In one breath Intel dramatically cut pricing on its Core 2 Quads. Intel’s swift response was even faster than NVIDIA’s after the RV770 launch. In the following breath however, Intel introduced new, lower power, and much higher priced Core 2 Quad CPUs. Enter the S-line.

TDP binning is something that AMD has done for quite a while on the desktop. The e-suffix parts (e.g. Phenom X4 9350e) are lower TDP parts, sold at a premium, to those users who need lower power consumption.

The Phenom X4 9350e and the 9150e are both 65W quad-core parts from AMD, while all of Intel’s quad-core CPUs have been 95W. Unwilling to allow AMD any sort of advantage, Intel has finally responded with 65W quad-core offerings of its own. The difference here is that while AMD’s 65W quad-cores are all significantly lower clocked Phenom processors, Intel’s 65W chips are available at up to 2.83GHz.

The Core 2 Quad Q9550S, Q9400S and Q8200S are all 65W TDP quad-core CPUs. They share the same specs as their non-S brethren. The only difference here is that instead of being 95W TDP parts, these CPUs can fit in a 65W thermal envelope.

Processor Clock Speed L2 Cache L3 Cache TDP Price
Intel Core i7-965 Extreme Edition 3.20GHz 1MB 8MB 130W $999
Intel Core i7-940 2.93GHz 1MB 8MB 130W $562
Intel Core i7-920 2.66GHz 1MB 8MB 130W $284
Intel Core 2 Quad Q9650 3.00GHz 12MB - 95W $316
Intel Core 2 Quad Q9550S 2.83GHz 12MB - 65W $369
Intel Core 2 Quad Q9550 2.83GHz 12MB - 95W $266
Intel Core 2 Quad Q9400S 2.66GHz 6MB - 65W $320
Intel Core 2 Quad Q9400 2.66GHz 6MB - 95W $213
Intel Core 2 Quad Q8300 2.50GHz 4MB - 95W $183
Intel Core 2 Quad Q8200S 2.33GHz 4MB - 65W $245
Intel Core 2 Quad Q8200 2.33GHz 4MB - 95W $163

 

The price premium for these new S-parts is huge. The Q9550S costs $103 more than the non-S, the Q9400S will set you back another $107 and the Q8200S is the most affordable with only an $82 premium. Note that in the case of the Q9550S and Q9400S you're actually more expensive than the entry level Core i7-920.

Intel achieves these lower TDPs by running at a lower core voltage. With a mature enough manufacturing process, which Intel’s 45nm process is, it’s quite possible to produce CPUs that run much cooler than average and on a lower voltage. CPU power varies with the square of the voltage, so any savings in voltage can result in a non-linear decrease in power consumption.

Don’t get too excited however. If you remember back to our review of the 9350e/9150e we found that the decrease in power wasn’t worth the added price. Even Intel has come forward and told us that these are primarily OEM parts and not intended for the high volume enthusiast community. With Intel being honest in its intended purpose for these S-class CPUs we don’t really have to do much to keep them honest, we just need to confirm the findings.

To do this we took a subset of our regular CPU performance tests and looked at performance, power consumption and power efficiency. We measured total system power consumption at the wall outlet, which does admittedly lessen the impact of a lower power CPU but it should give us an idea of the real world benefit of these processors. If you want to see how the Q9550/Q9550S performs across our entire suite of benchmarks take a look at AnandTech bench, our new publicly available benchmark database.

...and in case you’re wondering, no, they don’t overclock any better. Our Q9550S couldn’t get any further than the Q9550 we used in our Phenom II review.

The Test

CPU: AMD Phenom II X4 940 (3.0GHz)
AMD Phenom 9950 (2.6GHz)
Intel Core i7-965 (3.2GHz)
Intel Core i7-920 (2.66GHz)
Intel Core 2 Extreme QX9770 (3.2GHz/1600MHz)
Intel Core 2 Quad Q9650 (3.00GHz)
Intel Core 2 Quad Q9550S (2.83GHz)
Intel Core 2 Quad Q9450 (2.66GHz)
Intel Core 2 Quad Q9400 (2.66GHz)
Motherboard: Intel DX58SO (Intel X58)
Intel DX48BT2 (Intel X48)
MSI DKA790GX Platinum (AMD 790GX)
Chipset: Intel X48
Intel X58
AMD 790GX
Chipset Drivers: Intel 9.1.1.1010 (Intel)
AMD Catalyst 8.12
Hard Disk: Intel X25-M SSD (80GB)
Memory: G.Skill DDR2-800 2 x 2GB (4-4-4-12)
G.Skill DDR2-1066 2 x 2GB (5-5-5-15)
Qimonda DDR3-1066 4 x 1GB (7-7-7-20)
Video Card: eVGA GeForce GTX 280
Video Drivers: NVIDIA ForceWare 180.43 (Vista64)
Desktop Resolution: 1920 x 1200
OS: Windows Vista Ultimate 64-bit
Adobe Photoshop CS4 Performance
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  • JPForums - Wednesday, January 28, 2009 - link

    In general when people say average, they are talking about the Arithmetic mean.
    Arithmetic mean = 1/n*(X1 + ... + Xn)
    or in English a list of numbers divided by the number of items in the list.

    In your case this would mean summing the list of power measurements taken at one second intervals and dividing by the number of measurements which would be the integer number of seconds. You could then calculate joules by multiplying that average by the total time.

    The only way the sum of power measurements and calculation made by multiplying the average power by the time would be different is if the number of measurements for the sum and the average are different. In this scenario, the calculation made with more data points would be more accurate (think integration).

    So the question becomes: How did you calculate your average? It appears that your average has more data points given that the total test time is measured to 1/10 seconds and your summation was only one second intervals.

    That said, the difference between the summation and the result calculated from the average should be small as you stated. The Q9550S results, for instance, only differs by 113(3244-3131) joules and the Core i7-920 differs by a mere 86(2818-2732) joules. Even the Phenom X4 9950 only has a delta of 69(5474-5405) joules. However, the Phenom II 940 has a delta of 898(4697-3799) joules.

    This massive difference leads me to suspect that either the average power, the total time, or the total energy for this processor was reported incorrectly. If we assume the average power and the maximum power are the same, then the delta shrinks to 220(4697-4490) joules. Alternately, if we assume that the Phenom II 940 is the same speed as the Phenom 9950, the delta shrinks to 207(4697-4477) joules. Both of the assumptions seem unreasonable to me, and neither get the delta as small as it should be. So I ask, now that I've presented a reasonable case, please recheck your total energy numbers as Ryun suggested.
    Reply
  • harijan - Tuesday, January 27, 2009 - link

    It still it doesn't make sense. How can it use 4700 joules yet average 157 watts over 24 seconds? Or have a max of 188 Watts?

    4697 Joules / 24.2 seconds = 195 Watts average
    Reply
  • Anand Lal Shimpi - Wednesday, January 28, 2009 - link

    Woops, you're completely right :) The issue wasn't with the power measurement but with the performance. The performance data for the run that I measured power under was incorrect. A re-run fixes that problem. The Q9450 was also impacted slightly.

    It's worth mentioning that the performance and power data are taken at two different times. First the performance data, then the power data. The performance during the power run is close but not always identical to the performance during the performance run. There's going to be some variation depending on the test.

    Take care,
    Anand
    Reply
  • GourdFreeMan - Wednesday, January 28, 2009 - link

    I have to agree. There is something wrong with Anand's methodology. Also, look at his specious reasoning for the difference in processor ranking between his "average" power and the energy consumed in the Fallout 3 section, where the tests are run for the same time interval. He is measuring total system power, so the improved idle efficiency of the Nehalems should already be incorporated in those numbers. Average power draw is by definition total energy consumed divided by time interval over which it is consumed. Either taking instantaneous measurements and treating them as averages for each second or simple human error could be responsible for the discrepancy. Reply
  • JPForums - Wednesday, January 28, 2009 - link

    I wouldn't call it suspicious, just a flaw in the procedure. If you have a sine wave and a cosine wave at the same frequency, amplitude, and offset measured once per period, one will look much larger than the other even though they average out to be exactly the same. Likewise, if you have two computers drawing the same average power, but you happen to record one during mostly high fluctuations and the other during mostly low fluctuations, you'll get two very different results.
    You need more samples to get accurate results. The best method would be to record a power graph using the smallest period possible. Then, integrate the power under the curve. Convert the units to seconds to get energy in joules. Divide by the number of samples to get the average power.
    Reply
  • GourdFreeMan - Wednesday, January 28, 2009 - link

    JPForums, I appreciate your efforts to elucidate my remarks to Anand, but I must comment on two things. First, the word I used was "specious" not "suspicious". There is a difference, just as there is a difference between "average power draw" and "an average of periodically sampled power draws". These two are only guaranteed to coincide if the samples themselves are average powers or in the limit as the period they are sampled over approaches 0. (The latter remark is directed at your definition of average in the first paragraph of your other post). Reply
  • Ryun - Tuesday, January 27, 2009 - link

    I was expecting a much bigger delta compared to the 95W quads in wattage. Reply
  • Ryun - Tuesday, January 27, 2009 - link

    Meant to end with, "Thanks for the review."

    So, thanks. =)
    Reply
  • harijan - Tuesday, January 27, 2009 - link

    no idle power usage numbers? Reply
  • michael2k - Tuesday, January 27, 2009 - link

    Unfortunately 65W is still too hot for a 17" MacBook Pro.

    Reply

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