Am I correct in calculating the current draw of an amplifier?

Hello,

As you can see, I'm a newly registered user and a complete noob in a CarPC subject. I want to end up having a working system in my car (well, duh;)) and now starting to explore this subject in depth.

My question is about calculation of amplifier current draw (and choosing the appropriate PSU). If I take, for example, a TDA7850 amplifier, and look at its datasheet (or specs), I see the following:

Quote:

...

Specifications:

4 x 50 W/4 Ω max.

4 x 30 W/4 Ω @ 14.4 V, 1 kHz, 10 %

4 x 80 W/2 Ω max.

4 x 55 W/2 Ω @ 14.4V, 1 kHz, 10 %

...

As explained here, I should do the calculation as follows:

4x30=120W RMS output (for 4ohm impedance speakers). This is an output wattage for a single test tone (pure sine). As the class of this amp is AB, multiply it, roughly, by 2 to get the consumed power (50% efficiency). We have 240W RMS. Now divide it by the input voltage (~14V) and we get a drawn current of approx. 17A. Now divide it by 3 (As explained, music signal is never a "pure sine" and requires approx. 3 times less power) to get the drawn current at full volume. It comes to 6A.

Now, the approx. power that will be consumed by the amp (when the engine is running) is 6X14=84W.

The question is if this is the correct way for calculating power consumption? I found a couple of sites where there wasn't any division by 3. Does it mean that an 90W spare on PSU will power this amp without any problems? I've searched this site, but wasn't able to find a concrete answer to this.

Another question I want to ask is about the efficiency of this specific amplifier. If I look at its datasheet (link), I see from the efficiency vs. Pout graph (p. 12) that its efficiency is 85% at Pout=30W. Isn't it too high for class AB (as I saw in a couple of places, the efficiency for AB class ranges 50~70%)?

I want to apologize for the apparently "noob quetions" and would appreciate any help!

Thank you very much and sorry for the long post!:happy:

Alexei