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Even though there is a power supply unit in a computer case, why are there separated power circuits for CPU, GPU, and RAM on the motherboard?

I mean, why can not the CPU, RAM and GPU just take their energy from the power supply unit?

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    \$\begingroup\$ What are those "separated power circuits"? More than just for different supply voltage requirements for different components? \$\endgroup\$ Dec 3, 2022 at 14:24

6 Answers 6

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Because these chips require a lot of power, and that means a lot of current. A high-end CPU can pull over a hundred amps when running at full load! Having hundred-amp currents running all over your PCB isn't recommended if you can ever avoid it.

But fortunately, the CPU runs at a core voltage of only about 1 V. If you transfer the power across the board at a higher voltage, commonly 12 V, then step it down with a buck converter right where it's needed, you can cut the current requirements by a factor of 10 or more (ideally 12, in this example, but the converter isn't ideal)! It's the same reason mains power is transmitted at very high voltages and stepped down with transformers outside buildings; high transmission voltage means low transmission current means lower losses.

The use of local power regulation also makes it possible to adjust the voltage provided to the CPU--increasing the voltage slightly can allow for faster operation when needed, while decreasing it when idle can lower power consumption. Modern computers do all this automatically--the laptop I'm typing this on now currently has a CPU voltage that's going between about 0.6 V and 1.1 V* (in sync with a clock speed going between 900 and 4200 MHz) as more or less power is needed.

This goes for GPUs as well, which also consume a lot of power--often more than CPUs. RAM doesn't need as much power, but it does need to have its voltage stepped down from the system voltage, so it gets a converter too.


*As reported by the diagnostic tool CPU-Z.

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    \$\begingroup\$ Just like high voltage transmission lines versus the lower voltage AC in your house. \$\endgroup\$
    – DKNguyen
    Dec 3, 2022 at 8:09
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    \$\begingroup\$ It is entirely possible to run 100A through a PCB if you use enough copper. I have one board I designed recently that runs 500A through it. The main limitation in the case of an ATX supply is that there are only three 12V pins on the 24-pin connector, and each pin is rated to 8A. That's only 288W max, which is not enough for today's systems. If they want to keep backwards compatibility the simple solution was to add the 8 pin CPU connectors which each add an extra 24A @ 12V = 288W per connector. \$\endgroup\$
    – user4574
    Dec 4, 2022 at 3:56
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    \$\begingroup\$ @user4574 Yes, it's entirely possible. I said it's not recommended, not impossible. You're not going to be able to get a fine enough pitch for the dense signal lines of a motherboard if you use super-thick copper, though--I'm not familiar with motherboards in particular, but it's not uncommon for high-density boards to use 0.5 oz copper, because thinner copper can be etched more precisely. \$\endgroup\$
    – Hearth
    Dec 4, 2022 at 4:12
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    \$\begingroup\$ @Hearth Yes, 1oz and preferably 1/2 oz copper would be needed on the outer layers and some of the signal layers to breakout fine pitch BGAs using 3~4 mil traces. But that doesn't stop one from using thicker copper on some of the inner layers. The 500A board I designed had high density parts like an FPGA, a processor, an FMC connector, no lead chips, and thousands of other little parts. I just stuck like 6 layers of 3 oz copper on some of the inner layers and used that to route the high current stuff. But using the extra Molex is by far cheaper than using heavy copper layers when possible. \$\endgroup\$
    – user4574
    Dec 4, 2022 at 4:38
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    \$\begingroup\$ @Hearth I had the luxury of examining a dead motherboard the other day. Between the 4 so-called VRMs (i.e. the 12V->1V converters) that feed the CPU is basically solid copper. Basically one big polygon between the entire VRM section and that side of the CPU. And then above the CPU, another big polygon which feeds the integrated GPU. An entire half of the space around the CPU is occupied just by its power connections! I assume the ground is on an interior layer. \$\endgroup\$
    – user253751
    Dec 5, 2022 at 19:19
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What you're describing isn't really possible without silly amounts of decoupling.

A modern CPU might be drawing 100A, with 20A step changes in load, all at 1v DC, and wanting +-20mV. This is a nightmare to deal with at the best of times.

Ignoring what's happening inside the CPU itself (even though that's half the fun :-) ), let's look at what's going on outside.

You usually have a bunch of capacitors right around the CPU in the 1V domain. Their job is to buffer load changes just long enough for the local regulators to adapt. However, because these capacitors are so low-voltage, they cannot store much energy - energy stored in a capacitor is ½CV², and voltage is small, so we're on the wrong end of a quadratic. In addition, the CPU is relatively sensitive to voltage sags, so we can't draw down the capacitors all that much either. So a lot of capacitors here still only provide a modest amount of energy storage.

Then you get to a layer of local regulators, tuned to respond quickly. They do provide a certain amount of energy storage themselves, but they largely just pass through load changes to the next layer ASAP.

Then the board has a bunch of 12V power supply filtering. Each capacitor here is 144x as effective as a capacitor in the 1V domain! (More, because we can simply design our regulators to cope with a larger input droop, and be able to pull the power rail lower.) And yet you still require a fair amount to cover until the power supply can respond.

So, all told:

  1. System is in equilibrium at 80A.
  2. CPU does a step change from drawing 80A to 100A, right as mains power is at a zero crossing.
  3. Decoupling capacitors around the CPU start to voltage-droop under the load (80A in; 100A out)
  4. Local regulators start to respond to the additional load. Likely with a bit of overshoot.
  5. Decoupling capacitors start to recover from the droop, but the 12V board filtering starts to droop in turn.
  6. Power supply sense wires notice that the output is drooping, and start to ramp up the power supply 12V output.
  7. 12V board power starts to recover, but the internal power supply input DC rail starts to droop in turn.
  8. Power supply input DC rail starts to droop.
  9. Next cycle, the AC/DC conversion picks up the additional load.
  10. I could keep going into the wall socket, but let's stop here.

Now let's assume that instead I just tried to directly run a 1.0V rail from the PSU to the CPU. How much decoupling do I need? Well, I need enough to cover the entire time from the CPU step change to the power supply being able to respond. ...all at 1.0V... ...and all with only being able to draw 20mV from the capacitors.

So, to a first approximation, take all of the 12V filtering on the motherboard, and multiply it by 144x, and a little more because it needs to be better-regulated too. And try to squeeze that onto your motherboard, without going too far from the CPU. And try to make it cost-competitive with the current design. Have fun!

Oh, and GPUs are worse.

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    \$\begingroup\$ Fair warning, this is oversimplified to the point of being wrong in places. \$\endgroup\$
    – TLW
    Dec 3, 2022 at 19:36
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    \$\begingroup\$ Only 20 A step changes? I'd expect it to be zero to a hundred in a microsecond. \$\endgroup\$
    – Oskar Skog
    Dec 3, 2022 at 23:28
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    \$\begingroup\$ @OskarSkog - a modern cpu takes order-of-10us to ramp up its pll and such when switching frequencies (see e.g. travisdowns.github.io/blog/2020/01/17/avxfreq1.html ), and tends to idle at a far lower frequency, so it's not quite as much of a step change as you'd otherwise assume. \$\endgroup\$
    – TLW
    Dec 6, 2022 at 5:07
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    \$\begingroup\$ 20A is still probably a drastic underestimate though. \$\endgroup\$
    – TLW
    Dec 6, 2022 at 5:07
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    \$\begingroup\$ But you can lock it at a high frequency. \$\endgroup\$
    – Oskar Skog
    Dec 6, 2022 at 7:14
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The short answer is that the 24 pin ATX power supply connector pins are not rated to supply enough power to today's high-end CPUs.

Since they probably want to keep things backwards compatible with other ATX boards they can't just go changing the 24-pin connector. Therefore, the solution is to add extra connectors. The extra 8-pin CPU power connectors have three 12V pins on them, which gives an extra 24A per connector. You will often see that high end CPUs will have several of those 8-pin connectors.

DETAILS

An ATX 24-pin connector is the Molex 39-28-1243 equivalent. It has three 12V contacts, and those contacts are only rated to 8A each (see connector specification page 8). So, the main connector can only supply 24A max @ 12V = 288W.

As Hearth pointed out, the CPU and GPU consume a lot of power (potentially hundreds of amps at low voltage).

A high-end processor like an Intel i9-13900K consumes up to 253W intermittently. Given that the DC-DC converters that convert 12V to the CPU core voltage have efficiency loss, having a 288W source is marginal at best. Plus there might be other devices that want 12V also. For both those reasons extra 12V connector pins are needed.

As for the other part of your question,

I mean, why can not the CPU, RAM and GPU just take their energy from the power supply unit?

The CPU, RAM and GPU all are powered from the power supply. In a normal desktop computer, if you look at the cables, they should all be plugging into the ATX power supply at one end.

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To add to the existing answer, CPUs and GPUs will scale their frequency with load, and increasing frequency will require additional supply voltage to remain stable. Since the CPU and GPU loads are not well coupled, these frequencies, voltages and the supplies that provide the voltage must be independent.

As to why there is a frequency-voltage dependency:

  1. Propagation delay of a CMOS logic gate decreases with increasing voltage (https://web.mit.edu/6.012/www/SP07-L13.pdf pages 4-5), achievable CPU frequency will be inversely proportional to gate delays.
  2. Modern computer equipment implements Dynamic Frequency Scaling (https://en.wikipedia.org/wiki/Dynamic_frequency_scaling) to reduce power consumption.
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    \$\begingroup\$ Re "require additional supply voltage": Don't you mean "require additional supply current"? Similar for "provide the voltage". \$\endgroup\$ Dec 3, 2022 at 14:16
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    \$\begingroup\$ @PeterMortensen it’s my understanding that the current consumption is a byproduct of the increased voltage and frequency, and that more voltage is required for faster switching. I will try to find an authoritative reference for this today. \$\endgroup\$
    – Bryan
    Dec 3, 2022 at 17:12
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    \$\begingroup\$ @PeterMortensen: CPUs do change voltage slightly as they change frequency, letting power scale as f^3 instead of just linearly with frequency. (Also with number of transistors switching every cycle, and that depends on the workload. Running integer code with lots of stall, and/or only one core active, a CPU can spend more of its power/thermal budget in one place, running fewer gates at higher frequency and power.) DVFS (Dynamic Voltage and Frequency Scaling) not just DFS. Also lighterra.com/papers/modernmicroprocessors en.wikipedia.org/wiki/Dynamic_voltage_scaling \$\endgroup\$ Dec 4, 2022 at 16:44
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I will add another scenario: when the computer is in low power standby or sleep mode.

Energy legislation requires that computers be able to enter low-power sleep modes (Energy Star, EPEAT, EU ecodesign). Laptop users desire the ability to quickly put their computer to sleep and resume, while not consuming undue amounts of battery power.

This requires that most devices be powered down, while specific devices remain on. Generally, the CPU is shut down, while RAM goes into a low-power mode to preserve memory contents. Further, USB and network cards remain in a low-power mode to wake up the computer.

There can be even further power reductions. Modern PCs use a soft-off state, where a real-time clock alarm or network card can start the computer, but RAM is turned off. Another scenario is switchable graphics, where a laptop disables its GPU on battery power.

These combinations of low-power states requires segregation of power buses. In desktop power supplies, a separate standby supply is optimized for low current operation, and doesn't require the PSU fan to operate.

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Could add to existing answers, feed motherboard with 12V is very outdated solution, being used, because it's hard to change industry legacy.

In FAANG companies made r&d's on save power in their clouds, and they came to use 35V from power supply to motherboard, and claiming up to few tens percents better energy efficiency, than on currently wide used "classic" scheme (this is easy - 3 times more voltage, mean three times less current for same power, and copper is expensive now). Sure, these are custom designed motherboards, you should not try to feed off the shelf mb with non-standard voltage.

And 35V is very moderate enhancement. - In air-space widely used 127V 400Hz AC to power electronics, so savings are even more significant.

Electric vehicles, already made much more - for them typical to use 500-1500V for most consuming devices , but for so high voltages need much stronger isolation and significant precautions at production and service (mediasystem/etc, usually fed by separate network, 12V/24V/48V, and some gas auto producers like Jeep, offer generator addon option, instead of usual ~1.8KW generator, install about 3KW).

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