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JEE Main 202313 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
13 April
Shift
Morning Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2023 · 13 April · Morning Shift · Physics · Semiconductors

From the given transfer characteristic of a transistor in CE...

From the given transfer characteristic of a transistor in CECE configuration, the value of power gain of this configuration is 10x{10}^{x}, for RB=10kΩ{R}_{B} = 10kΩ, and RC=1kΩ{R}_{C} = 1kΩ. The value of xx is _____.

From the given transfer characteristic of a transistor in CE configuration, the value of power gain of this

Correct answer

0

Step-by-step solution

The formula to calculate the current gain (β)\left(\beta\right) in CE configuration is given by

β=∆IC∆IB...(1)\beta = \dfrac{∆ {I}_{C}}{∆ {I}_{B}} . . . \left(1\right)

Considering the third and the first points on the given graph, the current gain can be calculated as follows:

{β=(30−10)×10−3A(300−100)×10−6A=100\begin{cases}\beta & = & \dfrac{\left(30 - 10\right) \times {10}^{- 3}A}{\left(300 - 100\right) \times {10}^{- 6}A} \\ & = & 100\end{cases}

The formula to calculate the power gain \left(P\right)inCEmodecanbewrittenas</p><p>in CE mode can be written as</p><p>P = {\beta}^{2} \cdot \dfrac{{R}_{C}}{{R}_{B}} . . . \left(2\right)</p><p>Substitutetheknownvaluesoftheparametersintoequation(2)andsolvetocalculatetherequiredunknownparameter.</p><p></p><p>Substitute the known values of the parameters into equation (2) and solve to calculate the required unknown parameter.</p><p>\begin{cases}{10}^{x} & = & {100}^{2} \times \dfrac{1kΩ}{10kΩ} \\ & = & 1000 \\ & = & {10}^{3} \\ & ⇒ & x = 3\end{cases}

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