written 8.1 years ago by | modified 3.2 years ago by |
Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits
Marks: 10M, 5M
Year: May 2015, Dec 2015, 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, 5M
Year: May 2015, Dec 2015, Dec 2016
written 8.1 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= VoVo1 × Vo1Vin
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),
VoVo1 = hfb∗ie2∗Rcie28hib
AV2=VoVo1 = hfb∗Rchib
For first stage,
AV1=Vo1Vin……..(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= Vo1Vin = −hfe∗ib1∗hibhie∗ib1
AV1= −hfe∗hibhie…………..(8)
But,
hib=hie1+hfe
Substitute in equation (8)
AV1= −hfehie × hie1+hfe
AV1= −hfe1+hfe
AV1 ≈ -1
Multiply AV1 and AV2 to obtain AVT
AVT= AV1×AV2
AVT= −hfb∗Rchib
Application :