---
tags: [logic-gates, binary, memory]
---

# Creating memory with NAND gates

The
[logic circuit](Digital_circuits.md)
below demonstrates how memory can be created using
[NAND](Logic_gates.md#nand-gate)
gates. A single bit is stored in memory.

![](static/nand_latch_logic_circuit.png) Interactive version of circuit:

<iframe src="https://circuitverse.org/simulator/embed/nand-mem?theme=default&display_title=false&clock_time=true&fullscreen=true&zoom_in_out=true" style="border-width:; border-style: solid; border-color:;" name="myiframe" id="projectPreview" scrolling="no" frameborder="1" marginheight="0px" marginwidth="0px" height="500" width="500" allowFullScreen></iframe>

## 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

![](static/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

![](static/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`, **O** 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` since at Gate 1: `ON (I) + OFF (S) = ON` and `OFF (I) + OFF (S) = OFF`
and at Gate 2: `OFF (Gate 1) + OFF (S) = OFF`

This is illustrated in the diagram below. The space occupied by **A** and **B**
remains on (note it is illuminated) regardless of the state of **I**.

![](static/nand-mem-demonstrated.gif)