Basic Electronics Archives

Bohr’s Atomic model

Bohr’s Atomic model explain many of the atomic phenomena. He postulated three fundamental laws to overcome the inconsistency in the Rutherford model.

Postulate 1:- Electrons can not occupy states at all energy levels. Electrons can occupy states at only certain discrete energy levels. When electron occupies a discrete energy level, it does not emit radiation and is said to be Stationary (or Non-radiating state).

Postulate 2:-When an electron moves from Higher energy state (E₂) to lower energy state (E₁).It emitts radiation at a frequency given by

f= (E₂ – E₁)/h.

where ‘h’ is the plank’s constant h= 6.62 X 10^-34 Joules. E₁ and E₂ are Energies in Joules.

Postulate 3:- Any stationary state is determined by the condition that the angular momentum of the electron in this state is quantized and is a multiple of h/2π .

angular momentum = Integral multiple of h/2π .

mvr = nh/2π.

The total energy of electron in stationary state is W_n = (-13.6/n² ) eV.

Input and output characteristics of a transistor in Common Collector configuration

Common collector configuration, As we know collector is kept common to both input and output terminals.

To draw input and output characteristics of Common collector. The circuit diagram is given as follows

input characteristics:-

These are in between input current (I_B) Vs input voltage (V_BC) with output voltage (V_EC) as kept constant.

To obtain input characteristics , V_EC is kept constant and V_BC is increased in equal steps and the corresponding increase in I_B is noted and the same procedure is repeated for different values of V_EC and the characteristics are shown in the previous figure.

output characteristics:-

The output characteristics are obtained by keeping input current as constant (I_B = constant) and the output voltage V_EC is increased in suitable steps and the corresponding output current I_E is noted down.

This is repeated for different values of I_B .

Extrinsic Semi conductors (N-Type Vs P-Type)

Intrinsic semi conductors has a little current conduction capability.
This capability can be increased by many times by just adding a very small amount of impurity in (1 atom per 1 million pure atoms) the process of crystallization.
This process is called as doping
Si and Ge are tetravalent (4 electrons in Valance Band) atoms. So the impurity may be either tri valent (or) pentavalent.
Depending on the type of impurity added extrinsic semi conductors are divided into 2 types

i. Donor type (or) N-type.

ii. Acceptor type (or) P-type.

The detailed explanation about N-type and P-type semi conductors are given in the following table.

N-type Semi conductors	P-type Semi conductors
When a small amount of pentavalent impurity such as Arsenic, Antimony, Bismuth (or) Phosphorus is added to a pure semi conductor during the crystal growth. The resulting semi conductor is called N-type semi conductor.	When a small amount of trivalent impurity such as Boron, Gallium, Indium (or) Aluminium is added to a pure semi conductor in the resulting crystal . The resulting semi conductor is called P-type semi conductor.
Where N stands for negative.	Where P stands for positive.
When a pentavalent (or) a donor atom is added to Si the impurity atom forms 4 co-valent bonds with 4 Si atoms and fifth valance electron is left free. Which is loosely bound to the Antimony atom.	When a trivalent (or) a acceptor atom is added to Pure Si (or) Ge semi conductor. Boron has 3 valance electrons in its valance shell and pure Si has 4 valance electrons in its valance shell. The 3 electrons of Boron atoms forms 3 covalent bonds with 3 Si atoms and a whole is left free which is loosely bound to Boron.
One impurity atom provides one free electron yet an extremely small amount of impurity provides enough atoms to supply millions of free electrons.	simillarly a small amount of impurity provides millions of wholes.
In this type of semi conductor majority carriers are electrons which are responsible for conduction of current.	Majority carriers are holes and are responsible for conduction of current.
Number of holes are very less in N-type when compared to electrons. Hence holes are known as minority carriers.	Simillarly number of electrons less compared to number of holes in P-type. Minority carriers are electrons.

N-type is preferred over P-type , mobility of electron is high compared to hole mobility.

Early effect in Common Base Configuration

Early effect (or) Base-width modulation:-

In Common Base configuration in the Reverse Bias, As the voltage $V_{CC}$ increases, the space-charge width between collector and base tends to increase, with the result that the effective width of the base decreases. This dependency of Base-width on the Collector to emitter voltage is known as the early effect.

The early effect has three consequences:-

There is less chance for recombination with in the base region. Hence $\alpha$ increases with increasing $\left | V_{CB} \right |$ .
The charge gradient is increased with in the base and consequently, the current of minority carriers injected across the emitter junction increases.
For extremely large voltages, the effective Base-width may be reduced to zero, causing voltage break-down in the transistor. This phenomenon is called the Punch-through.

For higher values of $V_{CB}$ , due to early effect the value of $\alpha$ increases, for example $\alpha$ changes say from 0.98 to 0.985. Hence there is a very small positive slope in the CB output characteristics and hence the output resistance is not zero.

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Input and Output characteristics of transistor in Common Emitter Configuration

Input Characteristics:-

Input characteristics in Common Emitter configuration means input voltage Vs input current by keeping output voltage as constant.

i.e, $V_{EB}$ Vs $I_{E}$ by keeping $V_{CB}$ constant.

Therefore the curve between Emitter current $I_{E}$ and Emitter to Base voltage $V_{EB}$ for a given value of Collector to Base voltage $V_{CB}$ represents input characteristic.

common base input and output characteristics

for a given output voltage $V_{CB}$ , the input circuit acts as a PN-junction diode under Forward Bias.

from the curves there exists a cut-in (or) offset (or) threshold voltage $V_{EB}$ below which the emitter current is very small and a substantial amount of Emitter-current flows after cut-in voltage ( 0.7 V for Si and 0.3 V for Ge).

the emitter current $I_{E}$ increases rapidly with the small increase in $V_{EB}$ . with the low dynamic input resistance of a transistor.

i.e, $r_{i} =\frac{\Delta V_{EB}}{\Delta I_{E}}|_{V_{CB}\approx Constant}$

$input resistance =\frac{change in input voltage}{change in emitter current}|V_{CB}{\approx Constant}$

This is calculated by measuring the slope of the input characteristic.

i.e, input characteristic determines the input resistance $r_{i}$ .

The value of $r_{i}$ varies from point to point on the Non-linear portion of the characteristic and is about $100\Omega$ in the linear region.

Output Characteristics:-

Output Characteristics are in between output current Vs output voltage with input current as kept constant.

i.e, $f(I_{c},V_{CE})_{I_{E} = Constant}$

i.e, O/p characteristics are in between $V_{CB}$ Vs $I_{c}$ by keeping $I_{E}$ as constant.

basically it has 4 regions of operation Active region, saturation region,cut-off region and reach-through region.

active region:-

from the active region of operation $I_{c}$ is almost independent of $I_{E}$

i.e, $I_{c}\approx I_{E}$

when $V_{CB}$ increases, there is very small increase in $I_{c}$ .

This is because the increase in $V_{CB}$ expands the collector-base depletion region and shortens the distance between the two depletion regions.

with $I_{E}$ kept constant the increase in $I_{c}$ is so small. transistor operates in it’s normal operation mode in this region.

saturation region:-

here both junctions are Forward Biased.

Collector current $I_{c}$ flows even when $V_{CB}=0$ (left of origin) and this current reaches to zero when $V_{CB}$ is increased negatively.

cut-off region:-

the region below the curve $I_{E}=0$ ,transistor operates in this region when the two junctions are Reverse Biased.

$I_{c}\neq 0$ even though $I_{E}=0$ mA. this is because of collector leakage current (or) reverse-saturation current $I_{CO}$ (or) $I_{CBO}$ .

punch through/reach through region:-

$I_{c}$ is practically independent of $V_{CB}$ over certain transistor operating region of the transistor.

If $V_{CB}$ is increased beyond a certain value, $I_{c}$ eventually increases rapidly because of avalanche (or) zener effects (or) both this condition is known as punch through (or) reach through region.
If transistor is operated beyond the specified output voltage ( $V_{CB}$ ) transistor breakdown occurs.
If $V_{CB}$ is increased beyond certain limit, the depletion region( $J_{c}$ ) of o/p junction penetrates into the base until it makes contact with emitter-base depletion region. we call this condition as punch-through (or) reach-through effect.
In this region , the large collector current destroys the transistor.
To avoid this $V_{CB}$ should be kept in safe limits specified by the manufacturer

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Input and Output characteristics of transistor in Common Base Configuration

Input Characteristics:-

Input characteristics in Common Base configuration means input voltage Vs input current by keeping output voltage

\[ {\color{DarkGreen} V_{CE}} \]

as constant.

i.e, $V_{EB}$ Vs $I_{E}$ by keeping $V_{CB}$ constant.

Therefore the curve between Emitter current $I_{E}$ and Emitter to Base voltage $V_{EB}$ for a given value of Collector to Base voltage $V_{CB}$ represents input characteristic.

for a given output voltage $V_{CB}$ , the input circuit acts as a PN-junction diode under Forward Bias.

the emitter current $I_{E}$ increases rapidly with the small increase in $V_{EB}$ . with the low dynamic input resistance of a transistor.

i.e, $r_{i} =\frac{\Delta V_{EB}}{\Delta I_{E}}|_{V_{CB}\approx Constant}$

$input resistance =\frac{change in input voltage}{change in emitter current}|V_{CB}{\approx Constant}$

This is calculated by measuring the slope of the input characteristic.

i.e, input characteristic determines the input resistance $r_{i}$ .

The value of $r_{i}$ varies from point to point on the Non-linear portion of the characteristic and is about $100\Omega$ in the linear region.

Output Characteristics:-

Output Characteristics are in between output current Vs output voltage with input current as kept constant.

i.e, $f(I_{c},V_{CE})_{I_{E} = Constant}$

i.e, O/p characteristics are in between $V_{CB}$ Vs $I_{c}$ by keeping $I_{E}$ as constant.

basically it has 4 regions of operation Active region, saturation region,cut-off region and reach-through region.

active region:-

from the active region of operation $I_{c}$ is almost independent of $I_{E}$

i.e, $I_{c}\approx I_{E}$

when $V_{CB}$ increases, there is very small increase in $I_{c}$ .

This is because the increase in $V_{CB}$ expands the collector-base depletion region and shortens the distance between the two depletion regions.

with $I_{E}$ kept constant the increase in $I_{c}$ is so small. transistor operates in it’s normal operation mode in this region.

saturation region:-

here both junctions are Forward Biased.

Collector current $I_{c}$ flows even when $V_{CB}=0$ (left of origin) and this current reaches to zero when $V_{CB}$ is increased negatively.

cut-off region:-

the region below the curve $I_{E}=0$ ,transistor operates in this region when the two junctions are Reverse Biased.

$I_{c}\neq 0$ even though $I_{E}=0$ mA. this is because of collector leakage current (or) reverse-saturation current $I_{CO}$ (or) $I_{CBO}$ .

punch through/reach through region:-

$I_{c}$ is practically independent of $V_{CB}$ over certain transistor operating region of the transistor.

If $V_{CB}$ is increased beyond a certain value, $I_{c}$ eventually increases rapidly because of avalanche (or) zener effects (or) both this condition is known as punch through (or) reach through region.
If transistor is operated beyond the specified output voltage ( $V_{CB}$ ) transistor breakdown occurs.
If $V_{CB}$ is increased beyond certain limit, the depletion region( $J_{c}$ ) of o/p junction penetrates into the base until it makes contact with emitter-base depletion region. we call this condition as punch-through (or) reach-through effect.
In this region , the large collector current destroys the transistor.
To avoid this $V_{CB}$ should be kept in safe limits specified by the manufacturer

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Comparison of Transistor Characteristics in three Configurations

Comparison of Transistor Characteristics in three Configurations :-

Transistor in three Configurations Common Emitter, Common Base and Common Collector can be compared in terms of Input and output impedance’s ,voltage and current gains and some other parameters. Those are tabulated as follows

characteristic	Common Base (CB)	Common Emitter (CE)	Common Collector (CC)
Input Impedance/Resistance ( $Z_{i}$ )	low ( $\approx 100\Omega$ )	Medium ( $750\Omega$ )	Very High ( $750K\Omega$ )
Output Impedance ( $Z_{o}$ )	Very High ( $450K\Omega$ )	Moderate ( $45K\Omega$ )	low ( $25\Omega$ )
Current gain ( $A_{I}$ )	Unity	High	High
Voltage gain ( $A_{v}$ )	about $150$ High ( $\approx 131$ )	about $500$ High	less than $1$
Phase shift b/w i/p and o/p	$0^{o}/360^{o}$	$180^{o}$	$0^{o}/360^{o}$
Applications	High Frequency Circuits	For Audio Circuits	For impedance matching

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Current equation of the diode

The diode under Forward bias is as follows

The current equation related to the voltage V and current I is given by

$I=I_{o}(e^{\frac{V}{\eta v_{T}}}-1) Amperes$

I- Diode current

I_o– Reverse saturation current of the diode at room temperature.

V-applied External voltage

$\eta$ – constant = 1 For Ge

= 2 For Si.

$v_{T}=\frac{kT}{q}$ – volt equivalent temperature 26 mV at room temp.

where k-Boltzmann constant = 1.38 X 10^-23 J/K.

T-Temperature of Diode in kelvin ^oK = ^o C + 273.

q- charge of electron = 1.6X10^-19C.

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Reverse Saturation Current (I_o) of PN-Diode:-

Comparision table HWR, FWR and Bridge Rectifier

Parameter	Half-Wave Rectifier	Full-wave Rectifier	Bridge Rectifier
No of diodes	1	2	4
Maximum Efficiency ( η )	40.6%	81.2%	81.2%
$V_{dc}$	$\frac{V_{m}}{\pi }$	$\frac{2V_{m}}{\pi }$	$\frac{2V_{m}}{\pi }$
$I_{dc}$	$\frac{I_{m}}{\pi }$	$\frac{2I_{m}}{\pi }$	$\frac{2I_{m}}{\pi }$
Output RMS voltage	$\frac{V_{m}}{2}$	$\frac{V_{m}}{\sqrt{2}}$	$\frac{V_{m}}{\sqrt{2}}$
Average current I_dc	$I_{dc}$	$\frac{I_{dc}}{{2}}$	$\frac{I_{dc}}{{2}}$
Ripple Factor ( $\Gamma$ )	1.21	0.48	0.48
Peak Inverse Voltage (PIV)	$V_{m}$	$2V_{m}$	$V_{m}$
Output Frequency	f	2f	2f
TUF(Transformer Utilization Factor)	0.287	0.693	0.812
Form Factor	1.57	1.11	1.11
Peak factor	2	$\sqrt{2}$	$\sqrt{2}$

Drift and Diffusion currents

The flow of charge (or) current through a semi conductor material is of two types. Similarly the net current that flows through a PN diode is also of two types (i) Drift current and (ii) Diffusion current.

Drift current:-

When an Electric field is applied across the semi conductor, the charge carriers attains certain velocity known as drift velocity $v_{d}= \mu E$ with this velocity electrons move towards positive terminal and holes move towards negative terminal of the battery. This movement of charge carriers constitutes a current known as ‘Drift current’.

Drift current is defined as the flow of electric current due to the motion of the charge carriers under the influence of an external field.

Drift current density due to free electrons $J_{n} = qn\mu _{n}E$ atoms/Cm² and the Drift current density due to free holes $J_{p} = qp\mu _{p}E$ atoms/Cm².

The current densities are perpendicular to the direction of current flow.

Diffusion Current:-

It is possible for an electric current to flow in a semi conductor even in the absence of the applied Electric field (or) voltage provided there exists a concentration gradient.

concentration gradient exists if the number of electrons (or) holes is greater in one region than other region in a semi conductors.

Now the charge carriers move from higher concentration to that lower concentration of same type charged regions.

The resulting current is known as diffusion current.

Diffusion current density ( $J_{P}$ ) due to holes is $J_{p} = -q D_{p}\frac{dp}{dx}$ A/Cm² .

Diffusion current density ( $J_{P}$ ) due to holes is $J_{n} = q D_{n}\frac{dn}{dx}$ A/Cm² .

$\therefore$ Total current in a semi-conductor is the sum of drift and diffusion currents

In P-type total current density is $J= qp\mu _{p} E-qD_{p}\frac{dp}{dx}$ .

In N-type total current density is $J= qn\mu _{n} E+qD_{n}\frac{dn}{dx}$ .

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Crystal Oscillator

Crystal Oscillator:-

In LC oscillators the frequency of oscillation f_o depends on the tank circuit parameters L & C, whereas L & C values change with respect to time, temperature, aging etc. Therefore f_o does not remain constant so at high frequencies LC oscillators are unsuitable because of instability. Crystal oscillators are more suitable at high frequencies and uses crystal as oscillatory element.

Piezo-electric effect:-

It is the ability of certain materials to generate an electric charge when mechanical stress is applied and vice-versa ( vice-versa is called Reverse Piezo electric effect).

i.e, If mechanical pressure is applied across x-axis, the electric charges appear perpendicular to x-axis that is along y-direction. similarly if electric field is applied along x-direction mechanical strain is produced along y-direction.

working of quartz crystal:-

In this circuit crystal is placed between two metal plates then it acts as a capacitor with dielectric material as crystal between two metal plates.

i.e, when a.c voltage is applied across these plates the crystal vibrates at a frequency of the applied a.c voltage . when f_i = f_o resonance takes place and crystal vibrates with it’s natural frequency almost of constant value.

Equivalent circuit of crystal:-

when crystal is not vibrating it is equivalent to a capacitance C_m.

when it is vibrating it is equivalent to series R-L-C circuit as shown below

and the series resonant frequency is given by $\left | X_{L} \right | = \left |X_{C} \right |$

$\omega L =\frac{1}{\omega C}$

$f_{s}=\frac{1}{2\pi \sqrt{LC}}$

Parallel resonance $\left | X_{L}+X_{C} \right |=\left | X_{Cm} \right |$

$\omega L-\frac{1}{\omega _{C}}=\frac{1}{\omega _{Cm}}$

$\omega ^{2}=\frac{1}{L}\sqrt{\frac{1}{c}+\frac{1}{C_{m}}}$

$f_{p}=\frac{1}{2\pi }\sqrt{\frac{1}{L}(\frac{1}{c}+\frac{1}{C_{m}})}$ Crystal Oscillator has two modes of operation Series $f_{s}$ resonance mode and parallel resonance mode $f_{p}$ .

Crystal Oscillator with BJT:-

using crystal oscillator can be built up as follows

In this circuit crystal acts as a parallel tuned circuit at parallel resonance, the crystal impedance is maximum that is maximum voltage drop is there across C₁ this allows that maximum energy transfer through feedback network through f_p. BJT offers a phase shift of 180^o further 180^o is produced by the capacitor voltage. Oscillations are possible only through f_p, which provides stable oscillations.

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Hall effect

When a transverse magnetic field ‘B’is applied to a specimen (of metal (or) Semi conductor) carrying a current Ian Electric field E is induced perpendicular to both I and B. This phenomenon is known as Hall effect.

The figure shows the experimental arrangement to observe Hall effect Now

I $\rightarrow$ Current flowing in the semi conductor (x-direction)

B $\rightarrow$ Applied Magnetic field (z-direction)

E $\rightarrow$ Induced Electric field is along y-direction perpendicular to both I and B.

Now charge carrier electron is moving under the influence of two fields both electric field(E) and Magnetic field(B).

i.e, electron is under the influence of both E and B, E applies some force on electron similarly B.

under equilibrium $F_{E} = F_{B}$

$qE = Bqv_{d}------EQN(I)$ , where $v_{d}$ is the drift velocity

Electric field Intensity due to Hall effect is $E=\frac{V_{H}}{d}--------------EQN(II)$

$V_{H}$ is the Hall voltage between plates 1 and 2.

and d- is the distance between the two plates.

In an N-type Semi conductor, the current is due to electrons , plate 1 is negatively charged compared to plate 2.

The current density J related to charge density $\rho$ is $J = \rho v_{d}------------EQN(III)$

$J = \frac{Current}{Area}=\frac{I}{A}=\frac{I}{Wd}$

W- width of the specimen, d- height of the specimen.

From EQN(I) $E=Bv_{d}$ and From EQN(II) $V_{H}=Ed$

up on multiplying with ‘d’ on both sides $E d = Bd v_{d}$

$V_{H} = Bd v_{d}$

$V_{H} = B d \frac{J}{\rho }$ from EQN(III)

$V_{H} = B\frac{I}{Wd\rho }d$

$V_{H} = \frac{BI}{\rho W}$

$V_{H} = \frac{1}{\rho } . \frac{BI}{W}$ , let Hall coefficient $R_{H} = \frac{1}{\rho }$

$V_{H} = R_{H}. \frac{BI}{W}$ .

Uses of Hall effect (or) Applications of Hall effect:-

Hall effect specifies the type of semi conductor that is P-type (or) N-type.when $R_{H}$ is positive it’s a P-type semi conductor and $R_{H}$ negative means it’s N-type semi conductor.
It is used to find out carrier concentrations ‘n’ and ‘p’ , by using either $\rho = nq$ or $\rho =pq$ .
To find out mobilities $\mu _{n}$ and $\mu _{p}$ using the equation $\mu =\sigma R_{H}$ .
Some other applications of Hall effect are measurement of velocity, sorting,limit sensing etc.
used to measure a.c power and the strength of Magnetic field and also finds the angular position of static magnetic fields in a magnetic field meter.
used in Hall effect multiplier, which gives the output proportional to product of two input signals.

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Colpitt’s Oscillator

Colpitt’s Oscillator is an excellent circuit and is widely used in commercial signal generators upto 100MHz.

It consists of a single-stage inverting amplifier and an LC phase shift Network.

The two capacitors $C_{1}$ and $C_{2}$ provides potential divider used for providing $V_{f}$ . $C_{1}$ is the feedback element and which provides positive feedback required for sustained Oscillations.

The amplifier circuit is a self-Bias Circuit with $R_{1}$ , $R_{2}$ and parallel combination of $R_{E}$ with $C_{E}$ .

$V_{CC}$ is applied through a resistor $R_{C}$ (or) RFC choke some times. This RFC choke offers very high impedance to high frequency currents.

$R_{C}$ value has chosen in such a way that it offers high impedance. Two coupling Capacitors $C_{C1}$ and $C_{C2}$ are used to block d.c currents, that means they do not permit d.c currents into tank circuit.

These capacitors $C_{C1}$ and $C_{C2}$ provides a path from Collector to Base through LC Network.

when $V_{CC}$ is switched on , a transient current is produced in the tank circuit an consequently damped oscillations are setup in the circuit.

The oscillatory current in the tank circuit produces a.c voltages across $C_{1}$ and $C_{2}$ . If terminal 1 is more positive w.r. to 2 , then voltages across $C_{1}$ and $C_{2}$ are opposite thus providing a phase shift of $180^{o}$ between 1 and 2.

as the transistor is operating in CE mode , it provides a phase shift of $180^{o}$ .

Therefore the over all phase shift provided by the circuit results $360^{o}$ which is an essential condition for developing oscillations.

If the feedback is adjusted so that the loop gain $A\beta =1$ then then the circuit acts as an Oscillator.

The frequency of oscillation depends on the tank circuit and is varied by gang (or) group tuning of $C_{1}$ and $C_{2}$ means $C_{1}=C_{2}$ .

working:-

The capacitors $C_{1}$ and $C_{2}$ are charged by $V_{CC}$ and are discharged through the coil $L$ setting up of oscillations with frequency

$f_{o}=\frac{1}{2\pi }\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}$ .

these oscillations across $C_{1}$ are applied to the Base-Emitter junction and the amplified version of output is collected across Collector (the frequency of amplifier output is same as that of input of the amplifier) .

This amplified energy is given back to tank circuit to compensate losses.

therefore un damped oscillations results in the circuit.

Derivation for frequency of oscillations:-

chose $\left | A\beta \right |\geq 1$ for sustained oscillations.

Analysis(Qualitative):-

if $Z_{1}$ , $Z_{2}$ and $Z_{3}$ are pure reactive elements such that $Z_{1}=\frac{1}{j\omega C_{1}} =\frac{-j}{\omega C_{1}}$ , $Z_{2}=\frac{1}{j\omega C_{2}} =\frac{-j}{\omega C_{2}}$ and $Z_{3}=j\omega L$ .

from the general condition for an Oscillator

$\left | A\beta \right | =1$ $\Rightarrow h_{ie}(Z_{1}+Z_{2}+Z_{3})+Z_{1}Z_{2}(1+h_{fe})+Z_{1}Z_{3}=0$ .

$h_{ie}(-\frac{j}{\omega C_{1}}-\frac{j}{\omega C_{2}}+j\omega L)+\frac{j^{2}}{\omega ^{2}C_{1}C_{2}}(1+h_{fe})-\frac{j}{\omega C_{1}}.j\omega L=0$

find the real and imaginary parts,

$-j(\frac{1}{\omega C_{1}}+\frac{1}{\omega C_{2}}-\omega L)h_{ie}-\frac{1}{\omega ^{2}C_{1}C_{2}}(1+h_{fe})+\frac{L}{C_{1}}=0$

equating imaginary part to zero $(\frac{1}{\omega C_{1}}+\frac{1}{\omega C_{2}}-\omega L)=0$ , since $h_{ie}\neq 0$ .

$\frac{\omega C_{1}+\omega C_{2}}{\omega^{2} C_{1}C_{2}}=\omega L$ .

after simplification

$\omega ^{2}=\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}$ .

by substituting $\omega =2\pi f$ results $f_{o}=\frac{1}{2\pi }\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}$ .

substituting the value of $\omega ^{2}$ in the real part gives $h_{fe}=\frac{C_{2}}{C_{1}}$ . this is the condition for sustained oscillations.

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Current components of a PNP Transistor

The various Current components which flow across a PNP Transistor are as shown in the figure.

For Normal operation

Emitter Junction $J_{E}$ is Forward Biased.
collector Junction $J_{C}$ is Reverse Biased.

The current flows into Emitter is Emitter current $I_{E}$ , $I_{E} = I_{hE}+I_{eE}$ .

This current consists of two components

$I_{hE}$ or $I_{pE}$ – Current due to majority carriers(holes).
$I_{eE}$ or $I_{nE}$ – Current due to minority carriers(electrons).

since $I_{eE}$ is very small $I_{E} \simeq I_{hE}-----------Equation(1)$

All the holes crossing the Emitter junction $J_{E}$ do not reach the Collector junction because some of them combine with the electrons in the N-type Base.

$I_{hC}$ – is the hole current in the Collector.

∴ Base current = Total hole current in Emitter – hole current in Collector.

i.e, $I_{B} = I_{hE}-I_{hC}----------------Equation(2)$ .

If emitter were open circuited $I_{E} = 0$ Amperes which implies $I_{E} = I_{hE}$ from Equation(1) $I_{hE}\approx 0$ Amperes.

Under these conditions, Base-Collector junction acts as Reverse-Biased Diode and gives rise to a small reverse-Saturation current known as $I_{CO}$ .

when $I_{E} \neq 0$ , Total Collector current $I_{C}$ is the sum of current due to holes in the Collector and Reverse Saturation current $I_{CO}$ .

i.e, $I_{C} = I_{hC}+I_{CO}$ .

i.e, In a PNP Transistor $I_{CO}$ consists of holes moving across $J_{C}$ (from Base to Collector) that is $I_{hCO}$ and electrons crossing the junction $J_{C}$ (from Collector to Base) constitutes $I_{eCO}$ .

$I_{CO} = I_{hCO}+I_{eCO}$

i.e, $I_{E} = 0$ $\Rightarrow I_{C} = I_{CO}$ only

when $I_{E} \neq 0$ $\Rightarrow I_{C} = I_{hC}+I_{CO}$ .

$\therefore$ Total current in the transistor is given by $I_{E} = I_{B}+I_{C}$ .

$\therefore$ The general expression for Collector current is $I_{C} = \alpha I_{E}+I_{CO}$

$I_{C} =\frac{\alpha }{(1-\alpha )} I_{B}+\frac{1}{(1-\alpha )}I_{CO}$ .

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Working /Operation of NPN and PNP Transistor

NPN Transistor Working:-

For Normal operation of NPN Transistor Emitter junction J_E is Forward Biased and Collector junction J_C is Reverse Biased.

The applied Forward Biased at Emitter-Base junction injects a large number of electrons into the N-region and these electrons have enough energy to overcome the J_E junction and enter into the very thin lightly doped Base region.

Since Base is very lightly doped very few electrons recombine with the holes in the P-type Base region and constitutes a small Base current I_B in μA.

The electrons in the Emitter region are more when compared to electrons in the Collector. Only 5% (or) 1% of injected electrons combines with the holes in Base to produce I_Band remaining 95% (or0 99% of electrons diffuse into Collector region due to extremely small thickness of Base.

Since Collector junction is Reverse-Biased a strong Electro-static field develops between Base and Collector. The field immediately collects the diffused electrons which enters Collector junction and are collected by the Collector(Positive electrode).

Thus injected electrons from Emitter reaches Collector constituting a current known as $I_{E}=I_{B}+I_{C}$ Thus Emitter current is sum of Base current and Collector current. $I_{B}$ is very small in the Base region.

Current directions are always from negative to positive and Majority carriers are electrons in NPN Transistor.

NPN Transistor is preferred over PNP since the mobility of electron is more than that of hole that is electron moves faster than holes.

PNP Transistor Working:-

For Normal operation of PNP Transistor Emitter junction J_E is forward Biased and Collector junction J_C is reverse biased.

The applied FB at Emitter-Base junction injects a large number of holes in the P-type emitter region and these holes have enough energy to enter into very thin lightly doped Base region. Base is very lightly doped N-type region. Therefore very few holes combines with the Base region and constitutes a small Base current I_B (in Micro Amperes).

The holes in the Emitter region are more when compared to holes in the collector region.Only 5% or 1% of injected holes from Emitter combines with the electrons in the Base to produce I_B and remaining 95% (or) 99% of holes diffuse into Collector region due to extremely small thickness of Base.

Since Collector junction is Reverse-Biased a strong Electro-static field develops between Base and Collector. The field immediately collects the diffused holes which enters Collector junction and are collected by the Collector(negative electrode).

Thus injected holes from Emitter reaches Collector constituting a current known as $I_{E}=I_{B}+I_{C}$ , $I_{B}$ is very small in the Base region.

Majority carriers are holes in PNP Transistor.

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Working of PN-junction Diode under Forward and Reverse Bias Conditions

In order to consider the working of a diode,we shall consider the effect of forward and Reverse Bias across PN-junction.

Forward Bias:-

Forward Bias means the Positive terminal of the Battery has been connected to P-type and negative terminal to N-type in a PN-junction diode that is when an external voltage is applied to PN-junction in such a way that it cancels the barrier potential and permits the current flow such a bias is called as Forward-Bais.

Under No Bias voltage condition, Near the junction the holes moves towards the junction and electrons as well forms a region known as Depletion region, the region depleted with immobile ions .

when the applied voltage V establishes an electric field opposite to the potential barrier , as a result the width of the potential barrier is reduced as it is very small

0.3 Volts in Ge diode and

0.7 Volts in Si diode.

∴ a small voltage (V) is sufficient to completely eliminate the barrier that is the barrier is completely eliminated and the resistance at the junction becomes zero and the current flow across the diode can be explained as follows.

Now holes move towards junction simultaneously electrons since holes and electrons were repelled by the opposite terminals of the Battery, As the Battery voltage is sufficiently greater than barrier voltage electrons and holes gets sufficient energy to cross the barrier easily.

The continuous current in external circuit is due to electrons, the current in N-type material is due to movement of free electrons, when these electrons reaches the junction they combine with the holes at the junction and releases a new electron.Similarly, in the P-type region current is due to holes.

i.e, when an electron-hole combination takes place near the junction , A co-valent bond near positive terminal of the battery breaks down and it liberates an electron which moves towards positive terminal of the Battery as electron movement is towards positive terminal of the Battery this can be treated as hole movement in opposite direction.

therefore the constant movement of electrons and holes towards opposite terminals creates a high forward current in the external circuit.

PN-juction Diode in Reverse-Bias:-

When an External voltage V is applied to a PN-junction in such a way(direction) that it increases the Potential barrier is called as Reverse Bias that is Positive terminal of the Battery connected to N-type and negative terminal to P-type.

The applied voltage V acts in the Same direction to that of Potential Barrier.

that is when the PN-junction is Reverse Biased

The junction Potential Barrier width increases.
The junction offers higher resistance.
electrons and holes move away from the junction and a very small current flows through the junction because of minority carriers known as Reverse saturation current.

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Capacitance of a Co-axial cable

A Co-axial cable is a Transmission line, in which two conductors are placed co-axially and are separated by some dielectric material with dielectric constant (or) permittivity ( $\epsilon$ ).

a conductor is in the form of a cylinder with some radius, let the radius of inner conductor is ‘a’ meters and that of outer conductor be ‘b’ meters.

Now connect this co-axial conductor to a supply of ‘V’ volts , after applying ‘V’ assume positive charges are distributed on $M_{2}$ and negative charges on $M_{1}$ .

Now, a field is induced $\overrightarrow{E}$ between $M_{2}$ and $M_{1}$ because of flux lines, to find out $\overrightarrow{E}$ at any point P between these two conductors

location of P is out of the conductor $M_{2}$ an inside the conductor $M_{1}$ .

$\therefore \overrightarrow{E}_{at P} = \overrightarrow{E}_{\ due \ to \ inner \ conductor \ M_{1}}$ .

assume a cylindrical co-ordinate system $\rho ,\ \phi , \ z$ and axis of cable coincides with z-axis this is similar to a line charge distribution $\rho_{L}$ placed along the z-axis.

$\rho _{L}=\frac{Q}{L}$ .

$\therefore \overrightarrow{E}_{at P} = \frac{\rho_ {L}}{2\pi \epsilon _{o}\rho }\overrightarrow{a}_{\rho }$ .

$\therefore V = -\int_{1}^{2}\overrightarrow{E}.\overrightarrow{dl}$ .

$V = -\int_{1}^{2} \frac{\rho_ {L}}{2\pi \epsilon _{o}\rho }\overrightarrow{a}_{\rho }.(d\rho \overrightarrow{a}_{\rho }+d\phi \overrightarrow{a}_{\phi }+dz \overrightarrow{a}_{z })$ .

$V = -\int_{b}^{a} \frac{\rho_ {L}}{2\pi \epsilon _{o}\rho }\overrightarrow{a}_{\rho }.(d\rho \overrightarrow{a}_{\rho })$ .

$V = - \frac{\rho_ {L}}{2\pi \epsilon _{o} }(\ln a-\ln b)$ .

$V = \frac{\rho_ {L}}{2\pi \epsilon _{o} }(\ln b-\ln a)$ .

$V = \frac{\rho_ {L}}{2\pi \epsilon _{o} }\ln (\frac{b}{a})$ .

$V = \frac{Q}{2\pi \epsilon _{o}L }\ln (\frac{b}{a}) \ \because \ \rho _{L} = \frac{Q}{L}$ .

$\therefore C_{co-axial} =\frac{Q}{V} = \frac{2\pi \epsilon_{o}L}{\ln (\frac{b}{a})}$ .

L- length of the conductors.

b-radius of the outer conductor.

a- radius of the inner conductor.

$\epsilon$ – permittivity of the medium.

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Brewsters angle

index

Working of PN-junction Diode under Forward and Reverse Bias Conditions

In order to consider the working of a diode, we shall consider the effect of forward and Reverse Bias across PN-junction.

Forward Bias:-

Under No Bias voltage condition, Near the junction the holes moves towards the junction and electrons as well forms a region known as Depletion region, the region depleted with immobile ions .

when the applied voltage V establishes an electric field opposite to the potential barrier , as a result the width of the potential barrier is reduced as it is very small

0.3 Volts in Ge diode and

0.7 Volts in Si diode.

therefore the constant movement of electrons and holes towards opposite terminals creates a high forward current in the external circuit.

PN-juction Diode in Reverse-Bias:-

The applied voltage V acts in the Same direction to that of Potential Barrier.

that is when the PN-junction is Reverse Biased

The junction Potential Barrier width increases.
The junction offers higher resistance.
electrons and holes move away from the junction and a very small current flows through the junction because of minority carriers known as Reverse saturation current.

Working /Operation of NPN and PNP Transistor

NPN Transistor Working:-

For Normal operation of NPN Transistor Emitter junction J_E is Forward Biased and Collector junction J_C is Reverse Biased.

Since Base is very lightly doped very few electrons recombine with the holes in the P-type Base region and constitutes a small Base current I_B in μA.

Current directions are always from negative to positive and Majority carriers are electrons in NPN Transistor.

NPN Transistor is preferred over PNP since the mobility of electron is more than that of hole that is electron moves faster than holes.

PNP Transistor Working:-

For Normal operation of PNP Transistor Emitter junction J_E is forward Biased and Collector junction J_C is reverse biased.

Thus injected holes from Emitter reaches Collector constituting a current known as $I_{E}=I_{B}+I_{C}$ , $I_{B}$ is very small in the Base region.

Majority carriers are holes in PNP Transistor.

Conductivity of a Semi conductor

In a pure Semi conductor number of electrons = number of holes. Thermal agitation (increase in temperature) produces new electron-hole pairs and these electron-hole pair combines produces new charge particles.

one particle is of negative charge which is known as free electron with mobility $\mu _{n}$ another in with positive charge known as free hole with mobility $\mu _{p}$ .

two particles moves in opposite direction in an electric field $\overrightarrow{E}$ and constitutes a current.

The total current density (J) with in the semi conductor.

$\overrightarrow{J} = \overrightarrow{J_{n}} + \overrightarrow{J_{p}}$

Total conduction current density = conduction current density due to electrons + conduction current density due to holes.

$J_{n}= nq\mu _{n}E$ .

$J_{p}= pq\mu _{p}E$ .

n- number of electrons/Unit-Volume.

p-number of holes/Unit-Volume.

E- applied Electric field strength V/m.

q-charge of electron/hole $\approx 1.6X10^{-19}C.$

$J = nq\mu _{n}E +pq\mu _{p}E$ .

$J = (n\mu _n +p\mu _{p})qE$ .

$J=\sigma E$ .

where $\sigma = (n\mu _{n}+p\mu _{p})q$ is the conductivity of semi conductor.

Intrinsic Semi conductor:-

In an intrinsic semi conductor $n=p=n_{i}$

$\therefore$ conductivity $\sigma _{i}= (n_{i}\mu _{n}+ n_{i}\mu _{p})q$

$\sigma _{i}= n_{i}(\mu _{n}+ \mu _{p})q$

where $J_{i}$ is the current density in an intrinsic semi conductor $J_{i} = \sigma _{i} E$

Conductivity in N-type semi conductor:-

In N-type $n> > p$

number of electrons $> >$ number of holes

$\therefore \sigma _{N}\simeq n\mu _{n}q$

$J _{N}= n\mu _{n}q$ .

Conductivity in P-type semi conductor:-

In P-type $p> > n$

number of holes $> >$ number of electrons

$\therefore \sigma _{p} \approx p\mu _{p}q$ .

$J_{P}= p\mu _{p}q E$ .

Full Wave Rectifier

Full Wave Rectifier (FWR) contains two diodes $D_{1}$ and $D_{2}$ .

FWR converts a.c voltage into pulsating DC in two-half cycles of the applied input signal.

Here we use a Transformer, whose secondary winding has been split equally into two half waves with a common center tapped connection ‘c’.

This configuration results in each diode conducting in turn when it’s anode terminal is positive with respect to Center point ‘c’ of the Transformer.

Working of Full Wave Rectifier:-

During positive half cycle of applied i/p signal

point ‘P’ is more positive w.r.to ‘c’.
point ‘Q’ is more negative w.r.to ‘c’.

i.e, Diode $D_{1}$ is Forward Biased and $D_{2}$ is Reverse Biased , under this condition the equivalent circuit is as shown below

$\therefore V_{o} \approx V_{i} =i_{L}R_{L}$ , when there is no diode resistance.

Similarly the conditions of diodes will be reversed for the negative half cycle of i/p signal.

point ‘P’ is negative w.r.to ‘c’.
point ‘Q’ is positive w.r.to ‘c’.

i.e, Diode $D_{1}$ is Reverse Biased and $D_{2}$ is Forward Biased , under this condition the equivalent circuit is and output voltage is $V_{o} \approx i_{L}R_{L}$ .

the i/p and o/p wave forms are as shown below

FWR is advantageous compared to HWR in terms of its efficiency and ripple factor.

Ripple Factor ( $\Gamma$ ):-

$\Gamma = \frac{V_({ac})rms}{V_{dc}} = \sqrt{\frac{(V_{rms})^2}{V_{dc}}-1}$

to find out $V_{rms}$ and $V_{dc}$ of output signal

$\therefore V_{rms} = \sqrt{\frac{1}{\pi }\int_{0}^{\pi }V_{m}^{2} (\sin ^{2}\omega t ) d\omega t}$ , $\therefore V_{dc} = \frac{1}{\pi } \int_{0}^{\pi }V_{m} (\sin \omega t ) d\omega t$

$V_{rms} = \sqrt{\frac{1}{\pi }\int_{0}^{\pi }V_{m}^{2} (\frac{1-\cos 2\omega t}{2} ) d\omega t}$ , $V_{dc} = \frac{-V_{m}}{\pi } [-2 ]$

$V_{rms} = \sqrt{\frac{V_{m}^{2}}{\pi }\int_{0}^{\pi } }(\frac{1}{2}\pi )$ , $V_{dc} = \frac{2V_{m}}{\pi }$

$V_{rms} = \frac{V_{m}}{\sqrt{2}}$ .

$I_{rms} = \frac{V_{rms}}{R_{L}}=\frac{V_{m}}{\sqrt{2}R_{L}} = \frac{I_{m}}{\sqrt{2}}$ and $I_{dc} = \frac{2I_{m}}{\pi }$ .

now the ripple factor results to be $\Gamma = \sqrt{\frac{(\frac{V_{m}}{\sqrt{2}})^{2}}{(\frac{2V_{m}}{\pi })^{2}}-1}$

$\Gamma = \sqrt{\frac{V_{m}^{2}\pi ^{2}}{8V_{m}^{2}}-1}$

$\Gamma = \sqrt{\frac{\pi ^{2}}{8}-1}$

$\Gamma = 0.482$

Effect of negative feedback on Band width of an Amplifier

Let the Band Width of an amplifier without feedback is = BW. Band width of an amplifier with negative feed back is $BW_{f}= BW (1+A\beta )$ . Negative feedback increases Band width.

Proof:- Consider an amplifier with gain ‘A’

Now the frequency response of the amplifier is as shown in the figure. Frequency response curve means gain (dB) Vs frequency (Hz)

the frequency response of an amplifier consists of three regions

Low frequency region ( $< f_{1}$ -lower cut off frequency).
Mid frequency region ( between $f_{1}$ and $f_{2}$ ).
High frequency region ( the region $> f_{2}$ -upper cutoff frequency)

Gain in low- frequency region is given as $A_{vl} = \frac{A_{v}}{1-j\frac{f_{1}}{f}}---------EQN(I)$ ,

$A_{v}$ -open loop gain,

$f$ – frequency,

$f_{1}$ – lower cut off frequency, where Gain in constant region is $A_{v}$ .

Gain in High-frequency region is $A_{vh} = \frac{A_{v}}{1+j\frac{f}{f_{2}}}$ .

In low-frequency region:-

since open loop gain in low-frequency region is $A_{vl}$ and gain with feedback is $A_{vlf} = \frac{A_{vl}}{1+A_{vl}\beta }$

From EQN(I) $A_{vl} = \frac{A_{v}}{1-j\frac{f_{1}}{f}}$ after substituting $A_{vl}$ in the above equation

$A_{vlf} = \frac{\frac{A_{v}}{1-j\frac{f_{1}}{f}}}{1+\frac{A_{v}}{1-j\frac{f_{1}}{f}}\beta }$

$=\frac{A_{v}}{1-j\frac{f_{1}}{f}+A_{v}\beta }$

$= \frac{A_{v}}{1+A_{v}\beta-j\frac{f_{1}}{f} }$

Now by dividing the whole expression with $(1+A_{v}\beta )$

$A_{vlf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{\frac{1+A_{v}\beta-j\frac{f_{1}}{f}}{(1+A_{v}\beta )} }$

$A_{vlf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{1-j\frac{f_{1}}{f}\frac{1}{(1+A_{v}\beta )} }$

$A_{vlf} = \frac{A_{vf}}{1-j\frac{f_{1}^{'}}{f}}$ , where $A_{vf} = \frac{A_{v}}{1+A_{v}\beta }$ and $f_{1}^{'} = \frac{f_{1}}{1+A_{v}\beta }$

for example lower cut-off frequency $f_{1} =20 Hz$ implies $f_{1}^{'} = \frac{20}{1+A_{v}\beta }$ is decreasing with negative feedback.

In High-frequency region:-

Gain with out feed back in High frequency region is $A_{vh} = \frac{A_{v}}{1+j\frac{f}{f_{2}}}$

Now Gain with negative feed back is $A_{vhf} = \frac{A_{vh}}{1+A_{vh}\beta }$

Substituting $A_{vh}$ in the above equation

$A_{vhf} = \frac{\frac{A_{v}}{1+j\frac{f}{f_{2}}}}{1+\frac{A_{v}}{1+j\frac{f}{f_{2}}}\beta }$

$A_{vhf}=\frac{A_{v}}{1+j\frac{f}{f_{2}}+A_{v}\beta }$

$A_{vhf}= \frac{A_{v}}{1+A_{v}\beta+j\frac{f}{f_{2}} }$

Now by dividing the whole expression with $(1+A_{v}\beta )$

$A_{vhf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{\frac{1+A_{v}\beta+j\frac{f}{f_{2}}}{(1+A_{v}\beta )} }$

$A_{vhf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{1+j\frac{f}{f_{2}}\frac{1}{(1+A_{v}\beta )} }$

$A_{vhf} = \frac{A_{vf}}{1+j\frac{f}{f_{2}^{'}}}$ , where $A_{vf} = \frac{A_{v}}{1+A_{v}\beta }$ and $f_{2}^{'} = f_{2}(1+A_{v}\beta )$

for example lower cut-off frequency $f_{2} =20K Hz$ implies $f_{1}^{'} = 20 K (1+A_{v}\beta )$ is increasing with negative feedback.

In Mid-frequency region:-

Gain with out feed back is $A_{v}$

and the gain with negative feed back is $A_{vf} = \frac{A_{v}}{1+A_{v}\beta }$

With out feedback	With feedback
lower cut-off frequency is $f_{1}$	lower cut-off frequency $f_{1}^{'}= \frac{f_{1}}{1+A_{v}\beta }$ , increases
upper cut-off frequency is $f_{2}$	upper cut-off frequency is $f_{2}^{'} = f_{2}(1+A_{v}\beta )$
BW = $f_{2}-f_{1}$	$BW_{f} = f_{2}^{'}-f_{1}^{'}$ increases

Thus negative feedback decreases lower cut-off frequency and increases upper cut-off frequency.

$\therefore$ Over all gain decreases with negative feedback and Band Width increases.

Classification (or) topologies of feedback Amplifiers

There are 4 different combinations possible with negative feedback in Amplifiers as given below

Voltage-Series.
Current-Series.
Voltage-Shunt.
Current-Shunt.

The first part represents the type of sampling at the output .

i.e , Voltage- Shunt connection.
Current-Series connection.

and the second part represents the type of Mixing at the input

Series- Voltage is applied at the input.
Shunt-Current is applied at the input.

For any Amplifier circuit we require

High Gain
High Band Width
High Input Impedance
and Low Output Impedance.

Classification of feedback Amplifiers is also known as feedback Topologies.

Voltage-Series feedback Connection:-

at i/p side connection is Series and at o/p side connection used is Shunt since o/p is collected is voltage.

Series connection increases i/p impedance and Voltage at the o/p indicates a decrease in o/p impedance.

i.e, $Z_{if} = Z_{i}(1+A\beta )$ and $Z_{of} = \frac{Z_{o}}{(1+A\beta )}$ .

Current-Series feedback Connection:-

Series connection increases i/p impedance and Current at the o/p indicates an increase in o/p impedance.

i.e, $Z_{if} = Z_{i}(1+A\beta )$ and $Z_{of} = Z_{o}(1+A\beta )$ .

Voltage-Shunt feedback Connection:-

In this connection, both i/p and o/p impedance decreases .

i.e, $Z_{if} = \frac{Z_{i}}{(1+A\beta )}$ and $Z_{of} = \frac{Z_{o}}{(1+A\beta )}$ .

Current-Shunt feedback Connection:-

Shunt connection decreases i/p impedance and Current at the o/p indicates an increase in o/p impedance.

i.e, $Z_{if} = \frac{Z_{i}}{(1+A\beta )}$ and $Z_{of} = Z_{o}(1+A\beta )$ .

Effect of negative feedback on different topologies:-

Type of f/b	Voltage gain	Band Width with f/b	i/p impedance	o/p impedance
Voltage-Series	decreases	increases	increases	decreases
Current-Series	decreases	increases	increases	increases
Voltage-Shunt	decreases	increases	decreases	decreases
Current-Shunt	decreases	increases	decreases	increases

Similarly negative feedback decreases noise and harmonic distortion for all the four topologies.

p style=”text-align: justify;”>Note:- for any of the characteristics in the above table, increase ‘s shown by multiplying the original value with $(1+A\beta )$ and decrease ‘s shown by dividing with $(1+A\beta )$ .

Colpitts oscillator

Colpitt’s Oscillator is an excellent circuit and is widely used in commercial signal generators upto 100MHz.

It consists of a single-stage inverting amplifier and an LC phase shift Network.

The amplifier circuit is a self-Bias Circuit with $R_{1}$ , $R_{2}$ and parallel combination of $R_{E}$ with $C_{E}$ .

$V_{CC}$ is applied through a resistor $R_{C}$ (or) RFC choke some times. This RFC choke offers very high impedance to high frequency currents.

These capacitors $C_{C1}$ and $C_{C2}$ provides a path from Collector to Base through LC Network.

when $V_{CC}$ is switched on , a transient current is produced in the tank circuit an consequently damped oscillations are setup in the circuit.

as the transistor is operating in CE mode , it provides a phase shift of $180^{o}$ .

Therefore the over all phase shift provided by the circuit results $360^{o}$ which is an essential condition for developing oscillations.

If the feedback is adjusted so that the loop gain $A\beta =1$ then then the circuit acts as an Oscillator.

The frequency of oscillation depends on the tank circuit and is varied by gang (or) group tuning of $C_{1}$ and $C_{2}$ means $C_{1}=C_{2}$ .

working:-

The capacitors $C_{1}$ and $C_{2}$ are charged by $V_{CC}$ and are discharged through the coil $L$ setting up of oscillations with frequency

$f_{o}=\frac{1}{2\pi }\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}$ .

This amplified energy is given back to tank circuit to compensate losses.

therefore un damped oscillations results in the circuit.

Derivation for frequency of oscillations:-

chose $\left | A\beta \right |\geq 1$ for sustained oscillations.

Analysis(Qualitative):-

from the general condition for an Oscillator

$\left | A\beta \right | =1$ $\Rightarrow h_{ie}(Z_{1}+Z_{2}+Z_{3})+Z_{1}Z_{2}(1+h_{fe})+Z_{1}Z_{3}=0$ .

$h_{ie}(-\frac{j}{\omega C_{1}}-\frac{j}{\omega C_{2}}+j\omega L)+\frac{j^{2}}{\omega ^{2}C_{1}C_{2}}(1+h_{fe})-\frac{j}{\omega C_{1}}.j\omega L=0$

find the real and imaginary parts,

$-j(\frac{1}{\omega C_{1}}+\frac{1}{\omega C_{2}}-\omega L)h_{ie}-\frac{1}{\omega ^{2}C_{1}C_{2}}(1+h_{fe})+\frac{L}{C_{1}}=0$

equating imaginary part to zero $(\frac{1}{\omega C_{1}}+\frac{1}{\omega C_{2}}-\omega L)=0$ , since $h_{ie}\neq 0$ .

$\frac{\omega C_{1}+\omega C_{2}}{\omega^{2} C_{1}C_{2}}=\omega L$ .

after simplification

$\omega ^{2}=\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}$ .

by substituting $\omega =2\pi f$ results $f_{o}=\frac{1}{2\pi }\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}$ .

substituting the value of $\omega ^{2}$ in the real part gives $h_{fe}=\frac{C_{2}}{C_{1}}$ . this is the condition for sustained oscillations.

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Basic Electronics Lab Manual

Full Wave Rectifier with and without filter

Common Emitter Characteristics

FET CHARACTERISTICS

Applications of Cathode Ray Osciloscope

Applications of operational- Amplifier

Frequency-Response of CE Amplifier

Basic Electronics lab viva questions

Basic Electronics Lab Viva Questions

Experiment wise viva questions Basic Electronics Lab

PN-Junction Diode:

What is a diode?
What are the applications of diode?
Draw the symbol of PN junction diode and mark anode and cathode.
Define threshold voltage or cut in voltage of a diode.
define dynamic and static resistances of a diode.
Cut-in voltage of Si diode is…………………
Cut-in voltage of Ge diode is…………………
Draw the VI characteristics of PN junction diode.
What is forward bias of a junction diode? Explain with circuit diagram.
What is reverse bias of a junction diode? Explain with circuit diagram.
Define space charge or Depleted region of a diode?
Explain the working operation of a junction diode in forward and reverse bias.
p type material means………….. Give some examples of N-type materials
n type material means…………Give some examples of P-type materials
Ideal diode acts as short circuit under forward bias. Or draw the equivalent circuit of ideal diode in forward bias.
Ideal diode acts as open circuit under reverse bias. Or draw the equivalent circuit of ideal diode in reverse bias.
Draw the equivalent circuit of practical diode in forward bias and reverse bias.
What are the differences between avalanche and Zener break down.
What are intrinsic and extrinsic impurities?
Differentiate a conductor, semiconductor and dielectric material.
Write the diode current equation.

Zener Diode:

What is Zener diode? Define how it is different from ordinary PN diode.
What are the applications of Zener diode?
Draw the symbol of Zener diode and mark anode and cathode.
Define Zener breakdown.
What is the Zener breakdown voltage value———–.
How Zener diode is acting as a voltage regulator?
What is the cut-in voltage of a Zener diode————.
Draw the equivalent circuit of Zener diode?
Zener diode operates in reverse bias? Why?
Explain avalanche multiplication in detail.
Explain Zener multiplication in detail.
Give two differences between avalanche and Zener multiplication.

Half Wave rectifier and Full wave Rectifier (same questions are valid for FWR)

What is a rectifier? How diode is used as a rectifier?
Define ripple factor?
What is the need for filter at the output of a rectifier?
What is percentage of regulation?
Give the ripple factor values for HWR and FWR (read comparison table between HWR and FWR).
What is the efficiency of HWR and FWR? Give values.
What are the advantages of FWR over HWR?
What are the disadvantages of HWR?
What are the drawbacks of FWR?
Define PIV of a diode.
PIV of HWR and FWR are………..
why we need rectifiers?
where we use rectifiers ?
what are the differences between rectifier and converter?

CE/CB characteristics:

Draw the symbol of transistor.
Draw pnp and npn transistor symbols.
What are three configurations of a BJT?
Draw the circuits of CE, CB and CC configurations.
What is CE or Grounded Emitter configuration?
What is CB or Grounded Base configuration?
What is CC or Grounded Collector configuration?
What are input and output characteristics of BJT?
Draw input and output characteristics of CE configuration.
Draw input and output characteristics of CB configuration.
What are the applications of CE and CB transistors?
Comparison table of CE, CB and CC configurations in terms of input resistance, output resistance, voltage gain, current gain and applications.
What are the three terminals of a transistor?
Define doping levels of emitter, base and collector.
Define early effect in CB transistor.
The majority carriers in pnp transistor are………..
The majority carriers in npn transistor are………..
Why BJT is called as bipolar device?
Is BJT a uni polar or bipolar device?
What are the applications of transistors?
what is the phase difference between input and output waveforms of CE transistor( ans: 180^o)
What is the phase difference between input and output waveforms of CB transistor ( ans: 0^o or 360^o).
Why NPN transistor is preferred over PNP transistor?
Why CE is preferred over other combinations in voltage amplifiers?
Define saturation, cutoff, active regions of a transistor?
What is cutoff region of operation of a transistor?
Read about comparison tables of Si, Ge transistors for active cutoff and cutin and saturation voltages.

JFET:

What is FET or JFET?
What are the differences between BJT and FET?
Why BJT is called as Current controlled device?
Why FET is called as Voltage controlled device?
Draw the symbol of FET?
Symbols of N-channel and p channel FETs.
Differences between N- channel and P- channel FET’s?
Draw the input or output characteristics of JFET?
What are the meanings of source drain and gate?
What are the applications of FET?
What are the advantages of FET over BJT?
Draw h-parameter model of a BJT in CE/CB configurations?
Define pinchoff voltage of a FET?
Explain about linear saturation and breakdown regions of CE/CB output characteristics?
Explain about linear saturation and breakdown regions from drain or output characteristics of JFET
Define 4 h-parameters of BJT in CE configuration?
Draw the equivalent circuit of FET?
define µ, g_m , r_din FET? what is the relation between those three parameters?
define trans conductance and drain resistance of FET?
The input resistance of a JFET is very high in Mega ohms.
Working of JFET.
Applications of JFET.

Operational Amplifier:

What is Operational Amplifier?
What are the ideal characteristics of Op-amp?
What are the applications of op-amp?
What is inverting amplifier?
What is non inverting amplifier?
What is meant by open loop and closed loop op-amp?
Identify the feedback resistor in inverting and non inverting amplifiers (ans:R_f).
What are the advantages of op-amp over feedback amplifiers?
Define slew rate?
Draw circuits of integrator and differentiator using op-amp
Draw adder and subtractor circuits using op-amp.
What is virtual ground in op-amp circuits?
What is open loop gain of an ideal op amp?

Oscillators:

RC Phase Shift Oscillator:

Mention the two conditions required for oscillations in RC phase shift oscillator?
Give the formula for frequency of oscillations in RC phase shift oscillator?
The phase produced by a single RC network is RC phase shift oscillator?
RC phase shift oscillator uses positive feedback or negative feedback?
The phase produced by basic amplifier circuit in RC phase shift oscillator is?
What is the difference between damped oscillations and un damped oscillations?
What are the applications of RC phase shift oscillator?
How many resistors and capacitors are used in RC phase shift feedback network?
How the Barkhausen’s criterion is satisfied in RC phase shift oscillator
Mention the basic reason for any oscillations?
What is meant by Barkhausen’s criterion?
Audio frequency range is————
RC phase shift oscillator is ——–
Oscillator is a circuit operates on internal input power supply yes or no?
Define an oscillator?
Oscillator did not take any external input yes or no?
Type of feedback used in oscillators is——-
Radio frequency range is———–
Show that single RC section provides a phase shift of 60 degrees.
Positive feedback causes instability yes or no?
In oscillators loop gain must be————–
In oscillators the overall phase shift produced by the circuit is————
Feedback gain 𝛽 must be less than———–

COLPITTS AND HARTLEY’S OSCILLATORS:

What are the advantages of hartley’s and colpitts oscillators over RC phase shift oscillator?
Applications of Colpitts and Hartleys oscillators are—–.
Colpitts oscr uses ———- in its feedback network.
Hartleys oscr uses ———–in its feedback network.
Define mutual inductance—–
colpitts and hartleys ocsillators are used at ———— frequencies.
Give the formula for frequency of oscillations in Colpitts oscillator.
Give the formula for frequency of oscillations in Hartley’s oscillator?

SINGLE STAGE CE AMPLIFIER:

What is the phase difference between input and output waveforms of a CE amplifier?
What type of biasing is used in the given circuit?
What is the effect of emitter bypass capacitor on frequency response of a CE amplifier?
What is the effect or importance of coupling capacitor?
Why source resistance Rs is used in the input side?
What are the different regions of operation of a BJT?
The phase difference of input and output in CB amplifier?
CE amplifier is voltage or current amplifier?
Draw the equivalent h-parameter model of a CE amplifier?
Why NPN transistor is preferred over PNP transistor?
What is the effect of bypass capacitor over stability of the CE amplifier?
The CE amplifier is in voltage divider bias or in fixed bias configuration?
The values of hfe,hie,hoe and hre in CE configuration are—————-.
hfe,hie,hoe and hre are called as——————–.
What do you mean by loading effect?
What are the practical applications of single stage CE amplifier?
How do you know the amplifier is single stage CE amplifier?
Why this circuit is called RC coupled amplifier?
What is the operating point in CE amplifier from the design specifications?
Quiescent conditions means——–
D.C conditions are for———
What is the relationship between collector current and base current?
Does β and hfe are one and the same?
Why coupling capacitor Cb is connected in reverse (-,+) at the input side and Cc is connected in forward(+,-) at the output side?
How do u know a transistor is working in cut off region?
Write VBE(active),VBE(sat),VBE(cutoff) values of a Si and Ge transistors
Transistor meaning is———
Define frequency response of a CE amplifier?
Define voltage gain of a CE amplifier?
Units of gain as a ratio are———-
Units of gain in logarithim is———
Magnitude response and frequency response are one and the same say yes or no?
Define bandwidth?
What are 3 dB frequencies?
Define cutoff frequencies?
BW is approximately equal to f_H justify?
Gain is constant in————-
Semi logarithmic graph is linear or non linear graph?
Stability factor is a function of—–
In the circuit R1 and R2 are used for——–
The feedback type that Re and Ce introduces in the circuit is———–(Ans: negative feedback).
Negative feedback increases stability yes or no(Ans:yes).
CB amplifier provides more band width than CE amplifier justify?

Note: Read advantages and disadvantages and applications of every experiment.

ALL THE BEST

Prepared by P.Lakshmi Prasanna