Power and gain are not the same thing. Assuming sufficient current, a 100 watt amp with 20 dB gain has the same power as a 100 watt Amp with 29 dB gain--100 watts. It will have the same ability to control speaker movement and will possibly be better with speakers that are considered difficult loads. It just won't get as loud, again assuming that both amps have enough current. I think that might be the point you are trying to make, but it might confuse people the way it was worded. It is also worth noting that once you reach the max input level required for full power, applying more voltage using a preamp will make no difference whatsoever, other than possibly overloading the input circuitry. I'm not sure what the required level is for the TPA3255, but I would guess around 1v or less, and almost certainly not more than 2v.
I know what Power is, and I know what Gain is. The difference between a 100W output with 20dB gain vs 100W output with 29db gain is the on the first one your input signal needs to be almost 3 times higher to reach the same 100W(4ohms speakers?) So, Power=Voltage^2/Z (I know this method is not very accurate because Impedance changes with Frequency, but let's assume 4ohms) How much output voltage (don't confuse with DC input voltage) swing we need to produce 100W output power? (100W*4ohms)^(1/2)=V =>V= 20V rms.
I used this website to convert db Gain to Voltage Gain:
http://www.muzique.com/schem/gain.htm
First scenario:
20db Gain= 10x voltage ratio => 10= Vout/Vin => Vin=20V/10 = 2V
On the other scenario:
29db = 28.18x voltage ratio => 28.18 = Vout/Vin => Vin=20V/28.18 = 0.709V
Now you know the importance of Input signal voltage and Gain on an amplifier.
What do you think will happen if you feed a 0.2Vrms Audio(from real music, not pure tones) signal to both amplifiers?
On the first amplifier:
Gain=10=Vout/Vin => 10= Vout/0.2V => Vout=2Vrms. P=V^2/Z => P=2^2 / 4 = 1W
On the second amplifier:
Gain = 28.18 = Vout/Vin => 28.18=Vout/0.2V =>Vout=5.636Vrms. P=V^2/Z => P = 5.636^2/4 = 7.94W.
Remember how Amir used 0.157V and the Volume pot at 3 o' clock to reach 5W?
29.125db Gain = 28.59 Voltage Gain. Then, (0.157*28.59)^2 / 4 = 5.0369W.
29.216dB Gain = 28.89 Voltage Gain. Then, (0.157*28.89)^2 / 4 = 5.1432W.
It cannot get more accurate than that:
At maximum volume on the pot it would reach aprox 8W.
So you will ask how did he reach:
Well, he just increased the input signal to 2Vrms from his signal generator and used the volume pot to control output power until the amplifier clipped.
See the importance of the input signal? Of course there are other factors: DC input voltage is important because it dictates how much output voltage swing the amplifier can reach. Current is important because there is no power without current. Input voltage noise and ripple is important to reach low THD+N levels on the amplifier. Power supply output capacitors (capacitance) are important because it determines how much instant peak power the power supply can provide. Etc....
I mean, people can do whatever they please. If they want to buy a 1GW power supply then go on. But if your input signal remains the same (low) you won't see a benefit.
Again, my first question is how much output power are you really using from your amplifier. Did you measure it? If you tell me ok, I'm using 80-100W and I want more power, ok. In that case it is really power what is limiting you. But if you tell me that your power is 1-5Watts, then the power supply is not the limiting factor. In that case just increase the input signal voltage and you will get plenty of extra power.