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 | • modified 8.1 years ago |
Figure 1: Circuit Diagram of Common Source amplifier.
At mid frequency all connected capacitors acts as short circuit. Hence AC equivalent using JFET becomes,
Figure 2: Small signal equivalent circuit
Voltage Gain (AV):
VO=IO∗RD
But IO=−Ids
VO=−Ids∗RD ....(1)
Apply KCL at drain terminal,
Ids=I1+gmVgs
I1=VOrd
Ids=VOrd+gmVgs .....(2)
substitute equation (2) in (1),
VO=−(VOrd+gmVgs)RD
VO+VOrdRD=−gmVgsRD
VO(1+RDrd)=−gmVgsRD
VO(rd+RDrd)=−gmVgsRD
VO=−gmVgs(RDrdrd+RD)
VO=−gmVgs(RD||rd)
From input side we get vin = vgs,
hence, VO=−gmVin(RD||rd)
AV=VOVin=−gm(RD||rd)
Input Resistance (Rin):
Rin=VinIin
Since Iin passes only through R1||R2,
Rin=VinIin=R1||R2
Output Resistance (RO):
RO can be obtained using following steps:
i. Set Vin=0.
ii. Connect and imaginary voltage source VO that delivers current IO.
RO=VOIO
Figure 3 : AC equivalent for RO
RO=VOIO=RD||rd