Processors power dissipation and power consumption. losses in transistors?

Discussion in 'General Electronics Chat' started by BilColin, Feb 23, 2016.

  1. BilColin

    Thread Starter New Member

    Feb 23, 2016
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    Hello,

    Can anyone explain the energy efficiency of transistors in specific CPU. How much of the electricity consumed at a specific time is released as heat. Almost all energy consumed is converted into heat or not? I can't find the answer.
    The specified TDP does not answer. It's only tell what cooler is required to dissipate generated heat. Also, AMD and Intel measured TDP by a different methods and are not fully comparable, also TDP is valid for the entire family, not for a specific CPU. TDP is not the same as power consumption. But I found posts in many forums in the internet that claims that TDP = power consumption. Is this correct?
    Perhaps the question should be asked - where does power go in CMOS, how much is converted in to heat, how much is losses. If we use MOSFET transistror as a switch, I can find formulas for power dissipation, but for CPU? The Leakage current /power/ in transistors is converted into heat. But there are switching power, short-circuit power due to non-zero rise/fall times e.t.c. Can anyone help for find the answer?
    Is it right the claim that power consumption of one CPU at specific moment = 99.99% heat dissipation or almost all electrical energy is converted to heat.

    Thank you.
     
  2. Picbuster

    Member

    Dec 2, 2013
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    This is difficult:
    a: speed depended ( frequency and power are not linear connected)
    b:the used internal resistors, capacitors and coils (each wire can act like coil or as a capacitor all depending on frequency)
    c:the construction and used material in the junctions.
    d; environment ( temp, humidity airflow)
    no way that you are able to calculate without information as stated above.
     
  3. Papabravo

    Expert

    Feb 24, 2006
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    There are two general effects for CMOS processes.
    1. Power increases linearly with frequency
    2. Power increases quadratically with Vcc.
    #2 is the primary reason why you lower and lower Vcc values, starting with +5V and dropping to 0.8V.

    You can usually find this information in datasheets in the footnotes.
     
  4. BilColin

    Thread Starter New Member

    Feb 23, 2016
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    Yes, this is understandable. And the Leakage current is increases with temperature.
    But is it correct this statement - that almost 100% of all electric energy consumption /no matter how is it in specific time 10W or 100W/ in one CPU is converted to heat no matter frequency, temperature e.t.c. because the law of conservation of energy states that energy can neither be created nor destroyed. Just transforms from one form to another. And here /CPU/ we have energy consumption and conversion - electric energy to heat, there is no conversion to example mechanical energy, so almost all electrical energy 99.99% is converted to heat. And little magnetic field/electric field is generated in thansistors but is insignificant. Is this statement correct?
     
  5. BilColin

    Thread Starter New Member

    Feb 23, 2016
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    The question is in one specific time. All that - increase in temperature and voltage also increase power consumption is clear and everyopne know that. The question is - no matter 10W or 100W, CPU power consumption, how much of that energy is converted to heat.
     
  6. Picbuster

    Member

    Dec 2, 2013
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    If you move an electron or any mass you create speed and friction resulting into heat.
    Now you question is: is speed producing heat? no but the friction( banging all types of electrons through polluted material) is.(my point c)
    However; to achieve speed you need energy. total energy= friction energy + energy to create and maintain speed.
    therefore most of the energy is friction (heat).
     
  7. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Yes.
     
  8. Picbuster

    Member

    Dec 2, 2013
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    a lot of work is don in the second part of the last century. Look at Thermal Heat Transfer most of the universities will have this in their library.
     
  9. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    For any integrated circuit, some of the energy coming in the power pins goes out some I/O pins (or else, what good is the part?). Also, some energy is radiated as EMI. But most of the I/O power goes to another chip, and on and on. So your basic question/premise is correct. The vast majority of the energy entering a high performance CPU IC is dissipated locally as heat. For CMOS parts, that heat comes from charging and discharging capacitors through resistors. CMOS output stages are MOSFETS with Rdson specs, and CMOS input stages are MOSFETs with gate charge specs. Every time one gate drives another gate, it either charges or discharges the gate capacitance through its output impedance. We're talking very, very small capacitors and reasonably small resistors, but that times 10 billion adds up.

    Many people are aware that for CMOS processors, speed = power. This is because the gate structure of a MOSFET is a sheet of glass and draws almost exactly zero static current; all of the power dissipation comes from dynamic current, current associated with inputs and outputs changing state. But it wasn't always this way. The original Star Trek episode "The Ultimate Computer" was about a new supercomputer ("M5") taking over the operation of a starship, making the crew obsolete. The computer goes nuts and taps directly into the warp core to get more power because it wants to do more things. At the time this aired, that was seriously incorrect; computer power consumption and CPU performance were not directly linked. For TTL IC's, the static current through a device is a significant percentage if not the majority of the total current through the device when operating at speed. So early computer CPU boards did not draw significantly more power when running compared to when halted.

    ak
     
  10. dl324

    Distinguished Member

    Mar 30, 2015
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    Static dissipation wasn't zero but, for a time, it could be neglected. As I recall, around 90nm, static power dissipation in microprocessors became a huge issue.
     
  11. AnalogKid

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    Recall why?
     
  12. dl324

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    It was a combination of factors. VCC reductions to reduce dynamic power which required lower threshold voltages and increased static leakage (exponentially), gate length, oxide thickness. Essentially, the pursuit of Moore's Law.
     
  13. AnalogKid

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    So, even glass leaks when it's 5 atoms thick?
     
  14. dl324

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    Under the right conditions. EPROMs depended on electrons being able to jump the oxide barrier when it was around 500 angstroms.
     
  15. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
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    I can't give you any numbers....but all the heat dissipated in a transistor is generated during the TRANSITION period.....if every device went from 0 to 1 and back instantaneously, there would be no heat generated. Considering HOW MANY transistors are actually in a modern microprocessor, the efficiency is already astounding.
     
  16. crutschow

    Expert

    Mar 14, 2008
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    Not true. Most of the heat generated is from charging and discharging the stray circuit capacitances in the circuit.
    It does occur during the transition period but it's independent of the transition time (neglecting any power loss that might occur when both transistors are ON during the transition in a CMOS circuit).
    The dissipated energy is proportional to the voltage, capacitance, and circuit operating frequency, and equals fCV^2.
     
  17. dl324

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    More generally,  \small P_{tot} = P_{dyn} + P_{sc} + P_{leak}
     
  18. dl324

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    Forgot the most important reason... That was about the time when leakage power exceeded active power.
     
  19. BilColin

    Thread Starter New Member

    Feb 23, 2016
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    Thank you all. Now it is much more clear to me. This thread can be of help to many more people looking for answer.
     
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