How many machine cycles and T-states does ANI 4BH take?

2 machine cycles (opcode fetch + memory read), 7 T-states

1 machine cycle (opcode fetch), 4 T-states

2 machine cycles (opcode fetch + memory write), 7 T-states

Which flag effect does ANI 4BH have?

AC = 1, CY = 0, and S, Z, P are set by the result

AC = 0, CY = 0, and S, Z, P are set by the result

CY is set if the result is negative

No flags are affected, as with MOV

What happens in the second machine cycle of ANI 4BH (stored at 2000H-2001H)?

4BH is read from 2001H into a temporary register, then A AND 4BH goes to A

The result of A AND 4BH is written to memory location 2001H

Nothing new is read; the AND was already done in T4 of M1

The operand is read from the memory location whose address is 4BH

2075 Baisakh · 5 marks

Execution of ANI 4BH with timing

Short answer

ANI 4BH ANDs accumulator with immediate byte 4BH, result in A. It sets AC=1, CY=0, and S, Z, P per result. The 2-byte instruction has opcode fetch (4T) and memory read (3T), total 7 T-states.

Explain the execution of instruction ANI 4BH in 8085 with the help of timing diagram.

ANI 4BH (opcode E6H) ANDs the accumulator with the immediate byte 4BH and stores the result in A: A <- A AND 4BH. It sets AC = 1, clears CY, and sets S, Z, P by the result. It is a 2-byte instruction with 2 machine cycles: opcode fetch (4T) and memory read (3T) = 7 T-states. Assume it is stored at 2000H, 2001H.

M-cycle	T	A15-A8	AD7-AD0	ALE	Control	IO/M S1 S0
M1 (opcode fetch, 4T)	T1	20H	00H (PCL)	1	-	0 1 1
	T2	20H	Hi-Z (bus floats)	0	RD low	0 1 1
	T3	20H	E6H (opcode in)	0	RD low, rises at end	0 1 1
	T4	-	-	0	none (decode)	0 1 1
M2 (memory read, 3T)	T1	20H	01H	1	-	0 1 0
	T2	20H	Hi-Z (bus floats)	0	RD low	0 1 0
	T3	20H	4BH (data in)	0	RD low, rises at end	0 1 0

      |  M1 (OF)  | M2 (MR) |
      |T1 T2 T3 T4|T1 T2 T3 |
 ALE  |HI lo lo lo|HI lo lo |
 RD   |hi LO LO hi|hi LO LO |
 WR   |hi hi hi hi|hi hi hi |

Explanation

M1: PC = 2000H on the bus with ALE = 1 in T1. RD low in T2-T3; opcode E6H is read into IR. In T4 it is decoded and the CPU knows one more byte is needed.
M2: PC = 2001H on the bus; RD low; the data byte 4BH is read into a temporary register. At the end of this cycle the ALU performs A AND 4BH and the result goes back to A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Execution of ANI 4BH with timing

Explanation

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