Same watts at 8 and 4 ohms?


I'm in the market for an integrated amp and trying to sort through tech specs. My understanding of the tech aspects of hi-fi gear is limited. Looking for some clarity in regard to watts-per-channel specs.

It is my understanding that wpc at 4 ohms is typically 1.5x -2x the wpc at 8 ohms.

But I'm seeing a number of respectable mid-fi integrateds with the same wpc for both 8 and 4 ohms. The NAD 388 is one and I think this is true for several of the Cambridge Audio units at a similar price point ($1500-$2000).

The NAD features make a point of saying " 4-ohm stable for use with a wide range of speakers". 

Would appreciate any insight to what these specs mean and what 4 ohm stable really means to me. My speakers are 4 ohm speakers.

Thanks,

George
n80
     Two words:  Marginal power supply.  Compare a Superamp of its era, such as a 125 wpc Marantz, or a 125 wpc Audire or 200 wpc Brystom amp and you will hear. 
     Taking up space inside the Marantz is a pretty decent amp. for what it is, but it also has a tuner and dial, a preamp, all kinds of switch gear, etc. Inside the Bryston are two 375  watt transformers and 8 4,000 mf filter caps, plus 8 output transistors with heat sinks.  The Audire has 2 500 watt transformers, 8 26,000 mf filter caps, and 12 outputs. All newer Audire amps have many more outputs.  The sound is what you would expect, as is the actual power available.  Running 4 B&W's through these three, the first thing you notice is how much bass the Marantz loses at higher volumes, followed by the Bryston and the Audire.  The Audire is rated at 125/250/ 400 wpc at 8/4/2 ohms respectively, the Bryston about 300 at 4 ohms, and is not designed to run at 2 ohms. The Marantz just a wee bit more than 125 at 4 ohms, and shows it when pushed.
     Why do you think a 125watt x 5  A/V receiver (even Marantz) cannot pus a single set of speakers and a non-powered sub? 
     Weight can matter.  more stuff weights more, especially with big heat sinks.     Interestingly, my recently refurbished (At Bryston!) Bryston preamp runs hotter than my Audire amps did at full tilt pushing 4 inefficient. speakers. 
Danvignau what is your point none of those amps even come close to doubling down when impedance is halved. My krell has a 5kva transformer and 260,000 uf of capacitance and 56 output devices but it even can't technically double down on power with halved impedance.

Better amps have very robust power supplies that can provide the extra current required to double the power.
As I said before it’s not just the power supply, but the output stage as well. Especially the use of BJT (bi-polar) output stages in complimentary NPN/PNP push pull configuration over using complimentary Mosfets as the P channel is Mosfets weak one the N Channel is fine.

My krell has a 5kva transformer and 260,000 uf of capacitance and 56 output devices but it even can’t technically double down on power with halved impedance.
But it comes close

Cheers George
It does come close, but I need another one to vertically bi-amp my apogee duetta 2 speakers, they are very inefficient but sound great.
invalid,

Of course you can get more power.  It is ludicrous to think otherwise.  An amp is like (bit of an exaggeration to get the point across) your mains amplifying a 60 Hz signal.  If you connect a 120 ohm load to mains, you will get 120 watts, if you connect a 60 ohm load, you will get 240 watts, if you connect a 30 ohm load, you will get 480 watts, etc, until you blow your fuse.

AS LONG AS YOUR MAINS DOES NOT COLLAPSE AND YOU DO NOT EXCEED YOUR CURRENT LIMIT, you WILL get more power.  It is virtually the same in amps.  It is like supplying a 60 hz continuous signal.  As long as the power supply can supply the needed current, it will keep on supplying more power.  The limitation with amps is the heatsink/output stage design (transistor current limits), the maximum internal impedance and its stability at different frequencies.  The ideal amp (again, an asymptotic exaggeration) will even handle a short cirtcuit across its speaker where the current will be limited by only the internal impedance of the output stage.  This obviously will blow the transistors but you get my point.