If, instead of being a 2-dimensional grid of elements, a RAM chip was arranged as a single row 1 bit wide, how long would it be?
Supplementry question:
How does this compare to the length of the same data travelling down a DSL connection?
It doesn’t really make sense to talk about the “length” of data going down a DSL connection. Well, I take that back. You could talk about
So, let’s analyze that. Let’s say I’m downloading something from Singapore at a rate of 150 kb/s, which is slightly faster than most DSL connections. It takes about 115 ms for data to get to me from Singapore (judging by the ping times to www.eecs.nus.edu.sg), and Singapore is about 9400 miles away from me here in Boston. So, the data is travelling at a rate of 9400 / .115 = 81,739 miles per second, or a bit under half the speed of light.
Let’s say it takes you about 1.85 hours to download a 1-gigabyte file over your DSL connection. Then, in total, all the data you downloaded would fit on a 150kb/s internet-like network stretching 544,381,740 miles. That’s approximately the distance from earth to Jupiter.
Err…you could talk about how large a network it would take so that you receive the first bit of information at the destination at the moment the last bit leaves the source. Which is what I did.
As for your second question, we’ll look at a 1GB DIMM with chips along one side. For simplicity, we’ll compute the memory density per square millimeter for the entire DIMM, rather than each individual chip. This means that our number will be slightly larger than it should be, but the difference should be slight.
So, a DIMM is 134mm by 26mm. (134*26) mm^2 / one billion bytes = 3.484 * 10^-6 mm^2 / byte.
So, each byte requires a square of area 3.484 * 10^-6 mm^2 to exist. If we take the square root of that, we find that such an area is .00187mm on each side. Multiply that by one billion, and we get 1,870,000mm, or 1.870 km, or 1.16 miles.
Please double-check my math and reasoning before you do anything important with this data.
Thanks for the answers. I confess I am struggling to find any important application for this info. Perhaps that should be the next challenge.
But a gigabyte is actually 2^30 = 1,073,741,824 bytes.
Actually, the size of RAM chips are standard. So by the virtues of industrial standards, a 2GB RAM chip is the same size as a 1GB RAM chip. The internal circuity will of course differ, but as I am not a computer engineer, I have no idea what the make-up will be like.
I am not sure if it can be analysed as such…for one thing, a huge download is often broken up into packages and transmit through different servers and routers, so you may be reciving more than one packages at a time. Some packages may be lost, so some may be resent. It’s not a direct conenction from Boston to Singapore, but rather a round-about loop, which is why the speed might be less than 115 ms per bit.
IRIC, it also possible that the files you are accessing are actually cached by your ISP.
When asking “how long”, usually I’ll want to know the unit of measurement. Computer memory are just electrical signals of 1 and 0 being maintained by electricity. As mentioned, I am not sure of the physical make-up of RAM, but very likely it consists of super-small micro-size registers. Even if you take out those registers and lie them out, the final length might not even be impressive.
Hope this helps!
This depends. Do RAM makers always sell modules marked in base-2 units, or do they sell them marked in base-10 units with the actual capacity being the next lower/higher power of 2?
I honestly don’t know anymore, it’s been so long since I bought RAM, but I’m also making a point: Measuring the capacity of computer components (disk drives, RAM sticks, etc.) is a frustrating affair due to the obfuscating practice, common in the industry, of deliberately confusing base-10 units with base-2 units.
That’s a good start, but you neglected routing area. RAM words are selected by row and column addresses being set to 1 for the proper cell - when a RAM cell sees the and of these, it knows to read or write. The address is decoded outside the ram, and the select lines go between rows and columns. A linear ram would have to have 30 select lines running along side the cells, and you’d have to route at least two to each cell. It would be, as it is called in the trade, a routing nightmare.
Because of the large decoding logic and buffers on the dimm, I think you’re numbers are a bit high. But it would still be rather long. And slow, of course, since light travels a nsec a foot.
Still base two. Though most of the memory I work with is internal to chips.
If you’re only concerned with the length of the circuitry that actually stores the bits, then the answer ends up being quite bit a smaller. Basing my numbers off of the Dual Port SRAM Cell listed here, I get a mere 10.3079215 kilometers. (= 19.2um/cell * 1,073,741,824bits / 2 bits/cell ). That same site mentions that a DRAM memory cell would be even denser, but at the moment my google-fu is failing to turn up any links to what those dimensions would be.
Hayden Sikh, your calculation seems to be for one gigabit of RAM, not one gigabyte as specified in the OP. Also, the Dual Port SRAM cell described in your link has only one bit of memory per cell, not two (Dual-Port means that the cell can have simultaneous reads and writes). So, I think you need to multiply your answer by a factor of sixteen.
However, as you say, that’s for SRAM, not DRAM, the latter being much smaller. Here is an old Samsung paper from 1995 (the Dark Ages!) showing DRAM cell size, with the densest available being 0.4μm x 0.835μm per bit. More recent advances have brought the feature size down even more (I’m not finding easily-readable links with a quick search right now), but let’s run the numbers with each bit taking 0.40 μm of length of the OP’s “linear DRAM”: 0.40μm x 1,073,741,824 x 8 = 3.345km or just over 2 miles. This assumes, of course, that all read/write/decoding circuitry, select lines etc, are alongside the DRAM cells and therefore do not add to the length (a horrible waste of space, but that’s what the OP asked for!).
[The minimum dimension of a DRAM cell using current-technology 90nm processes is probably of the order of 200nm (=0.2μm), which would give a one-mile gigabyte DRAM.]
:smack: Thanks for the correction – I knew I should have been more skeptical of that number. Really should know better than neglecting the bit to byte conversion, though… I think I’ll take that as a hint that it’s past my bedtime. 