Narrow Band Pass Filter

 Narrow Band Pass Filter

Ans. 

                                                  

                                                  

Here , in this topic we are going to learn Narrow Band Pass Filter ( NBPF ) . So , the Narrow Band Pass Filter is also known by the other name called as or we can say that known as multiple feedback filter .

             So , question arises here , why we called or known by the name " multiple feedback filter " .

              Because this Narrow Band Pass Filter has two feedback path , Hence it is called by the name is multiple feedback path .

               In this Narrow Band Pass Filter , there is only one active component is taken in use is operational amplifier ( Op-amp ) in place of two active ( Op-amp ) .

                   This Narrow Band Pass Filter is work only in inverting configuration or we can say that in another words that inverting mode .

                 The Narrow Band Pass Filter or we can say that multiple feedback filter is design for particular or specifically for value of center frequency  ( fc ) , Q which is called as quality factor and Bandwidth .

                   So , we know that in Band Pass Filter there is two cut-off frequency , which is 

1) High cut-off frequency , which is denoted by ( fH ) .

2) Low cut-off frequency , which is denoted by ( fL ) .

          And here we also know that , High cut-off frequency is always greater than low cut-off frequency that is , fH > fL .

           The Q-factor , which is known as 

quality factor is greater than 10 that is ,Q >10

          Now here , if we see the relation between Q-factor , 3db bandwidth and the centre frequency ( fc ) is given by , 

                        Q = fc / BW

                         But   Band-width = fH - fL

             i.e.      Q = fc / fH - fL

       The center frequency is as , fc = √ fH . fL

Where as , 

Q = quality factor

fL = Low cut-off frequency 

fH = High cut-off frequency

BW= Band-width

fc = center frequency

So , Here in this Narrow Band Pass Filter there is two capacitor is there .

One capacitor C1 at the inverting input of the terminal and second capacitor C2 at the inverting terminal for feedback .

       So in this circuit , for specific calculation C1 = C2 = C it means that the capacitor C1 and C2 is both equal to C .

         So , in this circuit what is the value of Resistor (R1) , (R2) and (R3)

              R1 = Q / 2 .pi.fc.C.AF

              R2 = Q / 2 .pi.fc.C (2Q square - AF)

              R3 = Q / 2 .pi.fc.C

   This is required for calculation purpose .

               Gain is denoted by AF

Here , AF is the gain at the center frequency ( fc ) . The center frequency gain shows the 

                AF = R3 / 2R1

   The gain must be satisfied the following condition i.e. therefore ,

                      AF < 2Q square   , it means that gain must be less than 2 times quality factor square .

 

    If we see , the frequency response of the Narrow Band Pass Filter , so the frequency response draw in between frequency and gain .

                                        

                                                

So , here in this frequency response graph show that , ( fL ) show low cut-off frequency and ( fH ) show high cut-off frequency .  So in both low cut-off frequency and high cut-off frequency there is narrow band pass filter's bandwidth is exist .

             Here , in this center frequency ( fc ) , we can change with new frequency ( f'c ) without changing the gain or bandwidth by changing the R2 to R'2 so like that .

                  If we change the value of Resistor R2 with any value R'2 so it is given as ,

               R'2 = R2 ( fc / f'c ) whole square

In this way we can change the center frequency with new center frequency .

                   

Explain reason needs for modulation

 Explain reason needs for modulation 

Ans. 

Modulation is a message carrying signal , which is transmitted over a long distance and for it to establish a reliable communication , it needs to take the help of a high frequency signal which should not affect the original characteristics of the message signal .

1) The needs of modulation is to prevent against disturbance of noise .

2) When we transmitted our message signal , than after transmitting our message signal , it can attacked by unwanted signal in the channel , so here modulation help to protect.

3) Message signal can be modified or we can say that in other words that , it changes into unwanted signal by noise , So modulation help to not happen this type of issue .

4) So , for not cause problem because of noise i.e. therefore modulation process must be done before transmitting our message signal .

5) Here , if we not done modulation process , than our message signal gets weak and corrupted due to unwanted signal i.e. therefore signal receiving antenna required larger in size .

           Because of Modulation process our message signal's wavelength get shorter and frequency gets high and because of that signal receiving antennae size will be reduced .

            So modulation is prove very helpful in all branch of communication purposes .

Explain the application of elasticity

 Explain the application of  elasticity 

Ans:-

1) The application of elasticity, while designing concrete or steel structure.

2) The application of elasticity, to bear or we can say that in other words that, to sustain more load than actual load .

3) The application of elasticity, to determine strength of the material .

4) Elasticity play a vital role in the field of engineering, to determine the tensile strength of different - different materials.

5) The application of elasticity, the elastic property has to be taken into account in the construction of build buildings and fly over bridges etc .

Explain the different types of electric current effect

 Explain the different types of electric current effect

Ans. The different types of electric current effect i.e. there are three different types of Electric Current effect such as , given as 

1) Heating Effect

2) Magnetic Effect

3) Chemical Effect

So , let's see , and know about in complete and detail information of these three effects of electric current .

1) Heating Effect :

  In this type of effect , electric current play a very vital role . Heating effect of electricity is one of the widely most used effective in the world .

          When current is passed or we can say that in other words that , when current is flow through a conductor, it is generated heat , because of opposition of Electric Current by a given resistance .

         The heating effect of an electric current , depends upon these three factors such as , 

        i) Resistance (R)

       ii) Time (t)

      iii) The Amount Current (I)

                                                             

                a) Resistance ( R ) : Resistance of the  conductor , the lower the resistance , the lower heat generated by the electric current and the higher the resistance, the higher heat generated by the electric current .

                b) Time ( t ) : In time for which , current flow the larger , the amount of heat produce or we can say that in other words that , in a given time how many current (A) flow through the conductor in presence of given resistance . That much heat generated by the electric current .

               c) The Amount Of Current ( I ) : The Amount of current higher the larger the heat generated by the electric current and the amount of current lower the smaller the heat generated by the electric current .

         Thus , heating effect H = I(square).R.t      this given equation is called as or we can say that , known as joule heating equation .

Application of heating effect of electric current is given as ,

          i) Electric iron

         ii) Electric kettles

        iii) Electric cooker

        iv) In lighting appliances like as filaments lamp , fluorescent etc.

2) Magnetic Effect : 

When an electric current passes or we can say that in other words that when an electric current flow through the given conductor then electric current  produces magnetic field ,which is called as electromagnetic field .

               So all of you may ask , how to conclude this , it's really produces magnetic field or not . So for detecting this or we can say that , to proving this , we can take help of compass . To detecting flow or passes of electric current through the given conductor . To measure of how much electric current passes through the given conductor , we uses galvanometer . Galvanometer is very useful to determine to check or measure mili ampere electric current through the given conductor .

                The imaginary lines around a magnet are called as or we can say that known as field lines of magnet .

                When we drop iron fillings to settle around a bar magnet , they are arranged in a such way that, in a mimics the magnetic field of lines or we can say that in other words that , when we allowed iron fillings to settle around a bar magnet , they are arranged in a pattern which follows the magnetic field of lines .

                Magnetic field is a vector quantity that is , it has both direction and magnitude . If a conductor carrying a current is held by  the right hand rule . Keeping the thumb straight , it mean consider that , if the direction of electric current is in the direction of thumb , then the direction of wrapping of other four fingers , then it shows the direction of magnetic field .               

Applications of Magnetic effect of Electric Current given as ,

         i) Electric door bell

        ii) Electric fan 

       iii) Electric motor

       iv) Electric generators

        V) Electric transformer

       vi) Because of electromagnetic, lifting of  heavy loads of iron and also iron scrap etc.

3) Chemical Effect :

 Here in para , we learn complete about , what is chemical effect of electric current ? So, when electric current is passed or we can say that flow through a conducting solution , some chemical reaction take place ,  liquid that conducts electricity are solutions of acids , base and salts .

              You may be heard , the term electroplating ? the process of deposition a layer of any desired metal on another material by means of electricity is called as or we can say known as electroplating .

              When electric current is passed or we can say that , flow through the water , water get dissociates or we can say that in other words that separation of different molecules . Here in water, hydrogen and oxygen , so than , hydrogen molecule is deposited over negative pole and oxygen molecule is deposited on positive pole .

               So , the deposition of the hydrogen and the oxygen at a different pole is visible or in other words , see in the form of bubbles .

               Sometimes , the colour of the solution changes , when an electric current passes through it or we can say that in other words that, when an electric current flow through it .

                 The chemical reaction depends upon the types of different solution , through which the electric current is passed .

    Application Of Chemical Effect Of Electric Current is given as ,

                i) Electroplating

               ii) Electrolysis

          There are many several application of chemical effect of electric current . Instead of the above application , there are several many more application of these , chemical effect of electric current .

Explain working of positive series clipper with the help of diagram and waveform

 Explain working of positive series clipper with the help of diagram and waveform 

Ans. 

                                       

Positive half cycle 

When positive half cycle , apply at the input ( I/P ) terminal of the circuit , the diode in this circuit act as or we can say that , it behave like a open switch .

         And i.e. therefore , it does not allow or we can say that in other words , it not permit to flow of electric current through the load resistance ( RL ) .

         So , because of that , waveform of this operation region , When we give input signal sinusoidal wave , it clips or we can say that in other words that , it cut and remove the positive half signal waveform .

Negative half cycle

When negative half cycle , apply at the input ( I/P ) terminal of the circuit , the diode in this circuit act as or we can say that in other words that , it behave like as a close switch .

        And i.e. therefore , it allow or we can say that in other words that , it give permit to flow of electric current through the load resistance ( RL ) .

           So , because of that , waveform of this operation region , when we give input signal sinusoidal wave , it not clips or we can say that in other words that , it not cut and remove the negative  half signal waveform .

Explain construction of N-P-N transistor with the help of diagram

 Explain construction of N-P-N transistor with the help of diagram

Ans. 

                                                          

In this N-P-N transistor , it has basically three (3) regions i.e. emitter , base and collector .

         The junction which present in between Emitter and Base is called as or known as Emitter-Base junction .

            So that , similarly like as , the junction which present in between Base and Collector is called as or we can say that known as Base-Collector junction .

             In this N-P-N transistors , due to presence of two (2) junction in between three (3) region like as  Emitter , Base and Collector , it act like or we can say that in other words it behave like as two (2) PN junction diode .

             It is very interesting to know that , the doping level of all these three (3) regions are different .

              So in this , Emitter region are highly doped , the Base region is lightly doped and the  last region is collector region , which is come or fall in between Emitter and Base region or we can say that in other words that , the doping level of collector region is moderate .

             It is noteworthy or we can say that , it is notice-able that we cannot interchange or in other words , we cannot exchange the Emitter and Collector region .

              The reason for this , is that the thickness of the Collector region is some what greater than , the Emitter region . So , that it can dissipate more power or we can say that it can dissolve more power .

State the application of photo-diode

 State the application of photo-diode

Ans. 

1) The photo-diode is used in optical communication system .

2) The photo-diode is used in a automobile devices and also you called as automobile circuits .

3) The photo-diode is used in medical devices .

4) The photo-diode used in a solar cell panel , which is very very helpful to us .

5) The photo-diode is used in a logic circuit or called as logic device .

6) The photo-diode uses in a photo-detection circuit or we say that in other words that , photo-detection devices .

7) The photo-diode is used in a camera's light meter .

8) The photo-diode is used in a street light .

9) The photo-diode is used in , for exact measurement of the intensity of light in science and industry fields .

10) The photo-diode is used in a consumer electronics devices or we can called as electronics circuits like as smoke detectors , compact disc players , television and remote control in VCRs .

11) The photo-diode is used in a detection circuit or we can called as detection devices .

12) The photo-diode is used in a character recognised circuits or we can say that in other words that , character recognised devices .

13) The photo-diode is used in a demodulation process or we can say that in another words that , it is used as detectors .

14) The photo-diode is useful in , for detection of visible and invisible light rays .

15) The photo-diode is used in a Photography or we can say that in other words , is used in shooting photographs .

16) The Photo-diode is used in a spectroscopy .

17) The photo-diode is used in a Laser Range Finder ( LRF ) .

18) The photo-diode is used in a Analytical instrumentation .

19) The photo-diode is used in a Medical Imaging Instrument ( MII ) .

20) The photo-diode is used in a Beam Alignment .

21) The photo-diode is used in a obstacle detector .

22) The photo-diode is used in a Bar Code Scanner .

INVERTING OPERATIONAL AMPLIFIER

 INVERTING OPERATION AMPLIFIER

                                                  

To understanding Inverting operational amplifier , then you have to first understand the concept of Virtual Ground .

Now in this blog article we will go to learn about virtual ground in simple and precisely. This virtual ground concept is only , applicable when we are providing the negative feedback to this operational amplifier (Op-amp) , or we can say in other words this virtual ground concept is only valid when we providing negative feedback to this operational amplifier .

              Now , here for a given (op-amp) the open loop gain of this (op-amp) is 10 raise to the power 6 . And we know the output (o/p) voltage (Vout) of (op-amp) is given as A multiply with differential input voltage .

                                              

             This is the input voltage between these inverting and the non-inverting input terminals . Now , here let's we consider that through this negative feedback , we are controlling the output (o/p) voltage of this (op-amp) in a such a way that , the output voltage is always less than the saturation voltage. Or in other words we can say that we are operating this (op-amp) in a linear region . 

           So , let consider that the output (o/p) voltage is ten volt (10 v) . So, we can say that 10 v is equals to the 10 raise to the power 6 multiply with this differential input voltage . Or we can say that in other words that the differential input voltage is equal to 10 micro volts . 

           Now , here this differential input voltage is nothing , but the differential between this inverting and the non-inverting input terminals . So , we can write this differential input voltage as a (V plus minus V minus) ( V+ -V- ) this is equals to 10 micro volt . 

          Now , here this 10 micro volt is very very small signal and we can almost neglect it . So , we can write this (V plus minus V minus) (V+ - V-) as approximately equal to zero volts or we can say that in other words that Vplus , that is equal to Vminus .

           It means that the inverting and the Non-inverting input (I/P) terminals are at the same potential or we can say that in other words that there is virtually short between this inverting and non-inverting input (I/P) terminals .

             Now here , the term we use virtual it means that , these two terminals are not actually short circuited , but they are virtually short circuited . So, whatever voltage that appear at one terminal , the exact and same voltage will appear at another terminal . 

            So , now in this configuration , this Non-inverting input terminal is grounded . So, we can say that Vplus that is equals to zero (0) . So, accordingly to this result , Vminus should be equals to zero (0) . It means that this terminal is not actually grounded , but it will act like as a virtual ground . 

            So , this negative feedback will ensure that the difference between this inverting and the Non-inverting input (I/P) is very very small or we can say that in another words that it is almost negligible .

           And because of that , we can consider these both input (I/P) terminal at the same potential . So , this concept is known as the virtual ground concept .

          So , now let use this concept , if virtual ground and let's derive the expression between this Vout and Vin .

           And let's say that node is , node X . And let's say that the current , that is flowing through this resistor Rf is If . We had seen that the (Op-amp) has very high input (I/P) impedance or if we assume the ideal (Op-amp) is infinite . It means that no current is entering into this (Op-amp) or we can say that in other words that current ( I ) that is equal to zero (0) .

                                                  

So , as you can see here , just by changing the value of this ( Rf ) and ( R1 ) we can control the gain of this (Op-amp) . And we can use this (Op-amp) as a amplifier .

         Now , here the negative sign indicates that the output voltage is 180 degree out of phase with respect to input voltage .

          So , let's say that , if we have applied the sinusoidal signal at the input , then at the output we will get the amplified sinusoidal signal , which is having a 180 degree phase with respect to the input signal .

           And that is why this configuration of the (Op-amp) is known as the inverting configuration . Because the output (O/P) will be get inverted with respect to input (I/P) voltage .

Wide band pass filter

 Wide band pass filter

Ans. The wide band pass filter is formed by cascading high-pass section and low-pass section . 

Look here , the wide band pass filter is made by two order , First is high-pass filter and second is low-pass filter . So when we combine or we can say that cascading first order high-pass section and first order low-pass section is called as or we can say that known as wide band pass filter ( WBPF ) .

                                                             

  Here , question is arises , that how we can conclude first order high-pass section ? 

For concluding first order high-pass section is that , at non-inverting terminal in input ( I/P ) section of non-inverting terminal capacitor is placed . And with capacitor , resistor is connected in parallel that , combination is proved that this is first order high-pass section .

   And also question arises is that , how we can conclude first order low pass section ? 

For concluding first order low-pass section is that , at non-inverting terminal in input ( I/P ) section of non-inverting terminal Resistor is placed , and with resistor , capacitor is connected in parallel that combination is proved that , this is first order low-pass section .

In this , the product is plus and minus + or - 20 db ( decibel ) per decade . In which condition ? , when the first order low pass and high pass section cascade with each other .

  As like as , if we talk about second order high-pass section and second order low-pass section , the product is plus or minus + or - 40 db ( decibel ) per decade . In which condition ? , When the second order low-pass section and high pass section  cascade or we can say that combine with each other .

If we see the frequency response of wide band pass filter . Here , wide band pass filter has two cut off frequency . First is ( fH ) and second is ( fL ) . 

Whereas , ( fH ) stands for or we can say that it shows the high cut off frequency and ( fL ) stands for or we can say that it show low cut off frequency . 

In both of these cut off frequency ( fH ) is always greater than ( fL ) . In other words we can say that high cut off frequency is always greater than low cut off frequency. 

In wide band pass filter , if we talk about Q-factor or we can say that quality factor . It is denoted by capital Q .

Because of Q-factor we can show that wide band pass filter and narrow band pass filter. When Q-factor is less than 10 i.e.  Q < 10 than it shows wide band pass filter , and Q-factor is greater than 10 i.e.  Q > 10 , than it shows narrow band pass filter .

                                                             

                                                        

In wide band pass filter , if we see the relation between the Q , 3dB bandwidth and the center frequency fc is given by  

                                Q = fc / Bw

                          i.e. Q = fc / fH-fL

In wide band pass filter our, frequency response is given by in the above graph . Here in this graph fL is low cut off frequency and fH is represents high cut off frequency , and 0.707 is the gain magnitude .

In between high cut off frequency and low cut off frequency there is pass band is exist . Where as AFT is a total pass band gain .

At fL point low cut off frequency, there is plus +20 dB per decade is present . And if we move towards fL to fH , it means that if we move low cut off frequency to high cut off frequency +20 dB is constant for some given of time and after it goes minus -20 dB per decade -20 dB/decade at point high cut off frequency ( fH ) .

From fL to fH there is pass band is there and from outside of these fL and fH there is stop band is present .

And one thing always remember that fH is always greater than fL or we can say that high cut off frequency is always greater than low cut off frequency .

And after all of these , there is another point is center frequency . Center frequency is arises in between both fH and fL and it is denoted by ( fc ) .

                i.e.     fc = √ fH .fL

And , if we talk about gain of high pass low pass filter than it is given as ,