## Classification (or) topologies of feedback Amplifiers

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

1. Voltage-Series.
2. Current-Series.
3. Voltage-Shunt.
4. 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, $Z_{if}&space;=&space;Z_{i}(1+A\beta&space;)$    and    $Z_{of}&space;=&space;\frac{Z_{o}}{(1+A\beta&space;)}$.

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

Voltage-Shunt feedback Connection:-

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

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

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

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.

Note:-  for any of the characteristics in the above table, increase ‘s shown by multiplying the original value with $(1+A\beta&space;)$ and decrease ‘s shown by dividing with $(1+A\beta&space;)$.

(No Ratings Yet)

## 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 $BW_{f}=&space;BW&space;(1+A\beta&space;)$. 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

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

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

$A_{v}$ -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 $A_{vh}&space;=&space;\frac{A_{v}}{1+j\frac{f}{f_{2}}}$  .

In low-frequency region:-

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

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

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

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

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

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

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

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

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

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

In High-frequency region:-

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

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

Substituting $A_{vh}$ in the above equation

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

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

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

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

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

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

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

for example lower cut-off frequency $f_{2}&space;=20K&space;Hz$  implies $f_{1}^{'}&space;=&space;20&space;K&space;(1+A_{v}\beta&space;)$  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 $A_{vf}&space;=&space;\frac{A_{v}}{1+A_{v}\beta&space;}$

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

(No Ratings Yet)

## 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

$\therefore&space;V_{o}&space;\approx&space;V_{i}&space;=i_{L}R_{L}$, 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 $V_{o}&space;\approx&space;i_{L}R_{L}$.

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 ($\Gamma$):-

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

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

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

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

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

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

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

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

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

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

$\Gamma&space;=&space;0.482$

(No Ratings Yet)

## 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  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.

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, $r_{i}&space;=\frac{\Delta&space;V_{EB}}{\Delta&space;I_{E}}|_{V_{CB}\approx&space;Constant}$

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

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, $f(I_{c},V_{CE})_{I_{E}&space;=&space;Constant}$

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, $I_{c}\approx&space;I_{E}$

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.

$I_{c}\neq&space;0$ 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

(No Ratings Yet)

## 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:-

1. There is less chance for recombination with in the base region. Hence $\alpha$ increases with increasing $\left&space;|&space;V_{CB}&space;\right&space;|$.
2. The charge gradient is increased with in the base and consequently, the current of minority carriers injected across the emitter junction increases.
3. 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.

(No Ratings Yet)

## 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 $F_{E}&space;=&space;F_{B}$

$qE&space;=&space;Bqv_{d}------EQN(I)$, 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 $J&space;=&space;\rho&space;v_{d}------------EQN(III)$

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

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 $E&space;d&space;=&space;Bd&space;v_{d}$

$V_{H}&space;=&space;Bd&space;v_{d}$

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

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

$V_{H}&space;=&space;\frac{BI}{\rho&space;W}$

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

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

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 $\rho&space;=&space;nq$  or $\rho&space;=pq$.
• To find out mobilities $\mu&space;_{n}$ and $\mu&space;_{p}$ using the equation $\mu&space;=\sigma&space;R_{H}$.
• 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.

(No Ratings Yet)

## 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 $v_{d}=&space;\mu&space;E$ 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 $J_{n}&space;=&space;qn\mu&space;_{n}E$ atoms/Cm2  and the Drift current density due to free holes $J_{p}&space;=&space;qp\mu&space;_{p}E$ 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  $J_{p}&space;=&space;-q&space;D_{p}\frac{dp}{dx}$    A/Cm2 .

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

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

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

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

(No Ratings Yet)

## 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 $\mu&space;_{n}$ another in with positive charge known as free hole with mobility $\mu&space;_{p}$.

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.

$\overrightarrow{J}&space;=&space;\overrightarrow{J_{n}}&space;+&space;\overrightarrow{J_{p}}$

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

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

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

n- number of electrons/Unit-Volume.

p-number of holes/Unit-Volume.

E- applied Electric field strength V/m.

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

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

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

$J=\sigma&space;E$.

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

Intrinsic Semi conductor:-

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

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

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

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

Conductivity in N-type semi conductor:-

In N-type $n>&space;>&space;p$

number of electrons $>&space;>$ number of holes

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

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

Conductivity in P-type semi conductor:-

In P-type $p>&space;>&space;n$

number of holes $>&space;>$ number of electrons

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

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

(No Ratings Yet)

## 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 $I_{E}$,  $I_{E}&space;=&space;I_{hE}+I_{eE}$.

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 $I_{E}&space;\simeq&space;I_{hE}-----------Equation(1)$

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, $I_{B}&space;=&space;I_{hE}-I_{hC}----------------Equation(2)$.

If emitter were open circuited $I_{E}&space;=&space;0$ Amperes which implies  $I_{E}&space;=&space;I_{hE}$ from Equation(1) $I_{hE}\approx&space;0$ 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 $I_{E}&space;\neq&space;0$  , Total Collector current  $I_{C}$ is the sum of current due to holes in the Collector and Reverse Saturation current $I_{CO}$.

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

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 $I_{eCO}$.

$I_{CO}&space;=&space;I_{hCO}+I_{eCO}$

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

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

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

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

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

(No Ratings Yet)

## Working /Operation of NPN and PNP Transistor

NPN Transistor Working:-

For Normal operation of NPN Transistor Emitter junction JE is Forward Biased and Collector junction JC 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 JE 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 IB 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 Iand 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 JE is forward Biased and Collector junction JC 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 IB (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 IB 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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## 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% $V_{dc}$ $\frac{V_{m}}{\pi&space;}$ $\frac{2V_{m}}{\pi&space;}$ $\frac{2V_{m}}{\pi&space;}$ $I_{dc}$ $\frac{I_{m}}{\pi&space;}$ $\frac{2I_{m}}{\pi&space;}$ $\frac{2I_{m}}{\pi&space;}$ Output RMS voltage $\frac{V_{m}}{2}$ $\frac{V_{m}}{\sqrt{2}}$ $\frac{V_{m}}{\sqrt{2}}$ Average current Idc $I_{dc}$ $\frac{I_{dc}}{{2}}$ $\frac{I_{dc}}{{2}}$ 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 $\sqrt{2}$ $\sqrt{2}$
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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

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

I- Diode current

Io– 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  = 1.6X10-19 C.

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

## Mass Action law

Mass-Action law:-

Under thermal equilibrium for any semi conductor the product of number of holes and the no. of electrons is constant( and is independent of the amount of donor and acceptor impurity doping)

$\large&space;np&space;=&space;n_{i}^{2}$

n- electron concentration (or) number of electrons

p- hole concentration (or) number of holes

ni is the intrinsic carrier concentration

If a pure semi conductor is doped with N-type impurities, the no. of electrons in the conduction band(CB) increases above a level and no. of holes decreases below a level in the valance band (VB).

Similarly the addition of P-type impurities to a pure semi conductor increases holes in the valance band(VB) above a level and decreases the no. of electrons in conduction band(CB) below a level .

Charge Densities in N-type Semi conductor:-

Under equilibrium in a pure-Semi conductor $np=n_{i}^2$

when an N-type impurity is added then in N-type Semi-conductor $n_{N}p_{N}=n_{i}^2$ , the suffix N represents type of Semi conductor.

Number of electrons in N-type semi conductor= Number of donor atoms added+ number of holes in N-type semi conductor.

i.e, $n_{N}=&space;N_{D}+p_{N}$

since number of holes are very less in N-type material $n_{N}=&space;N_{D}$ ($\because&space;p_{N}$ is negligible)

$p_{N}&space;=&space;\frac{n_{i}^{2}}{n_{N}}$

$p_{N}&space;=&space;\frac{n_{i}^{2}}{N_{D}}$

$n_{i}$ – intrinsic  carrier concentration.

$p_{N}$ – number of hole concentration in N-type material.

$n_{N}$– number of electrons in N-type Semi conductor/ number of donor atoms.

Charge Densities in P-type Semi conductor:-

Under equilibrium in a pure-Semi conductor $np=n_{i}^2$

when an P-type impurity is added then in P-type Semi-conductor $n_{P}p_{P}=n_{i}^2$ , the suffix P represents type of Semi conductor.

Number of holes in P-type semi conductor= Number of acceptor atoms added+ number of electrons in P-type semi conductor.

i.e, $p_{P}=&space;N_{A}+n_{P}$

since number of electrons are very less in P-type material $p_{P}=&space;N_{A}$ ($\because&space;n_{P}$ is negligible)

$n_{P}&space;=&space;\frac{n_{i}^{2}}{p_{P}}$

$n_{P}&space;=&space;\frac{n_{i}^{2}}{N_{A}}$

$n_{i}$ – intrinsic  carrier concentration.

$n_{P}$ – number of electron concentration in P-type material.

$p_{P}$– number of holes in P-type Semi conductor/ number of acceptor atoms.

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## Crystal Oscillator

### Crystal Oscillator:-

In LC oscillators the frequency of oscillation fo depends on the tank circuit parameters  L & C, whereas L & C values change with respect to time, temperature, aging etc. Therefore fo 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 fi = fo 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 Cm.

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

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

$\omega&space;L&space;=\frac{1}{\omega&space;C}$

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

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

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

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

$f_{p}=\frac{1}{2\pi&space;}\sqrt{\frac{1}{L}(\frac{1}{c}+\frac{1}{C_{m}})}$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 fp. BJT offers a phase shift of 180o further 180o is produced by the capacitor voltage. Oscillations are possible only through fp, which provides stable oscillations.

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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                ($\approx&space;100\Omega$) 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               ( $\approx&space;131$) 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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## Basic Electronics Important Questions Unit 5 (FAQ)

UNIT-5

Data Acquisition systems: Study of transducer (LVDT, Strain gauge, Temperature, Force).

Photo Electric Devices and Industrial Devices: Photo diode, Photo Transistor, LED, LCD, SCR, UJT Construction and Characteristics only.

Display Systems:  Constructional details of C.R.O and Applications.

ASSIGNMENT- UNIT5

1. Explain with neat sketch, the working principle of LVDT.
2. Give the applications of CRO. Explain the Block diagram of Cathode ray tube in detail. 10M.
3. Write a short note on LED and LCD.

1. Distinguish between photodiode and LED.
2. Why is SCR known as negative resistance device?
3. What are the different types of transducers used for the measurement of temperature. Explain the principle of any of these.
4. Draw the symbol of a SCR and VI characteristics .explain its principle of operation.
5. What is meant by sensitivity in a CRO and explain the necessity of a saw tooth generation of a CRO.
6. What are the important characteristics of an LCD?
7. What is an LVDT? By means of a neat sketch explain how a LVDT is used in measurements.
8. Give the block diagram of a CRO, Explaining the importance of each block.
9. Explain about an instrumentation amplifier.
10. Write a note on UJT.
11. How are DIAC and TRIAC different from operational point of view.
12. What is a transducer?
13. List out the applications of a SCR.
14. Explain the principle involved in a strain gauge.
15. Draw the symbols of SCR, TRIAC, DIAC and UJT.

1. Explain the working of Cathode Ray Tube of CRO with neat block diagram in detail.
2. What are the applications of CRO?
3. Differentiate Photo Diode and Photo Transistor in detail in terms of operation and applications.
4. Differentiate LED and LCD in detail in terms of operation and applications.
5. Explain the working of SCR and draw the characteristics of it.
6. Explain the working of UJT and draw the characteristics of it.

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## Basic Electronics Important questions units 4 (FAQ)

Unit-4:  Operational Amplifiers – Introduction to OP Amp, characteristics and applications – Inverting and Non-inverting Amplifiers, Summer, Integrator, Differentiator, Instrumentation Amplifier.

Digital System: Basic Logic Gates, Half, Full Adder and Subtractors.

ASSIGNMENT- UNIT4

1. What are the four basic building blocks of Op-Amp explain each block in detail. 10M.
2. Write in detail about Instrumentation amplifier.
3. Design and realize Full Subtractor (Give the necessary expression and Truth table and Circuit diagram)

1. What are the ideal characteristics of OP-AMP?
2. What is an operational amplifier? Mention some of its applications.
3. A 5mV, 1 kHz sinusoidal signal is applied to the input of an OP-amp integrator for which R=100KΩ and C=1µF.Find the output voltage.
4. What are the different parameters of an OP-amp? Sketch the circuit of a summer using OP-amp 741to get Vo= -(-V1+2V2-3V3).
5. Write a note on basic logic gates.
6. Draw the truth table of an EX-OR gate.
7. State and prove De-Morgan’s theorems. Discuss the working of half adder and full adder and give their truth tables.
8. Write a note on universal gates.
9. Realize OR gate using NAND gates.
10. What is half Subtractor? Realize full Subtractor using using NAND gate only.
11. What is an op-amp? List the four building blocks of an op-amp.
12. Draw the schematic symbol of an op-amp in detail. Write the expression for output voltage of an op-amp?
13. What are the applications of op-amp?
14. What are the advantages of op-amp?
15. Why op-amp is named as op-amp?
16. What are universal gates and why those are called as universal gates?
17. Draw EX-OR and EX-NOR gates and give the truth tables of each gate.
18. Name some basic logic gates draw their symbols and give the corresponding truth tables.
19. Draw half adder and half Subtractor.

1. Draw and explain block diagram of op-amp in detail. Or what are the basic building blocks of an op-amp explain in detail.
4. Full Subtractor/ Half Subtractor.
5. Explain the operation of inverting and non-inverting amplifier circuits using op-amp and calculate gain of each circuit in closed loop configuration.
6. Define common mode rejection ratio (CMRR).a differential amplifier has a differential mode gain of 100 and common mode gain of 0.01 then find CMRR in dB?
7. Write in detail about instrumentation amplifier.
8. Explain the operation of the following circuits of op-amp in open loop configuration.    a. Non –inverting amplifier.        b.   Inverting amplifier  c.  Differential amplifier.
9. What are the characteristics of practical op-amp?
10. Define slew rate of op amp? Problem on calculation of Slew rate.
11. Explain how op-amp is used as  Summer and difference amplifier or Subtractor.

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## Basic Electronics Important Questions Unit 3 (FAQ)

UNIT – 3

Unit-3:  Feedback concepts: Properties of negative feedback amplifiers, Classification, Parameters. Oscillators: Barkhausen’s Criterion, LC Type and RC type oscillators and crystal oscillator. (Qualitative treatment only)

ASSIGNMENT- UNIT3

1. Draw the basic block diagram of feedback amplifier and explain each block in detail. 10M.
2. What is the effect of negative feedback on input and output impedances of a voltage series feedback amplifier? Give necessary expressions.
3. Write a short note about RC Phase Shift Oscillator.

1. What is meant by feedback?
2. What is tank circuit? How it is used in oscillators?
3. State Barkhausen’s criterion.
4. Differentiate positive feedback with negative feedback.
5. If open loop voltage gain of feedback amplifier is 200 and feedback factor is 1/5 then find
6. Gain with negative feedback.
7. BW with negative feedback if open loop BW is 100 KHz.
8. What are the advantages of negative feedback?
9. What is the effect of negative feedback on bandwidth?
10. Explain how gain stability is improved by negative feedback?
11. Draw block diagrams of voltage- series and current- shunt feedback topologies?
12. What are four topologies of feedback amplifiers?
13. What is positive feedback .Give the advantage of positive feedback?
14. Write the definition of oscillator?
15. Differentiate oscillators with the frequency of operation?
16. Draw under damped over damped and un-damped oscillatory signals.
17. .In the forward path of an oscillator, two amplifiers of equal gain are cascaded. If the feedback path transfer function is 1/81, find the gain of each amplifier.

1. What is the effect of negative feedback on input and output impedances in voltage-series feedback amplifier give necessary expressions?
2. Derive the general equation for LC oscillator circuits used in oscillators.
3. Distinguish between positive feedback and negative feedback in amplifiers. State the advantages of negative feedback in amplifiers.
4. Draw the circuit diagram of Hartley’s oscillator and explain its operation in detail. Derive the expression for frequency of oscillation.
5. Draw the circuit diagram of Colpitt’s oscillator and explain its operation in detail. Derive the expression for frequency of oscillation.
6. What is the effect of negative feedback on Bandwidth give explanation with necessary equations?
7. Write in detail about four topologies of feedback amplifiers and compare the topologies in terms of gain, BW, input impedance and output impedance.
8. Write the working operation of the following circuits and derive the expression of frequency of oscillation. i. RC-Phase shift oscillator. ii. Wein-bridge oscillator. iii. Crystal oscillator.
9. Draw a neat circuit diagram of an RC phase shift oscillator using BJT and explain its working principle
10. A Hartley oscillator is designed with L1=2mH, L2=20mH, and a variable capacitance. Determine the range of capacitance value, if the frequency of oscillation is varied between 950 kHz and 2050 kHz.

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