i7 920QM vs i7 620M?

I notice the i7 920QM has a low frequency of 2.0GHz, does this mean for usage not requiring more than 2 cores, the i7 620M is better?

 

Not really no, the i7-920 having four cores will drop cores and overclock the 1-2 cores required that the task will depend on,

But then again, 920 costs a massive amount more.

In general though the 920 is much better, but then it really depends on your task.

 

The 920XM will destroy the 620M in any situation, but that of course comes at a much higher cost.

 

also way higher TDP... ppl with it have pushed it so far that it pulled 105W under load... rather than standard 55W TDP... and yes 920XM will destroy 620M but even a 820QM is enough really speaking... IMO , 920XM is not worth way more.

 

Now how can you possibly measure that?

Yes, that's right. Of course, the 920XM does allow overclocking depending on the laptop, so it could have better performance when using only one or two threads.

 

Looking at the Turbo Boost numbers, the clock speeds they reach on 4/3/2/1 cores are:
i7-620M: NA/NA/3067/3333
i7-920XM: 2267/2267/3067/3200

I'd expect the two to be about the same when running on 1 or 2 cores, though the 920XM should have an advantage due to Clarksfield's lower memory latency.

 

The difference is cost. Because theres a ~$700 difference.

Cost aside you would be VERY hardpressed to make the 620 perform better in any given task.

 

Agreed. Arrandale kicked the memory controller off the die (I assume that's what you are alluding to). Moreover, Clarksfield has more L3 cache per core when running just 1 or 2 cores (since the L3 is inclusive). Both of these things would tend to make Clarksfield a bit faster. OTOH, if you happen to run an application that can benefit from SSE 4.2 instructions, Arrandale will be faster (Clarksfield isn't capable of SSE 4.2).

 

If what you are saying is true, then the 920XM would be pulling fairly close to a stock Core i7 975 Extreme.

 

no. it's not really comparable.
975 has a stock speed of 3300mhz iirc.

 

He's talking about power consumption and pointing out how sean473 is making stuff up.

 

i didn't make it up... Mandrake did some measuring with Asus G73 and 920XM.. go and ask him.. don't flame me...ask mandrake... that's what he wrote when he stressed benchmarked the Asus G73.

 

And just to remind you, Sean, TDP (_THERMAL_ Design Power) is not power consumption. Never has been. A perfect superconductor would have a TDP of 0, and power consumption of near infinity. So just because a processor pulled 105 watts (and is that the processor, or the entire notebook? Overclocked, or stock?) does not necessarily mean that it's TDP went over 55 watts.

 

Most of the energy consumed by a CPU becomes heat; what else could it become? Also, the perfect superconductor wouldn't consume any power at all. Sure, some applications of that perfect superconductor could involve transformation of energy, and might achieve perfect efficiency, but power consumption cannot be infinite without an infinite power supply.

 

Ok, but there is absolutely no way one can measure a processor's power consumption separate from the rest of the computer. The figure of 105 watts sounds like it was measured either with a kill-a-watt before the power adapter, or by estimating based on battery voltage draw, and for the entire system, not the processor itself.

 

Well I meant the TDP of the 920XM would've been close to the 975 Extreme's TDP.

i7-920XM (According to Sean473) - 105W
i7-975 Extreme - 130W

 

the 920 also has the same TDP.. http://ark.intel.com/Product.aspx?id=37147

 

Interesting...

 

this 920 is not laptop CPU...

1ku Bulk Budgetary Price 284.00

 

thanks, this thread is getting a bit upsetting.

i7-920XM - 55W TDP

all desktop i7-9xx (aka Bloomfield) and i7-9xxX (aka Gulftown) - 130W TDP

 

Mandrake overclocked it to desktop levels and it was on par pretty much exactly, but the 920 doesnt come close to the 975 stock,

All i know is 920 will eat 620 but in terms of cost, i' d assume the 620 is definately worth it.

 

The 920XM is alot better than the 620. And it is an unlocked processor, which means it can go higher then the turbo boost. but if you want a good laptop processor for a good price i would go for the 17 720QM or the I7 820QM, both are great but i personally would take the 820QM, more cache memory.

 

It can come back out as useful instructions. Part of the problem I think is that when people say "power consumption" they only measure what goes in, and not what comes out. That's largely what I meant by a perfect superconductor having a near infinite power consumption (meaning you can shove in as much power as you want) and still have a TDP of 0 (since all that power just passes right through, creating no heat). As I stated before when this topic came up, oh, months ago, I don't actually have an idea of how efficient a CPU is at converting the power it intakes into useful commands that it outputs (it'll vary depending on CPU model, and I'm sure the actual data is buried somewhere in some paper). Let's assume, however, for the sake of argument, that there's a CPU with, oh, 100 watts TDP that at maximum load turns 20% of the power it takes in into usable instructions with the rest as heat, compared to one with the same TDP that can only convert 10%, with the rest as heat. That means that at maximum load, the 20% CPU would be drawing 125 watts, while the 10% CPU would be drawing about 111 watts.

 

"Useful instructions" aren't a form of energy, and so you can't transform electrical energy into them. Otherwise, the energy from "useful instructions" would just be building up in the CPU over its lifetime, until at some point it exploded.
Also, the very term "power consumption" directly implies that the power is consumed. Electrical power does not "pass through", only current does that. Conservation of energy means that energy in = energy out, and since CPUs don't store significant amounts of energy, that pretty much means power in = power out. Indeed, a CPU transforms pretty much 100% of input power as heat, and just happens to do useful calculation as a fortunate byproduct of this.

The perfect CPU would have zero power consumption.

 

Sorry, by useful instructions I mean the electrical signals that come out of the CPU, and are passed on to other components to get them to "do things". I think part of the problem is that the right term should probably be power requirement as opposed to power consumption; the problem being, as I said before, that people only measure what goes in, and not what comes out. In this sense, what they're measuring is more correctly labeled power requirement as opposed to power consumption. Thus, your perfect CPU, while having 0 power consumption, would still have a power requirement; otherwise the electrical signals coming out of it that give the results of its calculations would be spontaneously generated out of nowhere...

Personally, I'm not as sure as you are that a CPU will transform pretty much 100% of input power as heat, but I can't seem to find any actual numbers to prove or disprove the issue one way or another.

 

Sure, a CPU has output signals, but the power that is consumed by the other components when they "do things" doesn't go through the CPU; it's supplied independently to those other components. Each component has its own power rating which would obviously be independent of the others, or individual power ratings for components would be useless. (I'd make the distinction that power is energy per unit time, but it doesn't matter that much here).

Because the signals travel through non-zero resistance there will also be energy lost (i.e. turned into heat) at the interconnections between the CPU and the motherboard, though this would be minimal. Past that point it is presumably the motherboard that draws the energy needed for communications, not the CPU. In any case, these communication signals are there to transmit data to other components, but almost no electrical energy (which should comes from the PSU), and the unfortunate energy losses associated with transmitting this data would, for the most part, be accounted for in the power consumption of the motherboard.

Including the CPU, almost all the components of a computer are devices that transform electrical energy mostly into heat, sometimes in novel and interesting ways. For example, a spinning hard drive has kinetic energy, but ultimately it uses up energy making heat due to frictional losses, and vibrations that spread to the rest of the computer (including audible sound). In essence, the only computer component that outputs useful energy is the screen, which produces light.

Though the unfortunate fact of the laws of thermodynamics (specifically entropy) means that a CPU cannot manipulate data while consuming no power, the "perfect CPU" would nonetheless need no power at all. Just because there's an output signal doesn't mean there needs to be output power; though it is impossible in practice, in theory the output electrical signals don't necessarily need to consume power, because there is no need to transfer energy between the CPU and other components, only data.

What other forms of energy could the electricity become? Monitors produce light, fans move some air, and hard drives make sound/vibrations, but the CPU doesn't do any of this. To some extent, there is also most likely an energy difference between a "1" and a "0" in storage, but this would also be minimal.

 

I don't actually disagree with most of what you say, I think that to a large extent we're just coming at the issue from different sides. For most of my comparisons, I'm not looking at the motherboard, or other components, just the CPU. To this end, what I see is this; power of X amount goes into a CPU, and some Y amount of power comes out of it as data (because, at our current level of technology, it is impossible to have data come out of the CPU without power, as the CPU can only output things as a modulated electrical current, which means we have voltage and amperage which is power), and the remaining amount of power should end up being dissipated as heat. This is, actually, why my original response made the point of asking whether or not that stated 105 watts was for the processor only, or for the entire notebook, which would then mean that other components would also have been using up some of that stated 105 watts.

I'm not certain how you mean the motherboard draws the power needed for communications. I would argue that the motherboard distributes the power from the PSU to the other components, but it doesn't draw a separate amount of power for communications; the power for communications between components is transmitted to the components themselves as part of their power requirements. When you have communication between components, the motherboard is simply a conduit that connects those components together (unless by motherboard you mean the components and chipset integrated onto many motherboards, which I tend to separate out as their own components).

I don't see how even in theory output electrical signals wouldn't mean there isn't output power. As stated in the first paragraph, to output an electrical signal in the first place, you need some amount of amperage, and some amount of voltage, or there wouldn't be anything coming out in the first place. And, if you have amperage and voltage, you have power, by definition. You may not have power _consumption_, in that the perfect CPU would consume no power, but you would still need to have a power requirement, in that you need to input the power that is modified by the CPU to come out.

As for what other forms of energy the electricity becomes, well, it doesn't become anything else. It comes back out as electricity, in the form of the modulated current that represents the outgoing data. After all, at a very, very, simplistic view, the CPU is just a big mass of transistors that simply modulate an incoming data current into a (probably different) outgoing data current.

 

I agree that there's X input power and Y output power, but my points are that Y is very small compared to X, and that pretty much all of Y also becomes heat anyway, so it's entirely sensible to take X as the power used by the CPU.

The statement about the motherboard was indeed related to the chipset; it makes perfect sense to consider them included because the motherboard's power rating accounts for the power consumed by the chispet. It was more true when the northbridge wasn't part of the CPU, and so I guess currently the CPU would draw the power it uses to communicate over PCIe and QPI.

In any case, the more important point is that the power used in communications also becomes heat, so in order to accurately represent the power used by the computer, it has to be attributed to one component or another.

If you look at superconductors, they can have current but effectively zero voltage. For the purposes of the theoretical "perfect CPU", then, it is reasonable to consider that there would be zero input voltage and hence zero power. If you look at Ohm's law, there is no theoretical problem with zero voltage and nonzero current - V = IR is satisfied when V = 0 and R = 0, and current is anything but infinity.

Sure, if you look at only the CPU in isolation, there is some amount of output electrical energy (which, again, would be very small), but as I said before, that energy still becomes heat down the line; it is obviously power consumed, and it might as well be included in the CPU's consumption.

 

Well, that was my point about the lack of numbers; you argue that Y is very small compared to X, but we have no real numbers to prove it either way. And while Y may become heat anyway, it becomes heat in another component (probably the chipset/motherboard), and thus shouldn't be counted against the processor's TDP, as the chipset/motherboard does have its own TDP rating (and that should, at least in theory, account for Y).

For superconductors, yes, they have zero voltage difference from input to output, but I think it would be incorrect to say that they have zero voltage at input; if they did, there would be no current flow at all. I think we're hitting again the semantic difference between power consumption and power requirement; the "perfect CPU" would consume no power itself, but without an input signal (the power requirement; the instructions you send into the CPU via an electrical current, or the power you apply to one end of the superconductor) you would get no output signal (the modulated electrical current that serves as output signal, or the power you get out of the other end of the superconductor, that is equal to the input you put in).

 

If we trust Intel on the issue, the TDP of the mobile chipsets is only 3.5W, so if some of it were attributed to the CPU, it shouldn't make much difference.
Looking at the actual numbers in the Arrandale Datasheet, I found that the memory I/O voltage VDDQ is ~1.5V, while the current has a maximum of 3A, with an average of 0.33A on standby. This would imply consumption between 0.5W to 4.5W, but doesn't really tell me what it would be on average. You're right in that because of the need for I/O some of the power input into a CPU will become heat elsewhere, but I doubt that on average it would be more than ~2W compared to the 35W TDP.

If it is possible for the perfect CPU to consume no power (and hence for the signals within it to consume no power), then it's also possible for the input and output signals to consume no power, and so the CPU will not need any power. Why bring practical considerations into one area and drop them in another?

 

Didn't feel like reading if anyone answered you or not but you can measure power in hwmonitor and I believe in hwinfo32 as well.

/facepalm

lol

You are digging yourself a hole....

 

Hm. From your link, I find table 45 (page 94) more telling, especially as compared to table 18 (page 55). Table 18 lists the TDP for the CPU core at 25 watts in HFM for SV (which I think means standard voltage). Table 45 says (if HFM_VID means the input voltage in HFM) that the maximum voltage is 1.4 volts, and the maximum current (Iccmax) is 48 amps for SV. This would give a power figure of 67.2 watts. What confuses me is the Icc_tdc... I suspect that might actually be the current that would support the "usual" TDP, as there's that big disclaimer up front that Intel's TDP is maximum value under "normal" conditions, but even using that listed value of 32 amps, we get a power figure of 44.8 watts, which compared to the CPU core TDP of 25 is still a rather substantial difference.

As for the "perfect CPU", well, I suppose I insist on an input and output power because otherwise, our CPU would just be sitting there, and as far as we could tell, be doing nothing, which wouldn't make it much of a "processing unit", would it?

 

TDC definitely does seem a little strange. Even at the minimum voltage of 0.8V, the TDC gives 25.6W.
On the other hand, AMD's datasheets make a lot more sense, because they actually explain what the terms mean for their CPUs.

Theoretically, it's a situation where an infinitesimal change in voltage can cause a finite change in current, hence it's possible to have "on" and "off" both occurring at 0V, and hence you can have data signals without any power consumption. In the real world, there would have to be a very small voltage involved, and hence some minute power consumption, but we were talking hypotheticals.

 

For Intel, that might just mean what I've been saying, all that "excess" power just flows through the CPU and comes back out, and isn't converted to heat. Thus, assuming my previous numbers of 44.8 watts input and a TDP of 25 are correct, that means that 25 watts of that 44.8 are converted to heat, while the other 19.8 "flows through" (I doubt it actually just flows through, it's probably converted and modulated and such-like, but it does come back out).

Those AMD tables are actually worse in the other direction... for table 2.3.1, the TDP is always well over the VID_VDD Max times the TDC (32 compared to 28.8 for the first entry, TMZM80DAM23GG). This would seem to mean that the CPU is generating heat out of nowhere, but more likely means that either the listed TDP is "wrong", or that there's some other component contributing to TDP that AMD hasn't listed.

But if there is an infinitesimal change in voltage, wouldn't that by definition make the voltage non-zero?

 

If that's the case, that would be an extra 20W of heat in the motherboard, or elsewhere; that doesn't really seem right either. Intel seems to state that its TDC is for the design of voltage regulators; it's entirely possible that the value is derived completely differently to the TDP, and it definitely seems like a red herring here.

As for the AMD tables, they quite specifically explain the terms on page 17:

i.e. the extra power is that associated with VDDNB , VDDIO, VLDT, VTT, and VDDA. Not entirely sure what those are, but one of them is obviously some form of I/O, and the total of all of these is ~3.5W for all of those CPUs.
AMD also has a figure for I/O power of 3W, but that's in the halt state.

The power of the I/O signal would be zero in the same way that the power consumed by the CPU would be zero, i.e. any measurement would give zero. We can argue the semantics of infinitesimal quantities, but an infinitesimal voltage would imply infinitesimal power consumption, which, while technically not zero, definitely wouldn't show up on your power bill, no matter how long you ran your computer for (well, unless you ran your computer for an infinite period of time...).

 

Possibly. That still leaves us with my original, even larger numbers, which doesn't really seem much better.

This also assumes that VDD0 and VDD1 are both covered by the VID_VDD Max I mentioned earlier. It'd just be nice if we could actually have all these numbers for good clarification; I understand the need for confidentiality to a certain extent, but I can't see how publishing the numbers we're asking for would be a danger.

Now you're bringing practical considerations into things. Either way, I think we're reaching (relative) congruence on the issue of a "perfect CPU".

 

Thought this post would help the original discussion.

http://forum.notebookreview.com/showpost.php?p=5892567&postcount=463

 

I think the assumptions there are reasonable. From page 10 of AMD's datasheet:

(which also sorts out the issue of what VDDNB is)
Table 29 also clearly identifies VID_VDD as "VID-Requested VDD0, VDD1 Supply
Level", but better still is Table 31. Sadly, it only gives maximum current values, which again aren't that useful for calculating average power usage over time. However, it's clear that VDDIO and VDDA are related to the DDR2 memory, and I also found that VLDT is used to power HyperTransport (which used to be LDT). VDDA seems to be related to stability, but the current of 40mA is tiny anyway.

Also, to conclude on the "perfect CPU":
You said earlier that the perfect superconductor circuit outputs all the energy that goes in; that's true if we're thinking about a power circuit, where the aim is to transfer or transform electrical energy with minimal losses, and efficiency can be measured as (power out) / (power in).
However, a CPU cannot be looked at in the same way, because the purpose of the CPU is not to output power, but data, though it is an unfortunate fact that power is necessary for this also. It makes no sense to consider the efficiency of a CPU the same way as the aforementioned power circuit; it would be like saying that I am 75% efficient at eating food because I threw three quarters of my meal in the trash. Ideally, the CPU would just have data in and data out, with no power input or output.

 

Well, according to AMD (note 3, at the bottom of table 7 on page 17), TDP is calculated for the max. The problem is, then, that with the previously calculated gap of about 3.2 watts TDP, even just the calculated wattage for VDDIO and IDDIO1 (1.9 * 2) goes over that amount. Adding in the wattage of VLDT and ILDT (1.26 * 1.5) just makes it worse, as well as the small addition of VTT and ITT1.

Well, see, I would view your efficiency in the food case as 25%; you were efficient enough to convert 25% of your food into something "useful", while the 75% that you threw away got thrown out as waste. Similarly, if you input 20 watts (this would be power and data both) into a CPU, and got out 5 watts of data, that would also be, in my mind, a 25% efficient CPU. I suppose a lot of it comes down to what you're measuring; it actually seems easier in a case like this to measure "loss" (heat thrown off) to calculate what wasn't wasted. If you wanted to extend this backward to the conductor example, it would be 100 - (power lost). It's just easier in the case of a conductor to measure what comes out power-wise than loss.

 

It says it's calculated at the maximum Tdie, which is temperature, and at standard frequency and VDD being whatever the CPU happens to require at the time (though VDD doesn't vary much for AMD anyway). There is no mention of current, and, more importantly, it's quite specifically said that "TDP is not the maximum
power of the processor." As for the communications, you should consider that if the signal was on and drawing max current 100% of the time, no useful data would be transmitted. If we take the typical voltages with the max. currents, we get 1.8 * 2 + 1.2 * 1.5 + 0.9 * 0.75 + 2.5 * 0.04 = 6.175W, which would imply that you get 1s a little more than half the time (assuming an active high signal; if it's active low then 0s draw current and 1s don't). If you allow for AC losses due to switching and such, the signal is probably active half the time or less.

Yeah, my example wasn't great and I knew it at the time, but I couldn't see how to improve it and have it sound good. In the case of the CPU, it's more like the food you eat is waste as well. Accordingly, there's no such thing as 5 watts of data. It's X bits/sec of data accompanied by 5 watts of waste heat. In any case, the efficiency of a CPU is, in the most sensible sense, some number of calculations performed per watt. Obviously the "calculation" could be many different things and that's why there's no single standard measure of CPU performance, but the point is that the less power needs to go in, the more efficient the CPU is.

 

Hm. You have a point on the communications, although TDC (which is the current I was using) has a footnote of 6, which says that TDC is the sustained current drawn by VDD0 and VDD1 as measured at the same conditions as TDP, so that's covered.

Well, data carried on 5 watts of power, which could, in theory, be used to power another component that the data is being carried to. Which, by your previous argument does mean, I suppose, that it would get turned into heat (maybe in one of those interesting ways? ). I did mean to put this in my last post, and forgot, but I do agree with you that the efficiency of a CPU should really be more based on calculations per watt as opposed to total power consumption, but it just seems that in a lot of cases people just want to focus on power consumption, probably for heat and battery reasons.

 

The thing is, data lines will basically always go to the switching input of transistors, where they can't really usefully provide power to the circuit. For example, even though USB supplies both data and power, this is done on separate electrical wires.

The reason there is output power is that a certain level of output voltage is necessary to allow for the resistance of the connection, and to ensure that 1s and 0s are properly recognised as 1s and 0s at the other end. If you use any more voltage than you need for a safe margin, you're just wasting power, and generating more heat over the connections and in the input transistors.

 

Well, I was also just pulling numbers out of the air, but it did occur to me to wonder exactly how many instructions/data/commands could be carried on 5 watts of power. If we assume (again just pulling numbers out of the air) that each command is carried on, say, 50 mA and 1.5V, then 5 watts would translate to 66 and 2/3 commands a second, which, come to think of it, seems rather small...

 

A single lane can only have one bit at a time, but the key to transmitting large amounts of data is to toggle that lane very quickly, and to have more than one lane. For example, Intel's QPI runs at a clock rate of 2.4GHz, 2.93GHz, or 3.2GHz; there's some intermediate calculations you can see on the Wiki article, but it works out to a maximum bandwidth of 25.6GB/s. Unfortunately, I can't tell you how much power it uses.

 

Well, if we could figure out/know how much current is required per byte, then if we can find the voltage of the QPI to get the power per byte, and take that and multiply it by that maximum bandwidth and we'd have our answer. I have no idea where to even start looking, though.

 

Well, according to the specs for HyperTransport, on page 368 of the 3.1 specs, the maximum power per output bit (I assume they mean the power consumption of each lane) at 2.8GT/s is about 80mW. Consequently, if you had 60 lanes, you would consume ~4.8W and be transmitting 21GB/s of data.
Nowadays AMD runs their HyperTransport at a higher clock speed, and so they would get more bandwidth, but unfortunately the spec only lists up to 2.8GT/s (i.e. running at 1.4GHz)

 

That is not accurate. It is only a rough guess. In some cases it has been shown to be wildly inaccurate. Like I said, there is absolutely no way in hell it is possible to measure the processor's power consumption.

 

Hm. Interesting. Looks like my random 5 watt number wasn't that far off after all...

Well, I'm sure if you custom wired a motherboard it would be possible. We're talking custom work, though.

 

It may not be accurate down to the last little electron, but I am sure it is plenty accurate to guage how much power the CPU is curremtly drawing.

Are the temp sensors inaccurate as well?

You also said "in some cases it was shown to be wildly inaccurate". Could you explain this further and maybe provide a link or two?

 

My calculation (21GB/s) was for a ballpark (i.e. within an order of magnitude) figure for the bandwidth you get for 5W. However, the problem is that that was only for HT at 1.4GHz or so, while AMD's latest Turion II chips run HT at 1.6GHz or 1.8GHz. Additionally, that calculation had 60 lanes of HT while AMD CPUs tend to use 16, and the assumption is also made that the HT link would be in full operation 100% of the time, which is unlikely to be the case.

16 lanes at 80mW each would be 1.28W, and I don't think it would be more than 2W consumed at 1.8GHz. Under typical operating conditions, 1W or so is probably to be expected for an HT link with 16 lanes at 1.8GHz. The bandwidth of that link is 7.2GB/s in one direction, or 14.4GB/s in two - but if you're counting bidirectional bandwidth, you also have to note that there would also be equivalent power consumed at the other end. Of course, the CPU also has to interface with memory, which would be another 1W or so.