Jump if minus (JM) in 8085 Microprocessor

In 8085 Instruction set, we are having one mnemonic JM a16, which stands for “Jump if Minus” 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 sign flag value is 1. If the sign flag value is 0, program flow continues sequentially. It is a3-Byte instruction.

Mnemonics, Operand	Opcode(in HEX)	Bytes
JM Label	FA	3

Let us consider one example of this instruction type JM 4000H. It is a 3-Byte instruction. The result of the execution of this instruction is shown below with an example.

Address	Hex Codes	Mnemonic	Comment
2000	3E	MVI A,30	A ← 30H
2001	30		8-bit operand 30H
2002	06	MVI B,40	B ← 40H
2003	40		8-bit operand 40H
2004	90	SUB B	A ← A – B= 30H – 40H = -10H = F0H. As the result is -10H and as the result is negative, so S flag bit will be 1
2005	FA	JM 4000	Jump Minus, i.e. Jump when S = 1, as the subtraction result is -10H, so S 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 JP 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 JM 4000H execution is as follows –

Summary − So this instruction JM requires 3-Bytes, 3-Machine Cycles (Opcode Fetch, Memory Read, MemoryRead) and 10 T-States for execution as shown in the timing diagram.