11 ms·

> it's consuming 200 W to provide its transistors with about 1 to 2 volts, which means the chip is drawing 100 to 200 amperes of current from the voltage regula

> it's consuming 200 W to provide its transistors with about 1 to 2 volts, which means the chip is drawing 100 to 200 amperes of current from the voltage regulators that supply it. Your typical refrigerator draws only 6 A. High-end mobile phones can draw a tenth as much power as data-center SoCs, but even so that's still about 10–20 A of current. That's up to three refrigerators, in your pocket! This feels out of place coming from IEEE.

amelius 5y ago
I don't know. I bet half of IEEE only worries about data/signal processing in their dayjobs and never thinks about power distribution. Such a comparison immediately makes clear what the problem is.
rzwitserloot 5y ago
Especially considering the fact that this is harping on about Ampere. Which is _not_ the number to be looking at here; that'd be watts. That fridge is chugging down 6A at 110 or 220V (assuming it's a new fridge, unless its absolutely gigantic or incredulously inefficient, sounds like that'd be a 110V model) - not at 1 to 2 volts. If someone can build a fridge that is so efficient, it can make do with 6A @ 2V, dang. Where can I buy me one of those? That's 12W total, I can power one of these for a full hour with 4 AA batteries.
- wheels 5y ago
  I thought the same initially, and do think the analogy is bad, but a few seconds later I wondered if the point that they were making was that the interconnects carry the same amperage: the required gauge for a connector (i.e. wire) is determined by amps, not watts. As a result you can send more power down smaller cables at higher voltages.
  - willis936 5y ago
    Yeah the ohmic losses in the power delivery networks are the killer and the topic of this article.
  - mindslight 5y ago
    That is the point they are making, but comparing it with "three refrigerators" is seemingly invoking the power of a fridge. To understand what they're saying, you need to understand current as distinct from power regardless of the scale. If you do understand this distinction, but you don't have a feel for what 100 amps is, perhaps a good comparison is starting a car.
  - deleted 5y ago
    [deleted]
- Dylan16807 5y ago
  > this is harping on about Ampere. Which is _not_ the number to be looking at here; that'd be watts. You say that like it's obvious. I don't see why.
  - dragontamer 5y ago
    Agreed. In my electrical engineering classes, we use amps to determine the gauge of wires. The important calculation here is watts = I^2 * R Where R is the resistance of your wire, and watts is the power wasted in your wires. And I is amps. That little squared sign is a bit intimidating. Under normal circumstances, you want to increase voltage to reduce wire loss. But computer chips only operate at low voltage.
    - unnouinceput 5y ago
      And the reason they operate at low voltages has to do with the micro-scale these SoC transistors are separated from each other. Increase the voltage and you get a shortcut, which will render your chip useless.
      - brennanpeterson 5y ago
        This is true but also wrong. The transistors themselves have a specific operating voltage.
        unnouinceput 5y ago
        On current technology, you are absolutely correct. But if you go and say something "let's do instead those high voltage transistors, so we can deliver to them more power" you can't, exactly because at this scale you'll get shortcuts. It's a physics limitation.
      - dheera 5y ago
        > Increase the voltage and you get a shortcut It's possible to fill them up with goo that has a much higher dielectric breakdown than air, so I don't believe this is the reason
- deepnotderp 5y ago
  Actually for power delivery networks current (amperes) is mostly what you care about
- throwaway9870 5y ago
  As someone who has designed many chips, amperage absolutely matters because it is not DC, it has very rapid transients based on workloads and that, combined with inductance, can make power delivery very difficult. Additionally, the high current requires careful design of the package and routing because of resistance and electromigration even in the DC case.
  - uberswe 5y ago
    I think the point is that the text makes it seem like you have something consuming the same amount of power as 3 refrigerators in your pocket.
  - awesomeusername 5y ago
    Can you comment on why is there always a drive towards lower voltage? And how this impacts design. A nieve take would have thought higher voltage would equate to less loss to heat. But then with a higher voltage I guess those transients become relatively bigger?
    - throwaway9870 5y ago
      A large part of modern chip design is accomplishing what you are trying to do inside a given power budget. Thus, anything that lowers power is usually a huge win. Switching power in CMOS is roughly proportional to V^2 (basic textbook, real design includes a vastly larger number of issues), so lowering V is a huge win. Also, reducing the parasitic cap with a better process is a big win for a multitude of reasons. But shrinking the process generally would require the voltage to scale also because the gate oxide was getting thinner. That is not really true any more as we reach supply voltages that are so close the Von of the devices, but used to be a clear trend. High voltages are good when you want to deliver power because that is what you need (motor HP, space heater, etc.). But we don't use an IC for power, we use it to do logic, and in that case, logic is just much more efficient at low voltages.
- bserge 5y ago
  A modern fridge compressor uses less than 300W (for a big one). They're actually surprisingly efficient.
  - userbinator 5y ago
    Even ancient ones don't go over 300W, but 6A at 120V is 720VA which even with a terrible power factor of 0.5 is still 360W. Fridge compressors are usually 2-3A, or 100-200W.
- ASalazarMX 5y ago
  "[..] Austin Wilde held up the source of power that had enabled a Disinto to chew up a mountain in half a second - two flashlight batteries!" It's amazing how well Asimov's robot stories have aged in these A.I. times.
- dheera 5y ago
  No, amperes is the number to look at. If you just want to deliver more power on the SAME wire, you can do so by increasing the voltage. This is why a pretty thin wire can deliver power to a high speed train, or even an entire town. Trains often use voltages in the 16-50 kV range, and power lines that power entire towns can be upwards of 100 kV; at that voltage the entire town's average power might be only a handful of amps. (Heat dissipation of a resistive wire is I^2*R, not dependent on voltage.) In the case of a CPU, higher voltages don't work with the semiconductors, so they have no choice but to use high amperes, and that becomes a problem.
- sandworm101 5y ago
  >>a fridge that is so efficient, it can make do with 6A @ 2V, dang That's 12watts. You just need some great insulation and enough time. If you are willing to never open the door and can wait days for your beer to chill, a 12-watt fridge is very doable.
whatshisface 5y ago
Voltage drop = current × resistance. Power lost to heat = current² × resistance. I think they are making a reasonable point that resistance losses are likely to be a much bigger problem for a CPU than for a large appliance with similar wattage. 10-20A is an enormous current even on household wires (most household circuits are rated for 15-20A), and while wires on CPUs are shorter, they're also a lot thinner. The wires in the refrigerator would likely be unable to handle 20A at 2V.
- bserge 5y ago
  No, they handle it fine.
konschubert 5y ago
why? Amperage determines the wire diameter. High amperage means very wide wires. I think their point is that this is what ultimately drives the need to power from below.
- ReactiveJelly 5y ago
  It's frustrating to see them not spare a couple sentences to clear up a misconception that _many_ laypeople suffer from. Sure, _we_ know the difference between amps, watts, and watt-hours, because we paid attention in science class, but most people still get them mixed up.
  - wheels 5y ago
    To be fair, this is not a publication for lay-people; it's obviously and explicitly a publication for electrical engineers, which would not need these things explained. But it's still a terrible analogy since the phrasing seems to imply that it's talking about power, when it's actually talking about current.
    - hinkley 5y ago
      And possibly a heads up to tech selection people about what’s coming soon. I think you can expect those people to have potentially taken EE 101.
- asddubs 5y ago
  because it would lead someone not already familiar with what those figures mean and how they relate to one another to come to the wrong conclusion. and someone who does know doesn't need the analogy. When I think "fridge", I don't think "what wire diameter do I need to deliver power", I think about a big old hunk of metal using a bunch of power
  - JumpCrisscross 5y ago
    > would lead someone not already familiar with what those figures mean and how they relate to one another to come to the wrong conclusion It's the IEEE. It's not designed for average consumption. That's almost OP's point, which make this counterpoint a bit comical.
Keyframe 5y ago
What is Ohm's law? Come on, IEEE!
marcosdumay 5y ago
I don't see anything wrong with it. That refrigerator will be a real constraint on the width of the power wires of any place it's installed on. And adding the current of your devices is exactly what you need to do to size your power lines. It being on IEEE, I can't imagine anybody on their target audience will be confused and imagine they are talking about power.
MayeulC 5y ago
Instantaneous power draw can be quite considerable too, when you have millions of transistors switching in a short lapse of time. Typically you cannot really include capacitors on the die, so those are close to it. It might have to do with it, but I haven't read TFA yet.
- bsder 5y ago
  Actually, inductive ringing on the power grid is generally a bigger problem than lack of capacitance. Generally, not all the transistors in your chip switch. The transistors that don't switch provide a charge reservoir to draw from for the transistors that do. The problem is then backfilling all that current that got lost and you have to do that within one clock cycle--which is the "lots of current" that this article is talking about. Because you have these pulses of current snapping from on to off at fairly high frequencies being fed over long distances with very little resistance to damp them, inductance kicks in and starts causing oscillations (LC tank). However, at this point Moore's Law about performance is dead (2x every 18 months), so this is not a very big deal. Moore's Law about cost is still alive (double the number of transistors/halve the cost every 18 months). So, the big deal currently is in the embedded space where leakage is more problematic because the die is mostly determined by RAM and flash sizes which goes directly to current leakage and die size.
  - MayeulC 5y ago
    Thank you for the insightful comment!
deleted 5y ago
[deleted]
maccolgan 5y ago
>That's up to three refrigerators, in your pocket! This is the part where it feels out of place
nicoburns 5y ago
Amps aren't really relevant here, I have a wrench that will consume considerably more amps than that if you're able to supply them.
doctor_eval 5y ago
Am I mistaken that those 200 amps are distributed among billions of transistors? Is there any point where there is a single conductor carrying 200A? I would have thought that the actual power distribution is done some other way. How does this 200A actually work in practice? Do they start with higher voltage and step down? This poor analogy means I don’t understand the problem that’s been solved. I would really have liked to learn this from the article, but instead all I can think of is the one noisy compressor in my 20yo fridge sucking down 6A (probably 3A where I live) over cables almost as thick as a CPU is wide.
- magicalhippo 5y ago
  Yes, the 200 amps are consumed by the billions of transistors. As you might have noticed, these power-hungry chips have a large number of pins, and quite a lot of them are dedicated to power, exactly to avoid having a single pin having to carry 200 amps. Here's[1] the pinout of the AM4 socket, where the pink and green squares represents power and ground respectively. As you can see they make up almost half of the pins. The motherboard is primarily supplied by 12V these days, which is then converted down to the 1-2V needed using multi-phase buck converters[2]. If you've ever seen motherboards boasting some number of VRM phases, this is what they're talking about. Gamers Nexus has an overview[3] of this as well. The problem they're talking about is similar to the AM4 socket. You have a bunch of signal pins, and they're connected to output transistors inside the package. You'd like to avoid long connections, so ideally the pin is close to the relevant output transistors on the chip. However you also got all this power that needs to be supplied, and you gotta spread that out over multiple pins. So it's a challenge to best arrange the power vs signal pins, minimizing the detours either have to take. Long connections means higher inductance and resistance, which is a problem for both signal and power. Imagine instead the CPU was mounted like a sandwich, with pins on both sides. Then it would be easy, as you could just place all the power pins on one side and signal pins on the other side. This is the solution they propose, except on the chip level. [1]: https://www.docdroid.net/6cDW11N/am4-pinout-diagram-pdf https://www.docdroid.net/6cDW11N/am4-pinout-diagram-pdf [2]: https://en.wikipedia.org/wiki/Buck_converter#Multiphase_buck https://en.wikipedia.org/wiki/Buck_converter#Multiphase_buck [3]: https://www.gamersnexus.net/guides/1229-anatomy-of-a-motherboard-what-is-a-vrm-mosfet?showall=1 https://www.gamersnexus.net/guides/1229-anatomy-of-a-motherb...

amelius 5y ago

I don't know. I bet half of IEEE only worries about data/signal processing in their dayjobs and never thinks about power distribution. Such a comparison immediately makes clear what the problem is.

rzwitserloot 5y ago

Especially considering the fact that this is harping on about Ampere. Which is _not_ the number to be looking at here; that'd be watts. That fridge is chugging down 6A at 110 or 220V (assuming it's a new fridge, unless its absolutely gigantic or incredulously inefficient, sounds like that'd be a 110V model) - not at 1 to 2 volts. If someone can build a fridge that is so efficient, it can make do with 6A @ 2V, dang. Where can I buy me one of those? That's 12W total, I can power one of these for a full hour with 4 AA batteries.

wheels 5y ago
I thought the same initially, and do think the analogy is bad, but a few seconds later I wondered if the point that they were making was that the interconnects carry the same amperage: the required gauge for a connector (i.e. wire) is determined by amps, not watts. As a result you can send more power down smaller cables at higher voltages.
- willis936 5y ago
  Yeah the ohmic losses in the power delivery networks are the killer and the topic of this article.
- mindslight 5y ago
  That is the point they are making, but comparing it with "three refrigerators" is seemingly invoking the power of a fridge. To understand what they're saying, you need to understand current as distinct from power regardless of the scale. If you do understand this distinction, but you don't have a feel for what 100 amps is, perhaps a good comparison is starting a car.
- deleted 5y ago
  [deleted]
Dylan16807 5y ago
> this is harping on about Ampere. Which is _not_ the number to be looking at here; that'd be watts. You say that like it's obvious. I don't see why.
- dragontamer 5y ago
  Agreed. In my electrical engineering classes, we use amps to determine the gauge of wires. The important calculation here is watts = I^2 * R Where R is the resistance of your wire, and watts is the power wasted in your wires. And I is amps. That little squared sign is a bit intimidating. Under normal circumstances, you want to increase voltage to reduce wire loss. But computer chips only operate at low voltage.
  - unnouinceput 5y ago
    And the reason they operate at low voltages has to do with the micro-scale these SoC transistors are separated from each other. Increase the voltage and you get a shortcut, which will render your chip useless.
    - brennanpeterson 5y ago
      This is true but also wrong. The transistors themselves have a specific operating voltage.
      - unnouinceput 5y ago
        On current technology, you are absolutely correct. But if you go and say something "let's do instead those high voltage transistors, so we can deliver to them more power" you can't, exactly because at this scale you'll get shortcuts. It's a physics limitation.
    - dheera 5y ago
      > Increase the voltage and you get a shortcut It's possible to fill them up with goo that has a much higher dielectric breakdown than air, so I don't believe this is the reason
deepnotderp 5y ago
Actually for power delivery networks current (amperes) is mostly what you care about
throwaway9870 5y ago
As someone who has designed many chips, amperage absolutely matters because it is not DC, it has very rapid transients based on workloads and that, combined with inductance, can make power delivery very difficult. Additionally, the high current requires careful design of the package and routing because of resistance and electromigration even in the DC case.
- uberswe 5y ago
  I think the point is that the text makes it seem like you have something consuming the same amount of power as 3 refrigerators in your pocket.
- awesomeusername 5y ago
  Can you comment on why is there always a drive towards lower voltage? And how this impacts design. A nieve take would have thought higher voltage would equate to less loss to heat. But then with a higher voltage I guess those transients become relatively bigger?
  - throwaway9870 5y ago
    A large part of modern chip design is accomplishing what you are trying to do inside a given power budget. Thus, anything that lowers power is usually a huge win. Switching power in CMOS is roughly proportional to V^2 (basic textbook, real design includes a vastly larger number of issues), so lowering V is a huge win. Also, reducing the parasitic cap with a better process is a big win for a multitude of reasons. But shrinking the process generally would require the voltage to scale also because the gate oxide was getting thinner. That is not really true any more as we reach supply voltages that are so close the Von of the devices, but used to be a clear trend. High voltages are good when you want to deliver power because that is what you need (motor HP, space heater, etc.). But we don't use an IC for power, we use it to do logic, and in that case, logic is just much more efficient at low voltages.
bserge 5y ago
A modern fridge compressor uses less than 300W (for a big one). They're actually surprisingly efficient.
- userbinator 5y ago
  Even ancient ones don't go over 300W, but 6A at 120V is 720VA which even with a terrible power factor of 0.5 is still 360W. Fridge compressors are usually 2-3A, or 100-200W.
ASalazarMX 5y ago
"[..] Austin Wilde held up the source of power that had enabled a Disinto to chew up a mountain in half a second - two flashlight batteries!" It's amazing how well Asimov's robot stories have aged in these A.I. times.
dheera 5y ago
No, amperes is the number to look at. If you just want to deliver more power on the SAME wire, you can do so by increasing the voltage. This is why a pretty thin wire can deliver power to a high speed train, or even an entire town. Trains often use voltages in the 16-50 kV range, and power lines that power entire towns can be upwards of 100 kV; at that voltage the entire town's average power might be only a handful of amps. (Heat dissipation of a resistive wire is I^2*R, not dependent on voltage.) In the case of a CPU, higher voltages don't work with the semiconductors, so they have no choice but to use high amperes, and that becomes a problem.
sandworm101 5y ago
>>a fridge that is so efficient, it can make do with 6A @ 2V, dang That's 12watts. You just need some great insulation and enough time. If you are willing to never open the door and can wait days for your beer to chill, a 12-watt fridge is very doable.

whatshisface 5y ago

Voltage drop = current × resistance. Power lost to heat = current² × resistance. I think they are making a reasonable point that resistance losses are likely to be a much bigger problem for a CPU than for a large appliance with similar wattage. 10-20A is an enormous current even on household wires (most household circuits are rated for 15-20A), and while wires on CPUs are shorter, they're also a lot thinner. The wires in the refrigerator would likely be unable to handle 20A at 2V.

bserge 5y ago
No, they handle it fine.

konschubert 5y ago

why? Amperage determines the wire diameter. High amperage means very wide wires. I think their point is that this is what ultimately drives the need to power from below.

ReactiveJelly 5y ago
It's frustrating to see them not spare a couple sentences to clear up a misconception that _many_ laypeople suffer from. Sure, _we_ know the difference between amps, watts, and watt-hours, because we paid attention in science class, but most people still get them mixed up.
- wheels 5y ago
  To be fair, this is not a publication for lay-people; it's obviously and explicitly a publication for electrical engineers, which would not need these things explained. But it's still a terrible analogy since the phrasing seems to imply that it's talking about power, when it's actually talking about current.
  - hinkley 5y ago
    And possibly a heads up to tech selection people about what’s coming soon. I think you can expect those people to have potentially taken EE 101.
asddubs 5y ago
because it would lead someone not already familiar with what those figures mean and how they relate to one another to come to the wrong conclusion. and someone who does know doesn't need the analogy. When I think "fridge", I don't think "what wire diameter do I need to deliver power", I think about a big old hunk of metal using a bunch of power
- JumpCrisscross 5y ago
  > would lead someone not already familiar with what those figures mean and how they relate to one another to come to the wrong conclusion It's the IEEE. It's not designed for average consumption. That's almost OP's point, which make this counterpoint a bit comical.

Keyframe 5y ago

What is Ohm's law? Come on, IEEE!

marcosdumay 5y ago

I don't see anything wrong with it. That refrigerator will be a real constraint on the width of the power wires of any place it's installed on. And adding the current of your devices is exactly what you need to do to size your power lines. It being on IEEE, I can't imagine anybody on their target audience will be confused and imagine they are talking about power.

MayeulC 5y ago

Instantaneous power draw can be quite considerable too, when you have millions of transistors switching in a short lapse of time. Typically you cannot really include capacitors on the die, so those are close to it. It might have to do with it, but I haven't read TFA yet.

bsder 5y ago
Actually, inductive ringing on the power grid is generally a bigger problem than lack of capacitance. Generally, not all the transistors in your chip switch. The transistors that don't switch provide a charge reservoir to draw from for the transistors that do. The problem is then backfilling all that current that got lost and you have to do that within one clock cycle--which is the "lots of current" that this article is talking about. Because you have these pulses of current snapping from on to off at fairly high frequencies being fed over long distances with very little resistance to damp them, inductance kicks in and starts causing oscillations (LC tank). However, at this point Moore's Law about performance is dead (2x every 18 months), so this is not a very big deal. Moore's Law about cost is still alive (double the number of transistors/halve the cost every 18 months). So, the big deal currently is in the embedded space where leakage is more problematic because the die is mostly determined by RAM and flash sizes which goes directly to current leakage and die size.
- MayeulC 5y ago
  Thank you for the insightful comment!

deleted 5y ago

[deleted]

maccolgan 5y ago

>That's up to three refrigerators, in your pocket! This is the part where it feels out of place

nicoburns 5y ago

Amps aren't really relevant here, I have a wrench that will consume considerably more amps than that if you're able to supply them.

doctor_eval 5y ago

Am I mistaken that those 200 amps are distributed among billions of transistors? Is there any point where there is a single conductor carrying 200A? I would have thought that the actual power distribution is done some other way. How does this 200A actually work in practice? Do they start with higher voltage and step down? This poor analogy means I don’t understand the problem that’s been solved. I would really have liked to learn this from the article, but instead all I can think of is the one noisy compressor in my 20yo fridge sucking down 6A (probably 3A where I live) over cables almost as thick as a CPU is wide.

magicalhippo 5y ago
Yes, the 200 amps are consumed by the billions of transistors. As you might have noticed, these power-hungry chips have a large number of pins, and quite a lot of them are dedicated to power, exactly to avoid having a single pin having to carry 200 amps. Here's[1] the pinout of the AM4 socket, where the pink and green squares represents power and ground respectively. As you can see they make up almost half of the pins. The motherboard is primarily supplied by 12V these days, which is then converted down to the 1-2V needed using multi-phase buck converters[2]. If you've ever seen motherboards boasting some number of VRM phases, this is what they're talking about. Gamers Nexus has an overview[3] of this as well. The problem they're talking about is similar to the AM4 socket. You have a bunch of signal pins, and they're connected to output transistors inside the package. You'd like to avoid long connections, so ideally the pin is close to the relevant output transistors on the chip. However you also got all this power that needs to be supplied, and you gotta spread that out over multiple pins. So it's a challenge to best arrange the power vs signal pins, minimizing the detours either have to take. Long connections means higher inductance and resistance, which is a problem for both signal and power. Imagine instead the CPU was mounted like a sandwich, with pins on both sides. Then it would be easy, as you could just place all the power pins on one side and signal pins on the other side. This is the solution they propose, except on the chip level. [1]: https://www.docdroid.net/6cDW11N/am4-pinout-diagram-pdf https://www.docdroid.net/6cDW11N/am4-pinout-diagram-pdf [2]: https://en.wikipedia.org/wiki/Buck_converter#Multiphase_buck https://en.wikipedia.org/wiki/Buck_converter#Multiphase_buck [3]: https://www.gamersnexus.net/guides/1229-anatomy-of-a-motherboard-what-is-a-vrm-mosfet?showall=1 https://www.gamersnexus.net/guides/1229-anatomy-of-a-motherb...