I find this statement quite amusing, especially when you have declared 14nm LPP superiority over Intel's processes, based on the high idle state voltage of a single Intel 14nm mobile product.
But no, really. The voltage is high even when compared to Bristol Ridge in general or even the same SKU itself. The ramp up in voltage beyond 3.5GHz is extremely rapid.
If voltage was what define exclusively power dissipation then Bristol Ridge would had higher TDP than a FX4300, but of course you didnt think about this, instead you are changing the goal post by bringing Intel completely irrelevantly, although there can be a usefull comparison since their 14nm use higher voltage than their 22nm, yet it has somewhat better efficency despite the higher voltage, isnt it...
As for GF 14nm efficency in respect of Intel s there s no CPU using that process that would allow a comparsion, but perhaps that you have numbers that suggest that i would be wrong..?.
A hint, Zen consumed less than BDW in the Blender test, so at equal throughput it looks like GF s 14nm is certainly not lacking efficency, and that s a fact, not a theory like your non substancied statement that is even contradicted by the few info available.
P = C.V^2....
I see two terms here that have a say in power comsumption of a capacitive switching device, if voltage is higher by a ratio x then it can be compensated by a capacitance that is smaller by a ratio 1/x^2.
@The Stilt:
Why are they volted so high? I know, Ohm's Law, current = voltage/resistance, but still...it seems like AMD silicon comes with a massive overvolt ex-works. I rather hope AM4 motherboards can run a sort of quick stability test suite and undervolt and/or overclock the CPU when installed.
I explained it above, voltage in isolation has no meaning, that s just a red hering used by some people to badmouth AMD.
As for current in function of voltage that s still the formulae current = Voltage/resistance but resistance is defined as being equal to 1/2pi.F.C with C being the capacitance of the circuit and F the frequency, so it s current = Voltage.2.pi.F.C