Last Sync: 2022-10-15 17:00:05

Hardware/Logic_Gates/Creating_memory_with_NAND.md

@@ -9,4 +9,48 @@ tags: [logic-gates, binary, memory]
 
 # Creating memory with NAND gates
 
+The [logic circuit](/Hardware/Logic_Gates/Logic_circuits.md) below demonstrates how memory can be created using [NAND](/Hardware/Logic_Gates/Nand_gate.md) gates. A single bit is stored in memory.
+
 ![](/img/nand-memory.svg)
+
+## Components
+
+- **I** is where we input the bit that we want to remember.
+- **O** is the output of the remembered bit
+- **S** is an input that tells these gates when to set the memory.
+- **A, B, C** are internal wires that represent the individual bit values which channel into each gate
+
+## State changes
+
+### First state: S `ON`, I `OFF`
+
+- Start with **S** `ON` and **I** `OFF`.
+- Given the logic of NAND, **A** will be on.
+- This means **Gate 2** is receiving `ON` from **A** and `ON` from **S** therefore `B` is `OFF`
+- Moving to **Gate 4** it is receiving `OFF` from **B** and it's other input is not activated so is therefore also `OFF`. Consequently **C** is `ON`
+- `C` is fed as the second input into **Gate 3**, thus we have both inputs (**A** and **C**) as `ON`, therefore the output will make `O` `OFF`
+
+> Upshot: With **S** `ON`, output is the same as input
+
+![](/img/nand-mem-first.gif)
+
+### Second state: both S and I `ON`
+
+- With **S** and **I** both on, Gate A will be `OFF`
+- Gate 2 is receiving `OFF` from **A** and `ON` from **S** therefore **B** is `ON`
+- **A** feeds into Gate 2 which is receiving `ON` from **S**, therefore the output (**B**) will be `ON`
+- **A** feeds into Gate 3 and the other input is `OFF`, therefore Gate 3 is outputting `ON`. Thus **O** is `ON`
+- **O** feeds back to Gate 4 which is receiving `ON` from **B** thus we have two `ON` inputs going into Gate 4, turning **C** off
+- This means the two inputs going into Gate 3 are both `OFF`, keeping **O** `ON`
+
+> Upshot: With **S** on, the output is again the same as the input
+
+![](/img/nand-mem-second.gif)
+
+> So far we have seen that when **S** is `ON` you can change **I** on and off and **O** will change with it.
+
+### Memory
+
+When **S** is `ON`, **0** will mirror whatever state **I** is in. However if you turn **S** `OFF`, **O** will remain in whatever state it was when **S** was turned `OFF`. You can toggle **I** as much as you like, **O** will remain in its previous state. Hence creating a memory store of the past value of **I**.
+
+The specific reason for this is that, if **S** is `OFF`, both **A** and **B** are `ON`