How much power does a hard drive use?

Question

This is not as straight forward as it sounds.

Specs from Western Digital's site for a WD 3TB Green Drive:

Read/Write 6.00 Watts
Idle 5.50 Watts

Looks fine right? Look at this part of the spec: "12 VDC" and "Read/Write 1.78 A".

It was a long time ago, but when I was in college that would mean the drive uses 21.36 Watts (12V x 1.78A). 21.36 Watts is a lot more than the claimed 6.00 Watts.

I want to put four of these in a RAID 10 array, so I want to know the actual max power requirement.

Thoughts? Is this a simple typo? Do I need to plan on ~85 Watts of power to support four drives?

That "spec sheet" is a joke and incomplete. There's no mention of current draw for the +5 volt supply. I remember when 5.25" HDDs had full product manuals with graphs for all operations for both voltages. — sawdust, Mar 14 '13 at 08:41
[Ohm's Law](https://www.fluke.com/en-us/learn/best-practices/measurement-basics/electricity/what-is-ohms-law): **E** _(voltage)_ **x I** _(current)_ **= P** _(wattage)_, however **I** and **P** are generally not consistent in electronics in terms of energy consumed, i.e. **P** will usually fluctuate because **I** fluctuates based upon current demand at any given time. — JW0914, Nov 18 '19 at 12:51
21.36 is the physical cap of wattage that that particular HD can ever utilize as I understand it, so yes, right folks? This post is for people doing PSU planning on hosts/servers and not an electricity post, so whether terms are correct or not, I think the calculations AND reasoning of the OP are both correct but that’s not clear from the accepted answer, which only refers to the calculation and misses the PSU determination point? Asking for a friend … ;) — oemb1905, Oct 25 '21 at 04:28

score 19 · Answer 1 · edited May 13 '16 at 23:30

Most of you are way off here. You are confusing peak and average draw. 1.78 A is the start-up current value.

When calculating power dissipation, both 5V and 12V are considered. All the power to a drive is converted to heat, with 99% being dissipated by the drive and a small amount dissipated over the interface. Friction in the bearings and airflow on the platters results in heat. Losses in the drive motor electronics and windings and by the chipset waste the rest.

A 6W drive through an 80% efficient power supply will draw about 7.5W at the wall.

ubiquibacon · Answer 2 · 2020-12-05T11:32:47.947

12

Your calculation is correct, but your understanding of the term power dissipation is lacking :)

Electrical Specifications
Current Requirements
    12 VDC
    Read/Write  1.78 A

Power Dissipation <-- Energy measured in watts lost as heat
    Read/Write  6.00 Watts
    Idle        5.50 Watts
    Standby     0.80 Watts
    Sleep       0.80 Watts

Update:

Lots of hate in the comments. I encourage commenters to read exactly what this answer says and don't read anything else into it. This answer makes exactly two claims:

1.) The OP's calculations were correct

2.) The OP didn't understand the term "power dissipation".

The answer does not claim that WD specs are correct or that they make sense. In fact I would guess that WD has used the term "power dissipation" incorrectly (as others have already suggested) and should have instead used "average power consumption", but that is just a guess.

Some commenters have mentioned that all energy is lost to heat. This is mostly true, but is technically incorrect because some energy is lost to magnetic and gravitational forces. Since the primary loss of energy is most definitely in the form of heat it does make us raise an eyebrow to WD's supplied specs which do not include any peak or average power consumption numbers (a fact that one commenter for some reason blamed on this answer) and may suggest that the HD in question breaks the law of energy conservation (a fact that another commenter for some reason blamed on this answer).

So again, this answer does two things. It confirms that 12V x 1.78A = 21.36W and it provides a definition for the term power dissipation. That's it!

edited Dec 05 '20 at 11:32

answered Mar 14 '13 at 03:54

ubiquibacon

8,098
3
29
36

Wow. Thanks! I had no idea a hard drive would use that much power. – Jay Wen Mar 14 '13 at 04:06
17

_All_ the energy a hard drive consumes is lost as heat. Every last erg of it. It's not as if you're winding up any springs or lifting any weights! So there is still a disparity between the claimed current requirement and the "power dissipation" numbers. I wonder if the "read/write 1.78A" might be a peak figure and the "power dissipation" figures long-term averages. – Jamie Hanrahan Aug 13 '15 at 05:22
7

That is incorrect. Some is lost as motion. The disc does weigh something so that is the "lifting weights" part, except the weight is being spun, not lifted. – ubiquibacon Aug 18 '15 at 16:23
@ubiquibacon is right, there is some friction that is encountered with the rotational axel the platters are attached to – edhurtig Nov 20 '15 at 03:49
19

But, other than the stored rotational energy (which won't continually increase unless you continually increase the speed of the drive), all the energy "lost to rotation" is, in fact, converted to heat. Through friction. Which is a process where work/kinetic energy is converted to heat/thermal energy. – jhaagsma Jan 18 '16 at 20:43
If you spun a solid object in the vacuum of space it could still eventually stop because of the forces of gravity and magnetism acting on it which to the best of my knowledge do not create heat. The same forces act on objects on Earth so you can't really say "all" energy is eventually converter to heat. – ubiquibacon Aug 08 '17 at 00:44
3

@ubiquibacon: If you want to be pedantic, a solid object spun in the vacuum of space will not lose any energy to gravity unless it is asymmetric along the axis that it is rotating around, which would cause gravitational waves. Otherwise, no energy is lost if a symmetric disk is just rotating. And the object would not lose energy due to magnetism acting on it unless it was also magnetic and its rotation induced an electric current in some object around it. – kloddant Nov 15 '18 at 19:11
@kloddant, everything is magnetic, some materials more so than others. Google levitating frog. Your comment about asymetery is also not correct. A force acting on an object in one plane can effect that object in another plane. This is easily demonstrated by trying to displace a spinning gyroscope. – ubiquibacon Jun 06 '19 at 02:41
This answer suggests that HDDs break conservation of energy, which is just silly. Come on guys, this is high school physics. – Navin Dec 04 '20 at 11:51
2

-1. This answer is totally wrong, misleading, The answerer fails to make a distinction between peak consumprion and average consumption. – Ale..chenski Dec 04 '20 at 12:16
@Navin Please re-read answer and update. I made exactly zero claims of breaking the law of energy conservation. – ubiquibacon Dec 04 '20 at 15:11
@Ale..chenski, Please re-read answer and update. I made no mention of peak or average power consumption because they were not included in the WD specs. If WD has used the term "Power Dissipation" to mean "average" then they have used it incorrectly. The short of it is that WD has done a poor job with there spec sheet. – ubiquibacon Dec 04 '20 at 15:12
Lemme try to be more constructive than I was in my last comment: The Current Requirements spec is there to help you size your power supply whereas the Power Dissipation spec helps you size your cooling solution. Power Dissipation is always equal to average power consumption for devices like HDDs so there's no need for them to provide a spec for "average power consumption". I have no idea what you mean by "energy is lost to magnetic and gravitational forces". Energy is never just "lost", it can only be converted from one form to another. – Navin Dec 05 '20 at 03:36
@Navin, some have said that all energy is eventually lost to heat, but that is not correct. Some energy is used to spin the platter. Most of that spinning motion is eventually converted to heat through friction in the bearing and the platter spinning in its nitrogen filled chamber, but some of the spinning motion is lost to magnetic and gravitational forces which do not generate heat. Power dissipation means energy lost as heat. Period. If WD or any other mfg is claiming that means average power consumption, they are wrong and doing their customers a disservice using that term. – ubiquibacon Dec 05 '20 at 11:38
@ubiquibacon If by "spinning motion" you mean kinetic energy, all of that is converted into heat. If you want to be pedantic, a tiny fraction of that energy is converted into sound and that's why you can hear it running. What do you mean by "nitrogen filled chamber"? The platters inside any normal HDD spin in air. Look for the vent hole. Newer HDDs are helium-filled. I don't understand why you keep bringing up "magnetic and gravitational forces". That is not a form of energy! – Navin Dec 05 '20 at 12:44
@Navin some are nitrogen filled I believe. I though those were more common than the heliem filled ones. Regardless of the gas, we at least agree on the concept that gas molecules will cause friction and thus heat. Now do you believe in gravity and magnetic forces? If you answer yes then imagine something spinning in a vacuum in space undisturbed by any force. It would spin forever. Now have a magnetic force act on it. That will cause a change in motion that will not create heat. Same with gravity. – ubiquibacon Dec 05 '20 at 15:59

score 3 · Answer 3 · answered Jul 26 '18 at 04:58

Electrical Specifications <-- absolute maximum values

Current Requirements 12 VDC Read/Write 1.78 A

Power Dissipation <-- average operating values

Read/Write  6.00 Watts
Idle        5.50 Watts
Standby     0.80 Watts
Sleep       0.80 Watts

If you start looking at all the appliances around the house, read their electrical specs, then plug them into a power meter and read their actual usage.. it's much lower than what's on the label. Very illuminating.

Are you saying my 1000 Watts computer doesn't consume 1000 Watts 24/7? :P — dognose, May 01 '23 at 21:15

score 3 · Answer 4 · answered Mar 14 '13 at 05:19

3

If 85W seems like a lot for the PSU you are planning to use for this system - do not forget that during start-up/spin-up the current drain could be almost twice as much (up to 3A per drive).

answered Mar 14 '13 at 05:19

Alex P.

2,713
12
15

Thanks. This is certainly an eye-opener. I haven't looked at a desktop/server hard-drive since ~2004. I knew I would be looking at more power than a laptop drive, but I didn't know it would be that much more. – Jay Wen Mar 14 '13 at 07:37

score 2 · Answer 5 · answered Mar 14 '13 at 03:51

2

21.36 watts sounds about right. You can use a general rule of thumb for about ~25 watts per 7200 rpm drive.

answered Mar 14 '13 at 03:51

user530873

154
8

4

This is absolutely not true. I've got 8x4TB drives, an i3, 16GB of RAM, an SSD, and 6 case fans and my server only consumes 78W at idle at the wall as measured with a kill-a-watt device. During disk load it goes up to ~115W tops. – haventchecked Jan 05 '17 at 18:41
2

Motors take more power when starting... – Eduard Florinescu May 05 '17 at 12:13
2

@haventchecked Yes, on average over at least a second or however often your measuring device updates. But for choosing a power supply, you need absolute peak values, even if that peak never shows up on a coarse measurement. Of course you can also just hope that those peaks never happen all at the same time... but that's risky. – Nobody May 09 '17 at 22:11
2

Considering they all power on at the same time, typically speaking, when the computer is powered on, the peaks WILL happen at the same time with some regularity, and therefore peak values are what need to be used when planning the power requirements of a system. – music2myear May 09 '17 at 22:18
3

Some drives support being started in sequence. e.g. staggered spin-up. This means that the drive will not start spinning until the system (BIOS or EFI) tells it to. Ergo the drives will not all start spinning at the same time IF they support staggered spin-up. – Waxhead Aug 07 '17 at 18:07
that's right - my RAID setup with 6 HDDs on Gigabyte GA-X58-UD3R always seem to spin up drives one by one – Vitas Dec 03 '19 at 22:23
This is the correct answer - the other posters nitpicked the OP’s knowledge of electricity and failed to understand that when determining which PSU to purchase for a physical host, you absolutely must do exactly this calculation and plan on it, or the host won’t boot and/or perform well. And/or you purchase a larger PSU, and subsequently pay more for hosting it. – oemb1905 Oct 25 '21 at 04:23

Gabriel · Answer 6 · 2020-12-04T11:54:41.770

Normal drives need 3-6W when not idle.

High-speed drives (I have a Barracuda) take 8W on average (the documentation says). This means that you need to calculate it as 10W (2W as reserve power).

But that's the avverave power, under "normal" conditions. HOWEVER, according to the same documentation (see link), they can take 30W at spin up. That's only the motor, so don't forget to add 1-2W for the digital part (the microcontroller + circuit board).

When I did the power calculations for my PC, I considered two drives at 30W. I measured the power consumption of my PC but it was some years ago and I forgot the numbers. I will do it again during gaming and boot.
But I remember that the numbers were not that close to the max power of my PSU (800W I think).

Anyway, the idea is that the power requirements might seem small until you get into one of those "abnormal" running conditions, when all peripherals start to draw the maximum amount of power, at the same time. If you see your computer rebooting "without reason", this might be the cause.

score 1 · Answer 7 · answered Dec 04 '20 at 12:12

The 1.78A at 12V is a start-up / spin-up peak power. After the drive reaches steady spinning, it consumes quite less, depending on load / seek pattern. That's why serious multi-drive SCSI subsystems have a standard option to turn disk power one-by one, to avoid the surge of current upon power-up.

So, for a 10-disk array, you either need to plan for a 18-20A on 12V rail, or somehow have a built-in option in hardware to turn drive's power in sequence, one-by-one.

score 1 · Answer 8 · edited Aug 13 '15 at 03:42

@sawdust is correct in that the 5 VDC line isn't truly accounted for...but in most spinning consumer hard drives, the 5 VDC and the 12 VDC lines are about evenly split, and nowhere near 1.75A. For example, the specs on a nine year old WD 500 GB drive (Caviar 16SE WD50000KS) I use in a RAID array are:

5 VDC = 0.70A = 3.5W
12 VDC = 0.75A = 9.0W

The 1.45A here is pretty close to the drive you quoted...but it isn't ALL at 12 VDC, so your calculation is off. This drive draws 12.5W. My array of four drives draws 50W, or 1.2 KwH per day. i.e. my array costs me about $0.20 per day in electricity at local rates.

Green drives may run lower. Newer drives probably run lower. Check both voltages and do the math as above, but my four-drive, ancient array adds about $6 a month to my electric bill. It isn't a HUGE concern. You're more concerned about the Thermal Design Power of your power supply. Always have a far bigger power supply than you need.

score 0 · Answer 9 · answered Feb 04 '18 at 07:10

One of the arguments for using SSD is that they do use more power but its a smaller average load. This might make all the difference in some setups where you're spinning up the drive regularly. As they can often support DPD mode (essentially sleep) and use system RAM only due to lower latency the battery life can go up. I've tested this experimentally and got maybe a 20% increase with identical system running 10 with Crucial M300 SSD versus 500GB slim 5400 HDD. Its possible to accurately measure this with a tool like "Battmon" or similar and you can actually see the energy usage with things like screen brightness.

Alex T · Answer 10 · 2023-06-02T20:52:23.740

Isn't the primary danger here that all this gets confused by other factors too. For instance I've yet to see a cheap power supply maintain its 12V output at 12V when delivering its specified maximum current. Also, its no-load volts won't necessarily be 12.0000V either. I'm thinking in particular about a PV array feeding through a DC-DC convertor, where the PV array's nominal spec output is (say) 20V, the typical no-current output is actually only 18V, and the voltage when delivering 50% of its theoretical maximum current could be as low as 12V-14V. So it's usually wrong to think of a PV array as delivering 18V * its peak amps as its "watt power rating", but nonetheless that's what PV vendors tend to do. So it would not surprise me at all if the "12V" and "5V" input lines on a hard drive are no-longer "12.0000V" and "5.0000V" when drawing full current (if they ever were that accurate). Such voltage and current specs should be taken with a hefty dose of salt, rather than being multiplied together to calculate watts.

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). — Community, Jun 02 '23 at 21:08

How much power does a hard drive use?

10 Answers10

Linked