CPU TDP

CPU TDP question:
Why are the e6300, e6400 and e6600 all rated at 65w TDP? As I understand it,
the heat output is based on the CPU architecture multiplied by the CPU clock
speed and CPU vCore squared (roughly). With these 3 chips (and other CPU
families) the only different is their speed (and cache levels), so the heat
should be lowest in the 6300, middle in the 6400 and top in the 6600?

I expect that the TDP is just a max rating to enable users to select the
correct cooler and the slower chips in reality produce less heat than the
faster ones, but can someone confirm this?

In short, if I take an e6400 and overclock it from its stock 2.16GHz up to
the 2.4GHz stock speed of the e6600, then would the heat output match that
of a stock e6600?

GT wrote:
CPU TDP question:
Why are the e6300, e6400 and e6600 all rated at 65w TDP? As I understand it,
the heat output is based on the CPU architecture multiplied by the CPU clock
speed and CPU vCore squared (roughly). With these 3 chips (and other CPU
families) the only different is their speed (and cache levels), so the heat
should be lowest in the 6300, middle in the 6400 and top in the 6600?

I expect that the TDP is just a max rating to enable users to select the
correct cooler and the slower chips in reality produce less heat than the
faster ones, but can someone confirm this?

In short, if I take an e6400 and overclock it from its stock 2.16GHz up to
the 2.4GHz stock speed of the e6600, then would the heat output match that
of a stock e6600?



CPUs are plopped into "bins", for the purposes of the TDP number.
Everything in the same bin, might benefit from the same retail cooler.

Say I was looking in the Intel catalog, and saw this.

2.1GHz TDP=65W
2.2GHz TDP=65W
2.3GHz TDP=65W

2.4GHz TDP=89W
2.5GHz TDP=89W

What that would tell me, is the one just before the transition,
is nearest to being really representative. So the 2.3GHz has a real
TDP rating of 65W, while the slower ones don't actually get that hot.

The 2.4GHz one is in the next set of bins, and draws much less than
89W. It could be drawing 65W or a little more. But guessing at it,
really doesn't tell us what is going on. Measurements are more
illuminating.

When they include a heatsink fan in the retail CPU box, the first
three will come with the same cooler. The last two, will have a
slightly beefier cooler.

The accuracy and intent of TDP has changed over the years.
On my Northwood processors, the actual power might be very close
to the TDP. Maybe even a couple watts above it.

My Core2 processors are on the low side. I think my 3GHz Core2
65W, might have measured 45W. Another Core2 I've got, measures
36W or so. And that is measured on the ATX12V cable, so the real
power as measured at the socket itself, is lower. (Vcore emits
some heat, and isn't 100% efficient.) Those measurements are
done with something like Prime95 running.

For small overclocks, it is probably sufficient to use a simple
ratio, or other rule, and assume nothing is going to burn up.

If you wanted a more accurate method, you'd

1) Measure the stock consumption like I did.

2) Use FCV**2 equation, to estimate the new power rating.

Frequency scaling is directly proportional. Taking the
ratio of your two frequencies, I get 1.11x, which means
11 percent more power due to frequency.

The voltage term is V_squared. If the original Vcore was
1.3V and to keep the processor stable, you needed 1.35V,
then take the ratio of 1.35/1.3 and square the result to
get 1.078. That means 7.8% more power, comes from the
extra voltage I used in the overclock.

Taking both factors into account = 1.11 * 1.078 = 1.197
or close to 20% more power.

So if the original power was 40W, and you overclocked with
a bit more Vcore (tested for stability and found it necessary),
the estimated power would be 40W * 1.2 = 48W roughly. You could
then clamp on the ammeter, and verify it.

So then the next natural question would be, "is the retail
heatsink/fan good enough for 48W ?". No data is available
to answer that :-) You can't link the third party coolers Intel
buys, to any data. No theta_R is available. We have to assume
the cooler is "good enough" in some sense, but we don't know
how good.

Overclocking does have its limits. The most extreme I've heard
of, is overclocking the D 805 to 4GHz. When the increased Vcore
is taken into account, power consumption is around 200W. This
exceeds the rating of the 2x2 power connector on the motherboard.
And has caused foam plastic to melt, in the vicinity of the
Vcore regulator. So there are a few overclocks of that nature,
where it pays to keep your calculator handy. Smokin good fun...

*******

The story on the AMD side is even more puzzling. In that AMD
announced some time ago, that it planned to cheat on power
numbers. I don't remember the details, and I'm not
interested in looking it up :-) That's the problem with TDP,
is it is politically charged, and a marketing brownie point.
Too much honesty, isn't good for business. Actual measurements
are the only way to add realism to the picture.

Paul

"Paul" wrote in message
...
GT wrote:
CPU TDP question:
Why are the e6300, e6400 and e6600 all rated at 65w TDP? As I understand
it, the heat output is based on the CPU architecture multiplied by the
CPU clock speed and CPU vCore squared (roughly). With these 3 chips (and
other CPU families) the only different is their speed (and cache levels),
so the heat should be lowest in the 6300, middle in the 6400 and top in
the 6600?

I expect that the TDP is just a max rating to enable users to select the
correct cooler and the slower chips in reality produce less heat than the
faster ones, but can someone confirm this?

In short, if I take an e6400 and overclock it from its stock 2.16GHz up
to the 2.4GHz stock speed of the e6600, then would the heat output match
that of a stock e6600?

CPUs are plopped into "bins", for the purposes of the TDP number.
Everything in the same bin, might benefit from the same retail cooler.

Say I was looking in the Intel catalog, and saw this.

2.1GHz TDP=65W
2.2GHz TDP=65W
2.3GHz TDP=65W

2.4GHz TDP=89W
2.5GHz TDP=89W

What that would tell me, is the one just before the transition,
is nearest to being really representative. So the 2.3GHz has a real
TDP rating of 65W, while the slower ones don't actually get that hot.

The 2.4GHz one is in the next set of bins, and draws much less than
89W. It could be drawing 65W or a little more. But guessing at it,
really doesn't tell us what is going on. Measurements are more
illuminating.

When they include a heatsink fan in the retail CPU box, the first
three will come with the same cooler. The last two, will have a
slightly beefier cooler.

The accuracy and intent of TDP has changed over the years.
On my Northwood processors, the actual power might be very close
to the TDP. Maybe even a couple watts above it.

My Core2 processors are on the low side. I think my 3GHz Core2
65W, might have measured 45W. Another Core2 I've got, measures
36W or so. And that is measured on the ATX12V cable, so the real
power as measured at the socket itself, is lower. (Vcore emits
some heat, and isn't 100% efficient.) Those measurements are
done with something like Prime95 running.

For small overclocks, it is probably sufficient to use a simple
ratio, or other rule, and assume nothing is going to burn up.

If you wanted a more accurate method, you'd

1) Measure the stock consumption like I did.

2) Use FCV**2 equation, to estimate the new power rating.

Frequency scaling is directly proportional. Taking the
ratio of your two frequencies, I get 1.11x, which means
11 percent more power due to frequency.

The voltage term is V_squared. If the original Vcore was
1.3V and to keep the processor stable, you needed 1.35V,
then take the ratio of 1.35/1.3 and square the result to
get 1.078. That means 7.8% more power, comes from the
extra voltage I used in the overclock.

Taking both factors into account = 1.11 * 1.078 = 1.197
or close to 20% more power.

So if the original power was 40W, and you overclocked with
a bit more Vcore (tested for stability and found it necessary),
the estimated power would be 40W * 1.2 = 48W roughly. You could
then clamp on the ammeter, and verify it.

So then the next natural question would be, "is the retail
heatsink/fan good enough for 48W ?". No data is available
to answer that :-) You can't link the third party coolers Intel
buys, to any data. No theta_R is available. We have to assume
the cooler is "good enough" in some sense, but we don't know
how good.

Overclocking does have its limits. The most extreme I've heard
of, is overclocking the D 805 to 4GHz. When the increased Vcore
is taken into account, power consumption is around 200W. This
exceeds the rating of the 2x2 power connector on the motherboard.
And has caused foam plastic to melt, in the vicinity of the
Vcore regulator. So there are a few overclocks of that nature,
where it pays to keep your calculator handy. Smokin good fun...

*******

The story on the AMD side is even more puzzling. In that AMD
announced some time ago, that it planned to cheat on power
numbers. I don't remember the details, and I'm not
interested in looking it up :-) That's the problem with TDP,
is it is politically charged, and a marketing brownie point.
Too much honesty, isn't good for business. Actual measurements
are the only way to add realism to the picture.

Paul

Thanks Paul. I was asking with a view to under-volting and possibly
under-clocking for a media centre PC. I have used the formula before to
calculate power/heat, but was just wondering about the TDP classification.
You have confirmed what I thought - the TDP is just a group thing for a
family of processors, the actualy power/heat varies based on vCore and
Frequency. Thanks.