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.