

Intel Core i “Haswell” CPUs May Require New Power Supply Units for PCs - fdm
http://www.xbitlabs.com/news/cpu/display/20130430230347_Intel_Core_i_Haswell_Microprocessors_May_Require_New_Power_Supply_Units_for_PCs.html

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kwantam
Perhaps someone will helpfully design a "PSU stability dongle" comprising a
26.1 Ohm resistor (standard 1% resistor value sufficient to ensure >450mA from
12V) and a Molex connector. Make it a passthrough.

Of course, it'll have to be a >5W resistor, but that's what I like to call an
opportunity to excel: just more room to add some rakish heat sink fins. You
could even spend some of your power wasting budget on LEDs instead, but
obviously we'll have to charge more for that model. The margins, you see.

~~~
MertsA
Technically it would have to be on the same rail as the 12V CPU power
connector so a Molex 8981 connector
([http://en.wikipedia.org/wiki/Molex_connector#Disk_drive_conn...](http://en.wikipedia.org/wiki/Molex_connector#Disk_drive_connector_.28Molex_8981_Series_Power_Connector.29))
wouldn't cut it.

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joenathan
Molex is a company that makes many different types of connectors, so a Molex
connector would cut it.

[http://www.molex.com/molex/search/keyword_select.jsp?channel...](http://www.molex.com/molex/search/keyword_select.jsp?channel=Products&chanName=Product%20Name)

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twoodfin
Only if you want the C6/C7 states, which are the really low power draw ones.
If you're the kind of person who builds your own PC, it's unlikely you really
care about saving a few watts while in sleep. At worst, you'll get Ivy Bridge-
level power consumption.

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dman
On the contrary in 2013 the kind of person who builds their own PC's is the
only one who likely cares about technical specs. If "good enough" is good
enough for you there is very little reason to build your own box these days.

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criley
On the contrary to your contrary: the kind of person who builds their PC is
the kind who customizes how it works _when they're using it_. There's no
automatic assumption that they also harbor green feelings and want lower power
consumption for idle/sleep.

Hell, many are proud of their beefy 1kW+ power supplies and the ridiculous
power load they draw.

Others are proud of their energy-destroying bit-coin mining setups.

You certainly can't assume that the home builder cares at all about power
consumption.

~~~
jkldotio
>Others are proud of their energy-destroying bit-coin mining setups.

I'd be proud of my bitcoin mining setup too if it could violate the laws of
thermodynamics.

~~~
solistice
On the opposite side, if you had an energy creating bitcoin setup, why would
you care about the bitcoins?

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coldtea
> _As it appears, end-users will either have disable low-power states of
> Haswell or get a new power-supply units compatible with the new Intel chip._

Errr, is not like many end users update their desktop PC with new CPUs anyway.

And those very few that do, they also buy a new motherboard, so to get a new
power supply is nothing significant in comparison.

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MikeKusold
Everyone will need to buy a new Haswell motherboard. Unless the reports I've
read months ago changed, Haswell is going to be on a new socket.

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nivloc
Correct. Haswell is LGA 1150, SB/IB were LGA 1155.

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batgaijin
WTF is up with this? Is Intel just taking as much money as they can? Or are
these legitimate issues that they waited to address?

~~~
wtallis
There weren't any legitimate technical reasons for switching from LGA1156 to
LGA1155, but Haswell and LGA1150 are introducing major changes to power
delivery and voltage regulation. Given that power delivery accounts for
something like half the pin count of a modern CPU, a new socket for Haswell is
quite justified. (Although, those power delivery changes only really benefit
the mobile market, but Intel's long been unwilling to produce different dies
for the desktop market and the mainstream mobile market.)

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astrodust
I'm still perplexed why Intel doesn't redesign their CPU socket to be more
efficient at delivering power. They're channelling a lot of juice through some
very tiny pins or pads, a fairly straight-up evolution of the old 8088 chip
that fit into a DIP socket.

Is it somehow not practical to have several bigger pins for power that can
handle more current than to have literally hundreds of pins dedicated to
power? A surprising percentage of the pins on a modern Intel chip do nothing
more than power the chip. The traditional pair of +Vcc and GND pins just can't
cut it, apparently, and no wonder with some chips drawing over 100W of power.

Not too long ago video cards started taking a direct feed from the PSU rather
than relying on the PCI or PCIe bus. It's surprising the same thing hasn't
happened to high-power integrated circuits.

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elwin
There are some interesting reasons for that.

When a lot of transistors in an integrated circuit all switch at the same
time, it can cause the chip's power and ground voltage levels to shift,
relative to the circuit board's power and ground levels. The size of the
difference depends on the inductance between the chip and board, i.e. the
inductance of the chip's power and ground pins. Inductance can be minimized by
connecting a lot of inductors in parallel. Lots of small pins are better than
a few big ones.

If the inductance is too high and the chip's power voltage falls below its
ground voltage, this will randomize every storage element on the chip. The
rule of thumb is that a third of the pins need to be power or ground.

~~~
astrodust
You're implying it could be done if whatever load distribution and power
condition that's done outside of the chip, which consists of a lot of analog
components to help manage rapid changes in power consumption, could be somehow
packaged inside the chip.

So, in rough terms, the internals of a large-scale chip are not one big
integrated circuit, but a large number of smaller modules that are massively
interconnected, then?

That rule of thumb seems to apply to only a particular class of chips.
Wouldn't the number of pins be somehow proportional to the power draw, as at
higher currents induction would become a more severe problem? It's just
usually the case that more power-hungry chips have more pins, as the 2011
socket is for Intel's flagship CPUs, the 1155 ones more commodity-oriented.

With the power voltage dropping below ground, that unless you had a floating
ground, that'd be implying reverse flow of current, negative voltage, right?
Or are you talking about a non-zero voltage ground? I'm not sure what the
presumption is in real-world CPU design.

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elwin
> That rule of thumb seems to apply to only a particular class of chips.
> Wouldn't the number of pins be somehow proportional to the power draw, as at
> higher currents induction would become a more severe problem?

I think it depends on the chip's speed. The problem is with rapid changes in
current. Of course, higher currents can also have higher fluctuations.

> With the power voltage dropping below ground, that unless you had a floating
> ground, that'd be implying reverse flow of current, negative voltage, right?
> Or are you talking about a non-zero voltage ground?

Suppose the board's ground rail is 0 V and the power rail is at 12 V. The
chip's ground voltage might bounce up to 9 and its power down to 8. It does
cause reverse currents and other bad effects.

~~~
astrodust
Ah, so the ground gets pulled up and the power driven down.

Thanks. This makes a lot more sense. I never thought Intel was doing something
for no reason, but the reasoning wasn't obvious.

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wtallis
Just how much of the system is getting shut off in these new C-states? A lot
of the best power supplies have a single 12V rail, and the 3.3V and 5V rails
are provided by stepping down the voltage from the 12V rail. It seems like any
PSU with that kind of design wouldn't be affected unless the total system
power consumption drops below 6W, which would require shutting off at least
the expansion cards, disk drives, any power-hungry USB devices, and
potentially the case fans. Even 4 DIMMs plus the keyboard and mouse necessary
to wake up the system might be sufficient to keep the power draw high enough.

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petermonsson
I'm going to make a wild guess that the C6/C7 states are intended to be used
in notebooks and other battery powered devices. For desktops the rest of the
system will easily loose 5W somewhere else.

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ajross
FTA, quoting Robert Pearce: _"I fully expect the [motherboard] companies to
disable C6/C7 in the BIOS (though consumers could enable it if they chose to)
as there are simply too many PSU's in the market space which might not work
correctly."_

... because the motherboard manufacturers can't figure out how to stuff a 240
ohm resistor across the 12v rail?

Edit to be fair: if the power savings from going from the legacy suspend to
the Haswell states is less than 600mW, then this isn't worth it and disabling
the Haswell states is the right solution. In my experience though, PC
motherboards almost never idle at less than 3-5W.

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stephengillie
It's like how motherboard manufacturers couldn't figure out how to keep the
voltage regulators on their boards from overheating in the summer and
desoldering themselves.

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msandford
It's surprising that these kinds of hiccups happen as infrequently as they do.

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gshubert17
The difference between a 0.5 amp minimum current on the 12 volt supply and the
0.05 amp minimum current on the CPU might be enough to power a CPU fan (such
as the Precision 690 from Dell, which is about 0.45 amp at 12V). I'm not
familiar with other hardware requirements or constraints, so it might not
work.

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venomsnake
After how many years of standby will the savings of the low power state
justify replacement on a 200$ PSU that is 80+ platinium rated? And anyway
power states must be disabled for decent overclock to be achieved.

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blappu
Couldn't Haswell motherboards automatically test the PSU on bootup?

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gvb
Not likely. The problem is that the PSU will shut down (self protection) if it
has less that its minimum current draw. It has to do this because the voltage
regulation loop goes unstable if it was designed to have a higher minimum
current draw. If the PSU didn't shut down, it would fry things, including
itself. Really bad.

To test the PSU works at 0.5A would end up shutting down the computer if the
PSU didn't work at 0.05A... exactly what happens today. Unfortunately, at that
point it is too late to tell the user "your PSU is old" or to quickly connect
a resistor across the +12v rails to work around the problem.

As an experiment, find an old PSU and plug it into the wall (nothing connected
to the power output connectors). It won't power on. If you connect a 2 ohm 10W
resistor (or two 1 ohm 5W resistors in _series_ ) into the "hard drive" Molex
connector +12v (yellow) to ground (black), it should start up and be happy.

Disclaimers:

1) The resistor will become quite warm quite quickly (6W), watch your fingers.

2) If you damage something, it is a learning experience, not my fault.

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CamperBob2
Ah, yes, the other side of the Halting Problem: not only can you not tell if
your program will ever halt, but you also can't tell if your program has been
_externally_ halted.

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UnFleshedOne
Computers were doing that since forever though. Remember all those "your
computer wasn't shutdown correctly, do try to do better next time would you"
messages? Just write a bit into CMOS before doing a test, then assume the test
failed if the bit is still set on next power up.

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ivybridge
Hmmm. Maybe if they change the socket interface the new mobos can deal with
this issue.

~~~
lmm
They'd have to change the ATX power socket on the motherboard, not the CPU
socket; not impossible but it would be a shame after it's stood for this long.

