16
\$\begingroup\$

I would like to understand how the computation process causes the processor to get hot. I understand that the heat is generated by the transistors.

  1. How does the transistors generate the heat exactly?
  2. Is the correlation between the number of chips and the heat generated linear?
  3. Do CPU manufacturers optimize the positions of single transistors in order to minimize the heat generated?
\$\endgroup\$
  • \$\begingroup\$ Switching cases heat and hence clock speed also has a direct relation with the heat generated. \$\endgroup\$ – nidhin Mar 17 '15 at 14:17
  • 4
    \$\begingroup\$ In practice, real chips heat up because of the movement of electricity through a resisting conductor, as noted in the answers. But if this subject interests you, you might want to read about how the creation of binary information itself necessarily requires creation of waste heat; A nice essay that doesn't require a lot of physics can be found here plato.stanford.edu/entries/information-entropy \$\endgroup\$ – Eric Lippert Mar 17 '15 at 21:12
  • \$\begingroup\$ I don't have the math to properly answer (and elaborating on Eric's comment), but there are implications for this question in quantum computing and reversible gates. There is "kT ln(2) energy dissipated per irreversible bit operation". Part of Landauer's principle. If you have (A & B), this has two inputs and one output. That information that is lost in this process has to go somewhere and becomes entropy (heat)... assuming I understood it enough to describe this. \$\endgroup\$ – user35419 Mar 18 '15 at 14:20
25
\$\begingroup\$

A transistor (FET, in modern ICs) never switches instantly from full OFF to full ON. There is a period while it's turning on or off where the FET acts like a resistor (even when fully ON it still has a resistance).

As you know, passing a current through a resistor generates heat (\$P=I^2R\$ or \$P=\frac{V^2}{R}\$).

The more the transistors switch the more time they spend in that resistive state, so the more heat they generate. So the amount of heat generated can be directly proportional to the number of transistors - but it is also dependent on which transistors are doing what and when, and that depends on what the chip is being instructed to do.

Yes, manufacturers may position specific blocks of their design (not individual transistors, but blocks that form a complete function) in certain areas depending on the heat that block could generate - either to place it in a location with better heat bonding, or to place it away from another block that may generate heat. They also have to take into account power distribution within the chip, so placing blocks arbitrarily may not always be possible, so they have to come to a compromise.

\$\endgroup\$
  • \$\begingroup\$ There are many many many contributing factors to heat production in a CPU. Just because this answer doesn't happen to mention your favourite one doesn't mean it is wrong. I suggest, instead of down-voting perfectly valid answers, you try writing your own answer. If it is better than this one then the community will decide and it will get more votes. \$\endgroup\$ – Majenko Mar 20 '15 at 19:43
  • 1
    \$\begingroup\$ So bias it by giving people a better answer to put their votes into instead of mine. \$\endgroup\$ – Majenko Mar 20 '15 at 19:54
  • \$\begingroup\$ I will, but I won't have time for a while (days). For now, I agree that heat is generated by resistive losses. However, I don't think your answer gives a very deep understanding. \$\endgroup\$ – HKOB Mar 20 '15 at 20:23
21
\$\begingroup\$

All current flow in anything that isn't a superconductor generates heat. In chips, it's mostly flowing in aluminium "metal" layers (why not copper? Nasty chemical interaction with other parts of the silicon, it turns out).

What causes current to flow? Every time a transistor changes state, this can be modeled as a capacitor (the FET gate of the driven logic gate plus parasitic wire capacitance) charging/discharging through the wire and output FET of the previous gate. This is "switching" or "dynamic" power. It's proportional to switching speed and the square of the voltage; hence the drive from 5V to 3.3V to 1.8V for better efficiency.

The insulators are not perfect, and in some places are very thin. Transistors may not be fully "off". If a FET has an off resistance of a megaohm, and you put a million of them in parallel, it looks like a 1 ohm resistor. This is "leakage" power. It's proportional to number of transistors.

I spent a decade working at a startup on power optimisation. :) There are a lot of techniques: speed/leakage tradeoffs ("high k metal gate"), turning off parts of the circuit entirely, clock gating, reduction of clock frequency, sizing and placement.

\$\endgroup\$
  • \$\begingroup\$ Today's high-performance microprocessors actually do use copper interconnects, with thin layers of other metals to prevent the copper from reacting with silicon. Aluminum is still used on less complex and bigger-process chips, though, as it's a whole lot less complicated to work with. \$\endgroup\$ – hobbs Mar 18 '15 at 1:24
0
\$\begingroup\$

1) Any time there is current flow, heat is generated by the collisions of the electrons. 2) Yes, generally, the correlation is linear. 3) It is very unlikely that CPU manufactures optimize the position of individual transistors, to minimize the heat generated (they are all inside the same casing).
When a CPU is "idle", although it uses a minimum amount of current, it generates heat. As the processor starts to "process" information, the individual transistors switch states. This switching also generates heat. In addition, the switching frequency affects the heat generation rate, the higher the frequency the higher the heat generation rate. Since the heat dissipation capacity of the chip is fixed, it can overheat if it's operated at a higher frequency than it was designed to operate.

\$\endgroup\$
0
\$\begingroup\$

it is simple we know that according to joules law that whenever the electron are flowing through the conductor the heat produced due to the resistance of the material because every conductor have some amount of resistance in it.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.