Fundamental of Electronics Engineering (Lecture 10)

Hello everyone. So, in previous class I was pitching you about rectifier circuit, then

the circuit. Now, in previous class you have seen suppose input, AC voltage, if it is

given to rectifier circuit, then the output of this rectifier circuit, it is consecutive DC.

Now, the ripple can be removed. In positive DC both AC as well as easy components, both

are present. So, in order to remove the ripple, we are again connecting one filter circuit. So, the

output of this filter circuit, it is DC in nature. So, here the output of this filter circuit,

it is very less ripple, very less ripple. So, here AC component is very nice or is removed. So,

how AC component is removed, AC component from this per second DC, it is removed by using filter

circuit. This is the DC output of the filter circuit. So, this DC output voltage, it varies, it varies

with this input, it input AC voltage as well as the value is in the load register. So, we can write

this filter circuit. Suppose this is R L and we are taking DC out because R L. Now, the

voltage, here is DC in nature, that is ripple free, very less ripple is present in the output of

this filter circuit. So, this DC is produced. Now, this DC, value of this DC voltage, this

output DC voltage, it varies with the variation in the AC input voltage as well as with the

AC in R L. But nowadays, we want a fixed, this is a plane. So, how we can achieve of

it's, this is supply. Thicks, this is supply in this V, V naught, this V output, it should not

vary with the changing output load or with the, with the variation in the AC input voltage. So, how

this can be achieved, it can be achieved by using voltage regulator. So, we have to connect

one voltage regulator. So, the output of this voltage regulator is always fixed. So, this

V output, it will not depend upon the load R with the change in R L as well as with the

variation in AC input voltage. So, this is the use of voltage regulator. So, nowadays, this fixed

DC supply is actually used in many applications. Means, the, this will not, it should not

change with the variation in load register and with the variation in AC input voltage. So, we will

study about voltage regulator in this class. So, before studying voltage regulation, we will study

about Z and R L. So, we have studied earlier about T n Johnson L. So, Z and R L, it is

similar to T n Johnson L, but the percent risk of operating atoms here, the percent risk

difference, percent risk in or broken level, broken level of infinity atom, infinity is very large.

Okay, Z and R L, it is also of T n Johnson L, it is also a Christian diode, but the concentration

or doping level of the infinity is very very high in case of Z and R L. Okay. So, this is a spatial purpose

that Z and R L, it is of a spatial purpose, spatial, means Z and R L, or the spatial

Christian diode, that Z and R L works in reverse bias condition. See, it is having application

in reverse bias condition, okay. So, we will see its characteristics in forward bias as well as

reverse bias case, okay. So, Z and R L, we can represent Z and R L like this,

this is a node, and this is catod, this is the symbol of Z and R L, okay. And the symbol of

normal phi in Jensen diode, it is like this, this is psi and psi, this is a node,

and this is catod, in case of normal T n Jensen diode, in case of normal T n Jensen diode,

this this is bar, okay. And this bar, in case of Z and R L, it is replaced with this symbol,

of Z, okay, let this Z, okay. So, this is the symbol of Z and R L, then we will see the V and

characteristics, suppose this Z and R L, if it is met up of silicon, semi-conductor, semi-conductor,

semi-conductor material, then it is forward characteristic, will be just like silicon,

T n Jensen diode, V f, this is V r, reverse bias, reverse bias, this is forward bias,

what is, this is I reverse side, and this is forward side, okay. Now,

okay, and this is the Z, as V Z, okay, and this is become voltage, okay. So, the V i characteristics

of a Z and R L diode, it is just similar as T n Jensen diode in forward bias case, okay. In reverse bias case,

see if V are increasing, if we are increasing the reverse voltage, then after this return voltage,

that is this is basic, it is represented as the return voltage. So, after the return voltage,

of sub-interest in current is observed. So, this is the sub-interest current, okay. So, Z and R L is Z diode,

it is a really used in reverse bias, why is used Z and R L in reverse bias is for making Z and R L as

being regulator, okay. So, this is the V i characteristics of Z and R L, V i characteristics,

of a Z and R, okay. Then, then we will study about, we will study about reverse bias,

break on mechanism, okay. See, a Z and R L is operating, it operates in reverse bias condition.

So, in reverse bias condition, break on occurs due to two phenomenon, first one is due to

effect and second is due to Z and R effect. So, due to these two effects due to these two effects,

break down occurs in any normal danger zone that are all Z and R, due to these two effects,

okay. So, after break down a sub-interest in current is observed in the reverse bias condition.

So, first of all, we will study about average effect,

suppose a p-interest in power, suppose this is p-side and this is n-side, okay and,

so in p-side, the majority carriers are holds in n-side, the majority carriers are electrons,

okay and say, this is depletion layer, this is the width of the depletion layer, okay. Now,

suppose due to the increase in temperature, magnetic carriers say electrons in this p-side

and holds in this n-side or in the depletion layer, okay. It, like this this is p-side and it is connected

to negative potential of the boundary, okay. So, this is in reverse bias, okay. So, with the increase

in, with the increase in temperature or at room temperature, some magnetic carriers, these are

magnetic carriers.

magnetic carriers suppose electrons and this is hold, these are x n-due to the increase in temperature,

okay. So, with the increase in temperature, some magnetic carriers are x n-due in the

the crystal layer ok. Now, with due to the reverse bath that is negative terminal of the

battery is connected to P to P type and positive terminal of the battery is connected to

n type. Now, due to this due to this the electrons electrons which are magnetic

area in in the crystal layer ok. So, it will be attracted it will be attracted towards the

positive terminal of the battery ok. So, it will gain a very high very high

energy. This electron will gain a very high amount of energy and this electron it will it will

will gain a very high amount of energy. So, this electron it will it will move very fast and it will

it will break the this this electron which achieves a very high amount of energy. So, this

electron it will it will move very fast and it will break it will break this electron balance

electron from the from the from the semiconductor atom and this electron it will be released free. So,

it will become free electron ok. So, once this electron is removed from the from the this balance

that. So, now this is free. So, again this free electron it will it will it will collide with another

another semiconductor atom ok and again it will it will release the balance electron

balance electron and make this balance electron as free electron. So, in this way

or average or a group of free carriers, free charge carriers are available ok. So, into the presence

of or into the average of free carriers due to the presence of a loves number of free carriers in

way the high current is also of just after the average electron. So, this is due to

average effect ok. So, we so average effect it occurs generally for less though for less

though semiconductor. When the doping level is less then we will then we are seeing

average effect in that die ok. So, I will do this average effect if very large amount of current

is observed just after the vector ok. Then second effect is zenn effect

zenn effect ok. So, because of zenn effect it will last amount of current is observed just after

the breakdown due to zenn breakdown effect. So, we will see again

this sub increase in k i suppose this is v r and this is v l this is i l and this is i r. So, just suppose

this is v z ok v z zenn breakdown voltage just after this just at this voltage a sub increase in

k i is also before this i i r it is approximately equal to it is equal to 0 it is in the range of

micro m b r or p k m b r h. So, it is approximately equal to 0 before this end but after after

the breakdown a sub increase in current is observed. So, this is due to ideal average effect or zenn effect.

So, we are going to be zenn effect. So, how this current increases subtly. So, due to zenn effect

we are saying for say and zenn effect this focus due to have we go say we can have that right

level of double concentration it is very high then we are saying that the breakdown it occurs

due to zenn breakdown effect. Suppose this is k side and again this is m side,

majority carriers are poles and here the majority carriers are electrons ok. Now,

suppose this is it is connected to can be that is the direction of the electric field is from positive

to negative and here that this is the direction of the electric field ok. Now, this zenn effect

it occurs in highly dope sending conductor. So, due to heavy dope say due to heavy dope the

depth of the decryption result the depth of the decryption result it decryzes ok. So,

the depth with the decryption the depth this depth of this distance that is the decryption result

this zenn thickness it decryzes it decryzes with the

depth increase gain bumping level. So, this reason this the distance the this this distance

the decryption result the decryption result it decryzes it decryzes due to increase in dopeing level

in case of semiconductor. So, in zennadow we are seeing that the looping level is very high

looping level is very high in case of zennadow ok. So, the thickness of the decryption result it decryzes.

So, with the decryption in this decryption layer decryption layer thickness this is decryption layer thickness

of very high elected field is observed and the direction of this electric field and the direction

of this electric field both are in the same direction. So, a very high electric field is created

when or reverse bias condition is achieved. When we are connecting this external

battery to this type of semiconductor then then a very high elected field is created across

this junction across this reason. So, because of this very high electric field it is very high

it is in the range of 10 to the power 7 volt per meter ok. So, if this this range of electric

field is created here then it it writes the power in bond due to very high due to very high

electric field due to very high electric field that is the it is in the range of 10 to the power 7

volt per meter the current bond it is broken due to this very high electric field. So, due to the

break in the overing bond, opening bond due to the break in the current bond a last amount of

free carriers a last amount of free carriers that is free charge carriers they are created ok. So,

with this last amount of free charge carriers a very high electric current is observed just

up to the break down. So, this is the mechanism this is Z and R effect because of this Z and R effect

a very high reverse current is also just up to the break down ok. So, if the electric field this electric

field and this electric field because of this electric field is the electric field is in the range

of 10 to the power 7 volt per meter that it will break the overing bond over and bond suppose this is

silicon atom silicon atoms see this is it is somewhere by 4 valence electrons again this

silicon atom it is already by 4 silicon 4 valence electrons ok. Now, this 2 this 2 and this 2

and again the silicon is 2 this this variance electrons are say these 2 valence electrons are

say to make a overing bond ok. But with the when E is approximately in the range of

10 to the power 7 volt per meter then this will break this will break this problem bond and make

the charge carriers free. So, due to this a very large current is observed in the reverse gas condition

after the break down the break down the break down voltage say it is busy. So, if we are is better than

or equal to design then a very large amount of reverse current very large value of reverse current

will be seen. So, this is due to Z electric then we will see the application of Zeta

application of Zeta. So, a Zeta diode can be used to make a regulator ok.

A Zeta diode is generally used for making a regulator circuit. So, what is the circuit atom for regulator?

Suppose this is unregulated this is the power supply ok. Then it is it is connected to

Zeta diode. So, Zeta diode and Zeta diode and Zeta diode ok like this. So, this is the circuit

diagram of a regulator circuit. So, I am directly power supply it can be replaced with this voltage.

Suppose V I V I this is unregulated DC voltage R is is series register this is series register

and this is Zeta diode it is it is specific as V Z and V Z and V Z and V Z and V Z and V Z this is R and

across R and V L output it is this this V L is regulated voltage. So, this is the this is the

diagram of a regulator circuit ok V I this is unregulated DC voltage and V L is regulated

V Z voltage this R S is series register register resistance and this is this is Zeta diode and

it is specified by Z and V Z is the this decon voltage it can be Z hand it with the Z the maximum power

that it can be stand ok. Now see this is the circuit diagram of a regulator circuit in this regulator

circuit this is Zeta diode ok. So, Zeta this is Zeta diode this Zeta diode it is always connected it is

reverse pass with this V I this is unregulated this voltage is unregulated DC voltage and the

output of a regulator circuit it is it is V L it is regulated voltage ok. So, the main aim of a regulator

circuit is to produce a regulated voltage by using Zeta and this Zeta diode it is always in reverse pass

condition why how it is in reverse pass condition. So, this this ah positive terminal it is always

connected to cathode this is cathode and this side is unknown ok. So, it is connected like this. So,

Zeta diode in reverse pass condition ok we are seeing this Zeta diode in reverse pass condition

cathode only wall came stabilizer or regulator now in order to analyze this circuit first of all

we have to we have to remove this we have to remove this Zeta diode and we have to

measure the voltage across 0.80 first of all how we are analyzing we are analyzing this regulator

circuit ok. So, in order to analyze this circuit first of all we have to remove remove this Zeta diode

and after removal we have to measure this voltage what is across A V. So, what is across A V if it is

equal to voltage across L this this voltage across R L and voltage across R L it is equal to

being L ok. So, it is equal to R L divided by R L plus R this R into this V i ok. Now,

if V A V which is equal to V L if it is greater than or equal to V Z that is specified V Z and V Z

this V Z is the vector voltage. So, if V A V this voltage across A V and if both are equal V

and both are equal if V A V is equal to V L and if it is greater than or equal to V Z then

we can say that the diode that is Zeta diode is in on condition what is mean by on condition

if this calculated if this calculated V L if it is more than or equal to this V Z then we can

say that this Zeta diode it is in on condition. So, it is of on condition this diode it will be

raised with V Z that is constant voltage source. Now, this is replaced with a constant voltage source

ok by V Z if it is the diode is in on condition this is the first case and second case is

if V A V second case is if V A V is less than V Z then what will happen then

that is the Zeta diode is in of state then the Zeta diode it is in of state. So, we can remove this

this is A V point A V if it cross A V if V A V V A V what is V A V V V V V V V V V V V is equal to V L

V L. So, if it is less than V Z this is the Zeta diode that is the break on voltage of the

Zeta. If V A V is less than the break on voltage of the Zeta diode then in that case Zeta diode

behaves as as it is in of state ok. So, it can be replaced with Opel. So, how you can say that this is in

of state you can from the V A characteristics suppose this is V R and this is I R ok in of

the state suppose this is V Z. So, when V I is more than when this V V S is more than V Z

then only it will be have as a as a constant voltage source otherwise otherwise we have seen that

at this place when V L is less than V Z suppose at this point at this point very less amount of

current in state. So, I is approximately equal to 0. So, in that case we can remove this diode.

So, this is in of state when V A V is less than V Z ok. So, in this way we can calculate this V A V

So, it is replaced with like this this is V Z this is in on state this is in on state ok.

Now, in on state in on state how we can write expressed I R I R it is equal to I Z plus I L

during on state during on state when V L is greater than or equal to V Z this is the

condition for on state of the diode of the generator then in that case I R it is equal to I Z plus I L

I L is 0 across R L I Z is the diode current this diode current I R is the current 0 across the

series resistor. So, you can write I Z is equal to I R minus I L now how we can express I R I R it is equal to

V R divided by R that is voltage drop across this R divided by R. So, V R how you can express V R it

is equal to V I minus V L or V Z V L or V Z both are same divided by R and how we can express I L

I L can write I R it is equal to V L divided by R L ok and how we can express power this

equation across the diode it is equal to I Z into V Z I Z into V Z and this power this

equation it should be less than V Z M this is it that is power this equation across the diode it is

equal to I Z into V Z and it should be less than V Z M this is the rating of this diode that is

the power this is V Z because this diode should be less than the maximum power rating of the Z and

diode ok now we will say this example this is the circuit diagram of a regulator circuit. So,

this is the regulator circuit ok now here V I is 16 volt this is an regulator V I under

the regulator DC voltage and this is series resistance series resistance is 1 kilo ohm and

and V Z of this Z and I L it is 10 volt and is at M that is the maximum power that it can withstand

it is equal to 30 milli watt and this is L ok. So, we have to determine V L V R I Z and V Z under

two conditions when R is equal to 1.2 k n kilo ohm and V when R is equal to 3 kilo ohm. So,

first of all we will see when R L is equal to 1.2 kilo ohm then in that case whatever

that is of V L V R I Z and V Z. So, just now we told you that till we had discussed that first of all

we have to remove this we have to remove this in order to in order to obtain in order to obtain

voltage across A V we have to remove this Z and at that only we can determine the state. So, first of all

we have to determine the state of the power determining the state of the Z and at that.

So, how this can be achieved how the state of the Z and at that can be known. So, V A V it is equal to

how it can be calculated it is equal to R L divided by R L plus this series register into V I that is

16 volt. So, R L is equal to say in this case it is equal to 1.2 kilo ohm divided by 1.2 plus 1 that is

2.2 kilo ohm into multiplied by 64. So, finally equal to 8.73 this is equal to 8.17 volt.

Now, we are getting V A V it is equal to 8.73 8.73 volt. Now, this V A V voltage across A V which is equal

to V L that is what is across R L again it is less than this is less than V Z that is the

break on voltage of the power then under this case you can see the drop V I drop this is I this is V R

and this is I R I F and for our voltage can see the drop.

This is V A V V R giving V A V is at this point and V Z is at this point.

So, in this case we have not achieved V V V L equal to or voltage Z. So, it is not in

state but it is in of state. So, if it is in of state I will be equal to 0 ok. So, we have to

represent this in of state. So, in of state we can represent like this. So, it will be equal to

it will be open it will be open. So, the diode the diode that is Z and a diode

is in of state ok. So, how we can determine V as V as V as V is equal to just we have calculated

it is equal to V A V it is equal to 8.7V it is equal to 8.7V then how we can calculate V R V R it is equal

to V I minus V L. So, it is equal to 16 more minus V L it is 8.7V volt 7.7V volt ok. Then we have to

determine I Z what will be I Z see this is open. So, I Z will be equal to 0. So, I Z it is equal

to 0 and V L ok. Now, volt 1 is how to calculate V Z now if I Z is 0. So, basically V Z it is also equal

to 0 because we said it is equal to I Z into volt is volt this V A V V A V V A V that is equal to

V L. So, basically it is equal to I Z into V L. So, it is equal to 0 multiplied by V L. So, it is equal to 0

1 ok. So, when R L is equal to 1.2 0 when R is equal to 1.2 0 then in that is this Z and I

have it is in of state of state means I Z is equal to 0, V Z is equal to 0 ok and V L V L is the volt is

of a cross this R L. So, volt is not because R L is equal to R L divided by R L plus R into V L and

volt is not across this R it is equal to this volt L minus this volt L ok. So, in this way we can calculate

then we have to calculate all the values when R L is equal to 3 kilo.

When R L is equal to 3 kilo volt we have to determine again G L V R I Z and T Z ok. So, when R L is equal to 3 kilo

volt. So, again we have to determine V A V ok. Why we have determining V A V in order to determine

the state determine the state of the Z and I have we are determining the state of the Z and I have

ok. So, V A V is equal to R L divided by R L is equal to 3 kilo volt ok. So, R L is equal to R L divided by R L

plus R into this V I. So, it is equal to 3 divided by 3 plus 1 that is 4 into 6 3. So, it is equal

to 12 volt. Now V A V which is equal to V L this is greater than this is greater than this

is equal to 12 volt and 12 volt is greater than 10 volt ok that is it is greater than this Z. So,

the diode is in on state. So, we can write. So, we can write the diode is in on state.

So, if it is in on state then we can replace this with this.

So, we can replace this diode with this Z this is 10 volt ok and this is the direction of I Z.

So, we have to determine V L V L it is equal to it is equal to V Z because the diode is in on condition

V L this is V L V L across R L it is V L. So, V L is equal to V Z because the diode is in on condition.

So, it is equal to 10 volt ok. Then second we have to say how to calculate V L?

V R it is equal to this V I minus V L. So, it is equal to 16 volt minus V L is 10 volt. So, it is equal

to 6 volt. Then third is how to calculate I Z? Third one is we have to determine I Z and I L ok.

So, first of all we can calculate I L. I L it is equal to V L divided by R L. So, V L is 10 volt I is V

kilo ohm ok 10 volt divided by 6 kilo ohm it is equal to 3.6V V L ok. So, in this way we can

determine I L ok. Now, I have I have we can determine I L I R it is equal to V R divided by this

R R. So, it is equal to V R is 6 volt divided by 1 kilo ohm. So, how much it is equal to it is equal to

6V V L. So, how we can determine I Z? I say as I R it is equal to I Z plus I L. So, from this equation

we can determine I Z it is equal to I R minus I L. So, I R I L is 6V L minus 3.3.3V L. So, I R it is equal

to 2.67V L 6V L 6V L minus 3.6V L L it is equal to 2.67V L L. Then we can calculate P Z it is equal

to I Z into B Z into B Z. I Z is I Z we have calculated it is 2.67V L L into B Z it is equal to 10 volt.

So, in this way it is equal to 22 it is equal to 26.7V L L which is less than 13V L L.