0 + (10’’/60)’ ]
         = (π / 180) X [400 + (20’/ 60)0 + (10’’/60 X 60)0 ]
         = (π / 180) X [40 + 1/3 + 1/360]0
         = (π / 180) X [40 + 0.33 + 0.003]
         = (π / 180) X [40.333]
         = (0.22 π)c
Hence 40020’10’’ = (0.22 π)c

Tricks

Halogen Derivatives and Alcohol, Phenol, Ether

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