It's supposed to do that. The voltages help to make the jump from no load to load less of a power jump for the processor. Let's say the proc is recieving 1.31V, but when load is placed on the processor, there is a small amount(not noticeable to human's) where the processor would take in the voltage and since this isn't recognized by the system until after it happens then voltage is pumped through to the processor( this would be a drop in voltage until the system can compensate).
Then there is the other way: load to idle. A smaller voltage(1.25V) allows the system to stop supplying voltage to the processor when it stops using the voltage is has(thus a spike in voltage occurs).
I hope this isn't too confusing, I'm usually bad at explaining ideas even if I know them well
I understand, I am pretty good with electrical stuff. It is just I have never overclock before, so I don't know if it is supposed to drop this much. This is 0.06V, which means I am using close to 40Amps.
That's a lot more than what I thought a CPU can draw.
I have no idea how you translate 0.06 VDC drop to a 40 amp load. Somehow you are assuming a load resistance of 0.0015 ohms. Why?
On my older system I saw something like this when set up years ago. Is is an ASUS mobo. At the time I found on ASUS forums that they routinely set their systems to supply to the CPU under near-no-load conditions a core voltage about 0.05 VDC higher than you actually specify in your BIOS settings. This is in recognition that the NORMAL behavior is that Vcore will actually drop by about 0.05 volts when the CPU is under load. I presume this is related to some minor resistance between the point where the mobo supply voltage is measured and regulated, and the point within the CPU where Vcore is measured.