Why does increasing voltage make an overclock more stable?

[Giscardo]

I have been overclock since the days of the Cyrix 6x86 PR166+ (ran at 133). I was reading tom's hardware before anandtech existed, and i read anand since his first reviews. Despite all this, I'm still not sure as to why increasing the voltage would make an overclocked cpu more stable? It seems counterintuitive, becasue wouldn't more voltage make teh cpu run hotter, and therefore less stable?

 

[jobberd]

correct me if im wrong, but when you overclock, it doesn't make the cpu run any hotter itself. However, the cpu cannot handle all the extra input/output it has to deal with, which is why it might crash. So you up the voltage, and the cpu has more power to work with, and therefore becomes more stable (although the added heat decays its lifespan)

 

[skull]

There are gaps in the circuits that the electricity goes though with lower freqency it can go around the gaps when you increase the clock speed it doesn't have time to go around so increased voltage makes it kind of arc accross.

 

[L3Guy]

Giscardo;

I am going to guess here.
If the CPU is thermally limited, you are absolutely correct.
However, if one section of the chip is RC drive limited,
a bit of extra voltage can increase speed in the areas that are marginal at default voltages.

Why not keep increasing voltage? Because after some limit, the voltage will either cause thermal limits, or start to punch through the fet oxide.

Just my two cents.

Doug

 

[jeremy806]

ok dudes. Here is my thoughts on the subject.

Power stored in a capicitor is 1/2 x CV^2.

Power dis by a CPU is fCV^2.

Increasing f or V increases the CPU power dis.

CPU temp will increase until the watts can be dissipated. That is, a HSF solution dissipates so many Watts/Kelvin, etc.

Reason voltage adds to stability: The chip has a high voltage, a low voltage and a middle (for example 2.0V, 0, and 1.0V). A '1' is when the voltage gets around the high voltage. A '0' is when the voltage gets around the low voltage (noise margins). When you increase f, the circuit is switching faster. Errors occur if the voltage does not get within the margin fast enough. Increasing voltage causes faster switching, assuming that the '0' and the '1' are a fixed distance from the middle voltage. (Example, you want to go 25 mph in your car, you can push the petal down half way or all the way, both will get you there but flooring it will get you there faster.)

So, you increase f to speed things up, you increase V to avoid errors, you add a better HSF to dissipate the extra heat that you just added to the system.


jeremy806

 

[Maverick]

man I learn so much useful information in this forum. What Jeremy said makes lots of sense.

 

[Demon-Xanth]

Basically it helps drive the signal over the threashold voltage since all circuits have some RLC effects.

 

[br0wn]

The equation for maximum frequency (fmax) relates to voltage (V) is:

fmax =~ (V - Vthreshold)^2 / V


So increase voltage, will increase the limit of a circuit to its maximum
frequency (linearly).

Unfortunately, increasing voltage has effect of increasing power consumption
(doubling voltage, roughly quadruple the power consumption).

 

[Freeze]

I think another reason is because of a feature know as current drive capability. The current in a transistor is non linear (to a degree, it has it's moments ). It's proportional to the Voltage squared (grossly simplified). This increased current will charge/discharge the interenal capacitors faster, thus increasing circuit speed (decreased delay times). It also increases stability by increasing the Noise Margins (basically how high a low can be until it's no longer a low and how low a high can be before it's no longer a high). This may only effect the high Noise margin though. This comes at a price though, namely higher power consumption and more heat. Plus there is a limit on how fast a chip can go based on the manufacturing process (as we all know). I could be off a little here, but I think I got the basics down.