Question:

The input signal given to a CE amplifier having a voltage gain of 150 is $v_i = 2 \cos \Bigg(15t+\frac{\pi}{3}\Bigg)$ The corresponding output signal will be

Updated On: Jul 18, 2024
  • $2 \cos \Bigg(15t+\frac{5\pi}{6}\Bigg)$
  • $300 \cos \Bigg(15t+\frac{4\pi}{3}\Bigg)$
  • $300 \cos \Bigg(15t+\frac{\pi}{3}\Bigg)$
  • $75 \cos \Bigg(15t+\frac{2\pi}{3}\Bigg)$
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The Correct Option is B

Solution and Explanation

Here,
Input signal, $v_1=2 \cos \Bigg(15t+\frac{\pi}{3}\Bigg)$ and voltage gain, $A_v$ = 150
As $A_v=\frac{V_0}{V_i}$
$\therefore$ Output signal, $V_0=A_v V_i$
Since CE amplifier gives a phase difference of $\pi (=180^\circ)$ between input and output signals.
$\therefore V_0=150\Bigg[2 cos \Bigg(15t+\frac{\pi}{3}+\pi\Bigg)\Bigg]$
$=300 cos \Bigg(15t+\frac{4\pi}{3})$
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Concepts Used:

P-n Junction

A P-N junction is an interface or a boundary between two semiconductor material types, namely the p-type and the n-type, inside a semiconductor.

Biasing conditions for the p-n Junction Diode:

in p-n junction diode two operating regions are there:

  • P-type
  • N-type

There are three biasing conditions for p-n junction diode are as follows:

  • Zero bias: When there is no external voltage applied to the p-n junction diode.
  • Forward bias: P-type is connected to positive terminal of the voltage potential while n-type is connected to the negative terminal.
  • Reverse bias: P-type is connected to negative terminal of the voltage potential while n-type is connected to the positive terminal.