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Jump if zero (JZ) result in 8085 Microprocessor
In 8085 Instruction set, we are having one mnemonic JZ a16, which stands for “Jump Zero” and “a16” stands for any 16-bit address. This instruction is used to jump to the address a16 as provided in the instruction. But as it is a conditional jump so it will happen if and only if the present zero flag value is 1. If the zero flag value is 0, program flow continues sequentially. It is a 3-Byte instruction.
| Mnemonics, Operand | Opcode(in HEX) | Bytes |
|---|---|---|
| JZ Label | CA | 3 |
Let us consider one example of this instruction type JZ 4000H. It is a 3-Byte instruction. The result of execution of this instruction is shown below with an example.
| Address | Hex Codes | Mnemonic | Comment |
|---|---|---|---|
| 2000 | 3E | MVI A,40 | A ← 40H |
| 2001 | 40 | 8-bit operand 40H | |
| 2002 | 06 | MVI B,40 | B ← 40H |
| 2003 | 40 | 8-bit operand 40H | |
| 2004 | 90 | SUB B | A ← A – B= 40H – 40H = 00H |
| 2005 | CA | JZ 4000 | Jump Zero, i.e. Jump when Z = 1, as the subtraction result is 00H, i.e. zero, so Z flag bit will remain with value 1 |
| 2006 | 00 | Low order Byte of the target address | |
| 2007 | 40 | High order Byte of the target address PC ← 4000H, So the program control will be transferred to the address 4000H | |
| 2008 | 78 | MOV A, B | This instruction will not get control now as JZ will transfer the control to the memory address 4000H |
| …. | …. | …. | …. |
| 4000 | 41 | MOV B, C | Next instruction at address 4000H will get the control |
The timing diagram against this instruction JZ 4000H execution is as follows –
Summary − So this instruction JZ requires 3-Bytes, 3-Machine Cycles (Opcode Fetch, Memory Read, MemoryRead) and 10 T-States for execution as shown in the timing diagram.
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