Can someone explain voltage to me?

V

Valdin

New Member
#1
I've been toying with o'clocking my PII 450 and am trying to get a feel for what is all involved and what components are affected.

I understand how the FBS and multiplier work to get the MHz. I don't get what roll the voltage plays in all of this. How do you set it? How does it affect the FBS or MHz or anything else? What, if any, are the dangers of changing the voltage?

Also, I have an ASUS P2B-L motherboard, 100MHz bus, PC100 RAM, CL TNT AGP card. If I increase the bus speed, how does that affect the RAM and AGP card? Do I need to do something to counter the effects of the bus increase? I thought I read that you would need to set something to 2/3 bus speed. What is that something? Any help is greatly apprecitated. Thanks!
 
S

Splatt

New Member
#2
Well, for starters, it's FSB - Front Side Bus. This is the speed at which the bus is working (basically the backbone of the mobo). The voltage is referring to the core voltage of the cpu. A PIII550 (for instance) is made to run at 100fsb and 2.0v. If you increase the fsb, then it may require more juice (volts) to run. So if you go to 124 you may need 2.05v or so... However, a cpu is a resistor. The more juice, the more resistamce, the more heat.... That's why when o/c'ing cooling is needed.


When o/c go up in the smallest increment you can on the fsb. If it runs, but is slightly unstable (Freezes, stutters, etc...) then you can try to up the voltage slightly...


Good luck
 
J

JL_in_Vail

New Member
#3
The mobo also plays a part in the voltage req's - every time the processor cycles, it takes a burst of current. More cycles = more current (and as Splatt said, more heat). Current = voltage/resistance (btw Splatt, if resistance was increased, current draw would drop, not increase). If the mobo can't reliably supply enough current for the CPU at 2.2v, it may at 2.3. (That's what all those capacitors around the CPU are for).

------------------
...hope this helps...
 
S

Sterling_Aug

Guest
#4
The actual quality of the processor also plays a BIG role in overclocking.

Some CPU's will not overclock well no matter HOW much voltage you apply to them.

My PII-450 runs ROCK STABLE at 527 MHz @ 1.90 volts and if the air temp in my house is cooler than 80°F, then I can run 568 MHz @ 2.20 volts. All of this is while pushing the processor at 100% CPU load for long periods of time.
 
S

Soul Apathy

New Member
#5
But here is the basic question...On an Asus board, how do you increase the voltage?? where would you measure the voltage using a multimeter to confirm it? I have not seen any voltage jumpers on Asus boards...I have an Asus P2BVE with a PIII 550..o/c at 633MHz 115 FSB..I have a cool processor..but my system locks up if I pass 115...I believe my other components can handle the increased bus speed...
 
StuTheWise

StuTheWise

Southpaw'ed Member
#6
Since no one else has been sarcastic (and seeing as how his question has been sufficiently answered), I shall take upon myself.
Voltage is the force that pushes power (amperage). Ohms is the measurement used to record the resistance involved. It takes one volt to push one amp through one ohm.
 
S

Soul Apathy

New Member
#7
Ok...can anyone answer the question, though...HOW to change the voltage on an ASUs P2BVE board...?????
 
Z

Zeke

New Member
#8
God, I feel such a fool! I've been using ammeters (Ampere-meters) to measure electrical CURRENT for the past 25 years!
 
J

JL_in_Vail

New Member
#9
Yo! -

Basic electrical theory:

V (voltage) = IR (current * resistance)

P (power, watts) = IV (current * voltage)

Electricity is not "pushed" - ever. It tries to get from high potential to low potential. Voltage is the difference in potential, and can be measured whether there is current flowing or not. When a circut is completed, current flows. Resistance (aka impedance) can be viewed as something which "blocks" the flow of current - the higher the resistance (measured in ohms), the lower the current. That's why you can put your hands on the terminals of a car battery and feel almost
nothing, as your body has a fairly high resistance (roughly 10,000 ohms). Drop a wrench (zero ohms)
across the terminals and you'll get all the current the sucka' can deliver! (V/zero = infinity, or for a car battery, about 500 amps - arc welder supreme!)
So, what does this mean to an overclocker? Simply this: as the power (wattage) demands from a CPU increase (raising the MHZ effectively lowers the resistance), the mobo must deliver more current. As the mobo PS (which is stepping the 12v/5v power IT receives down to the stable 2.2 ~ 3.5v a CPU needs) has only a limited amount of current it can deliver, raising the voltage lowers the amperage demand. (P = IV) However, here's the kicker. Remember, voltage is a potential, in this case, the potential to overcome resistance. That's why the 12v car battery won't give you a shock, but the 110v light socket will. Inside your computer, it means that the dielectrics inside those millions of transistors which make up your CPU are being increasingly stressed - voltage reaches a certain level, and the current will "jump" the gap - voila, no POST (or erratic behavior), and potential to fry your CPU. Having the voltage too low can cause damage too: voltage low, current high - too much current will damage any circut unable to carry the load (and in a CPU, we're talking micro-amps). OK, there's my piece - now overclock away!

------------------
...hope this helps...

[This message has been edited by JL_in_Vail (edited 07-31-99).]
 
R

Rogue Gunman

New Member
#10
JL_in_Vail, that was a great explanation of electrical fundamentals. I believe what StuTheWise was trying to do, however, was to convey a more simple cause and effect relationship by using something that people are often more familliar with- hydraulics. Quite often in basic electricity classes the teacher or instructor will use a students preexisting understanding of the properties of hydraulics to build a basic understanding of the properties of electricity. They are by no means identical, but on a fundamental level they are very similar.
 
J

JL_in_Vail

New Member
#11
ya, Rogue

...my teacher used just that analogy, but in the area of resistance/current it gets a little soft. Here's a question, if MY computer is on a spaceship going .99 the speed of light, and YOUR computer...

keep it coming!
 
D

diablovision

New Member
#12
from a stationary observer's view, time dialates so that the cpu runs only 2% as fast. This is the Einstein factor (upsilon?) that is sqrt(1-(v/c)^2).
=P

Oh, and the last statement about too low of voltage killing the cpu isn't valid. From the eq V = IR, it's clear that current is proportional to voltage when resistance is kept constant.

[This message has been edited by diablovision (edited 08-03-99).]
 
D

diablovision

New Member
#13
The thing i don't understand about relativity is this:

Let's say Mr. A is on a ship going .99c relative to a point occupied by Mr. B.
Mr. A's (relative to Mr. B) time dialation is 50.25, which means his clock should run 50 times slower than Mr. B's.
From his perspective, Mr. B would see A moving at .99c, and should expect A's clock to be 50.25 times slower.
But if A were really fixed and B were moving, B would see that A's clock was faster (because B's clock was dialated).

But relativity states that neither one can be authoritatively ruled the fixed point, so both men's clocks would appear to run faster than the other man's.

AAAAAAAAh!
What im trying to say is that the guy who is time dialated looks at the outside clock and thinks it is going fast (because his clock is dialated). But this can't be possible?


Which is the slower clock? Are they both?
Does anyone understand this?


AAAAAAAAHHHH!
 
M

Machupo

New Member
#14
So wait... if you overclock your puter on a spaceship going .99c, everyone who looks inside (physicists do all sorts of wacko things -- people seeing inside a ship going .99c, frictionless planes, where will it all end?) will so the speed of the CPU a whole lot faster, right? hahaha, physics suck...

-machupo
 
S

Soul Apathy

New Member
#15
They are both operating at the same speed...Your perception is being affected by your body's inability to compansate for....splattt...Just out of curiosity...is anyone going to answer the original question....-HOW TO CHANGE THE VOLTAGE ON AN ASUS P2BVE...AND #2...HOW TO CHECK THAT VOLTAGE WITH A MULTIMETER
 
J

JL_in_Vail

New Member
#16
-> Go to the Asus website, download the manual, read the manual.

-> Go to the Intel website - there you will find all the white papers on pin assignments for all Pentium CPU's. How you pull the voltage off of the socket is the poser - you might just be able to run the board w/o the CPU.

-> A CPU uses the same amount of power no matter what the voltage is. Power=voltage*current. Voltage goes down, current goes up. A 100 watt bulb at 120 volts uses around 0.83 amps. A 100 watt bulb at 12 volts uses 8.3 amps. That's why a table lamp can use 18-ga wire, but your KC Daylighters need 12-ga wire and a 20A fuse.
 
S

Soul Apathy

New Member
#17
...there is not a manual for the P2BVE...the P2B-99...is the closest match to ut...one problem...the manual link does not work on the asus site for that model
 
M

Machupo

New Member
#18
what happens when you copy the link into your address bar and step back a directory (and/or remove the file name), do you get an index, or just another file? what's the name of the file? if it's listed at .htm, try changing it to .html... whatever, just fool around with it
 
Dimarini

Dimarini

New Member
#19
JL_in_Vail

Voltage = Current x Resistance

Since the resistance of your cpu stays the same when you raise the voltage the amperage must also increase.

Since Power = Voltage X Current when you increase the voltage the power must go up.

This is why when you keep the same heatsink and you increase the voltage to your cpu your temperature will go up.
 
M

MGP

New Member
#20
JL_in_Vail,

-> A CPU uses the same amount of power no matter what the voltage is.
Dimarini is closer to the truth. Actually, the power is not constant with voltage either. It's a dynamic property of the CPU, that is it's also frequency dependent. The reason is that every time a signal changes state in the CPU, the transistor used in the logic cell (and there may be up to 20+ MILLION of them!) conducts current from the processor core supply to ground. It's a relatively small current, but there are lots of them that add up on each clock cycle. The power requirements of modern CPU's rise with clock frequency because of this.

Here's a general equation for power consumption in high speed CMOS circuits. Not all CPU's are pure HSCMOS, but the general equation is similar for different fabrication processes.

Pd = C * f * V^2

where:

Pd = Power Dissipation in CPU
C = Sum of Junction Capacitances in CPU
f = Operating Frequency
V = Core Voltage

Note: C is a constant for a given CPU/Fabrication Process. You can estimate the power increase in your CPU due to increased clock frequency or core voltage without knowing it's exact value, since it doesn't change.

There is also a constant current draw from the CPU with no clock applied, but it is very low in comparison to the operating (dynamic) current draw when the CPU is clocked at full speed.

So in general, power increases linearly with an increase in frequency, and is a square of the increase in voltage. If you bump your frequency 10%, then the power will go up 10%. If you increase your core voltage by 10%, your power will increase 21%!

That is why there is a HUGE drive by CPU designers to lower the core voltages and shrink the geometry of new CPU's. Smaller geometry lets you clock faster and run at lower voltages at the same time. This makes power manageable with reasonable heatsink/fan solutions. It also lets them get more dies/wafer in manufacturing and that translates to $$$ -- lower cost/die for the manufacturer and cheaper CPU's for us!




[This message has been edited by MGP (edited 08-05-99).]