written 7.8 years ago by | modified 2.8 years ago by |
Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits
Marks: 10M, 5M
Year: May 2015, Dec 2015, Dec 2016
written 7.8 years ago by | modified 2.8 years ago by |
Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits
Marks: 10M, 5M
Year: May 2015, Dec 2015, Dec 2016
written 7.8 years ago by |
Figure 1 : Circuit Diagram of CASCODE Amplifier
Cascode amplifier has following features :
i. Large input resistance.
ii. Large output resistance.
iii. Bandwidth is large.
iv. Voltage Gain is low compare to CE-CE Cascade.
In this above circuit Vcc, R1, R2, R3, Re are used to bias transistor Q1 and Q2 in active region. Re is used to make Q-point stable against temperature for both the transistor.
Analysis of Cascode Amplifier :
Figure 2 : AC equivalent using h-model
Figure 2 shows the AC equivalent using h-model. At mid frequency all connected capacitor act as a short circuit.
AVT= $\frac{Vo}{Vo1}$ × $\frac{Vo1}{Vin}$
Vo=io×Rc……………..(1)
io= hfb ie2
Substitue io in equation (1),
Vo=hfb ie2 Rc…………(2)
Apply KVL at the input of CB configuration,
Vo1= ie2 hib………….(3)
Divide (2) by (3),
$\frac{Vo}{Vo1}$ = $\frac{hfb*ie2*Rc}{ie28hib}$
AV2=$\frac{Vo}{Vo1}$ = $\frac{hfb*Rc}{hib}$
For first stage,
AV1=$\frac{Vo1}{Vin}$……..(4)
Vo1= ie2 hib………(5)
Since ie2 and hfe ib1 are opposite in direction,
ie2= -hfe ib1
Substitue in equation (5),
Vo1=-hfe ib1 hib…….(6)
Apply KVL at the input,
Vin – hie ib1=0
Vin = hie ib1………..(7)
Divide (6) and (7)
AV1= $\frac{Vo1}{Vin}$ = $\frac{-hfe*ib1*hib}{hie*ib1}$
AV1= $\frac{-hfe*hib}{hie}$…………..(8)
But,
hib=$\frac{hie}{1+hfe}$
Substitute in equation (8)
AV1= $\frac{-hfe}{hie}$ × $\frac{hie}{1+hfe}$
AV1= $\frac{-hfe}{1+hfe}$
AV1 ≈ -1
Multiply AV1 and AV2 to obtain AVT
AVT= AV1×AV2
AVT= $\frac{-hfb*Rc}{hib}$
Application :