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Cluster
12-18-02, 07:31 PM
I recently had a question asked that I thought would be most beneficial in the forums.

I just read that post about "basics of overclocking out of intrest. One little question that has been bugging me for quite a long time: the voltage (vcore, etc)- is heat the only thing that limits it? Because i thought, having something with very high voltage going through it would surely blow the circuit? I.e. if heat wasn't a problem, could you have the processor running at like 80v or something, or would it just blow the electric paths?

I thought i had an answer to his question, but when i worked it out, my theory didn't work. Started to contradict some of what i was saying. So i'll state where i had a problem also.

When i was calculating where the heat would increase, i had thought that an increase in voltage would cause a larger amperage draw. But that isn't true since voltage and amperage are inverly proportional to each other. If this is so, then increasing the voltage should decraease the amperage for a given wattage.

Or am i wrong in presuming that the wattage ncessary for a CPU is dictated by the amount of work it's doing, and not directly effected by what the voltage is set at?

A great question, any takers?

NookieN
12-19-02, 12:31 PM
If you're talking pure resistors, amperage and voltage are inversely proportional. However, transistors have both resistive and capacitive properties (not to mention the parasitic resistance and capacitance found in CMOS circuits).

An increase in voltage should not cause a noticeably larger amperage draw if the chip's clock speed remains the same. However, since a certain amount of energy is required for each clock cycle, increasing the clock speed will cause an increase in current draw. There are no hard and fast rules about this, but I might guestimate that raising the clock speed 10% will lead to a 5-6% increase in current demand.

Raising the voltage alone will increase wattage (and thus heat dissipation). Raising the voltage AND the clockspeed will doubly impact power requirements.

ArBiTaL 24
12-21-02, 10:51 AM
THX Cluster ;)

A little on the complex side though, :D

nihili
12-21-02, 11:16 AM
Originally posted by ArBiTaL 24
THX Cluster ;)

A little on the complex side though, :D

Please remember this is the Technical Discussion forum. It is assumed that discussions will be quite technical and may require advanced training to follow. If you want a less technical answer, please use the other forums. Also, please keep thread in this area free from extraneous posts. The idea is to foment advanced discussion by reducing chatter and the need to make everything widely accessible. If you have questions about this, please read the rules sticky. I will be deleting excess posts on sight and at my disgression.

nihili

KickFlip
12-24-02, 05:53 AM
If you're talking pure resistors, amperage and voltage are inversely proportional.

I don't get how thats true. I always thought that voltage and current have a proportional relationship.

Say you have a resistor of 5ohms. Put 5V across it and 1A of current runs through it. 5V / 5ohm = 1A

Take the same resistor and put 10V across it.
10V / 5ohm = 2A

The voltage has increased and thus the current has increased due to Ohm's Law: V = IR

Unless I'm missing something.......

Since87
12-24-02, 09:01 AM
Originally posted by nihili
I will be deleting excess posts on sight and at my disgression.

nihili

NookieN
12-24-02, 01:42 PM
Originally posted by KickFlip

The voltage has increased and thus the current has increased due to Ohm's Law: V = IR

Unless I'm missing something.......

Oh, yes you're correct. For some reason I was thinking of current and resistance when writing that.

Movax
01-09-03, 10:00 PM
A curcuit will only allow a fixed amount of amps to flow at a given voltage, so I don't think increasing the clock will increase current draw (..much, depending on how you look at it). But it will increase the voltage required (as we all know) as there is more power being drawn. And Power = Volts x Amps

Of course, if you up the voltage your amps will increase as well, and a conductor will only handle so many amps til it blows, much like a fuse. Also, the more voltage you put across the CPU the more likely the electrons are going to take the shortest path through it, and try to skip the connections on the die altogether.

Electricity will flow through just about anything if there is enough of a potential difference.

Just my take on this, not to be read as fact.

Since87
01-09-03, 10:23 PM
Originally posted by Movax

Just my take on this, not to be read as fact.

I take it that you have not read the sticky for this group or Nihili's post in this thread.

The current draw will go up as the frequency goes up even though the voltage stays the same. The reason for this is that every transistor switch that occurs inside the CPU injects a certain amount of electrical charge into parasitic capacitances.

The unit of electrical charge is the Coulomb.

1 Amp = 1 Coulomb / Second

The faster the transistors in the CPU are switching, the more Coulombs of charge are being injected per second, and thefore the current (Amps) goes up.

Movax
01-10-03, 07:16 AM
Originally posted by Movax
(..much, depending on how you look at it). .

Which is why I said this. Believe me I know electronics, I just haven't applied my knowledge to computer hardware much before this. That is why the disclaimer. Your explaination for current increase is exactly along the lines of what I was thinking.

I also would like to add, that as the clock speed goes up, the CPU is using more of it's muliple curcuits in a given period of time and everyone knows that adding another curcuit in parrallel increases current draw. I think I am correct in that statement.

Any comments for the rest of my earlier post? Was I right in thinking that?

(And yes, I have read the sticky. I'm not a high school dropout, just trying to expand my knowledge. If it seems I don't grasp the concepts and should not be here, I'm sure I will be told as much by a mod.)

NookieN
01-10-03, 10:29 AM
Originally posted by Movax
Also, the more voltage you put across the CPU the more likely the electrons are going to take the shortest path through it, and try to skip the connections on the die altogether.

Since you asked, yes, that's more or less correct. But for the transistors in processors, it's unlikely you'll blow them like a fuse (unless you run 10V through it... ). The most common transistor failure resulting from increased voltage is gate-oxide decay.

Gates switch a transistor on when an appropriate voltage is applied, however current is never supposed to flow through the gate. When you raise the voltage, current starts to leak across the gate, which in turn disrupts the gate oxide and shortens the life of the transistor.

bluce ree
01-10-03, 11:24 AM
Current is not always inversely proportional to voltage.

remember Ohm's law I= V/R.
It's only inversely proportional when the resistance doesn't change.

There is a Gate Bias current for FETS, though it's usually in the nanoamps to pico amps(10^-9 amps.)

think of it this way: A Fet is a voltage controlled switch. So, the more voltage you apply, the more force you apply to the switch to turn a circuit on or off.

Now for an analogy:
If you slam a light switch on/off as hard as you can, you'll probably break the switch and it will get stuck in either on or off position. FETs (the most basic part of a cmos CPU like a pentium etc.) work the same way.

The opposite is also true.. If you use very little force to move a switch, it will probably only go halfway & may go either on or off. Fets also have a minimum ON voltage.

As far as overclocking is concerned, the only thing preventing you from raising your VCore higher is the fact that the cpu will literally burn up if you raise it above a certain point. FETs have a maximum gate voltage (Vgs) rating. If you exceed it, you will likely blow the FET up. Now that we've talked about 1 FET, immagine 130 million of them in a .7 by .5 rectangle only .01 tall. Imagine how fast heat must build up. say each Fet gets only slightly hotter with a Vcore increase, now multiply that slightly hotter by 130,000,000. (Obviously the math isn't exact, but I'm trying to make a point.)

When you clock your CPU frequencies higher,

Normal FSB clock: _/¯\_/¯\_/¯\_/¯\_/¯\_/¯\_
Overclocked FSB: /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/
___________________________________> Time

you can see that the FSB is stable an HIGH / LOW states are clearly defined. The change from LOW to HIGH is a slant, because the Gate of each FET has a capacitance that must be charged/discharged before the FET turns on or turns off. This capacitance becomes a problem when you increase your FSB. As you can see from the bottom, the FET isn't On or OFF for very long, It spends more time in the middle (inbetween on/off.) When a FET is in the middle stage, it gets very hot, very fast.

By increasing the FSB, you effectively turn the FET on/off more times in 1 second than if it were running normally. As a result, the FET spends more time each second in the middle (not on or off) stage getting hotter.

Now for some more esoteric electronic theory:
The reason upping Vcore makes a CPU more stable is this:

Upping the Vcore makes the Gate of the FET switch faster. That is, if you increase the Vcore, you make the Gate capacitance charge faster. This means that your CPU will spend more time in either the ON or OFF state and less time in the middle state. Theoretically this means the CPU should produce less heat, but we know that's not true. Why not? Glad you asked. The other part of Ohm's law: Forumla for Power in DC circuits:

Power = Voltage^2 / Resistance. or Voltage x Voltage / Resistance.

because resistance won't change in this circuit, you can easily see that increasing the voltage will increase the Power (by a lot.) This power must be disipated. The CPU disipates it in the form of heat.

I hope this clears up some questions. If you have a question pertaining to MOSFETS, FETS, or Bipolar Transistors I strongly reccomend you read up at Http://www.howstuffworks.com or just get an electronics book (to really understand the theories/laws, you really need to work out some text book problems and see the math behind the theory.)

Without the math, we are all just guessing. the math is the proof.

--Bluce Ree

Electrical Engineer

bluce ree
01-10-03, 11:42 AM
just to clear it up:

Gate capacitance greatly increases the amount of Heat a CPU produces when you increase the FSB. Gate capacitance also causes bad data to go to/from the CPU because of the following:

lets say the gate capacitance takes approximately 1 picosecond to charge up at a normal Vcore.

Each Hz takes 2 cycles ( Do a search on Nyquist frequency.)

1 cycle = going from on to off or from off to on.

so, 1Hz = 2 cycles. each cycle needs to charge or to discharge the gate capacitance depending on whether you're going from off to on or from on to off respectively. This means that in order to run right on the edge of stability, you can't run the cpu faster than 1/2 the speed of Gate capacitance charge/discharge frequency.

Frequency = 1/ Time. So because at a standard Vcore, our gate capacitance charges at 1picosecond, we calculate our charge frequency like this:
1/ 1pico second or 1/10^-12. The answer is 1GHz, or 1,000,000,000,000 Hz.

Because the CPU uses 2 cycles per HZ we can't run the CPU faster than 1/2 the speed of the Gate Capacitance of each FET. This means that we can't run faster than 1/2 * 1GHz = 500MHz at stock Vcore. If you try to run faster than 500MHz, the FET will not be able to go from LOW to HIGH or vice versa, fast enough.

If you increase your Vcore, you make the Gate capacitance charge faster, which decreases the 1picosecond charge/discharge time. This allows the FSB to operate at a faster speed.

Since87
01-10-03, 11:55 AM
Very good job bluce ree.

I neglected shoot thru current, because I thought my explanation was going to be difficult enough to follow as it was, but yeah, it's an important issue and may well be more important.

Do you know if shoot thru tends to consume more power than that consumed in charging gate and interconnect capacitances?

I don't have any idea which is the larger contributor to power consumption in today's CPU's.

Edit: Just noticed that bluce ree and I were posting at the same time.

NookieN
01-10-03, 12:02 PM
Originally posted by bluce ree
Gate capacitance greatly increases the amount of Heat a CPU produces when you increase the FSB.

Do you mean when you increase the core frequency by increasing the FSB? The FSB, in and of itself, doesn't dissipate much heat on the die.

Originally posted by bluce ree
each cycle needs to charge or to discharge the gate capacitance depending on whether you're going from off to on or from on to off respectively.

Assuming, of course, that the transistor's input changes with the clock cycle. If input has not changed from one clock to the next, the transistor won't switch and the charge/discharge states do not occur. Unless the input is the clock itself, then yeah obviously it will switch.

bluce ree
01-10-03, 12:17 PM
Shoot through current is a function of the voltage being switch divided by the FET's Rdson (ON state resistance.)

I design switching power supplies & switching audio amplifers. I typicially switch voltages as low as 12volts to as high as 200V.

I have had Mosfets litterally explode off of the board and melt into my safety goggles because of design problems.

also gate capacitance is rarely an issue with circuit designers unless you specifically make Switchmode Audio amplifiers or high power RF communication. typical Audio amps switch mosfets around 250kHz (20kHz = highest audio signal, * 12 = 240KHz minimum--the *12 factor is something I hear of a long time ago to use as a general rule, but got no explanation.) and RF obviously has to switch EXTREMELY FAST, atleast 2 to 8 * faster than the operating frequency ie, 802.11a WiFi operates at 5GHz, so its driving MOSFETS must operate at atleast 10GHz.

RF = black magic.

Honestly there is no way for us to tell how much shoot-through current affects a CPU without knowing the exacty perameteres for each individual FET used. Hell, I honestly don't even know if they only use 1 type of FET, or if they use mixed P, N channel FETs. I only say FET because I have no idea if the process they use even has metal oxide (I think they use Gallium Arsenide or something brandy new.) Older P2's were biCMOS i think, but I have no clue what they are now.

I didn't post the forumla for capacitance charging / discharging either, but its almost not relevant because as the temperature of the CPU increases or decreases ALL THE FET'S PARAMETERS CHANGE. It's really one big dynamic system where everything is a variable and a nightmare to try and simulate.

Also, because the gate capacitance will have a tolerance, even if we had the parameters the tolerance would throw off our results (some would work, others wouldn't.) The only thing knowing gate capacitance would help is finding CPUs like the 1.8GHz rev C Northwood, where it would be VERY OBVIOUS that the CPU wasn't being run anywhere near its maximum potential at stock FSB/Vcore.

--bluce

bluce ree
01-10-03, 12:27 PM
The FSB indirectly affects CPU temperature.

The CPU will process the data it receives on the FSB. When it finishes calculating and returning a result, it will stay idle and cool. If you increase the work load of the CPU by giving it the same amount of work to do in less time, the CPU will heat up because it will spend more time computing per second instead of sitting idle.

Of course the probably 1024 mosfets involved in the actual reading/writing of data to & from the FSB will produce so little heat its negligble--its the fact that as you increase FSB you decrease the amount of time the CPU has to sit idle.

The CPU will always have some switches switch with the FSB, reguardless of what its doing.

bluce ree
01-10-03, 12:31 PM
The main reason you draw more current with a higher FSB, is because you're asking the CPU to do the same job in less time than it normally takes. Time is realtive, if you do a job in 10 seconds and you burn 100 calories, then you're asked to do the same job twice in 10 seconds you will burn atleast 200 calories if not more.

NookieN
01-10-03, 12:50 PM
Raising the FSB increases the amount of data the cpu can send to and receive from memory. And, when you have a lock bus multiplier, raising the FSB will force an increase in the core clock rate.

But the cpu is _always_ doing something regardless of whether there's data on the FSB or not. Instructions take more than one cycle to complete, so even if the bus were idle for a few cycles you might still have some processing going on.

Moreover, you have the cache. Not only can you be executing instructions and data entirely from the cache, but even if you're not executing anything the cache _must_ be switching in order to save data. The cache is quite power hungry, even when the FSB isn't doing anything.

Not disagreeing with you... just saying the the FSB is but a small piece of the puzzle in such complex circuits.

Movax
01-10-03, 04:23 PM
As far as my statement about electrons just blowing through the die, and skipping the paths altogether, I was refferring to the original post that said "could you have the processor running at like 80v or something".

Even if you had liquid nitrogen cooling I doubt that you could hit 80v... this is why I said what I said about amps and volts and conductors shorting/blowing like a fuse...

Some good posts here.

Movax
01-10-03, 04:27 PM
Nookie..

Also - different CPU instuctions use different parts of the cpu and the faster you run the CPU the more instructions it will execute in a given time, resulting in more heat.

Data paths etc are also being used more often, so they don't have the chance to cool down.

shane46and2
02-01-03, 04:33 PM
Hello all,

I have a few notes on CPU power and voltage. There are three parts to power consumption in a CMOS logic circuit.
1) Switching Power
2) Short Circuit Power
3) Leakage Power

All CMOS logic gates have two paths for current to flow during this transition, 1) to the load capacitance(next gate/gates in design) 2) Strait to ground. The short circuit power is type 2, and the switching power is type 1. The leakage power is power disipated during the states where no transitions occur. This power is generally small, but can be big when considering 50Million gates.

___/^^^^^^^^^^^^\___________/^^^^^^^^^^^\______ -> Output Voltage Vs. Time
1&2 3 1&2 3 1&2 3 1&2 3 -> Power disipation Types

Switching power follows the charging and discharging of a capacitor at a given frequency. It follows this Eq: Pow=A*C*F*(V^2). The C is the effective capacitance of the gate, A is a scale factor that is the number of gates switching at frequncy F, and of course V is supply voltage. As you can see if frequency is doubled the power is doubled, as voltage is doubled power is quadroopled. This power is a the largest part of the power disipated in a logic circuit running when the rise and fall times of the input signals are <<the pulse width of the signal.

The short circuit power as I have said above is the "wasted current" that is lost because there exists current path from power to ground during this short time. This power is more dependant on the input voltage, ie how long it takes to get from low to high or high to low. This time is approximetely constant vs. frequency, because the rise and fall time of a gate does not change. Now how could you express power consumed by this? Well it will double when frequncy doubles, and it will increase when voltage increases. Pow ~F*V.

The Leakage power is basically the amount of power disipated while the gates are not in a state of switching. Basically even MOS transistor draws some current when it is turned off. So this current increases with the number of gates, and the voltage.