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
    Commented Mar 14, 2013 at 8:41
  • 2
    Ohm's 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
    Commented Nov 18, 2019 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
    Commented Oct 25, 2021 at 4:28
  • *highest posted answer
    – oemb1905
    Commented Oct 25, 2021 at 4:34

11 Answers 11


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.


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


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!

  • 19
    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. Commented Aug 13, 2015 at 5:22
  • 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. Commented Aug 18, 2015 at 16:23
  • 21
    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
    Commented Jan 18, 2016 at 20:43
  • 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
    Commented Nov 15, 2018 at 19:11
  • 2
    -1. This answer is totally wrong, misleading, The answerer fails to make a distinction between peak consumprion and average consumption. Commented Dec 4, 2020 at 12:16

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).

  • 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
    Commented Mar 14, 2013 at 7:37

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
    Commented May 1, 2023 at 21:15

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

  • 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. Commented Jan 5, 2017 at 18:41
  • 2
    Motors take more power when starting... Commented May 5, 2017 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
    Commented May 9, 2017 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. Commented May 9, 2017 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
    Commented Aug 7, 2017 at 18:07

@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.


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.


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.

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To calculate the power consumption of HDD-SSD devices, it's very important to know how the Power-Unit dispense the power to devices. Thats because Disks and Power Units designed and depended each other natively. As you can see below, some Disks (HDDs) uses both different Voltages but depended on "conditions" and "states".
If take a look at power-unit wires, the basic are:
Red=5V, Orange=3.3V, Yellow=12V, Black=Ground(-).
Please take a look at the table below: Example-Table of Power distribution of a PC Power Supply
We notice that e.g. 130W MAX (out of 500W) can be 'consumed' equally for +3.3V and +5V together. Note also, combinations like for example (+12V)+(-12V) = 24V output (for advanced devices like some 'advanced' audio controllers etc.) can be achieved.

Hard drives label tells you the MAXIMUM consumption of:
eg: 5vDC : 0.6A | 12vDC : 0.45A (450mA)
( This HDD will need (5v)3W+(12v)5.4W = 8.4W MAX but in reality consumes about ~20W depended on circuit's Electrical Resistance (Ωhm) but mostly note the totally difference between, let's say 'Referenced' Power and 'Consumed' Power.)

  • rule: Volts x Amperes = Watts (unit of power)
    1W = 1V x 1A
    1W = 1V² / Ω = 1A² x Ω (Ωhm electrical resistance)
    i (Note that in a circuit, when the V increases, the A decreases, and vice versus.)

  • (5V x ?A) + (12V x ?A) + (3.3V x ?A) = TotalPower(W)
    (Red)W + (Yell)W + (Orange)W = TotalW ('consumption').


  • HDDs (mechanical drive) 7200rpm or 5400rpm (more rounds per minute the more power consumption.)
  • 3.5" HDD will typically use 12V for the motor and 5V for the circuit board.
  • 2.5" HDD drive will typically use only 5V, for both motor and circuit board.
  • SSDs do not use any 'mechanical parts', but just writes to 'memory chips' (flash memory).
  • M.2 SSDs are powered using 3.3V

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.


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.

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    – Community Bot
    Commented Jun 2, 2023 at 21:08
  • Actually I think he makes a good point. Although admittedly it would make more sense as a comment. Commented 22 hours ago

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