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The definition of voltage gain is \$V_{\text{out}}/V_{\text{in}}\$.

However, I read some articles about the gain in decibels, and I have a confusion now.

Here is an article about it: https://en.wikipedia.org/wiki/Gain#Voltage_gain

Here, I understand the definition of Power gain in decibels, which is $$ \text{Gain} = 10 \log \left( {P_{\text{out}} \over P_{\text{in}}} \right)\text{ dB} $$

However, I can't understand why Voltage gain in decibels is $$ 20 \log \left( {V_{\text{out}} \over V_{\text{in}}} \right)\text{ dB} $$

If \$ 20 \log \left( {V_{\text{out}} \over V_{\text{in}}} \right)\text{ dB} \$ is derived from $$ 10 \log {\left( {V_{\text{out}}^2 \over R_{\text{out}}}\right) \over \left({V_{\text{in}}^2 \over R_{\text{in}}}\right) }\text{ dB} $$ then this is the power gain, not the voltage gain, isn't it? However, the Wikipedia says it is a formula for the Voltage gain in decibels. I thought the voltage gain in decibels would be \$ 10 \log \left( {V_{\text{out}} \over V_{\text{in}}} \right)\text{ dB} \$. Actually, the example section in that linked page uses voltage gain \$ V_{\text{out}} \over V_{\text{in}} \$.

Why did \$V^2/R\$ suddenly come out from the voltage gain in decibels?

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4 Answers 4

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As you say, the decibel is a unit of power ratio.

\$ G\ [\mathrm{dB}] = 10 \log_{10}\left(\dfrac{P_1}{P_2}\right)\$.

When the input and output impedances are equal and then we can express the gain in terms of voltage as

\$ G\ [\mathrm{dB}] = 20 \log_{10}\left(\dfrac{V_1}{V_2}\right) \$

I wouldn't call this the "voltage gain in decibels." I'd rather say it's the decibel gain, calculated from the voltage gain.

Sometimes, you will see a voltage gain expressed in decibel according to this formula even when the input and output impedances are different. There is no technical justification for this --- it's simply a shorthand practice that's become common through usage.

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I'm going to say something that you might initially think of as totally wrong:

If the voltage gain of a circuit is 6dB, the power gain is also 6dB

  • To produce 6dB voltage gain requires a voltage gain of 2 and 20log(2) = 6.02dB
  • To produce 6dB power gain requires a power gain of 4 and 10log(4) = 6.02dB

This means you can talk about gain and not worry whether it is power or voltage gain - it is either or both.

For the same input and output impedance in non-dB terms, if the voltage gain is G, the power gain is \$ G^2 \$. The square term within the log part of the formula becomes a "multiply-by-2" term outside the log part hence 10dB becomes 20dB.

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This isn't more than a definition. For power quantities, you use 10. For field quantities, you use 20.

From that Wikipedia:

The equivalence of \$10 \log_{10} \frac{a^2}{b^2} \$ and \$20 \log_{10} \frac{a}{b}\$ is one of the standard properties of logarithms.

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This is an important convention, because at the end of the signal chain, in areas such as communication or audio, voltage ultimately turns to power. Decibels are a relative power measure that comes from audio: the measure of audio level loss in long distance telephone circuits.

The bel and decibel are not intended to be ways of reducing arbitrary relative measurements to a log scale.

For instance, we don't say that a 10 kΩ resistor is 3 decibels more resistive than a 5 kΩ resistor! That would be a joke, like the one in The Art of Electronics (Horowitz and Hill):

[A] gold-plated op-amp for this application is the ultra-low-noise LT1028, which is 13 dB quieter and only 10 dB more expensive [...]

So the \$10 \log_{10}\$ scale of a relative voltage measurement may be valid in some situation, but it's just not called decibels, any more than something that costs $10 being called ten decibels more expensive than a $1 part.

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  • \$\begingroup\$ Would $10 be ten decibels more than $1? Or did we lose the logarithmic relationship somewhere? :-) \$\endgroup\$ Jun 3, 2013 at 6:53
  • \$\begingroup\$ Actually yes, 10*log(10) = 10. Sorry for commenting all these years later. :-p \$\endgroup\$
    – anon
    May 4, 2020 at 9:47

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