State the need of biasing of transistor

 State the need of biasing of transistor 

Ans. 

                                                       

To understanding the need of biasing of transistor , first of all we need to understand the term what is biasing ? 

          Biasing means to supply a power supply to the circuit or we can say that apply D.C or A.C signal to the given circuit for start operating .

              So , in electronics and in electricals all devices or we can say that all circuits are work or we can say that start operating only by supplying power supply .

              So like that, here transistor work in a biasing mode . A biasing is required to operate the transistor or we can say that to start working of the transistor . Without biasing , it is not more able to obtain an output ( O/P ) .

              A biasing is required to activate the transistor and prevents it to either saturation mode or cut-off mode . A biasing is a phenomenon or a concept of getting a proper D.C collector current at a certain D.C voltage by setting up a proper point .

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 .

Explain how transistor works as a switch with i/p and o/p waveform

 Explain how transistor works as a switch with i/p and o/p waveform

                                               

                                                

Ans. First of all we have to start the transistor or we can say that activate the transistor . So for activating transistor , voltage ( Vbe ) should always in a forward bias and voltage ( Vce ) should always in reverse bias .

If we apply the voltage less than the cut off voltage or we can say that knee voltage , then transistor acts or behave as a open switch .

And if we apply the voltage more than the cut off or we can say that knee voltage , then transistor acts or behave as a close switch .

So , the derivation of these circuit is !

                         Input section of the circuit 

                   Vbb - Vb . Rb = 0

        i.e.     Vbe = Vbb - Ib . Rb    ....equation (1)

            Output section of the circuit 

                   Vcc - Ic . Rc - Vce = 0

                   Vce = Vcc - Ic . Rl

            i.e.   Vo = Vcc - Ic . Rl      ....equation (2)

     

             Therefore case (1)

        i.e.    Vin <  cut in voltage

                  Ib = 0

                  Ic = 0

               From equation (2)

                Vo = Vcc - Ic . Rl

                Vo = Vcc - 0 . Rl

       i.e.    Vo = Vcc       [ Off State ]

             

                      Case (2)

                 Vin  > cut in voltage 

                 Base current ( Ib ) is greater 

                 Collector current ( Ic ) is greater

                 Vo = Vcc - Ic . Rl

              But  ( Rl = Vcc )

        i.e.    Vo = Vcc - Vcc

                  Vo = 0              [ ON State ]