Please help solve this Bipolar Junction Transistors (BJTs) problem. +9 V100 kn10 knR;I kn1 knFig. P4.914.91 For the emitter-follower circuit shown in Fig.P4.91 the BJT used is specified to have B valuesin the range of 20 to 200 (a distressing situationfor the circuit designer). For the two extreme val-ues of B (B(a) lĘ, VE, and Vg.(b) the input resistance R;.(c) the voltage gain v,/v,.20 аnd B200), find:8

Given the BJT as common collector circuit we need to determine the following for each value of current gain β=20 & β=200: a) IE,VE and VB. b) the input resistance Ri. c) the voltage gain V0Vs . a) To find the the current IE,VE and VB: To find the d.c currents IE,VE and VB we first of all need to do d.c analysis. D.C Analysis: In d.c analysis capacitors acts as open circuit and a.c voltage sources are deactivated. Applying KVL: Vcc-IBRB-VBE-(β+1)IBRE=0For β=20: 9 -IB×100-0.7-21×IB×1=0 IB=9-0.7100+21 =8.3121 =0.068 mA IE=(β+1)IB =1.428 mA VE=IERE =1.428 V VB=VBE+VE =2.128 VFor β=200: 9 -IB×100-0.7-201×IB×1=0 IB=9-0.7100+201 =8.3301 =0.027 mA IE=(β+1)IB =5.54 mA VE=IERE =5.54 V VB=VBE+VE =0.7+5.54 =6.24 VThus, for β=20: IE=1.428 mA,VE=1.428 V & VB=2.128 V for β=200: IE=5.54 mA,VE=5.54 V & VB=6.24 Vb) To find the input resistance Ri : To find the input resistance we need to do a.c analysis. The ac equivalent circuit is: For β=20rπ=VTIB =26 mV0.068 mA =0.382 kΩFor β=200rπ=VTIB =26 mV0.027 mA =0.963 kΩRi =RB||[rπ+(1+β)(1||1)] kΩ =100||[rπ+(1+β)×0.5] kΩFor β=20:Ri =100||[0.382+21×0.5] =9.81 kΩFor β=200:Ri =100||[0.963+201×0.5] =50.36 kΩThus, for β=20:Ri=9.81 kΩ for β=200:Ri=50.36 kΩ .c) To find voltage gain V0Vs : From the a.c equivalent circuit: V0=(1+β)ib(1 kΩ||1 kΩ) ....(1) ib=Vinrπ+(1+β)(1 kΩ||1 kΩ) ......(2)By voltage division law:Vin=RinRin+100 kΩVs .....(3)Putting the value of Vin from eqn. (3) in eqn. (2), we get: ib=Rin[rπ+(1+β)(1 kΩ||1 kΩ)](Rin+100 kΩ)VsPutting this value of ib in eqn. (1) we get: V0=(1+β)(1 kΩ||1 kΩ) Rin[rπ+(1+β)(1 kΩ||1 kΩ)](Rin+100 kΩ)VsV0Vs=(1+β)(1 kΩ||1 kΩ)Rin[rπ+(1+β)(1 kΩ||1 kΩ)](Rin+100 kΩ) =(1+β)×0.5Rin[rπ+(1+β)×0.5](Rin+100 kΩ)For β=20:V0Vs=0.086For β=20:V0Vs=0.33Thus, for β=20:V0Vs=0.086 for β=20:V0Vs=0.33 .