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Old 01-11-2008, 07:04 PM   #1
kaz2100
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CPU clock and heat


Hya,

I am simply curious and have not found answer.

Question is:
Why does CPU generate more heat as clock goes up?

Happy Penguins!
 
Old 01-11-2008, 07:50 PM   #2
PhenuxRizing
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The faster the clock speed; the more electricity it draws in. The electrons moving at a faster pace into the conduits cause more friction thereby generating more heat.

This was a gross over-simplified explanation but that is the main idea.
 
Old 01-11-2008, 08:21 PM   #3
armanox
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Think of it as a resistor. When more current is sent through the resistor (clock gets faster, needs more juice), the power (watts) increases. As the resistor uses more watts, it generates more heat.

Another thought would be a light bulb. The brighter the bulb, the hotter it gets (more wattage used).
 
Old 01-12-2008, 10:55 AM   #4
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http://www.iue.tuwien.ac.at/phd/schrom/node41.html
 
Old 01-13-2008, 09:06 AM   #5
kaz2100
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Hya,

Thanks for replies.

Quote:
The electrons moving at a faster pace into the conduits cause more friction thereby generating more heat.
Why "speed" of electron gets faster as clock goes up? Also, as friction goes up, (under same voltage), current goes down.

Ohm's law (E =IR) then the wattage (defined as EI), is R(I^2) or (E^2)/R. Friction goes up means R gets bigger; so wattage goes down.

I also think R is intrinsic to circuit, (so clock does not affect..)

Another point I do not understand is: How electron gets higher speed? (under same voltage)

I still need to find missing link.

Happy Penguins!
 
Old 01-13-2008, 09:26 AM   #6
pixellany
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Electrons always move at the same speed......If the current increases, it simply means that more electrons are moving.

Current is defined as the rate of movement of charge: 1 ampere = 1 coulomb/second, and 1 coulomb = 6.241E18 electrons. In a computer chip, something like nano-amps might be more meaningful.....1 nano-amp ~ 6 billion electrons/second.

In re the orginal question---another perspective:
Think of each operation in a CPU as needing a specific amount of energy. Do the operations faster and the rate of energy consumption (AKA power) goes up.
 
Old 01-13-2008, 09:52 AM   #7
kaz2100
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Hya,

PMorph: Thanks for the link.

The second (Pstat) and third (Psc) equations are somewhat intuitive. (Ohm's law)

The last sentence
Quote:
In a typical digital CMOS VLSI system, like a microprocessor or a cache memory module, the dynamic power consumption dominates by far.
sounds like a fact, whether or not I understand.

So, I need to figure out the first equation (Pdyn)

I need to figure out if I understand correct: Following is my interpretation.

1. "due to the diabatic charging and discharging of the circuit's internal and external capacitances when the circuit is operating." To make life simple, there is only one capacitor, (charged = 1, discharged =0; its capacity is 1pF) then this state can be changed as clock speed.
2. If clock is 1 Hz, this means that this capacitor stays either 1 or 0 for one second.
3. Capacitor can store energy (but not consume.)
4.1 If this state stays on, no heat generation.
4.2 If this state changes, energy has to be discharged, -> this will generate heat, if there is any resistor around.

So, this capacitor can generate 1 unit of heat per 1 second at most. Now, if clock is 2Hz, 2 unit at most.
This sounds like the first equation (Pdyn), but Fig 3.3 does not show any resistor.

Where is the resistor?

Happy Penguins!
 
Old 01-13-2008, 10:00 AM   #8
kaz2100
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Smile

Hya

Quote:
Originally Posted by pixellany View Post
In re the orginal question---another perspective:
Think of each operation in a CPU as needing a specific amount of energy. Do the operations faster and the rate of energy consumption (AKA power) goes up.
Thanks, I miss your post while I was trying to figure out the Pdyn.

I think your re-phrasing is well describing.

I guess my confusion started when I understand CPU chip as a resistor load.

Cheers

Happy Penguins!
 
Old 01-13-2008, 10:27 AM   #9
pixellany
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Back to your comments on capacitors and resistors. Unless you are dealing with superconductors, every circuit has resistance---ie a current flow always results in a voltage drop (no matter how small)

If you like puzzles, try this:

2 identical capacitors--each 1 farad, one of which is charged to 10 volts, and the other is discharged. Connect them thru a resistor.
Before connecting, the energy in the first one is 0.5*CV^2 = 0.5*1*100 = 50 units
After the connection, charge is conserved (Q=CV). With twice the capacitance, the voltage is now 5 volts, and the stored energy is now 0.5*1*25 = 12.5 units. The missing energy has been dissipated in the resistor (as heat).

Now, let the value of the resistor be zero. Charge must still be conserved, so the voltage after connecting will be the same as before. Where does the energy go?
 
Old 01-13-2008, 05:10 PM   #10
PatrickMay16
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When you increment the CLOCK, the CLOCK increases. Now, because the processor is CLOCKIN' so much, it gets very HOT...

They call this the Gallo effect, as it was discovered by Professor Deven Gallo of harvard university.
 
Old 01-16-2008, 08:36 PM   #11
kaz2100
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Unhappy

Hya

Quote:
Originally Posted by PatrickMay16 View Post
Now, because the processor is CLOCKIN' so much, it gets very HOT...

They call this the Gallo effect, as it was discovered by Professor Deven Gallo of harvard university.
Yes, it is well known, but question is why "clockin" results in heat.

Quote:
Originally Posted by pixellany
2 identical capacitors--each 1 farad, one of which is charged to 10 volts, and the other is discharged. Connect them thru a resistor.
Before connecting, the energy in the first one is 0.5*CV^2 = 0.5*1*100 = 50 units
After the connection, charge is conserved (Q=CV). With twice the capacitance, the voltage is now 5 volts, and the stored energy is now 0.5*1*25 = 12.5 units. The missing energy has been dissipated in the resistor (as heat).
It was a while ago since I attended to physics class. I still have not figured out, but I do not think capacitor has to be "fully" charged, -- it can be partially charged. As soon as I solve this puzzle, I will post again.

By the way, I do not think the logic I followed at post #7 works when logic circuit is made of flip flop (two transistors). They are resistor load (not capacitor load), so to keep either 1 or 0, some sort of power is necessary,
and I do not think that changing state consumes power.

1 Hz (1 second of load) and 2Hz (0.5second load times 2) which is same amount.

Happy Penguins!
 
  


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