Guest Eugene

Testbed4 and Power Consumption

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Tygrus, take a look at a bunch of hard drives. Their exteriors are the same. I don't think your arguments hold much water. Before you get offended, take another, hard look at all the hard drives you have ;). If you don't have a number of disks available to compare, take a look at SR's article photos. Please don't die of boredom, I'd feel guilty.

The only area of practical difference, as has already been noted above, is in the coupling of internal heat generating components to the exterior. The exterior shells of any hard drives will have practically identical thermal resistance.

P.S.

Here's a Fujitsu 15K disk:

MAU3147NP_top.jpg

And here's a Seagate 7200RPM disk:

ST3400832AS_top.jpg

You can clearly see how the 15K disk is a server drive and how it has been designed to dissipate heat in a high airflow environment. In contrast the Seagate is obviously designed to be tremendously efficient in low airflow situations like in a Comcrap desktop. :unsure:

P.P.S Please forgive my facetiousness, but come on... look at them. They're all the same! I think watts will be quite adequate without any super-cold-still-air/wind tunnel aerodynamics tests...

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Now, frisbee tests of the platters... that would be practical! You have to test for vibrational instabilities, especially with server disks. If the platter doesn't fly straight and true, I don't think SR should recommend it. <_<

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I could be wrong, but I suspect you won't be able to detec any problems (even if they do exist) with a frisbee test with a drive that spins 15k RPM (or even 7200RPM) :-P

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I'm still going to say that I'm more concerned with heat generation, than actual current draw,

212585[/snapback]

The two are equal.

212588[/snapback]

Even if one drive is louder than the other? Is energy dissipation in the form of sound significant?

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Even if one drive is louder than the other?  Is energy dissipation in the form of sound significant?

Not really. Look how loud a piezo buzzer using tenths of a watt can be, for example. Even a 5 watt speaker is painfully loud to my ears. The amount of sound energy radiated by a hard drive can probably be measured in microwatts.

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Eugene and jtr1962,

Very nice test tool. :)

We started doing power measurements on HDDs at silentpcreview.com in May this year precisely because temperature testing is difficult to do reliably. There's no question that power draw and heat generation are equal. The differences caused by varying internal approaches to cooling and any external "thermal dissipation design" are too small to be significant.

Here's the page from our HDD Test Methodology article that discusses this & related issues: http://www.silentpcreview.com/article242-page5.html

Our crude-looking HDD power test rig is shown at the bottom of this page -- http://www.silentpcreview.com/article242-page3.html It's a simple circuit interrupt that inserts 0.25 ohms of resistance in the +12V line and 0.2 ohms of resistance in the +5V line. The voltage drop is measured, and we use Ohm's law to calculate power for each voltage line from that. Add the 2 power numbers and we get total power draw for the drive. We measure during idle and seek (with and without AAM).

Despite the simplicity of our test method, comparing some of the same HDDs SR tested, it looks like our resuilts are generally close, within a watt for both idle and seek. Most of our measurements are lower except for the seek power of the Hitachi 7K400, which we reported as 15.5W compared to SR's 13.8W. It would be nice to find out whether the differences are due to the different test setups or to sample variances. We'll never know -- unless SR & SPCR decide to start swapping drives... unlikley given the hassles and minor dis crepancies...

Anyway, just thought the SR community would be interested. We always refer our readers to SR for HDD performance testing results -- we focus entirely on acoustics, vibration and heat.

Mike Chin, SPCR

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Yes, manufacturer vs. real-world startup current requirements are still of interest. Maybe not on every drive, but some makers don't readily list individual current on each rail for startup, and a reality check once in a while would be useful.

Oh. Good to see you here, Mike. (we see Ralf regularly on Ars :) ).

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It would be nice to find out whether the differences are due to the different test setups or to sample variances.

Probably both but maybe more due to differences in the test setup. Resistors always have a tolerance to them, and this can affect measurements. Same thing with multimeters. In the case of the tester I calibrated the tolerances out to the limits of the circuit design and my ability to take measurements. I have a 5.5 digit bench multimeter which is accurate to 0.005%, at least for measuring voltage. For measuring current I've used 0.1% tolerance resistors as a reference on which I can calculate the errors in the usual 1% to 5% tolerance low-ohm power resistors needed for measuring higher current draw. With both of these things I was able to get the tester's power measurement accuracy to 0.5% or better under most circumstances. The only time it's worse than that is when you have currents of 2 amps or more flowing through my current sense resistors. The resistors heat up, and due to the temperature coefficienct they change in resistance slightly. By the time you're into 30 watt power levels on the 12 volt line the power readings might be about 1% off. However, that scenario is unlikely to happen on a steady-state basis when testing hard drives. The peaks lasting a few tenths of a second won't heat the sense resistors enough to affect the results.

All things considered Mike, I think you're doing quite well with your power measurements. Nice to see you here at SR!

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Welcome Mike Chin :). I'd just to like to say that you guys do a great job, and that you run a terrific site. I'm a frequent browser and respect the quality of your work. Cheers.

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Thanks for the welcome, gents. :) We keep trying.

Based on your comments, jtr1962, I think I might try using a bank of lower power but higher precision resistors in a parallel circuit. The higher precision + averaging aspect of parallel circuitry should mean better accuracy. I don't have a 5.5 digit bench multimeter is accurate to 0.005%...

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My Fluke 87 has a measurement mode that will measure over time and give me Max, Min and Avg readings. I think that measuring these over time during a specific test would give a more accurate picture of the power the drive is using.

For example, run the drive in idle for 1 minute and report min, max and avg readings, then run the drive through seeks (or other loads) for 1 minute and report min, max and avg.

Regards,

Wayne Sherman

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The only area of practical difference, as has already been noted above, is in the coupling of internal heat generating components to the exterior.  The exterior shells of any hard drives will have practically identical thermal resistance.

Not true. To a fair degree, the thermal properties of the stuff used to make HDDs is pretty much the same for everyone. But from one drive to the next, the thermal properties of the primary constituents, Aluminum and stainless steel, will vary as a function of how they're alloyed, how thick they are, the shape they take, and how they're constructed.

Most 7200RPM and slower drives have slab-sided base castings. Many 10K and 15K RPM drives have finned surfaces (which improves dissipation). Some drives have single-layer stainless steel top covers. Others have two layer covers with a sandwitched damper materail (which insulates) between the layers. Some drives have large PCBAs which cover (and therefore, insulate) a significant percentage of the HDD base casting, others have smaller PCBAs. Some drives have acoustical damper material between the PCBA and basecasting (which insulates). Even the packaging (BGA is quite different from LCC) used for the ASICs effects dissipation of the components. It also matters how the heat-dissipating components inside (spindle motor, voice coil motor, and Preamp) are mounted to the external components (the thermal convection path) and how air flows around these components inside the HDD.

Dissipation can and will be different from design to design. Assuming environmental factors stay constant, each design will attain it's own operating temperature as a function how it dissipates the heat it generates.

Now, that doesn't change the fact that a drive that consumes more power will generate more heat. You cannot change the laws of physics.

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2)  Similarly, pretend you had two drives, one with perfect thermal transfer (pretend it's solid diamond) while the other completely isolates the guts from the outside of the case with a layer of styrofoam.  The styrofoam-encased drive would appear incredibly cool on the outside, since almost all the heat generated would be held internally, even though it'd rapidly bake itself to death, while the cool drive would appear comparitively warm on the outside surface.

As such, it's quite possible for an apparently cooler drive to actually be running hotter where it counts--in fact, if you give me two drives with the same power dissipation, I'd wager the one that felt hotter is the better design, as it probaly does a better job of dumping the heat to the surface where it can be carried away.

I think that is true only in short term. Lets keep assuming both drives use same amount of power, say 10W, and one has good heat conductivity, and the other has heat-insulating layers. The one with insulation will indeed first feel cooler from outside surface, but its temperature *will keep rising* as long as it does not dissipate heat to outside air/whatever it has around it as fast as it generates new heat (uses power). So it will heat up until it dissipates that 10W *out* of the drive, at which point the cover will be as warm as the cover of the drive which conducts heat properly. (Needless to say, the styrofoam-drive would boil the internal components at that point...)

Anyway, great to know we're getting power usage figures in the future.

Just my 2 cents :)

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I think that is true only in short term. Lets keep assuming both drives use same amount of power, say 10W, and one has good heat conductivity, and the other has heat-insulating layers. The one with insulation will indeed first feel cooler from outside surface, but its temperature *will keep rising* as long as it does not dissipate heat to outside air/whatever it has around it as fast as it generates new heat (uses power). So it will heat up until it dissipates that 10W *out* of the drive, at which point the cover will be as warm as the cover of the drive which conducts heat properly. (Needless to say, the styrofoam-drive would boil the internal components at that point...)

Anyway, great to know we're getting power usage figures in the future.

Just my 2 cents :)

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Technically true, but only in the very, very extreme; that is, if you put, say, 10 watt-hours of heat (a hot cup of coffee) into a container, a solid aluminium container, the outside will get very hot, but will quickly dump all that heat to the environment. If you put it in a styrofoam container, the outside will never get more than warm, and while it will eventually disappate all that heat, it'll do so over a very long period of time.

In the case I think you're thinking of, it's true that this theoretical insulated hard drive will eventually have to start dumping all the heat input to the outside surface, but depending on how good the insulation is, by the time it's getting that much heat through the insulation the internal temperature could be hot enough to melt the internal components. Which is why I was implying that the drive would fail before you were dumping all the heat input to the surface.

In reality, as it applies to real hard drives, if you run the drive for long enough at a constant power level to reach total thermal equilibrium, then the total outside temperature will reach a point consistent with the power input. That doen't change the fact that it's where that heat ends up that matters, and top-plate measurements don't necessarily reflect that in a meaningful way--the drive with better thermal transfer to the top will "look" the hottest, whether it is or not.

Edited by Makosuke

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In the case I think you're thinking of, it's true that this theoretical insulated hard drive will eventually have to start dumping all the heat input to the outside surface, but depending on how good the insulation is, by the time it's getting that much heat through the insulation the internal temperature could be hot enough to melt the internal components.  Which is why I was implying that the drive would fail before you were dumping all the heat input to the surface.

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True, it can fail before that and stop producing more heat. (But the coffee cup is quite different story, as it is just a certain amount of heat energy in a box, not something that produces more of it - but yea the whole styrofoam idea is of course quite extreme)

In reality, as it applies to real hard drives, if you run the drive for long enough at a constant power level to reach total thermal equilibrium, then the total outside temperature will reach a point consistent with the power input.  That doen't change the fact that it's where that heat ends up that matters, and top-plate measurements don't necessarily reflect that in a meaningful way--the drive with better thermal transfer to the top will "look" the hottest, whether it is or not.

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I believe that differencies between real world hard drives are not that big, and it is possible to give them enough time so that temperature stabilizes (after all they represent the metal version, not styrofoam). So i think temperature measurements do indicate how much the drives generate heat, just less accurately than measuring the actual power input. Although i think sides would be better temperature measuring points than the top plate (which is quite bad imo).

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If you want a cooler drive, just replace all your data with 1s. Being long and thin, they are much more aerodynamic than 0s, so the air will flow by them more easily. And with their fins (serifs,) it provides more surface area for the heat to escape.

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If you want a cooler drive, just replace all your data with 1s.  Being long and thin, they are much more aerodynamic than 0s, so the air will flow by them more easily.  And with their fins (serifs,) it provides more surface area for the heat to escape.

213332[/snapback]

LOL :lol:

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the two are equal? Um, ok. So I guess none of the energy the drive is drawing is being used to spin the platters or move the heads or power the electronics eh? It's all just converted directly to heat with 100% effeciency? If you say so. Be a pretty useless "drive" then...

I'm still going to say that I'm more concerned with heat generation, than actual current draw,

212585[/snapback]

The two are equal.

212588[/snapback]

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comparing the amount of sound output from devices designed to create sound and the by-product sound of a mechanical device doesn't make any sense. A speaker is designed to make sound with relative effeciency, while I can pretty safely guarantee that a HD isn't designed to intentionally make any sound at all. In fact I can also pretty safely guarantee that the reduction of its sound output is a design concern. You're right, though, that the amount of energy converted to sound in a HD probably isn't very relevant. But comparing that to a speaker is kind of silly. (Not to mention a "5 watt speaker" doesn't really mean anything with regard to how loudly a speaker is playing. An amplifier outputting 5 watts to a speaker is a little more meaningful, but a speaker's 5 watt power handling rating doesn't tell you anything about how loud it is or is capable of being. That's just a rating of how much power the voice coil can safely handle without melting.)

Even if one drive is louder than the other?  Is energy dissipation in the form of sound significant?

Not really. Look how loud a piezo buzzer using tenths of a watt can be, for example. Even a 5 watt speaker is painfully loud to my ears. The amount of sound energy radiated by a hard drive can probably be measured in microwatts.

212782[/snapback]

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the two are equal? Um, ok. So I guess none of the energy the drive is drawing is being used to spin the platters or move the heads or power the electronics eh?  It's all just converted directly to heat with 100% effeciency? If you say so.  Be a pretty useless "drive" then...

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You put energy into a "black box" (in this case a HDD) in form of electricity. Yes, that energy is meant to spin the platters and move the heads and power the electronics. But have you ever thought what happens to that energy?

You spin the platters: some of the energy is turned directly to heat because spindle motors aren't 100% effective. Some of the energy actually spin the platters. For the first seconds when the power is turned on, the rpm of the platters increase but then suddenly the platters stop accelerating at a rated rpm (5400 or 7200 or 10000 or whatever). Still some power is required. Why? It isn't used for accelerating the platters, it is used to maintain the speed because some of the energy stored as kinetic energy of the revolving platters is turned into into heat every second the drive is operational. The causes of this are, for example, friction in spindle motor bearings and air circulating inside the drive due to being contact with both the spinning disc and stationary HDD casing and top cover. Because there is differences in these velocities, there is differences between air molecules and they tend to HEAT up. So tend the bearings.

So, the only power that is not turned into heat instantly is the one that was initially supplied to spin-up the drive. Even that energy is turned when the power is cut of and the platters start to decelerate due to friction. (There's even some added load, "dynamic braking", and some power is used from spindle motor in form of back-EMF and spindle motor functioning as a dynamo. This energy taken from spindle motor to supply head actuator movement after external power supply is cut-off actually causes the discs to lose their rpm faster. Energy does NOT disappear, NOR does it appear out of nowhere!)

Same laws of physics apply to head actuator and electronics on the PCB.

So even though producing heat is not the primary purpose of using HDDs, they produce it, as an inevitable side effect.

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the two are equal? Um, ok. So I guess none of the energy the drive is drawing is being used to spin the platters or move the heads or power the electronics eh?  It's all just converted directly to heat with 100% effeciency? If you say so.  Be a pretty useless "drive" then...

213625[/snapback]

It's converted into heat indirectly and indeed with 100% efficiency. And it's not just HDDs that do that. Cars do the same.

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AFAIK the power consumption in seek mode is significantly higher when AAM ist turned off, especially on Maxtor drives, IIRC. So I think you should consider this.

I appreciate the idea of power consumption measurements. And I think idle/seek measurements are more important to most readers than the spinup power consumption. Anyone who uses large drives arrays should be able to afford a decent controller that can stagger the spinup of the drives. With a lot of drives spinnung up at the same time, I think you could beat any conventional PSU.

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