Disadvantages of Junction Field Effect Transistors (JFETs)

-

 

  • It has high drain resistance
  • Moderate input impedance
  • It is slower in operation

Comments

Post a Comment

Popular posts from this blog

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

An electrostatic CRT has parallel deflection plates 1.5 cm long and 6mm apart. The screen is 55cm from the centre of the deflecting plates. The deflecting and accelerating potentials are 150 V and 2.5 kilovolts respectively. Determine the:

-
Image
 Ld = length of deflecting plates, m  d = distance between deflecting plates,m m = mass of the electron ,kg L = distance between screen and the centre of the deflecting plates, m Ea = accelerating voltage to the plates, volts Ed = deflecting potential of the plates (I) Velocity of the electron Soln Velocity of electron,  Vox = √{(2eEa)/(m)} Ea = 2500 v Ed = 150v Ld = 1.5cm L = 55cm d = 6 mm m = 9.1094 x 10^-31 kg e = 1.6022 x 10^ -19 Coulombs = {√ (2 x 1.6022 x 10^-19 x 2500)/(9.1094x 10^-31)} = √ 8.794x 10^ 14 = 29655037.64 m/s (II) Amount of deflection D = {((L x Ld x Ed)/(2 x d x Ea)) x (m)}     = {((0.55 x 0.015 x 150)/(2 x 6 x 10^-3 x 2500))}     = 0.04125 m (III) Deflection sensitivity              = { ( Ld x L )/(2 x d x Ea)}           = { ( 1.5 x 10^ -2 x 0.55)/(2 x 6 x 10^ -3 x 2500)}           = 2.75 x 10^ -4 m/ volt (IV) Deflection factor     ...
Read more

With the aid of a labelled diagram, describe the operation of a p-channel Junction Field Effect transistors

-
Image
 When a drain to source voltage Vds is applied between the drain and source terminals and the voltage on the gate is zero, the two PN junctions at the sides of the bar tend to establish a depletion layer. The holes will flow from the source to the drain through a channel between the depletion layers. The size of the depletion layer determines the width of the channel and hence the current conduction through the channel. When a reverse voltage Vgs is applied between the gate and source terminals , the width of the depletiom layer is increased this reduces the width of the conducting channel hence imcreasing the resistance of the p-type  bar. Hence reducing the current flowing from the drain to source. When the reverse bias voltage Vgs on the gate is reduced the width of the depletion also decreases and hence increasing the crossectional area of the Channel and hence flow of current by holes increases.
Read more