Fujitsu creates readable, writeable 'nanohole' hard drive
Step toward 1TB per square inch data density
Fujitsu's scheme to produce hard drives that can hold a terabyte of data in each square inch of recording surface has taken a step closer to realisation. It has made a 2.5in disk made of its proposed 'patterned medium' and verified the disc's read/write capability.
Fujitsu's approach uses anodised aluminium to create a pattern of "nanoholes", each holding a portion of magnetic material used to store a single bit of data. The aluminium-oxide surrounding these so-called 'nanoholes' helps magnetically insulate each bit from all the others, preventing one from affecting another, which might lead to data corruption.
Fujitsu began work on patterned media in June 2005. In January 2007, it said it had brought the technique forward to the point at which it could pledge it will be able to one day produce 1TB per square inch storage-density drives.
That density requires nanoholes of diameter 13nm. Fujitsu currently can get them down to 25nm. That said, it used 100nm holes for its latest effort: to make a 2.5in patterned media rotating disk capable of being read and written using a "currently available" flying read/write head.
Fujitsu said it had also created for the first time an "ideally ordered" arrangement of nanoholes that that drives are going to need to be able to function efficiently.
While the parts are coming together, the 1TB per square inch drive is still some way off: Fujitsu has said in the past it expects such products to appear in the 2010 timeframe.
COMMENTS
RE: Current capacity (lots of math)
Wow! - you made hard work of that. How's this:
(a) they hope to fit 1TB/sq in using 13nm holes. They can currently make 100nm holes.
(b) A 100nm hole is (100/13)^2 ~= 59 times the size of a 13nm hole (forget all the meaningless additional digits which)
(c) If 13nm holes allow 1TB/sq in, we can assume that 100nm holes give roughly 1/59 as much capacity, or 17GB/sq in.
Roughly speaking, a 2.5" platter gives a ring with inner diameter 1.25", outer diameter 2.5" of usable space. The area is thus PI x ((2.5/2)^2-(1.25/2)^2) ~= 3.7 sq in.
For a 3.5" disk this increases to 7.2 sq in per platter-side.
So each 2.5" platter should be able to fit 3.7 x 17 = 63 GB per side at current densities, increasing to 3.7 TB ultimately.
A 5 platter 3.5" drive using both sides should yield 5 (platters) * 2 (sides) * 7.2 (sq in / side) * 1 (TB/sq in) = 72 TB.
Title
Jesus.
Guys... You know that there's more than 1sq/in on a drive. You know that the stated (new density) is 1TB/sqin. So don't tell me that the capacity will be 16GB.
Also. Brett. Mate. I think you just multiplied the current density, by the area, to get current capacity. I like your thinking though. So lets try again
density * area = capacity
1 (TB per sq in) * 3.8 (sq in) = 3.8 TB. Better I think.
Now - for a multi-platter, multi side arrangement, we might expect four times that.
So what was the capacity?
I may just be being dense, but I'm still waiting for a figure, e.g. If they released a 3.5" IDE drive (for example's sake) at 13nm, you could store 84TB on your computer, etc.
Am I being dense or is there a figure like that?
Another advantage not mentioned
Because the magnetic domains are regularly arranged and all very close to the same size, magnetization is more quantized. In traditional drives, magnetic domains vary in size, and so the net magnetization per unit area can be quite variable, resulting in readout noise, which limits density.
Carnegie Mellon University, I believe with funding from IBM mostly, is working on a similar technology to this one, using magnetic nanoparticles, rather than nano-holes. Both are promising.
Current capacity (lots of math)
1 Inches (in) = 25400000 Nanometers (nm)
so... 1 square inch is 645160000000000 square nm
so now divide that by capacity they're shooting for (1TB - which is 1 followed by a bunch of zeros as far as the hard drive companies are concerned)
so... 645160000000000 / 1000000000000 = 645.16 square nm
now take the square root to find the length that the 13nm pits take up.
this is 25.4nm (which is probably correct since the 13nm pits need some area around that isn't a pit to support the structure and provide that magnetic shielding)
now... do some algebra... 25.4 / 13 = N / 100
13 is the target, 100 is the current working prototype.
N is 195.384615... lots more decimal places... anyhoo...
square this...
we get this...
38175.147928994... again... lots more decimal places... now divide the nm area by this and we have the capacity in bytes.
645160000000000 / 38175.147928994... = 16900000000
16.9 GB. Far cry from 1 TB, but it's getting there.
and hopefully I didn't miss copying a zero somewhere.
