Output at 1 = AB
Output at 2 = \(\overline{A+B}\)
Output at y  = AB +  \(\overline{A+B}\)

A	B	A.B	A + B	\(\overline{A+B}\)	AB + \(\overline{A+B}\)
0	0	0	0	1	1
0	1	0	1	0	0
1	0	0	1	0	0
1	1	1	1	0	1

Voltage across Zener diode = 15 V
Therefore, current in 3kΩ = \(\frac{V}{R}=\frac{15}{3 \times 10^3}\)
= 5 x 10^-3 = 5 mA
Voltage across 500 Ω = 25 - 15 = 10 V
Therefore, current in 500 Ω = \(\frac{10}{500}\) = 20 mA
Therefore, current in Zener diode = 20 - 5 = 15 mA

A	B	\(\bar{A}\)	\(\bar{A} B\)	\(A+\bar{A} B\)	A + B
0	0	1	0	0	0
0	1	1	1	1	1
1	0	0	0	1	1
1	1	0	0	1	1

(i) \((A+B)(A+B)=AA+AB+BA+BB\)     \((\because AA=A)\) 
\(=A+AB+B\) 
= A+ AB +B = A(1 + B) + B    \((\because AB+AB=AB)\)
= A + B   \((\because 1+B=1)\) 
(ii) \(A(\bar{A}+B)\)\(=A \bar{A}+A B=AB\)      \((\because A\bar{A}=0)\)
(iii) (A + B) (A + C) =  AA + AC + BA + BC
= A + AC + BA + BC
= A (1 + C) + BA + BC
= A + BA + BC
= A(1 + B) + BC   \((\because 1+C=1)\)
= A + BC  \((\because 1+B=1)\)

Diode D₂ does not conduct current as it is reverse biased.
Therefore, current flows through 3Ω is Zero.
The total resistance in the part of the circuit in which current flows = 4 + 2 = 6Ω
Potential ditference = 12 V
Therefore, current I = \(\frac{V}{R}=\frac{12}{6}= 2A\)

(i) In military, it is used for locating and detecting the targets.
(ii) It is used in navigation systems such as ship borne surface search, air search and missile guidance systems.
(iii) Radars are used to measure precipitation rate and wind speed in meteorological observations.
(iv) It is employed to locate and rescue people in emergency situations.

The process of superimposing low frequency baseband signal onto a high frequency radio signal is called modulation.

It is also refered as an IC. It consists of thousands to millions of transistors, resistors, capacitors etc. integrated on a small flat piece of semi conductor material that is normally silicon.

A solar cell, also known as photovoltaic cell, works on the principle of photovoltaic effect. Accordingly, the p-n junction of the solar cell generates emf when solar radiation falls on it.

(i) Indicator lamps on the front panel of the scientific and laboratory equipment.
(ii) Seven - segment displays
(iii) Traffic signals, emergency vehicle lighting etc.
(iv) Remote control of television, air conditioner etc.

The difference in potential across depletion layer is called the barrier potential.

NAND and NOR gates are known as universal gates because any other logic gate can be made from NAND or NOR gates.

Because of the differing size and the amount of doping, the emitter and collector cannot be interchanged.

The diffusion of the majority charge carriers across the junction gives rise to an electric current, called diffusion current.

The logic gates are considered as the basic building blocks of most of the digital systems. They have one output with one or more inputs.

(i) The loop phase shift must be 0⁰ or integral multiples of \(2 \pi\)
(ii) The loop gain must be unity That is IA\(\beta\)I = 1
Here, A → voltage gain of the amplifier
\(\beta\) →  feed back ratio
(iii) There should be a positive feedback.

Resistance decreases with increase in temperature. Hence, semiconductors are said to have negative temperature coefficient of resistance

The energy gap between the valence band and conduction band is called the forbidden energy gap.

Mobile communication is used to communicate with others in different locations without the use of any physical connection like wires or cables.

The shortest distance between the transmitter and the point of reception of the sky wave along the surface is called as the skip distance.

The range of frequencies required to transmit a piece of specified information in a particular channel is called channel bandwidth or the bandwidth of the transmission system.

The application of suitable DC voltages across the transistor terminals is called biasing
Types: i) Forward bias ii) Reverse bias

The process in which alternating voltage or alternating current is converted direct voltage or direct current is called rectification.

Step 1: AC (1 + B) = AC.1 [OR law-2]
Step 2: AC . 1 = AC [AND law – 2]
Therefore, AC + ABC = AC
Thus the Boolean identity is proved.
Circuit Description

The output at the 1^st AND gate : A\(\overline { B } \)
The output at the 2^nd AND gate : ĀB
The output at the OR gate: Y = A. \(\overline { B } \) + Ā .B

A	B	C	X = A.B	Y=\(\overline { X.C } \)
0	0	0	0	1
1	1	1	1	0

\({ I }_{ B }=\frac { { V }_{ i } }{ { R }_{ B } } =\frac { 20V }{ 500k\Omega } =40\mu A\)  [∵ V_BE = 0V]
\({ I }_{ C }=\frac { { V }_{ CC } }{ { R }_{ C } } =\frac { 20V }{ 4k\Omega } =5mA\) [∵ V_CE = 0V]
\(\beta =\frac { { I }_{ C } }{ { I }_{ B } } =\frac { 5mA }{ 40\mu A } =125\)

α = \(\frac{I_C}{I_E}\)
I_C = α I_E = 0.95 x 1 = 0.95 mA
I_E = I_B + I_C
∴ I_B = I_E - I_C = 1 - 0.95 = 0.05 mA

\({ E }_{ g }=\frac { hc }{ \lambda } \)
Therefore,
\(\lambda =\frac { hc }{ { E }_{ g } } =\frac { 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 1.875\times 1.6\times { 10 }^{ -19 } } \)
= 660 nm
The wavelength 660 nm corresponds to red colour light

The P.D. between A and B is given by
\(V =\left[V_{\mathrm{A}}-V_{\mathrm{B}}\right]-V_{\mathrm{b}}(\mathrm{Si}) \)
\(=[3.3-(-7.4)]-0.7 \)
\(=10.7-0.7=10 \mathrm{~V} \)
The value of current flowing through AB can be obtained by using Ohm’s law
\(I=\frac { V }{ R } =\frac { 10}{ 1\times { 10 }^{ 3 } } ={ 10 }^{ -2 }A=10mA\)

The diode is forward biased and it is an ideal one. Hence, it acts like a closed switch with no barrier voltage. Therefore, current that flows through the diode can be calculated using Ohm’s law.
V = IR
\(I=\frac{V}{R}=\frac{15}{5}\) = 3A

V= 10V,  R₁ = 2 Ω, R₃ = 2 Ω
Diode D₁ is reverse biased so it will block the current and diode D₂ is forward biased, so it will pass the current.
\(\mathrm{R} =\mathrm{R}_{1}+\mathrm{R}_{2} \)
\(=2+2=4 \Omega \)
\(\mathrm{I} =\frac{\mathrm{V}}{\mathrm{R}}=\frac{10}{4} \)
I = 2.5 A

(i) If the portion of the output fed to the input is in phase with the input, then the magnitude of the input signal increases
(ii) It is necessary for sustained oscillations.

(i) The emitter-base junction is forward biased by a dc power supply Vm and the collector-based junction is reverse biased by the bias power supply V_cB
(ii) The forward bias across the emitter base junction causes the majority charge carriers electrons in the emitter region to flow towards the base region and constitutes the emitter current (I_E).
(iii) Since the base region is very narrow, most of the electrons reach the collector region.
(iv) The electrons that reach the collector region will be attracted by the collector terminal as it has positive potential and flows through the external circuit. This constitutes the collector current(I_c)
(v) The holes that are lost due to recombination in the base region ate replaced by the positive potential of the bias voltage V_EE and constitute the base current (I_B)
\( \mathrm{I}_{\mathrm{E}}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}} \)
\(\mathrm{I}_{\mathrm{E}} \approx \mathrm{I}_{\mathrm{C}} \)     \((\because \mathrm{I}_{\mathrm{B}} is\ very\ small )\)

S.No	Avalanche breakdown	Zener breakdown
(i)	Heavily doped p-n junctions have narrow depletion layers of the order of <10-6 m.	It occurs in lightly doped junctions  Which have wide depletion layers.
(ii)	Electric field produced is strong in nature.	Weak electric field is produced.
(iii)	When a reverse voltage across the junction is increased to the breakdown limit, a very strong electric field is set up. It ruptures the covalent bonds in the lattice and thereby generating electronic-hole pairs. This effect is called Zener effect	Thermally generated minority charge carriers accelerated by the electric field gains sufficient kinetic energy, collide with the semiconductor atoms while passing through the depletion region. This leads to the breaking of covalent bonds and in turn covalent bonds and in turn generates electron-hole pairs
(iv)	Even a small further increases in reverse voltage produces a large number of charge carriers. Hence the junction has very low resistance in the breakdown region.	The newly generated charged carriers are also accelerated by the electric field resulting in more collisions and further production of charged carriers.
(v)	This process of emission of electrons due to the strong electric field is known as internal field emission or field ionization.	This cumulative process leads to an avalanche of charge carriers across the junction and consequently reduces the reverse resistance the diode current increases sharply.

The leakage current is the current that the diode will leak when a reverse bias is applied to it.

When a PN junction diode is forward biased, the depletion region decreases and the diode conduct once after the barrier potential is crossed, when it is reverse biased the depletion region increases and the diode does not conduct sci it is called as unidirectional device.

The process of adding impurities to the instrinsic semiconductor is called doping.

S.No Intrinsic semiconductor Extrinsic semiconductor (i)  A semiconductor in its pure form without impurity is called an intrinsic
semiconductor. When small amount of impurity is doped in pure semiconductors it becomes extrinsic semiconductors.
  (ii) The number of free electrons in conduction band is almost equal to number of holes in valence band. The number of free electrons and holes is never equal. (iii) Electrical conductivity is a function of temperature along. Its electrical conductivity depends upon the temperature as well as on the quantity of impurity atoms doped the structure.

RADAR stands for RAdio Detection And Ranging.

When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency. This is called centre frequency or resting frequency.

The total distance d between the transmitting and receiving antennas will be the sum of the individual distances of coverage.
d = d₁ + d₂
\(=\sqrt { 2R{ h }_{ 1 } } +\sqrt { 2{ Rh }_{ 2 } } \)
\(=\sqrt { 2R } \left( \sqrt { { h }_{ 1 } } +\sqrt { { h }_{ 2 } } \right) \)
\(=\sqrt { 2\times 6.4\times { 10 }^{ 6 } } \times (\sqrt { 40 } +\sqrt { 30 } )\)
\(=16\times { 10 }^{ 2 }\sqrt { 5 } \times (6.32+5.48)\)
= 42217 m = 42.217 km