The Figure below shows a circuit diagram of common emitter amplifier. Taking Vbe = 0.7 v and Ic = Ie, determine the following bias voltages and currents.

-


 (I) Base voltage Vb

Soln 

 Base voltage Vb = Voltage drop across resistor R2

 Base voltage Vb = { (R2/(R1+R2)) x Vcc}

                = { ( 4000/(4000+ 40000)) X 22 v}

                 = 2 Volts

(II) Emitter voltage Ve

Emitter voltage Ve = Vb - Vbe

                                  = 2 - 0.7

                                   = 1.3 Volts

(III) Emitter current Ie

        Emitter current Ie = Ve/Re

                                         = 1.3V/ 1500 ohms

                                          = 8.667 x 10^ -4 A

(III) Collector voltage Vc

       Collector voltage Vc = Ic Rc

                         = (8.667 x 10^ -4 x 10000)

                          = 8.667 Volts

(V) Collector emitter voltage Vce

Collector emitter voltage Vce = Vcc -Ic(Rc+Re)

= 22 - 8.667x 10^-4(10000+1500)

= 22- 9.967

= 12 Volts

     


Comments

Post a Comment

Popular posts from this blog

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

Properties of a cathode used in thermionic valves.

-
 It should have a low work function. A low work function helps to emit electrons from the cathode surface in a comparatively lower temperature It should have a high melting point. The temperature required to emit an electron from the cathode surface is quite high compared to the melting point of normal metals. A lower melting point causes vaporization of metals before they emit electrons. It should have high mechanical strength.
Read more