| Micro Instruction | Operation | Description |
| HLT | Halt | Halt Clock |
| MI | Memory Adress Register IN | Get the Memory Address (4-bit) from the Bus |
| RI | RAM IN | Get RAM content (8-bit) from the Bus |
| RO | RAM OUT | Put RAM content (8-bit) on the Bus |
| IO | Instruction Register OUT | Put Instruction Register argument (4-bit) on the Bus |
| II | Instruction Register IN | Get Instruction Register value (8-bit) from the Bus |
| AI | 'A' Register IN | Get 'A' Register value (8-bit) from the Bus |
| AO | 'A' Register OUT | Put 'A' Register value (8-bit) on the Bus |
| EO | SUM Register OUT | Put SUM Register value (8-bit) on the Bus |
| SU | Substraction | Perform a substract operation |
| BI | 'B' Register IN | Get 'B' Register value (8-bit) from the Bus |
| OI | Output IN | Get value (8-bit) from the Bus to display it on the LED |
| CE | Counter ENABLE | Increase Program Counter by one |
| CO | Counter OUT | Put Program Counter content (4-bit) on the bus |
| J | Jump | Jump to Memory Address |
| FI | Flag Register IN | Get Carry/Zero Flag from SUM Register |
Every OpCode requires two Micro Instruction Steps to be fetched from the current Memory Address in to the Instruction Register. We call this operation Fetch.
| OpCode # | OpCode # << 4 | OpCode | Syntax | Micro Instruction Steps | Operation | Description |
| - | - | - | - | Step 0: CO MI Step 1: RO II CE |
Fetch | Fetch OpCode from current Memory Address |
| 0 | 0 (0000 0000) | NOP | NOP | Step 2: - Step 3: - Step 4: - |
No Operation | Do nothing |
| 1 | 16 (0001 0000) | LDA | LDA [memory address] | Step 2: IO MI Step 3: RO AI Step 4: - |
Load value in to 'A' Register | Load value from RAM in to 'A' Register |
| 2 | 32 (0010 0000) | ADD | ADD [memory address] | Step 2: IO MI Step 3: RO BI Step 4: EO AI FI |
Addition | Load value from RAM in to 'B' Register and add it with 'A' Register. The result will be in the SUM Register and stored in to the 'A' Register |
| 3 | 48 (0011 0000) | SUB | SUB [memory address] | Step 2: IO MI Step 3: RO BI Step 4: EO AI FI SU |
Substraction | Load value from RAM in to 'B' Register and substract it from 'A' Register. The result will be in the SUM Register and stored in to the 'A' Register |
| 4 | 64 (0100 0000) | STA | STA [memory address] | Step 2: IO MI Step 3: AO RI Step 4: - |
Store value in to RAM | Store value from 'A' Register in to RAM |
| 5 | 80 (0101 0000) | LDI | LDI [4-bit value] | Step 2: IO MI Step 3: - Step 4: - |
Load value in to 'A' Register | Load value from Instruction Register in to 'A' Register |
| 6 | 96 (0110 0000) | JMP | JMP [memory address] | Step 2: IO J Step 3: - Step 4: - |
Jump | Jump to Memory Address |
| 7 | 112 (0111 0000) | JC | JC [memory address] | Step 2: CF=0 - Step 3: CF=1 IO J Step 4: - Step 5: - |
Jump on Carry | Jump to Memory Address if Carry Flag is set |
| 8 | 128 (1000 0000) | JZ | JZ [memory address] | Step 2: ZF=0 - Step 3: ZF=1 IO J Step 4: - Step 5: - |
Jump on Zero | Jump to Memory Address if Zero Flag is set |
| 14 | 224 (1110 0000) | OUT | OUT | Step 2: AO OI Step 3: - Step 4: - |
Display value | Display value from 'A' Register to the LED |
| 15 | 240 (1111 0000) | HLT | HLT | Step 2: HLT Step 3: - Step 4: - |
Halt | Halt Clock |