Consider the circuit shown in the figure below-The transistor has - - 100 and the external amplifier has A - Av - -103 and - VfV0 - 0-01- If small signal input base resistance for the BJT is r- - 1 k- then the value of overall voltage gain i-e - Avf - V0Vs is equal to kV-V-

A_f = A 1+ Aβ and A ∝μ ∴ A gt; gt;1 Thus, A_f = AA β = 1β Where, β nbsp;= V_fV_0 = R_1R_1+ R_2 = 11+1000= 11001 ∴ nbsp;A_f = 1β = 1001 V/V = 1.001 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Consider the ...

Question

Consider the circuit shown in the figure below.

The transistor has $β = 100$ and the external amplifier has $A = A_{v} = - 10^{3} a n d β = \frac{V_{f}}{V_{0}} = 0.01$ . If small signal input base resistance for the BJT is $r_{π} = 1 k Ω$ , then the value of overall voltage gain i.e ., $A_{v f} = \frac{V_{0}}{V_{s}}$ is equal to _______ $k V / V$ .
1.001

Open in App

Solution

The correct option is A 1.001
$A_{f} = \frac{A}{1 + A β}$
$a n d A \propto μ$
$∴ A >> 1$
$T h u s, A_{f} = \frac{A}{A β} = \frac{1}{β}$
$W h e r e, β = \frac{V_{f}}{V_{0}} = \frac{R_{1}}{R_{1} + R_{2}} = \frac{1}{1 + 1000} = \frac{1}{1001}$
$∴ A_{f} = \frac{1}{β} = 1001 V / V = 1.001 k V / V$

Suggest Corrections

Similar questions

Q. Consider the voltage divider circuit shown in the figure below.

The value of cut in voltage for the amplifier circuit

V_{B E (o n)} = 0.7 V

, and the value of

β = 140

, then the value of power dissipated by the transistor is equal to

Q. Consider the circuit shown in the figure below :

The value of

g_{m} = 68 m A / V, r_{π} = 1.45 k Ω a n d β = 100, t h e n t h e v a l u e o f \frac{V_{o}}{V_{m}}

is equal to _____. (Assume that the small signal current through the feedback resistor of 10 k

Ω

is negligible).

Q. Consider the circuit shown in the figure below:

Assume that the transistor has

β = 99, V_{A} = 100 V

and

V_{T} = 25 m V,

then the small-signal voltage gain

A_{v} = \frac{v_{0}}{v_{s}}

is equal to

Q. In the amplifier circuit shown below,

The value of the voltage gain

| \frac{V_{0}}{V_{s}} |

is equal to _____.
(Assume

V_{T} = 25 m V

and

β

is very large )

Q. Consider the transistor circuit shown in the figure below:

Assume that the biasing details are left out and the transistors are working in active region with

T_{1}

having transconductance

g_{m 1}

and input base-emitter small signal resistance

r_{π 1}

and for transistor

T_{2}

the transconductance

g_{m 2}

and input small signal base-emitter resistance

r_{π 2}

. Then the voltage gain

| A_{v} |

is given as

Join BYJU'S Learning Program

Consider the circuit shown in the figure below. The transistor has β=100 and the external amplifier has A=Av=−103 and β=VfV0=0.01. If small signal input base resistance for the BJT is rπ=1 kΩ, then the value of overall voltage gain i.e ., Avf=V0Vs is equal to _______ kV/V.1.001

The correct option is A 1.001Af=A1+Aβ and A∝μ ∴A>>1 Thus,Af=AAβ=1β Where,β=VfV0=R1R1+R2=11+1000=11001 ∴Af=1β=1001 V/V=1.001 kV/V