The difference is that you would now have two (posibly more effiecient) supplies and inefficient wiring, rather than one possibly inefficient one supply.
An example I have a device that draws ~6W (it is a d-link wireless router). The suplpy that came with it draws 21W from the mains supply. A waste of 15W, a better supply that replaced it draws 7W (85% efficient). Say for instance that I have a household supply to supply 12V from 240V. If it is supplying just this router (and I know I'm simplifiying here) and it is 90% efficient (only likely to be true near the full rated load - but I'll ignore that) then the supply in the device would need to be 93% efficient just to use the same amount of power (this is completely neglecting wiring losses)
Say that you did have a regulator on board (and I'm assuming a switching regulator) At low input voltages (12V) things like diode voltage drops (0.4V for a schottky) become important.
The voltage drop across the main switching transistor is another loss.
At 240V the 1.2V of diode (non schottky) is less of a concern (both because it is a smaller percentage of the supply and because it is carrying less current)
I don't deny that the idea sounds great, but I'm reasonably certain the math doesn't work.
I think the math still applies even to low power devices.
Most devices don't run at 12V so you wouod still need a wall wart of sorts to transform this into 5V or what ever your device requires.
Assuming that the two wall warts are both switching power supplies they are going to be similarly efficient (at least if they are well designed)
If you use 12V you will lose more power (given the same size wire) than if you use 240V
This is in addition to the fact that you have to have an 'extra' supply to give you the 12V in the first place - which adds another efficiency loss.
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The difference is that you would now have two (posibly more effiecient) supplies and inefficient wiring, rather than one possibly inefficient one supply.
An example I have a device that draws ~6W (it is a d-link wireless router). The suplpy that came with it draws 21W from the mains supply. A waste of 15W, a better supply that replaced it draws 7W (85% efficient). Say for instance that I have a household supply to supply 12V from 240V. If it is supplying just this router (and I know I'm simplifiying here) and it is 90% efficient (only likely to be true near the full rated load - but I'll ignore that) then the supply in the device would need to be 93% efficient just to use the same amount of power
(this is completely neglecting wiring losses)
At low input voltages (12V) things like diode voltage drops (0.4V for a schottky) become important. The voltage drop across the main switching transistor is another loss.
At 240V the 1.2V of diode (non schottky) is less of a concern (both because it is a smaller percentage of the supply and because it is carrying less current)
I don't deny that the idea sounds great, but I'm reasonably certain the math doesn't work.
I work for a power system company
A linear regulator throws energy away - it is less efficient than a switching supply
12V in 5V out 0.5A is a drop of 7V at 0.5A
3.5W wasted
I think the math still applies even to low power devices. Most devices don't run at 12V so you wouod still need a wall wart of sorts to transform this into 5V or what ever your device requires. Assuming that the two wall warts are both switching power supplies they are going to be similarly efficient (at least if they are well designed) If you use 12V you will lose more power (given the same size wire) than if you use 240V This is in addition to the fact that you have to have an 'extra' supply to give you the 12V in the first place - which adds another efficiency loss.