Figure below shows a curcuit diagram of JFET amplifier. Taking drain current (Id) =2.5 mA, Determine the:

-

 

(a) Gate voltage, Vg

Soln

Gate voltage Vg = Voltage drop across resistor R2

Vg = {R2/(R1+R2) } x Vcc

      = {280000/(2000000+280000)} × 16 volts

      = 1.9649 volts

(b) Source Voltage Vs

Soln

Source voltage, Vs = Id ×Rs

         = (0.0025 x 1500)

         = 3.75 volts

(c) Gate to Source Voltage Vgs

Soln

Vgs = Vg - Id Rs

        = 1.9649 - 3.75

        = - 1.79 volts

(d) Drain Voltage, Vd

Soln

Drain voltage Vd = Id Rd

                               = (0.0025 x 2500)

                               = 6.25 volts

(e) Drain to source voltage ,Vds

Vds = Vdd - Id(Rd + Rs)

        = 16 - 0.0025( 2500 + 1500)

         = 6 volts

Comments

Post a Comment

Popular posts from this blog

Explain Rutherford atomic model

-
 In his study on atoms ,Rutherford proposed Rutherford atomic model( Rutherford planetary model of an atom). The following are the salient features his model: (I) Every atom consist of tiny central core called nucleus which is entirely positively charged and it is a zone in which almost whole mass of the atom is concentrated. (II) Most of an atom is empty space (III) In free space around the nucleus there are electrons which move in orbits just like the planets around the sun and the centripetal force  needed for orbital motion of electrons is provided by electrostatic forces due to positively charged nucleus. (III) An atom as a whole is electrically neutral hence the total posively charge of the nucleus is exactly equal to total negative charge of all the electrons orbiting in an atom
Read more

Electron released from the cathode CRT is accelerated to the anode by accelerating potential of 2kV in time 0.25 microseconds. Taking the mass of the electron m = 9.1 x 10^-31kg and electron charge e = 1.6 x 10^-19C .Determine:

-
 (I) Velocity of the electrons Soln Velocity of electrons V = {√(2eEa)/(m)}      = {√ (2 x 1.6 x10^-19 x 2000)/(9.1 x 10-31)}       = 26519741.77 m/s (II) Acceleration of the electron Soln   Acceleration = change in velocity/time                 = (26519741.77 - 0)/(0.26x 10-6)                  = 1.01999 x 10^14 m/s^2 (III) Loss in potential energy     Loss in potential energy P.E = eEa                = (1.6 x10^-19) x ( 2000)                 = 3.2 x 10^-16 Joules (III) Gain in Kinetic energy       K.E = 1/2(m x V^2)      = 1/2( 9.1 x 10^-31 x 26519741.77^2)       = 3.2 x 10^-16 Joules
Read more