Multipliers in a nutshell, basically decide what speed your chip is going to run. Here's an example:
My processor runs at 3Ghz clock speed. To determine that, you would set the multiplier up as in my case anyway 15X200=3000 or 3Ghz.
To overclock the FSB, you advance the multiplier such as:
I think you get the point by now. So these little steps take the clock speed up on the processor.
As far as advantages go, it really depends on what your doing with the machine. Most people nowadays buy a lower clocked processor and overclock it to run at speeds equivalent of their higher clocked counterparts in essence, saving you money and gaining more performance.
Multipliers are one factor in determining how fast your processor runs. For example (and this is a simple one)
You have a processor with a 5 multipler.
The system bus is 100 Mhz
5 X 100 = 500 Mhz processor
It gets more complicated the newer machines get. Dual Data Rate and so on.
Most processors come locked these days, meaning you can't adjust the multipliers and only overclock them by changing the bus through the motherboard. I.E. changing it from 100 Mhz to 115 or 133 etc.
Some processors come unlocked like the ones mentioned earlier as well as the Athlon XP mobiles. Some of the first Athlons/Durons allowed you to unlock with them infamous pencil trick, but I wouldn't try that these days. You'll have an expensive paperweight.
Now I have a question. Obviously, increasing the voltage increases thermal output.
Does increasing the multiplier alone increase heat? For instance, I have my 3000+ @ 2.2 on 1.4 volts. If the proc was at 100% and I dropped the multiplier from 9 to 6, while maintaining the voltage and workload, would temps go down?
My gut feeling is "no", but if I knew, I wouldn't ask.
If the proc was at 100% and I dropped the multiplier from 9 to 6, while maintaining the voltage and workload, would temps go down?
Since the work load would be only 2/3s, with the speed reduced to 2/3s, yes, temps would go down. I would point out that it would take longer to get the same work done.
If however, you were increasing the base bus speed, to keep the workload constant, you would end up with a slightly higher temp (depending on OCability of your ram) and would probably have to raise the chip speed slightly to maintain perf.
I guess the laws of physics apply, so you have conservation of energy at work here.
So you are saying that the amount of work is responsible for the heat??
I'm not sure I understand.
If I have 100% workload @ 2.2 @1.5 volts, then I change the Vcore to 1.4 heat drops. I'm sure we'd both agree here.
At a multiplier of 6 my chip is stable at 1.1 Vcore (probably less).
I realize that less work gets done at a lower multiplier, but what I'm asking is if less heat is produced at the same voltage.
As far as I know, if you underclock, there is less heat. However, the main advantage of underclocking is that because you are stressing the hardware less you can also undervolt. The best heat reductions are gained by doing both. But if you wanted to only do one, either one, you should still get less heat.
Think of the clock itself as cycle of how many times electricity travels through the CPU per second. And the voltage itself is how much electricity goes through each cycle. Increasing either the number of the cycles per second or the amount of voltage per cycle will increase the total amount of electricity per second. Granted, this is a very generic way of looking at it.
And, of course, the inverse is true. That's why overclocking requires a good heatsink. :wink: