# DB to voltage gain conversion

#### Wendy

Joined Mar 24, 2008
23,320
It has been a very long time since I have done this kind of math. I have a tube type preamplifier for a friend. I want to make a solid-state equivalent just to test it against.
This amplifier has a gain of 39 BB, 42 DB, 45DB. I think without knowing that if I divide these numbers by 20 DB I will end up with a voltage gain, correct?
39DB=1.95 Gain
42DB= 2.1 gain
45 DB= 2.25 gain
is this correct or is my brain fart smellier than usual?

#### Papabravo

Joined Feb 24, 2006
20,626
The division by 20 is for voltage gain or current gain. For power gain you would use a factor of 10 because power is proportional to voltage or current squared.

$$dB(\text{Power})\;=\;10 *log_{10}\left( \cfrac{P_{out}}{P_{in}}\right)$$

$$Gain(\text {Power})\;=\; 10^{\left ({\cfrac{dB(\text {Power})}{10}}\right)}$$

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#### crutschow

Joined Mar 14, 2008
33,370
Note that, if it's not clear, where it says *Decibel values are with respect to zero reference, that means referenced to a gain of 1 which is 0dB.

#### crutschow

Joined Mar 14, 2008
33,370
The division by 20 is for voltage gain or current gain. For power gain you would use a factor of 10 because power is proportional to voltage or current squared.
The definition of dB is for relative power levels, where the division by 10 is used, and division by 20 is correct if you are measuring the difference in two volage or current levels at the same node.
It is also commonly used for voltage and current gain measurements from the input to output of a device, although that's only technically correct for equal input and output impedances of the device (but it's understood that requirement is mostly ignored).