If an amplifier has a power gain of 100, then its dB gain is

Concept:

Power gain is the ratio of the output power (P0) to the input power (Pin­), i.e.

\({A_p} = \frac{{{P_0}}}{{{P_{in}}}}\)

In terms of dB, the power gain is expressed as:

\(P(dB) = 10\log_{10}{\frac{P_{out}}{P_{in}}}\)

Property: log_x x = 1

Calculation:

With A_P = 100, we can write:

\(P(dB) = 10\log_{10}{A_P}=10\log_{10}{100}\)

\(P(dB) =10\log_{10}{10^2}\)

\(P(dB) =2\times10\log_{10}{10}\)

Since log₁₀10 = 1, the power gain in dB will be:

\(P(dB) =20 ~dB\)

Important Note:

It is the voltage gain or the current gain that is calculated as:

\(A_V=20log(\frac{V_2}{V_1})\)

\(A_I=20log(\frac{I_2}{I_1})\)