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Question: The input signal given to a CE amplifier having a voltage gain of 150 is \(2\text{cos(15t+}\dfrac{\p...

The input signal given to a CE amplifier having a voltage gain of 150 is 2cos(15t+π3)2\text{cos(15t+}\dfrac{\pi }{3}). The corresponding output signal will be
A. 300cos(15t+4π3)300\cos (15t+\dfrac{4\pi }{3})
B. 300cos(15t+π3)300\cos (15t+\dfrac{\pi }{3})
C. 75cos(15t+2π3)75\cos (15t+\dfrac{2\pi }{3})
D. 2cos(15t+5π6)2\cos (15t+\dfrac{5\pi }{6})

Explanation

Solution

Hint: In a CE amplifier, the phase difference between input voltage and output voltage is π. Voltage gain is the ratio of output voltage to input voltage. Output voltage is made the subject and calculated accordingly.

Formula used:
Av=V0Vi{{A}_{v}}=\dfrac{{{V}_{0}}}{{{V}_{i}}}

Complete step by step answer:
The common emitter amplifier is a type of bipolar junction transistor which is used as a voltage amplifier. The output of this amplifier is collected from the collector terminal and the input is provided from base terminal while the emitter terminal is common to both. The picture below represents a CE amplifier:

The following numerical data is given to us: V+
2cos(15t+π3)2\text{cos(15t+}\dfrac{\pi }{3}), Av=150 where Vi is voltage of the input signal and Av is voltage gain.
As we all know,
Voltage gain: Av=V0Vi{{A}_{v}}=\dfrac{{{V}_{0}}}{{{V}_{i}}} where V0V_0 is the output voltage
We can also write it as V0=Av×Vi{{V}_{0}}={{A}_{v}}\times {{V}_{i}} ……… (1)
So on substituting the given values in equation (1),
V0=150×2cos(15t+π3)=300cos(15t+π3){{V}_{0}}=150\times 2\text{cos(15t+}\dfrac{\pi }{3})=300cos(15t+\dfrac{\pi }{3})
In a CE amplifier, the phase difference between input voltage and output voltage is π. So we have to add π to the phase angle.
V0=300cos(15t+π3+π)=300cos(15t+4π3){{V}_{0}}=300cos(15t+\dfrac{\pi }{3}+\pi )=300cos(15t+\dfrac{4\pi }{3}).
Hence the correct option is (a).

Note: The possibility of the mistake is that you may choose option (b) because you may forget to take into account the phase difference between input and output voltage. In order to solve this problem or other problems like these, it is important to remember the characteristics of different transistors and amplifiers.