Consider the circuit shown in the figure below -The value of gm- 68 mA-V- r- 1-45 k- and - 100- then the value of VoVm is equal to - Assume that the small signal current through the feedback resistor of 10 k- is negligible-

Drawing the small signal equivalent circuit we get R_q= 20lΩ || 20kΩ || 2 kΩ = 1.66 kΩ V_0= 1.66 kΩ× nbsp; ( V_π14.5 )=114.48 V_π 9∵ Contribution by i_1 i ...

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Consider the ...

Question

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).

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Solution

Drawing the small signal equivalent circuit we get

$R_{q} = 20 l Ω | | 20 k Ω | | 2 k Ω = 1.66 k Ω V_{0} = 1.66 k Ω \times (\frac{V_{π}}{14.5}) = 114.48 V_{π} 9 ∵ C o n t r i b u t i o n b y i_{1} i s n e g l i g a b l e) \frac{V_{π}}{14.5} = I_{m} + 68 \times 10^{- 3} \times V_{π} V_{π} (\frac{1}{14.5} - 68 \times 10^{- 3}) = I_{m} I_{m} = \frac{V_{π}}{1.035 k Ω} V_{b} = V_{π} + V_{0} = 115.48 V_{p i} R_{i} = \frac{V_{π}}{i_{m}} = \frac{115.48 v_{π}}{V_{π} / 1.035 k} = 120 k Ω \frac{V_{0}}{V_{s}} = \frac{V_{b}}{V_{s}} \times \frac{V_{0}}{v_{b}} = \frac{R_{i}}{R_{i} + R_{s}} \times \frac{114.48}{115.48} . \frac{V_{π}}{V_{π}} = \frac{120}{120 + 10} \times \frac{114.48}{115.48} = 0.92 V / V$

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Consider the circuit shown in the figure below : The value of gm=68 mA/V, rπ=1.45 kΩ and β=100, then the value of VoVm is equal to _____. (Assume that the small signal current through the feedback resistor of 10 kΩ is negligible).

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).