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Write short note on cascode amplifier using BJT.
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The cascode amplifier is the two stage amplifier in which common emitter stage is connected to common base stage.

The CE-CB cascode connection is as shown in the figure:enter image description here

The input signal is applied at Q1 i.e at common emitter stage and output is obtained at Q2. 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 variation.

AC output voltage is obtained at RC collector .coupling capacitor are used to block dc signal pass a signals.

. AC signal is applied at base of Q1 which amplifies it with unity gain, and voltage V01 appears across collector of Q1. V01 acts as input to Q2 which further amplifies the signal and voltage Vo appears across collector of CB Configuration.

To perform small signal analysis we need to draw ac equivalent circuit of the given amplifier.To draw the ac equivalent circuit all capacitors must be replaced by short circuit and the DC sources connected to ground.

The AC equivalent circuit is as shown in the figure:

enter image description here

The overall voltage gain is the product of first stage gain to second stage gain.

AVT = AV2 * AV1

AVT= (Vo/V1) * (V1/Vin)

From the above figure we can see that Vo = Io*Rc.....1

The output current is Io = hfb ie2

hence subsituteing in equation 1 we get Vo = hfb ie2 * Rc....2

Now to determine Vo1 we need to apply KVL to the input side of common base connection

and we get Vo1 = ie2 * hib.....3

divide equation 2 by 3 we get.

Vo/V1 =(hfb ie2 * Rc)/( ie2 * hib )

Av2 =(hfb * Rc)/(hib)....................4

Now determine the gain for first stage,

Av1 = Vo1/Vin

from the above figure Vo1 = ie2 * hib...........5

Since ie2 and hfe ib1 are opposite in direction,

we can write ie2 = -hfe *ib2

Subsituteing in above equation 5

Vo1 = -hfeib2 * hib....................6

To determine Vin we need to apply KVL at the input side .

Vin – hie ib1=0

Vin = hie ib1………..(7)

Now divide (6) and (7)

AV1= Vo1/Vin = (−hfe∗ib1)∗(hib*hie∗ib1)

AV1= −hfe∗(hib*hie)…………..(8)

But WE know

hib=hie/(1+hfe)

Substitute in equation (8)

AV1= (−hfe/hie) ×( hie/1+hfe)

AV1= −hfe/(1+hfe)

AV1 ≈ -1 .......................9

Multiply AV1 and AV2 to obtain AVT

AVT= AV1×AV2

AVT= (−hfb∗Rc)/(hib).

The negative sign indicates the 180 degree phase shift provided by CE stage.

Input impedance : The input impedance is parallel combination of resistors at input side

Rin = R2 II R3 II hie......................10

Output Impedance : The output impedance is given by the output resistance of the second stage

Ro = Rc............11

In broadband amplifiers cascode configuration is used.

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