Relationship of opamp gain factor to dB?

[dpresley58]

I've used the search, but found nothing that relates directly to this question...

How does a Gain factor equate to dB?

From what I've been able to find, the dBs double for every factor of Gain. For instance, a Gain of 1<unity> = 0dB increase. Gain of 2 = 3dB, so a Gain of 3 would be 6dB. Is this correct?

[brett]

Hi
dB is a non-proportional scale.
a voltage gain of 10 = 10 dB
100 = 20 dB
1000 = 30 dB

The formula is: Take log10(voltage gain) x 10, where log10 is log in base 10.  Your scientific calculator or spreadsheet will have those functions.
cheers
PS sometimes you'll see power or signal to noise ratio or other things in dB instead of watts.  Even when it doesn't say so, this is the ratio of two signals.  Because power gain is the square of voltage gain, the formula above changes to log10(square(voltage gain)) x 10 or log(voltage gain) x 20.

[R.G.]

Gain *IS* db, just expressed a different way.

Actually, strictly speaking, db is the ratio of two powers. One of the powers is the reference power, the other is the power being measured.

For situations like that, brett is correct. If you have an input power and an output power, the power gain is Gp = Pout/Pin. So if Pout = 100W and Pin = 1W, then Gp = 100.

In radio work, the ratio of powers is very wide. You may speak of transmit power in MW and received power in nW, a ratio of 1: 1000,000,000,000,000. It's much more convenient to express the ratio as the EXPONENT of the ratio. So log10 (1,000,000,000,000,000)  = 15. This was defined as a "BEL". That turned out to be unwieldily large, so people started using "DECI-bels" or 1/10 of a bel.

So the ratio of 1MW to 1nW is 1/10E15, or in Bels, 15 Bel, or 150db, db standing for "decibels" and the way you calculate them is ten times the log to base 10 of the power ratio.

But we do audio, not RF. We usually want to know VOLTAGE ratio, not POWER ratio.

It turns out that if you have the same impedance level for one voltage and another, the ratio of the voltages can be accurately expressed as twenty times the log to base ten of the voltage ratios.  So if you have an input voltage of 0.1V from a 600 ohm source, and you get an output power of 32V into a 600 ohms load, the voltage ratio is
20 *Log10 (32/0.1) = 20 * log10 (320) = 20 * 2.505 = 50db.

Although it is only correct to speak of decibels in instances where the input and output impedances are the same, the audio world has ignored that, and we speak of voltage ratios without regard to the impedance. So a power amp which needs an input of 1.5V from a 10K source to give an output of 30V into 8 ohms is spoken of as having a gain of
20*log10 (30/1.5) = 20*log10(20) = 20*1.3 = 26db.

So "db" is simply another way to write "ratio of".

Some easy things to remember:
Ratio of 2 = 6db
Ratio of 3 = 9.54db
Ratio of 4 = 12 db (that is, ratio of 4 is the same as ratio of 2 times ratio of 2, or 6db + 6db because exponents add)
Ratio of 10 = 20 db

From this, we can easily state many different gains.

Gain of 100 is gain of 10 times gain of 10, or 20 +20 = 40db.
Gain of 300 is gain of 10 times gain of 10 times gain of 3; or 20 +20 +9.54 = 49.54db.