5 ms·

tl;dr New materials can help, but "resistive losses" aren't really the driving factor. The energy is a mix of leakage current and active current. Leakage curre

tl;dr New materials can help, but "resistive losses" aren't really the driving factor. The energy is a mix of leakage current and active current. Leakage current can be thought of as resistance - it's how much current flows through a transistor that's off. This can be better based on the material, but gets harder with smaller transistors. (Thinking about quantum tunneling as a resistance is good to get intuition, but not good enough to help solve the problem. A material with a lower bulk resistivity will not help here.) Active current is based on capacitance. Each FET has a little capacitor that needs to be charged and discharged every time the logic is switched - that adds up. Lowering the capacitance of each FET would reduce the energy required to switch it, but generally comes with bad tradeoffs. High-k dielectrics increase the capacitance, all other things being equal. But all other things are not equal, and they are used to create better performing FETs with lower power leakage.

saltcured 3y ago
I thought leakage current would be the "DC" loss that is independent of frequency, like we had in old bipolar logic. Isn't it fair to characterize the cmos/fet switching losses as resistance to moving the charges around? I understand leakage will go up if we increase voltages to support higher switching speeds, but aren't there still a lot of losses that happen with logic transitions and reduce when the states are stable, even if voltages are held constant? I realize it we can't move charges around for free, but in some fantasy superconducting-fet logic circuit, wouldn't the power consumption be reduced? I.e. much of the waste is resistive losses while charging and discharging those gates.
- timerol 3y ago
  > Isn't it fair to characterize the cmos/fet switching losses as resistance to moving the charges around? Not really. It makes more sense to think about it as filling and emptying capacitors. You are charging the gate capacitance up to the supply voltage, then dumping that charge to discharge the gate to 0 again. The energy of each capacitance that gets charged and dumped is CV^2/2, which happens for each logic transition. > I realize it we can't move charges around for free, but in some fantasy superconducting-fet logic circuit, wouldn't the power consumption be reduced? If there was no resistance when distributing charge, it would help a bit, but not enough to change the clock frequency by more than 20%, assuming that the fantasy superconducting-fet had normal leakage and gate capacitance.
  - saltcured 3y ago
    So the charge is work and the discharge is waste? I guess I am entertaining the idea of an idealized Maxwell-demon CMOS circuit, if we could bounce the charge between gates with very little work to just pump the charge back and forth.
    - timerol 3y ago
      That's a reasonable way to think about it - you take energy from the supply voltage to charge the gate capacitor when the logic line goes high, then dump it when the logic line goes low. If you had a lossless bidirectional voltage converter circuit for each gate capacitance, then you could charge the capacitor from the supply and discharge it back into the supply, removing any switching losses.
    - Dylan16807 3y ago
      They're both waste. Charging a capacitor to 1 volt means your average input voltage is .5 and half your energy goes to heat. Discharging to a ground line wastes the other half. As the sibling comment says, you would need voltage converters running both ways to avoid this waste.
moffkalast 3y ago
From what I understand while these two do contribute a lot to power usage, they don't really contribute that much to heating by themselves? Leakage should happen all the same in a processor that's completely idle and those typically don't heat up much. For higher clock speeds specifically I still don't see how lower resistance isn't key.
- thfuran 3y ago
  >while these two do contribute a lot to power usage, they don't really contribute that much to heating Those are the same thing. Or at least close enough as makes no practical difference. Only an extremely tiny fraction of the power used but a CPU is becoming anything other than heat.
  - moffkalast 3y ago
    Then we actually agree? You don't get heating without resistance, ergo resistance is the main problem. MRIs don't have any problems sending a thousand amps through their coils.
    - kayson 3y ago
      You're right that the resistance is where the heat is dissipated, but lowering the resistance does not actually change the amount of heat. Transistor switching can be modeled as a step input to an RC circuit [1]. If you integrate the power through the resistor to infinity, you'll see that the value of the resistor drops out. Intuitively, you might think of it like this: to charge a capacitor (or transistor) up to a certain voltage, you need a fixed number of electrons. That number of electrons will always pass through the resistor and generate heat based on their energy. Even if you change the resistor value, its still the same number of electrons, and the same amount of energy. What does change with resistance, though, is the time over which the power is dissipated. In practice, you have to make sure the resistors are small enough such that you can achieve your desired clock speed. There are actual resistive losses too, but they're mainly related to power delivery. [1] https://en.wikipedia.org/wiki/RC_circuit#Time-domain_considerations https://en.wikipedia.org/wiki/RC_circuit#Time-domain_conside...
vlovich123 3y ago
I think two major revolutions would be optical and reversible computing. The former would significantly shrink the heat generated which is a huge bottleneck but is very hard to build generic computing out of and expensive. The latter would basically result in computing obtaining a new theoretical lower bound on energy required but is purely research with no known approaches for actually building the things. Asynchronous clockless designs might also drastically cut the power budget but those have failed to find adoption for some reason.
- orbital-decay 3y ago
  Clockless designs did find their use, just not for the entire chips. Certain parts of modern CPUs are asynchronous.
  - vlovich123 3y ago
    Yeah, I'm just a bit surprised it didn't go further. Do you know what the reasons were that they couldn't make the entire thing clockless?
    - verall 3y ago
      I don't think clockless makes anything easier
    - AgentOrange1234 3y ago
      It makes reasoning about behavior very, very difficult. There is no tooling support for it.
  - infinite8s 3y ago
    What parts are clockless?
- tbrownaw 3y ago
  Reversible computing needs a place to store waste entropy, ie a rather large memory that's initialized to known values and filled with junk as the computation runs. Venting entropy out of a system is not reversible. Clockless designs mean that you compute readiness information on the fly instead of having it precomputed at design time. This additional run-time computation is not free, and tooling for clocked designs is good enough that the extra slack they need is often cheaper.

saltcured 3y ago

I thought leakage current would be the "DC" loss that is independent of frequency, like we had in old bipolar logic. Isn't it fair to characterize the cmos/fet switching losses as resistance to moving the charges around? I understand leakage will go up if we increase voltages to support higher switching speeds, but aren't there still a lot of losses that happen with logic transitions and reduce when the states are stable, even if voltages are held constant? I realize it we can't move charges around for free, but in some fantasy superconducting-fet logic circuit, wouldn't the power consumption be reduced? I.e. much of the waste is resistive losses while charging and discharging those gates.

timerol 3y ago
> Isn't it fair to characterize the cmos/fet switching losses as resistance to moving the charges around? Not really. It makes more sense to think about it as filling and emptying capacitors. You are charging the gate capacitance up to the supply voltage, then dumping that charge to discharge the gate to 0 again. The energy of each capacitance that gets charged and dumped is CV^2/2, which happens for each logic transition. > I realize it we can't move charges around for free, but in some fantasy superconducting-fet logic circuit, wouldn't the power consumption be reduced? If there was no resistance when distributing charge, it would help a bit, but not enough to change the clock frequency by more than 20%, assuming that the fantasy superconducting-fet had normal leakage and gate capacitance.
- saltcured 3y ago
  So the charge is work and the discharge is waste? I guess I am entertaining the idea of an idealized Maxwell-demon CMOS circuit, if we could bounce the charge between gates with very little work to just pump the charge back and forth.
  - timerol 3y ago
    That's a reasonable way to think about it - you take energy from the supply voltage to charge the gate capacitor when the logic line goes high, then dump it when the logic line goes low. If you had a lossless bidirectional voltage converter circuit for each gate capacitance, then you could charge the capacitor from the supply and discharge it back into the supply, removing any switching losses.
  - Dylan16807 3y ago
    They're both waste. Charging a capacitor to 1 volt means your average input voltage is .5 and half your energy goes to heat. Discharging to a ground line wastes the other half. As the sibling comment says, you would need voltage converters running both ways to avoid this waste.

moffkalast 3y ago

From what I understand while these two do contribute a lot to power usage, they don't really contribute that much to heating by themselves? Leakage should happen all the same in a processor that's completely idle and those typically don't heat up much. For higher clock speeds specifically I still don't see how lower resistance isn't key.

thfuran 3y ago
>while these two do contribute a lot to power usage, they don't really contribute that much to heating Those are the same thing. Or at least close enough as makes no practical difference. Only an extremely tiny fraction of the power used but a CPU is becoming anything other than heat.
- moffkalast 3y ago
  Then we actually agree? You don't get heating without resistance, ergo resistance is the main problem. MRIs don't have any problems sending a thousand amps through their coils.
  - kayson 3y ago
    You're right that the resistance is where the heat is dissipated, but lowering the resistance does not actually change the amount of heat. Transistor switching can be modeled as a step input to an RC circuit [1]. If you integrate the power through the resistor to infinity, you'll see that the value of the resistor drops out. Intuitively, you might think of it like this: to charge a capacitor (or transistor) up to a certain voltage, you need a fixed number of electrons. That number of electrons will always pass through the resistor and generate heat based on their energy. Even if you change the resistor value, its still the same number of electrons, and the same amount of energy. What does change with resistance, though, is the time over which the power is dissipated. In practice, you have to make sure the resistors are small enough such that you can achieve your desired clock speed. There are actual resistive losses too, but they're mainly related to power delivery. [1] https://en.wikipedia.org/wiki/RC_circuit#Time-domain_considerations https://en.wikipedia.org/wiki/RC_circuit#Time-domain_conside...

vlovich123 3y ago

I think two major revolutions would be optical and reversible computing. The former would significantly shrink the heat generated which is a huge bottleneck but is very hard to build generic computing out of and expensive. The latter would basically result in computing obtaining a new theoretical lower bound on energy required but is purely research with no known approaches for actually building the things. Asynchronous clockless designs might also drastically cut the power budget but those have failed to find adoption for some reason.

orbital-decay 3y ago
Clockless designs did find their use, just not for the entire chips. Certain parts of modern CPUs are asynchronous.
- vlovich123 3y ago
  Yeah, I'm just a bit surprised it didn't go further. Do you know what the reasons were that they couldn't make the entire thing clockless?
  - verall 3y ago
    I don't think clockless makes anything easier
  - AgentOrange1234 3y ago
    It makes reasoning about behavior very, very difficult. There is no tooling support for it.
- infinite8s 3y ago
  What parts are clockless?
tbrownaw 3y ago
Reversible computing needs a place to store waste entropy, ie a rather large memory that's initialized to known values and filled with junk as the computation runs. Venting entropy out of a system is not reversible. Clockless designs mean that you compute readiness information on the fly instead of having it precomputed at design time. This additional run-time computation is not free, and tooling for clocked designs is good enough that the extra slack they need is often cheaper.