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Question: An amplifier has a voltage gain \[{A_{\text{v}}} = 1000\] . The voltage gain in \[{\text{dB}}\] is: ...

An amplifier has a voltage gain Av=1000{A_{\text{v}}} = 1000 . The voltage gain in dB{\text{dB}} is:
A. 30 dB{\text{30}}\,{\text{dB}}
B. 60 dB{\text{60}}\,{\text{dB}}
C. 3 dB{\text{3}}\,{\text{dB}}
D. 20 dB{\text{20}}\,{\text{dB}}

Explanation

Solution

First of all, we will use the formula which relates voltage gain in dB{\text{dB}} to the voltage gain Av{A_{\text{v}}} . We will substitute the required values and manipulate accordingly to obtain the result.

Complete step by step answer: In the given question, we are supplied with the following data:
The voltage gain of an amplifier is 10001000 i.e. Av=1000{A_{\text{v}}} = 1000 .
We are asked to find the voltage gain in dB{\text{dB}} .

To begin with, we know a formula, which gives the voltage gain in dB{\text{dB}} , which is given below:
A=10log⁡AvA = 10\log {A_{\text{v}}} …… (1)
Where,
AA indicates the voltage gain in dB{\text{dB}} .
Av{A_{\text{v}}} indicates the voltage gain.

Now, by substituting the required values in the equation (1), we get:

⇒A=10log⁡Av ⇒A=10×log⁡1000 ⇒A=10×log⁡103 ⇒A=10×3 \Rightarrow A = 10\log {A_{\text{v}}} \\\ \Rightarrow A = 10 \times \log 1000 \\\ \Rightarrow A = 10 \times \log {10^3} \\\ \Rightarrow A = 10 \times 3 \\\

Again, multiplying we get the answer:
⇒A=30 dB\Rightarrow A = 30\,{\text{dB}}

Hence, the voltage gain in dB{\text{dB}} is 30 dB30\,{\text{dB}}. The correct option is A.

Additional information:
An amplifier, an electronic amplifier or a (informal) amplifier is an electronic system which can increase the signal power (a voltage or current that changes over time). The amount of amplification that an amplifier produces is determined by its gain: the ratio of output voltage, current, or input power. A circuit that has a power gain greater than one is an amplifier. Either a different piece of equipment or an electric circuit located inside another system may be an amplifier.

Note: While solving this problem, many students seem to have a confusion regarding the logarithm part, which clearly states that it is of base 1010 , rather than base ee . Taking the logarithm with base ee may deviate the calculated result from the actual one.