Jump if carry (JC) in 8085 Microprocessor

In 8085 Instruction set, we are having one mnemonic JC a16, which stands for “Jump if Carry” 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 carry flag value is 1.If carry flag value is 0, program flow continues sequentially. It is a 3-Byte instruction.

Mnemonics, Operand	Opcode(in HEX)	Bytes
JC Label	DA	3

Let us consider one example of this instruction type JC 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	37	STC	Set Cy flag bit. So Cy = 1
2001	D2	JC 4000	Jump Carry, i.e. Jump when Cy = 1
2002	00		Low order Byte of the target address
2003	40		High order Byte of the target address PC ← 4000H, So the program control will be transferred to the address 4000H
2004	78	MOV A, B	This instruction will not get control now as JC 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 JC 4000H execution is as follows –

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

Chandu yadav

Published on 27-Dec-2018 11:28:13

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