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?


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.


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.

  • \$\begingroup\$ -1. Not because it's wrong - it isn't - but because it doesn't really answer the question. It doesn't explain why it is this way, it just adds some background information. \$\endgroup\$ – Keelan Jun 2 '13 at 16:45
  • \$\begingroup\$ @CamilStaps OK see ya point - I've added details I think the OP is looking for to help him \$\endgroup\$ – Andy aka Jun 2 '13 at 17:00

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.


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.

  • \$\begingroup\$ Would $10 be ten decibels more than $1? Or did we lose the logarithmic relationship somewhere? :-) \$\endgroup\$ – Anindo Ghosh Jun 3 '13 at 6:53

protected by W5VO Jun 2 '13 at 20:13

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