What’s the relationship between gain (dB) and power (watts)?

This is where my technical ignorance seems significant.  Maybe I’d need a textbook to understand this:


Well, think about how much an amp can do.  Amps can't have infinite power, or output voltage, right?  They are limited devices.  Every one of them. What limits them?  Among many things, the voltage available to the output circuits. That is the absolute limit of output.

Your gain is 20x (28 dB) for instance.  OK, but your maximum voltage out is 10 Volts peak to peak.

1 volt pk-pk in x 20 = 20 Volts pk-pk. 

With this amp, 20 volts out is not possible.   You need a bigger amp.  What you'll get instead is a clipped signal that goes up to 10 and hangs there until the input voltage goes back down.

@erik_squires - Thank you. That’s helpful.

Well, think about how much an amp can do. Amps can't have infinite power, or output voltage, right? They are limited devices. Every one of them. What limits them? Among many things, the voltage available to the output circuits. That is the absolute limit of output.

Your gain is 20x (28 dB) for instance. OK, but your maximum voltage out is 10 Volts peak to peak.

1 volt pk-pk in x 20 = 20 Volts pk-pk.

With this amp, 20 volts out is not possible.   You need a bigger amp. What you'll get instead is a clipped signal that goes up to 10 and hangs there until the input voltage goes back down.

Look at it this way.

A Designer has design requirements.

Design an amp that produces 250 watts per channel with an input of x mV into an 8 ohm load that will produce the 250 watts per channel max without clipping.  an input impedance of y Ohms, output impedance of P Ohms, frequency range of 20 Hz to 20 kHz, etc.

This gives you the gain of the amp.  As MillerCarbon and many other have told you, it then is a straight forward calculation.

There are typical standards in many amp designs.   They call for an input voltage of x MV in order to produce max (before clipping) output of xxx watts per channel into an 8 ohm load.  So,  most audio amps have a set gain with the input sensitivity voltage of x mV to produce that max output.

This is electronics engineering 101.

The amp specs will tell you the amps gain based on the designed for input voltage to get that max power output.

The amp in question, has a gain switch that changes the amps gain.  They are either adding another gain stage in the amp via the switch to allow for lower input sensitivity or they are adjusting the input sensitivity and adding or removing a gain stage or just adjusting the input sensitivity.

either way, what MillerCarbon and some others have described is pretty accurate.

Its not rocket science, but it does require some knowledge of electronic design.

If you start with the required input voltage sensitivity, the required output power rating, the required load, input impedance requirement, output impedance requirements, the amp will have a calculated gain.

If you start with a set gain requirement and power output, into a set load, you still need to know what the input sensitivity is.

Pre-amps are pretty similar, except that there are standard pre-amp input voltage sensitivities, and max output voltage requirements (so you don't blow up your amp and speakers). 

anyway, I hope this didn't confuse you further.

enjoy 

To further confuse the issue,

Changes in amp output, measured in dB volts are directly related to SPL dB.
That is, increase amp output by 3dB and the SPL will increase the same.  Like with amps though, speakers have physical limits which cannot be exceeded

The practical difference is about reducing noise thats generated from the pre - amp feeding the amp. A pre-amp with a high output signal may deliver unwanted hum or hiss when the input sensitivity switch is at the higher setting. Along the same lines, if really efficient speakers are being used residual noise from the pre-amp may be audible; flipping the amp input switch to the lower setting will tend to reduce the noise.