What should my CPU Cache Voltage be set to?

AzurePaladin77

Honorable
Aug 31, 2013
3
0
10,510
Hi,

I have my i5-4670k working at 4.7GHZ with a Core Voltage of 1.150. It is set to adaptive and so is the CPU Cache Voltage, but in Asus Suite III it says the Cache OC Voltage is set to 1.275. The Multiplier for the Core and Cache is 47. Should I lower the Cache OC Voltage, or leave it as it is? I am using:

Corsair H50 CPU Water Cooler
Asus Z87-A Motherboard
Intel i5-4670K CPU
16GB 2400MHZ Memory
500GB Samsung SSD
2TB Toshiba HDD
(2) Geforce GTX 660s in SLI
750w OCZ Power Supply
Windows 8 Pro 64-bit

Thanks in advance for any help/advice.
 
Solution
Normally you don't need to play around with that setting. That CPU is around its wall. 4670k gets 4.0 - 4.8ish. 5ghz is rare with out good cooling. Your Core voltage can go up to 1.4 depending on your cooling solution. I recommend raising your vcore a little at a time, but if it doesn't stabilize then you probably hit your wall. Remember to actively check your temps.

imkvn

Distinguished
Jun 29, 2008
154
0
18,760
Normally you don't need to play around with that setting. That CPU is around its wall. 4670k gets 4.0 - 4.8ish. 5ghz is rare with out good cooling. Your Core voltage can go up to 1.4 depending on your cooling solution. I recommend raising your vcore a little at a time, but if it doesn't stabilize then you probably hit your wall. Remember to actively check your temps.
 
Solution

AzurePaladin77

Honorable
Aug 31, 2013
3
0
10,510


Thanks for the info. I forgot to mention I'm using a Corsair H50 CPU Water Cooler. I'm getting great temps, I just wanted to know if the CPU Cache OC Voltage was too high compared to my low Core Voltage. If you say I can just leave it as it is without it reducing the length of life of the chip then that's great. I tried to go up to 4.8GHZ but It froze in windows at around 1.28 and I didn't want to try 1.30 because it just seemed to high to be worth it. I want this chip to last for a while and for a 100Mhz increase it didn't seem worth it to have to increase the voltage by so much more.