This feels out of place coming from IEEE.
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.
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.
You say that like it's obvious. I don't see why.
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.
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
Watts are a function of Volts * Amps. So them talking about a fridge using 6 amps at 120 volts is really quite a silly comparison to using 200 amps at 1 volt. The fridge is using 720 watts of power compared to 200 watts from the chip.
In no way is it like having three fridges in your pocket.
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.
It's amazing how well Asimov's robot stories have aged in these A.I. times.
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.
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.
The wires in the refrigerator would likely be unable to handle 20A at 2V.
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.
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.
It's the IEEE. It's not designed for average consumption. That's almost OP's point, which make this counterpoint a bit comical.
It being on IEEE, I can't imagine anybody on their target audience will be confused and imagine they are talking about power.
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.
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.
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
[2]: https://en.wikipedia.org/wiki/Buck_converter#Multiphase_buck
[3]: https://www.gamersnexus.net/guides/1229-anatomy-of-a-motherb...
So would I be right to say that the problem is that the way power is delivered to individual transistors is getting too complex while the features are getting too small? The article started out by focussing on the total current at transistor voltages, but at 12 volts to the chip it’s just ~16 amps (still a lot!), spread across say 100 pins = 160mA per pin. Which isn’t as impressive as 200A. 200A is clearly not a wire you see going into a CPU.
It’s a shame that the article seems to have chosen to oversimplify the problem at the expense of helping us understand. The complexity of power delivery from the top of the chip looks fascinating, but it lost me at the outset with claims that wouldn’t pass muster with a high school electronics enthusiast.
It's the chip itself that needs 1-2V, so the buck converter is on the motherboard. So indeed 200A enters the chip. The linked GN article has a nice overview of the motherboard VRM layout.
The AM4 socket has 1331 pins though, so with roughly a quarter of those sourcing current that's 0.6A per pin (the other half of the power pins sinks that current).