Kumar, you can disregard this post, as your eyes will glaze over and you'll fall asleep. This is for anyone else who's curious as to how computer memory works. It's just the basics, this is not the only way to make any of the components.
Let's start with a
transistor. There are several transistor types, but that basic one will do for now. To make a transistor, you need 2 n-p junctions. That is, 2 boundaries between a n-doped region and a p-doped region.
An n-doped region is a region that has been doped with an electron donor element. That is, it has more electrons that the substrate. Typically, if Silicon is the substrate, the n-doped region will have an element with 5 electrons on its shell as a dopant.
A p-doped region is a region that has been doped with an element that will donate holes. That is, it has less electrons than the substrate. Typically, if Silicon is the substrate, the p-doped region wil have an element with 3 electrons on its shell as a dopant.
A transistor looks something like
this.
Logic circuits are made of logic gates. Each gate implements a logic operation, such as AND, OR, NOT, etc. The easiest way to describe them is through their truth tables. For instance, for an AND gate
(symbol)
:
Code:
v1 v2 | Output
0 0 | 0
0 1 | 0
1 0 | 0
1 1 | 1
v1 and v2 are the inputs. The gate only has an output of '1' when both inputs are '1'. The gate that I'll use to describe memory is a NAND gate
(symbol)
It is a negated AND. You'll notice from the symbol that the negation is signified by the circle. The truth table is:
Code:
v1 v2 | Output
0 0 | 1
0 1 | 1
1 0 | 1
1 1 | 0
One way to construct this gate is
this. It has 2 transistors, 3 resistors, a voltage source terminal (Vcc, at 6Volts), a Ground terminal (at the bottom of the image, the three horizontal lines), two input terminals (VA and VB), and one output terminal (Vout).
The way it works is this: Vout will be at 6V (approx) which signifies a '1', unless the connection with the ground terminal is established. If the connection with ground is established, Vout will be close to 0V, signifying a '0'.
The only way to establish a connection to the ground terminal is to activate both transistors, and that only happens when both inputs are '1'. So there we have a NAND gate.
Now onto the memory circuit. I'll describe one of the simplest ways to do this, an SR latch. There are other ways, but one is enough for out purposes.
The diagram for a SR latch is
this. Both inputs (S and R) are negated (they have a bar on top). That means that they are active when the input is '0'. S stands for Set, R for Reset. Q is the output.
This is a sequential circuit, which means it's a bit more complicated to explain, but I'll give it a shot.
The normal input for the circuit is S=1 and R=1, that means both off. Q and Q' (Q negated) are fed back, so we need to give them startup values too. Let's start with Q=0 and Q'=1.
If S is changed to '0' (it is activated), the NAND on top has '0' and '0' as its inputs, so It turns to '1'. The NAND on the botton now has '1' and '1', so it changes to '0'. The NAND on top now has '0' and '0', so it stays at '1'. The S can now be changed back to '1', and it won't matter. The circuit has "memorized" the '1'.
To turn it back to '0', we need to use the RESET. So we leave S to '1', and we change R to '0'. The NAND on bottom has '0' and '1', so it changes to '1'. The NAND on top now has '1' and '1', so it changes to '0'. The NAND on bottom now has '0' and '0' so it stays at '1'. The R can now be set back to '1' and it won't matter. The circuit has "memorized the '0'.
And that is one way to memorize a bit using semiconductors. Now looking at Kumar's theories, what problems are there?
1) Shaking some sand in water will not create transistors. There will be no substrate, no n-p junctions, no terminals, no transistors.
2) Shaking some sand in water will not create 4 transistors in the proper configurations. There will not be 2
NAND gates properly connected.
3) Shaking some sand in water will not supply enough voltage to feed the circuit. Opening a p-n junction takes a certain ammount of voltage. That will not come from shaking water.
4) The circuit requires a continual source of voltage, otherwise the circuit forgets the information stored.
5) Storing information in binary requires encoding. To encode only lower case letters, you need at least 5 bits, giving you 32 options.That means 10 NAND gates. Typically, computers use 8 bits to store each text element. Whatever information Kumar wants to encode will not magically be transformed into the proper code, and passed to the proper input terminals.
6) Reading the information will require decoding. Cells do not decode binary information from semiconductors.
7) There are other problems, but that's enough for right now.
Kumar, in case you read all this, did you understand why your current idea is wrong? No? Too bad.
Hmm, looking at the preview I can tell that was a long and boring post. I wouldn't read it, I recomend you don't either. Maybe I should put that recomendation on top.