written 8.1 years ago by | modified 3.2 years ago by |
Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits
Marks: 10M
Year: Dec 2016
written 8.1 years ago by | modified 3.2 years ago by |
Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits
Marks: 10M
Year: Dec 2016
written 8.1 years ago by |
Given: β = 100, Vbe = 0.7V
For DC analysis, all the coupling capacitor acts as open circuit as shown in below circuit diagram. We need to find equivalent circuit at base terminal (Rth & Vth).
Figure 1: Circuit diagram without coupling capacitor and equivalent circuit diagram
Vth can be found using voltage divider method:
Vth=10×12.2k12.2k+56k
Vth=1.78V
Rth=12..2k||56k=10.01kΩ
Apply KVL from base to emitter junction,
Vth−RthIB−Vbe−REIE=0 ----(1)
where IE=IC+IB=βIB+IB=IB(1+β)
Substitute in equation (1)
Vth−RthIB−Vbe−REIB(1+β)=0
IB=Vth−VbeRth+RE(1+β)
IB=1.78−0.710.01k+(1+100)0.4k
IB=21.42μA
Icq=βIB=100×21.42μA
Icq=2.142mA
Apply KVL from Vcc to Ground through collector to emitter,
Vcc−ICRC−Vceq−IERE=0 ----(2)
IE=IB+IC=2.142mA+21.42μA
IE=2.16mA
Substitute IE value in equation (2),
Vcc−ICRC−IERE=Vceq
10−2kΩ×2.142mA−2.16mA×0.4kΩ=Vceq
Vceq=4.852V
VC=Vcc−ICRC
VC=10−2.142mA×2kΩ=5.716V
VE=IERE=2.16mA×0.4kΩ=0.864
Answers: