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I found these formulas for different gains of an amplifier/filter on logarithmic scale. The topmost is power gain, simply 10 times the logarithm of the ratio of output to input power. The next one is voltage gain. But isn't the voltage gain here the exact same as the power gain, since $$P = \frac{V^2}{R}$$?

Aren't the two gains equal?

Why isn't power gain defined as $$10 log\frac{V_{out}}{V_{in}}$$?

EDIT:

Here is what my textbook says:

The ratio of input and output amplitudes $$|T(s)|=\frac{|u_o|}{|u_i|}$$ is usually expressed in decibels $$|T(s)|_{dB}=20*log*|T(s)|$$

So why does the factor of 20 make sense here? Doesn't this give the ratio of the squares of the voltages in dB rather than the voltage gain in dB? Isn't "voltage gain" using that definition still power gain

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    \$\begingroup\$ Remember this from high school math? log a² = 2*log a \$\endgroup\$ – Bart Oct 5 '20 at 12:45
  • \$\begingroup\$ I suggest you remove "But isn't the voltage gain here the exact same as the power gain since..." since it greatly obfuscates what your question is. It makes it sound like you don't understand the relation ship between squared and 20. \$\endgroup\$ – DKNguyen Oct 5 '20 at 14:57
  • \$\begingroup\$ Furthermore, for "Why isn't power gain defined as: $$10 log\frac{V_{out}}{V_{in}}$$ do you actually mean voltage gain? \$\endgroup\$ – DKNguyen Oct 5 '20 at 15:00
  • \$\begingroup\$ I think your voltage and current gain definitions are incorrect. The voltage gain is defined as: \$ 20log_{10}(\frac{V_{out}}{V_{in}) \$. It does not contain the ratio of the input and output impedance as you have defined. Similarly for the current gain. You have just defined power gain in three different ways. \$\endgroup\$ – sarthak Oct 5 '20 at 15:12
  • \$\begingroup\$ @sarthak No, the OP is correct. There is no such thing as voltage or current gain in dB. dB is always power gain. What you can do, however, is write power on the dB scale in terms of voltages and currents. \$\endgroup\$ – DKNguyen Oct 5 '20 at 15:23
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dB is never voltage or current gain. It is always power gain. Your equations that find dB using voltage or current are not finding voltage or current gain. They are finding power gain using voltage or current.

If I say something increased by 10dB, I am saying it increased by 10dB in power. Not voltage, current, nor any other amplitude. You can determine how much an amplitude increased from the 10dB increase in power, but the 10dB itself is always talking about the power increase.

I don't know why it was defined this way, but it is.

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  • \$\begingroup\$ So basically my textbook made an error? They define T(s) as the ratio of input and output voltage, and then use the factor of 20 before the logarithm on dB scale? That is, they define T(s) first in normal scale as a ratio of voltages, and then in log scale as a ratio of powers? \$\endgroup\$ – S. Rotos Oct 5 '20 at 18:06
  • \$\begingroup\$ @S.Rotos No it's not an error but a subtle interpretation that can be confusing if you're not used to it. Whenever you convert an amplitude gain on the linear scale into dB, you are converting it to a power gain. Similarly, whenever you convert a power gain in dB to an amplitude gain on the linear scale, you are converting it to an amplitude gain. \$T(s)_{dB}\$ is the power gain in dB that corresponds to the amplitude gain of T(s) on the linear scale. \$\endgroup\$ – DKNguyen Oct 5 '20 at 19:41
  • \$\begingroup\$ For example, if \$T(s)\$ is a linear voltage gain of 2, then \$T(s)_{dB}\$ is a power gain of 6dB, which is a linear power gain of 4. Since dB is always power you never say that a linear voltage gain of 2 is a gain of 3dB, since dB is always a power gain. Saying the gain is 3dB means the power doubled (not the amplitude) which means that the amplitude had to increase by \$\sqrt 2\$ \$\endgroup\$ – DKNguyen Oct 5 '20 at 19:47
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Because Rin may not be the same as Rout.

These formulae take that into account, essentially by translating the voltage gain and current gain into power gain.

Consider a buffer amplifier with a voltage gain of 1, an input impedance of 10000R, and a load impedance of 10R. The power increases 1000 times (power gain = 30dB) but the voltage gain is still 0dB. because Rout = Rin/1000.

The factor of 20 (not 10) in voltage gain does indeed reflect the fact that power is proportional to voltage squared.

If you pick a single value for R you can see that all three expressions for gain yield the same value : it makes no difference whether you compute 20log(V) or 10log(V^2) but the former is more convenient.

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    \$\begingroup\$ I rather think his confusion is the fact that the log of a number squared is twice the log of the number unsquared. \$\endgroup\$ – Bart Oct 5 '20 at 12:48
  • \$\begingroup\$ Bart is right, but I don't see why either answer addresses my question. Isn't "voltage gain" using that definition still power gain? \$\endgroup\$ – S. Rotos Oct 5 '20 at 14:22
  • \$\begingroup\$ I edited my question. \$\endgroup\$ – S. Rotos Oct 5 '20 at 14:30
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    \$\begingroup\$ @S.Rotos dB is always a measure of power gain. It's a little confusing when you're first learning it. \$\endgroup\$ – Hearth Oct 5 '20 at 14:54
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Equations below for power gain and voltage gain.

dB equations

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