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

Explain avalanche breakdown with respect to PN junctions.

-
 Avalanche breakdown is a phenomenon which occurs in lightly doped junctions which occur when a high reverse voltage is applied across the junction which tend to cause an increase in electric field across the junction. This generated electric field exerts a force on the electrons at junction and frees them from the covalent bonds. These free electrons will gain acceleration and it will start moving across the junction with high velocity. This results in collosion with neighbouring atoms. These collisions in high velocity tend to generate more free electrons. These free electrons will start drifting and electrons -hole pairs recombination occurs across the junction which results in a rapid increase in current uncontrollably. <a href="https://kol.jumia.com/api/click/link/f0647e0d-9d90-4183-b025-e2cf08db8f34/42461daf-6d6a-4cf2-b0ca-7e066511e571"><img src="https://kol.jumia.com/banners/OK6RTn7jyXBJ8ORsrvFNC6o0L8wUYB5zNJOYzNEI.jpg" alt="PHONES CATEGORY"
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            = 1/deflection sensitivity             = 1/ ( 2.75 x 10^ -4)             = 3636.3636 volt/meter
Read more