3D NAND in Numbers: Is it Economical?

As with all new semiconductor technologies, the one big question is whether it is economical. There are a ton of memory technologies that have better characteristics than 2D NAND (MRAM and ReRAM to name a couple) but none of them is currently scalable enough to challenge NAND in cost. IMFT's 16nm node is the smallest node shipping in volume, so let's compare Samsung's second generation V-NAND to that to see how it stacks up.

I am basing my analysis on Andrew Walker's blog post from 3DInCites. He is a respected analyst in 3D technologies and his blog post is based on Samsung's and Micron's presentations at the 2014 International Solid-State Circuits Conference, so the data should be as accurate as it can be.

Update 7/8: I was able to find some more accurate data regarding the die size, so I've updated this section with the new data.

  Samsung 2nd Gen V-NAND Micron 16nm NAND
Process Node 40nm 16nm
# of Layers 32 -
Die Capacity 86Gbit 128Gbit
Die Size 95.4mm2 173mm2
Cell Size 40,300nm2 1,312nm2

The only downside of Walker's data is that it only covers Samsung's first generation V-NAND, which was a 24-layer design with a die size of 133mm2 and capacity of 128Gbit. Fortunately, the second generation V-NAND is using the same 40nm node, so the only substantial difference between the first and second generation is the number of layers. The 86Gbit die capacity is certainly a bit odd but it actually makes sense as it is simply the first generation die cut in half with the added layers [(128Gbit / 2) * (32 / 24)]. As such, it is relatively safe to assume that the other variables (cell size etc.) are the same as with the first generation and that is what the table above is based on.

Update 7/8: Actually, the peripheral circuitry does not scale with the memory array, meaning that the die size is actually quite a bit larger than I originally thought.

Before we go deeper with the density analysis, I want to explain how cell size is calculated. As you can see, the figures are way too big to make any sense if you just look at the 40nm and 16nm figures because no matter how you try to multiply them, the results do not make sense. 

Oftentimes when cell size is discussed, it is only the actual size of the cell that is taken into account, which leaves the distance between cells out of the conclusion. However, the so called empty parts (they are not really empty as you saw in the X-ray a couple of pages back) take a part of the die area similar to the cells, meaning that they cannot be excluded. The appropriate way to measure cell size is from the inner corner of a cell to the outer corners of neighbouring cells in both X and Y axes. This is demonstrated by the black square on the above graph.

With the proper measurements, this is how V-NAND compares to 16nm NAND when just looking at cell size alone (i.e. excluding how the layers impact density). Doesn't look too good, huh?

If you are wondering where the 16nm comes from, it is mostly just marketing. 16nm refers to the smallest length (or resolution as it is often called) in the die, meaning that it is the finest line that can be printed. In the case of NAND, that can either be the wordline or bitline, or the distance between them. With Micron's 16nm NAND, that is likely the length of the wordline and the distance between two wordlines as the two are 32nm when combined (i.e. 16nm each). 

The actual cell size did not make justice to V-NAND because the whole idea behind the move to 3D NAND is to increase the cell size and distances between cells to get rid of the issues 2D NAND has. In the graph above, I took the amount of layers into account because you essentially need 32 2D NAND cells to achieve the same density as with 32-layer V-NAND and the game totally changes. 

The math behind that graph is just a couple of simple arithmetic equations. The actual cell area is 40,300 (155nm*260nm), meaning that the relative cell area is simply the actual cell area divided by the number of layers. That gives us 1,259nm2(40,300nm/ 32). To get the relative dimensions, the actual dimensions are divided by the square root of the number of layers (e.g. 155nm / √32 = 27nm).

NAND Cell Size - Relative

In the end, the relative cell size turns out to be smaller than Micron's 16nm NAND. The difference is not huge (~4%) but when the performance, power consumption and endurance advantages are taken into account, V-NAND is a clear winner. 

Bit Density

Another way to look at cost efficiency is the bit density. While cell size is a good measure, it does not take peripheral circuitry and ECC/spare bytes into account, which take a part of the die as well. Bit density is simply die capacity divided by die size and it gives us a number for quick and easy comparison. 

In this comparison, V-NAND is the leader without any doubts. The bit density is as much as 73% higher, resulting in a much more cost efficient design. The reason why the difference between cell size and bit density is so large is that 2D NAND requires a lot more die area for ECC bytes because it is more error prone. With V-NAND there is less ECC needed thanks to its higher reliability. In addition, Micron's peripheral circuitry design is relatively die consuming, so I wonder how Toshiba's/SanDisk's 15nm stacks up with V-NAND as they are claiming to have a more efficient peripheral circuitry design.

Update 7/8: V-NAND is still denser than the latest 2D NAND nodes but the difference is not overwhelming. 

All in all, there is a lot more than just cell area and bit density when it comes to cost efficiency. Yield and equipment cost are two major factors, which are both better for 2D NAND as it is a well known technology and some of the machinery can be reused when moving from one node to another. 

3D NAND: Hitting The Reset Button on Scaling RAPID 2.0: Support For More RAM & Updated Caching Algorithm


View All Comments

  • Cerb - Tuesday, July 1, 2014 - link

    The SSD just quitting could brick the drive, and could hose up the FS. Like ECC RAM, if you need power loss protection, you need it regardless of file system. IMO, they should all at least be equipped with enough to gracefully finish in-progress writes and shut down (not necessarily empty buffers, but set up a state that they can be guaranteed to be able to roll back from). Reply
  • sonicology - Tuesday, July 1, 2014 - link


    "Scaling below 20nm was seemed as a major obstacle but the industry was able to cross that..."

    seemed a major obstacle or seen as a major obstacle, but not seemed as a major obstacle
  • Bladen - Tuesday, July 1, 2014 - link

    Is it just me, or does the "Performance Consistency" page have interactive charts, all of which are entitled "(SSD name and capacity) - 4KB Random Write (QD32) Performance", but with different results in each of the graphs for the same SSDs and capacities? Also, descriptive text is missing below the last two.

    I'm presuming they are supposed to be a read and another read or write one, presumably at a lower queue depth.
  • MrSpadge - Tuesday, July 1, 2014 - link

    They're the same data in different scales. Reply
  • paesan - Tuesday, July 1, 2014 - link

    I purchased a 512GB MX100 when they first came out for $199. No way the extra speed from the 850 pro is worth over twice the price as the MX100. Nobody is going to keep that drive for 10 years anyway. In 10 years the drive will be obsolete. Most users won't even notice the difference in speed in their every day usage. Reply
  • MrSpadge - Tuesday, July 1, 2014 - link

    BTW: that image is surely not an X-Ray but "just" an ordinary SEM image (scanning electron microscope). Reply
  • Spatty - Tuesday, July 1, 2014 - link

    This is correct. SEM image after a FIB. Reply
  • drwho9437 - Wednesday, July 2, 2014 - link

    As I mentioned elsewhere, it is probably a STEM bright field image after a FIB liftout of the gate stack. Certainly not an X-ray; he should be correct the article... Reply
  • Cerb - Tuesday, July 1, 2014 - link

    "The wear out causes the insulators to lose their insulating characters" - shouldn't that be characteristic(s)? Reply
  • jann5s - Tuesday, July 1, 2014 - link

    I'm wondering If this technology will also end up in SD (XD) memory cards? Reply

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