The first spec is your gain in dB (The figure that makes the most sense to me the non-technically inclined).

The second seems to be the amp's sensitivity.

Below is a response to a question that I posted about amp sensitivity & its relation to gain on Audiogon. Almarg who responded is someone I regard as extremely knowledgeable.

From Almarg on Audiogon:

Roscoeiii: Am I understanding input sensitivity correctly, as follows?: The higher the input sensitivity (say 2V to reach full power, as opposed to 1V to reach full power), the more travel I will have in my preamp volume control. So someone without sufficient fine tuning in volume control or who wants to move out of the 8-10 range and more into the 10-1 o'clock range should try to find an amp with higher input sensitivity.

Do I have this right?

Almarg: Yes, provided that "full power" is the same in both cases.

But just to confuse the issue further, your statement would be more precise if it referred to a higher input sensitivity NUMBER, rather than to a higher input sensitivity. A higher input sensitivity number, properly speaking, corresponds to lower sensitivity. In other words, a more sensitive amp is one that requires less input voltage to reach full power.

Roscoeiii: I am not clear on if input sensitivity can give an idea of the amount of gain in dB (and what other specs might be needed to calculate dB gain from input sensitivity).

Almarg: If gain is not explicitly specified, and it often is not, it can be calculated to a reasonable approximation from the specified sensitivity and the specified maximum power capability.

As you realize, gain is the ratio of output voltage to input voltage, expressed in db.

The ratio of two voltages, V1 and V2, are converted to db based on the formula (20 x logarithm(V1/V2)).

The rated maximum output power into 8 ohms can be converted into voltage based on the equation P = (Vsquared)/R, where P is power in watts, V is volts, and R is resistance in ohms.

So for example in the case of an amplifier rated at 200 watts into 8 ohms and having 2 volt sensitivity, if we represent the output voltage corresponding to the 200 watts into 8 ohms as "Vout" and the 2 volt sensitivity as "Vin," we have:

200 = ((Vout)squared)/8

From which it can be calculated that Vout = 40 volts

Therefore the gain in db is

20 x log(Vout/Vin) = 20 x log(40/2) = 26 db

That is an approximation, as I indicated, in part because it does not reflect margin that may be built into the maximum power specification. But it will generally be a reasonable approximation for most purposes.

Regards,

-- Al