It will not draw 2A continuous if it dont carry any signal. But of you ask it ro produce 30W, it will need 2A to do that. It will also need 2A to produce a shorter burst of 30W... So, 2A ;-) So if you crank it you will need 2A. At 15W output, half the current will be needed if still fed 24V. Its all tied tighter like:
P=UxI (Power equals voltage times current)
I=P/U
I=30/24=1,25A... amp efficiency say 75% -> 1,7A...
//
P=UxI (Power equals voltage times current)
I=P/U
I=30/24=1,25A... amp efficiency say 75% -> 1,7A...
//
It is not relevant because the input current of a class-D amplifier, irrespective of the load, is discontinuous and therefore louder sounds mean wider (vs. taller) current pulses that are mostly supplied by the bus capacitors (due to their pulsed nature).
Of course, in an average sense, the load current would still be supplied from the DC bus, but it being chopped up in very small pieces (high fs) really makes things easier. For example, consider a buck converter that supplies 5A at 20V from 100V. The input current would still be 5A pulses, but at duty ratio of only 0.2, giving an average input current of 5A x 0.2 = 1A only. However, at a higher duty (typical for a D-amp), all you'd need a larger bypass capacitor to prevent the DC bus from sagging with the battery supplying only the average input current.
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