Laws of Boolean algebra:
Complement law:

A	Y=Ā
0	Y = \(\bar { 0 } \) = 1
0	Y=\(\bar { 1 } \)=0

The complement law can be realised as Ā = A
OR laws:

A	B	Y=A+B
0	0	Y = 0 + 0 = 0
0	1	Y = 0 + 1 = 1
1	0	Y = 1 + 0 = 1
1	1	Y = 1 + 1 = 1

The OR laws can be realised as:

1st law	A+0=A
2st law	A+1=1
3st law	A+A=A
4st law	A+Ā=1

AND law:

A	B	Y=A.B
0	0	Y=0.0=0
0	1	Y=0.1=0
1	0	Y=1.0=0
1	1	Y=1.1=1

The AND laws can be realised as:

1st law	A.0=0
2st law	A.1=A
3st law	A.A=A
4st law	A.Ā=0

The Boolean operations obey the folloWing laws:
Communtative laws:
A+B =B+A
A.B =B.A
A sociate laws:
A + (B + C) = (A + B) + C
A. (B.C) = (A.B).C
D stributive laws:
A (B + C) = AB + BC
A + BC = (A + B) (A + C) 
The above laws are used to simplify complicated expressions and to simplify the logic circuitry.

(i) The transistor in saturation region acts as a closed switch while in cut-off region, it acts as an open switch.
(ii) It functions like an electronic switch that helps to turn ON or OFF a given circuit by a small control signal which keeps the transistor either in saturation region or in cut-off region.

When the input is low:
(i) When the input is low (say 0 V), the base current is zero and transistor is not properly forward biased.
(ii) It is in cut off region, As a result, the collector current is zero and correspondingly the voltage drop across R_C, also becomes nearly zero, The output voltage is high and Is equal to V_CC. 
(iii) It means that the no current flows through the transistor and it is said to be switched off. The transistor acts as an open switch.
When the input is high:
(i) When input voltage is increased to a certain high value (say +5 V), the base current (I_B) increases and in turn decreases the collector current to its maximum.
(ii) The transistor will move into the saturation region. The increase in collector current (I_C). increases the voltage drop across R_C, thereby thereby lowering the output voltage, close to zero (since V_o = V_CC -I_CR_C). It means that maximum current flows through the transistor and it is said to be switched on. 
(iii)The transistor acts as a closed switch.

Construction:
(a) The amplification of an electrical signal is explained with a single-stage transistor amplifier as shown in figure.
(b) Single stage indicate that the circuit consists of one transistor with the allied components.

(i) An NPN transistor is connected in the common-emitter configuration
(ii) To start with, the Q point or the operating point of the transistor is fixed, so as to get the maximum signal swing at the output (neither towards saturation point nor towards cut-off).
(iii) A load resistance, R_C is connected in series with the collector circuit to measure the output voltage.
The resistance R₁, R₂, and R_E, form the biasing and stabilization circuit.
(iv) The capacitor C, allows only the AC signal to pass through.
(v) The emitter by pass capacitor C_E provides a low reactance path to the amplified AC signal
(vi) The coupling capacitor C_C is used to couple one stage of the amplifier with the next stage, while constructing multistage amplifiers
Vs is the sinusoidal input signal source applied across the base-emitter. The output is taken across the collector-emitter.
Collector current I_C = βI_B   [∵β = I_C/I_B]
Applying Kirchhoff's voltage law to the output loop, the collector-emitter voltage is given by
V_CE = V_CC - I_CR_C
Working of the amplifier :
During the positive half cycle :
(i) Input signal (V_s) increases the forward voltage across the emitter base. As a result, the base current (I_B in μA) increases. consequently the collector current (I_Cin mA) increases β times.
(ii) This increase the voltage drop across R_C(I_CR_C) which in turn decreases the collector-emitter voltage (v_CE). Therefore, the input signal in the positive direction produces an amplified signal in the negative direction at the output. Hence the output signal is reversed by 180⁰ as shown in figure

During the negative half cycle:
(i) Input signal (V_s) decreases the forward voltage across the emitter base. As a result base current (I_B in μA) decreases and in tum increases the collector current (I_B in μA).
(ii) The increase in collector current (I_C) decreases the potential drop across R_C and increases the collector - emitter voltage (V_CE).
(iii) Thus the input signal in the negative. direction produces an amplified signal in the positive direction at the output.
(iv) Therefore, 180 phase reverse is observed during the negative half cycle of the input signal as well as shown in figure.

The static characteristics of the BJT are
(i) Input characteristics
(ii) Output characteristics
(iii) Transfer characteristics.
(i) Input characteristics:
Input characteristics curves give the relationship between the base current (I_B) and base to emitter voltage (v_BE) at constant collector to emitter voltage (v_CE) and are shown in figure

(i) Initially, the collector to emitter voltage (V_CE) is set to a particular value (above 0.7 V to reverse bias the junction).
(ii) Then the base-emitter voltage V_BE, is increased in suitable steps and the corresponding base-current I_B is recorded.
(iii) A graph is plotted with V_BE along the x axis and I_B along the Y - axis
(iv) The procedure is repeated for different values of V_cE
The following observations are made from the graph :
(i) The Curve looks like the forward characteristics of an ordinary p - n junction diode.
(ii) There exists a threshold voltage (or) knee voltage (V_knee) below which the base current (I_b) is very small. This value is 0.7 V for silicon and 0.3 V for germanium transistors. Beyond the knee voltage, the base current increases with the increase in base-emitter voltage.
(iii) It is also noted the increase in V_CE, decreases the I_B. This shifts the curve outward.
(iv) This is because the increase in collector-emitter voltage increases the width of the depletion region which in turn, reduces the effective base width and thereby the base current.
Input resistance :
The ratio of the change in base-emitter voltage (∆V_BE) to the change in base current (∆L_B) at a constant collector-emitter voltage (V_CE) is called the input resistance (r_i)
\(\mathrm{R}_{\mathrm{i}}=\left[\frac{\Delta \mathrm{V}_{\mathrm{BE}}}{\Delta \mathrm{I}_{\mathrm{B}}}\right]_{\dot{\mathrm{V} C \mathrm{E}}}\)
The input impedance is high for a transistor in common emitter configuration.
Output characteristics :
The output characteristics give the relationship between the collector current (I_c) and the collector -emitter voltage (V_CE) at constant input current (I_B)
as shown in figure

(i) Initially I_B is set to a particular voltage. V_CE is increased in suitable steps and I_C is recorded
(ii) A graph is plotted with the V_CE along the x-axis and I_C along the y - axis
(iii) This procedure is repeated for different values I_B
The four important regions ln the output characteristics
(i) Saturation region
(ii) Cut-off region
(iii) Active region
(iv) Breakdown region
Output Resistance :
The ratio of the change in the collector emitter voltage (∆V_CE) to the corresponding change in the collector current (∆l_C) at constant base current (I_B) is called output resistance (r_o).
\(\mathrm{R}_{\mathrm{o}}=\left[\frac{\Delta \mathrm{V}_{\mathrm{BE}}}{\Delta \mathrm{I}_{\mathrm{C}}}\right]_{\mathrm{l}_{\mathrm{B}}}\)
The output impedance for transistor in common emitter configuration is very low.

FuIl wave rectifier :
The positive and negative half cycles of the AC input signal pass through the full wave rectifier circuit and hence it is called the full wave rectifier
Construction:

(i) It consists of two p-n junction diodes, a center-tapped transformer, and a load resistor (R₁)
(ii) The centre is usually taken as the ground or zero voltage reference point.
(iii) Due to the centre tap transformer, the output voltage rectified by each diode is only one-half of the total secondary voltage.
Working:
During positive half cycle :
(i) When the positive half cycle of the ac input signal passes through the circuit, terminal M is positive, G is at zero potential and N is at negative potential.
(ii) This forward biases diode D₁ and reverse biases diode D₂.
(iii) Hence, being forward biased, diode D₁ conducts and current flows along the path MD₁AGC.
During negative half cycle:
(i) When the negative half cycle of the AC input signal passes through the circuit, terminal N becomes positive, C is at zero potential and M is at negative potential.
(ii) This forward biases diode D₂ and reverse biases diode D₁.
(iii) Hence, being forward biased, diode D₂ conducts and current flows along the path ND₂BGC.
(iii) During both positive and negative half cycles of the input signal, the current flows through the load in same direction.

(iv) The output signal corresponding to the input signal is shown in Figure. Though both half cycles of AC input are rectified, the output is still pulsating in nature.
(v) The efficiency (η) of full wave rectifier is twice that of a half wave rectifier and is found to be 81.2 %.

HaIf wave rectifier:
Only one half of the input wave reaches the output. Therefore it is called half wave rectifier.
Construction:

(i) The circuit consists of a transformer, a p-n junction diode and a resistor
(ii) In a half wave rectifier circuit, either a positive half or the negative half of the AC input is passed through by the diode while the other half is blocked
(iii) It acts as a rectifier diode.

(iv) Efficiency (η) is the ratio of the output DC power to the AC input power circuit. supplied to the circuit.
(v) The efficiency (η) of a half wave rectifier is found to be 40.6 %.

i) Formation of depletion layer
(i) A single piece of semiconductor crystal is suitably doped such that its one side is p-type semiconductor and the other side is n-type semiconductor.
(ii) The contact surface between the two sides is called p-n junction. Whenever p-n junction is formed, some of the free electrons diffuse from the n-side to the p-side while the holes from the P-Side to the n-side.
(iii) The diffusion of charge carriers happens due to the fact that the n-side has higher electron concentration and the p-side has higher hole concentration.
(iv) The diffusion of the majority charge carriers across the junction gives rise to an electric current, called diffusion current.
(v) When an electron leaves the n-side, a pentavalent atom in the n-side becomes a positive ion.
(vi) The free electron migrating into p-side recombines with a hole present in a trivalent atom near the junction and the trivalent atom becomes a negative ion. Since such ions are bonded to the neighbouring atoms in the crystal lattice, they are unable to move.

(vii) As the diffusion process continues, a laver of positive ions and a layer of negative ions are created on either side of the junction accordingly.
(viii) The thin region near the junction which is free from charge carriers (free electrons and holes) is called depletion region.
(ix) An electric field is set up between the positively charged layer in the n-side and the negatively charged layer in the p-side in the depletion region.
(x) This electric field makes electrons in the p-side drift into the n-side and the holes in the n-side into the p-side.
(xi) The electric current produced due to the motion of the minority charge carriers by the electric field is known as drift current. The diffusion current and drift current flow in opposite directions.
(xii) Though drift current is less than diffusion current initially, equilibrium is reached between them at a particular time.
(xiii) With each electron (or hole) diffusing across the junction, the strength of the electric field increases thereby increasing the drift current till the two currents become equal
(xiv)Hence at equilibrium, there is no net electric current across the junction. Thus, a p-n junction is formed.
ii) Junction potential or barrier potential

(i) The movement of charge carriers across the junction takes place only to a certain point beyond which the depletion layer acts like a barrier to further diffusion of free charges across the junction.
(ii) This is due to the fact that the immobile ions on both sides establish an electric potential difference across the junction.
(iii) Therefore, an electron trying to diffuse into the interior of the depletion region encounters a wall of negative ions repelling it backwards.
(iv) If the free electron has enough energy, it can break through the wall and enter into the p-region, where it can recombine with a hole and create another negative ion.
(v) The strength of the electric potential difference across the depletion region keeps on increasing with the crossing of each electron until equilibrium is reached, at this point, the internal repulsion of the depletion layer stops further diffusion of free electrons across the junction.
(vi) The difference in potential across the depletion layer is called the barrier potential (V_b).
(vii) At 25°C, this barrier potential is approximately 0.7 V for silicon and 0.3 V for germanium.

When V_CE = 15 V, I_C = 1.5 μA
When V_CE is changed to 10 V, I_C = 1.4 μA

Voltage across AB, is V_Z = 9V
Voltage drop across R_s = 15 - 9 = 6V
Therefore current through the resistor R_s,
I = \(\frac { 6 }{ 1\times { 10 }^{ 3 } } \) = 6 mA
Voltage across the load resistor, = V_AB = 9V
Current through load resistor,
\({ I }_{ L }=\frac { { V }_{ AB } }{ { R }_{ L } } =\frac { 9 }{ 2\times { 10 }^{ 3 } } =4.5mA\)
The current through the Zener diode,
I_Z = I - I_L = 6 mA - 4.5mA = 1.5 mA

Photo diode:
A p -n junction diode which converts an optical signal into electrical current is known as photodiode.
(i) The operation of photodiode is exactly inverse to that of an LED. Photodiode works in reverse bias condition.

(ii) The direction of arrows indicates that the light is incident on the photo diode.
(iii) The device consists of a p-n junction semiconductor made of photosensitive material kept safely inside a plastic case as shown in fig.
(iv) It has a small transparent window that allows light to be incident on the p-n junction.
(v) Photodiodes car generate current when the p - n junction is exposed to light and hence are called as light sensors.
(vi) When a photon of sufficient energy (hv) strikes the depletion region of the diode, some of the valence band electron are elevated into conduction band, in turn holes are developed in the valence band. This creates electron-hole pairs. The amount of electron - hole pairs generated depends on the intensity of light incident on the p - n junction.
(vii) These electron and holes are swept across the p-n junction by the electric field created by reverse voltage before recombination takes place. Thus, holes move towards the n - side and electrons towards the P - side - when the external circuit is made, the electrons flow through the external circuit and constitute the photo current.
(viii) When the incident light is zero, there exists a reverse current which is negligible. This reverse current in the absence of any. incident light is called dark current and is. due to the thermally generated minority carriers.

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