# Decibels with Voltage Gain

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Decibels are also used to provide a comparison between voltage levels. Substituting the basic power equations $P_{2}=V_{2}^{2} / R_{2}$ and $P_{1}=V_{1}^{2} / R_{1}$ into Eq. (1) will result in
\begin{aligned}\mathrm{dB} &=10 \log _{10} \frac{P_{2}}{P_{1}}=10 \log _{10} \frac{V_{2}^{2} / R_{2}}{V_{1}^{2} / R_{1}} \\&=10 \log _{10} \frac{V_{2}^{2} / V_{1}^{2}}{R_{2} / R_{1}}=10 \log _{10}\left(\frac{V_{2}}{V_{1}}\right)^{2}-10 \log _{10}\left(\frac{R_{2}}{R_{1}}\right) \\ \text { and } \quad \mathrm{dB}&=20 \log _{10} \frac{V_{2}}{V_{1}}-10 \log _{10} \frac{R_{2}}{R_{1}}\end{aligned}
For the situation where $R_{2}=R_{1}$, a condition normally assumed when comparing voltage levels on a decibel basis, the second term of the preceding equation will drop out $\left(\log _{10} 1=0\right)$, and
$$\mathrm{dB}_{v}=20 \log _{10} \frac{V_{2}}{V_{1}}$$
Note the subscript $V$ to define the decibel level obtained.
Example 1: Find the voltage gain in dB of a system where the applied signal is 2 mV and the output voltage is 1.2 V.
Solution:
$$\mathrm{dB}_{v}=20 \log _{10} \frac{V_{o}}{V_{i}}=20 \log _{10} \frac{1.2 \mathrm{~V}}{2 \mathrm{mV}}=20 \log _{10} 600 = \mathbf{55.56} \mathbf{d B}$$
for a voltage gain $A_{V}=V_{o} / V_{i}$ of 600 .
Example 2: If a system has a voltage gain of 36 dB, find the applied voltage if the output voltage is 6.8 V.
Solution:
\begin{aligned}\mathrm{dB}_{v} &=20 \log _{10} \frac{V_{o}}{V_{i}} \\36 &=20 \log _{10} \frac{V_{o}}{V_{i}} \\1.8 &=\log _{10} \frac{V_{o}}{V_{i}}\end{aligned}
From the antilogarithm: and
$$\begin{array}{c}\frac{V_{o}}{V_{i}}=63.096 \\ V_{i}=\frac{V_{o}}{63.096}=\frac{6.8 \mathrm{~V}}{63.096}=\mathbf{1 0 7 . 7 7} \mathbf{~ m V}\end{array}$$

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