Jump if positive (JP) in 8085 Microprocessor

Microprocessor8085

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

Mnemonics, Operand
Opcode(in HEX)
Bytes
JPLabel
F2
3


Let us consider one example of this instruction type JP 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,30
B ← 30H
2003
30


8-bit operand 30H
2004
90
SUB B
A ← A – B= 40H – 30H = 10H. As the result is 10H and as the result is positive, so S flag bit will be 0
2005
F2
JP 4000
Jump Positive, i.e. Jump when S = 0, as the subtraction result is 10H, so S flag bit will remain with value 0
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 JP 4000H execution is as follows –


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

raja
Published on 27-Dec-2018 12:29:16
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