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Explain CASCODE amplifier with its application. OR Short Note on CASCODE amplifier. OR Draw and explain working of CASCODE amplifier.

Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits

Marks: 10M, 5M

Year: May 2015, Dec 2015, Dec 2016

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Figure 1 : Circuit Diagram of CASCODE Amplifier

  1. Figure 1 shows the circuit diagram of CASCODE amplifier. CE-CB cascade configuration is known as cascode amplifier.
  2. 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.

  3. 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.

  4. AC voltage is obtained across Rc. Cb, Cc, Ca block dc and passes ac. AC signal is applied at base of Q1 which amplifies it with unity gain, and voltage V01 appears across collector of Q1.
  5. V01 acts as input to Q2 which further amplifies the signal and voltage Vo appears across collector of CB Configuration.

Analysis of Cascode Amplifier :

enter image description here

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 :

  1. It is used in RF tunner.
  2. Used in Amplitude Modulation.
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