InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain, A_V of the amplifier is given by A_v = - ( beta_(ac) R_L )/(I_i)where beta_(Ac) is the current gain, R_L is the load resistance and r_1 is the input resistance of the transistor. What is the significance of the negative sign in the expression for the voltage gain? |
|
Answer» Solution :When an a.c. input SIGNAL `V_i` is superimposed on the bias `v_(B E)` the OUTPUT, which is measured between collector and ground, increases `V_(C C) = V_(CE)+I_C R_L` ` V_(B B) = V_(B E) +I_B R_B` when`v_i`is notzerowe have ` V_(BE )+V_i = V_(BE)+I_BR_B+DeltaI_B(R_B +R_i)` `impliesV_i =Delta I_B(R_B +R_I)` ` V_i= r DeltaI_B` changein `I_B ` causesa changein ` I_C` HENCE `beta_(a.c) = (DeltaI_C)/( DeltaI_B ) = (I_C)/(I_B)` As ` DeltaV_(C C)= DeltaV_(CE )+R_LDelta I_C = 0` ` impliesDelta V_(CE ) =- R_LDelta L_C` `impliesV_0=- R_L Delta I_C` ` = beta _(ac )DeltaI_B R_L` ` implies `Voltagegainof theamplifer ` A_0= (V_0)/( V_i )= (Delta V_(CE ))/( rDelta I_B) = ( - beta_(ac) Delta I_bR_l)/( rDelta I_E)` ` = beta_(ac )(R_L )/(r)` negative signin tehexperessionshowsthat ouputvoltageand inputvoltagehaephasedifferenceof `pi` |
|