Which is the correct definition of a machine cycle?

The time needed to complete one basic memory or I/O operation, made of 3 to 6 T-states

One period of the processor clock

The time needed to fetch and execute a whole instruction

The time needed to decode an opcode in T4

One clock period of the processor, the smallest unit of time (T = 1/f)

One complete opcode fetch cycle

The time for which ALE stays high

Which machine cycles does ANI 55H need?

Opcode fetch only (4T), because ANI is a logical instruction

Opcode fetch (4T) and memory write (3T) = 7T

Opcode fetch (4T), memory read (3T), memory read (3T) = 10T

If the 8085 operates at a 2 MHz clock, how long does ANI 55H take?

7 x 1 microsecond = 7 microseconds

4 x 0.5 microsecond = 2 microseconds

7 x 0.25 microsecond = 1.75 microseconds

2074 Bhadra · 2+5+1 marks

Machine cycle, T-states, ANI 55H timing

Short answer

A machine cycle is the time for one memory or I/O operation (3–6 T-states); a T-state is one clock period. ANI 55H requires opcode fetch (4T) and memory read (3T) = 7 T-states. At 2 MHz, T = 0.5 µs, so execution time = 3.5 µs.

What do you mean by Machine cycle and T-states? Draw a bus timing diagram for an instruction ANI 55H of 8085 microprocessor. Calculate the time required to execute such instruction, if microprocessor is operating at 2 MHz clock frequency.

Machine cycle

A machine cycle is the time needed to complete one basic memory or I/O operation: opcode fetch, memory read, memory write, I/O read or I/O write. It consists of 3 to 6 T-states.

T-state

A T-state is one clock period of the processor clock; it is the smallest unit of time (T = 1 / f). An instruction cycle is made of several machine cycles, and each machine cycle of several T-states.

Timing diagram of ANI 55H

ANI 55H (opcode E6H, data 55H) is a 2-byte instruction: M1 opcode fetch (4T) and M2 memory read (3T) = 7 T-states. Assume it is stored at 2000H.

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	55H (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 |

M1 reads E6H from 2000H, M2 reads 55H from 2001H. The AND operation is done after the data byte is received.

Execution time

Clock frequency f = 2 MHz, so T = 1 / 2 MHz = 0.5 microsecond.

Time = 7 x 0.5 microsecond = 3.5 microseconds.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Machine cycle, T-states, ANI 55H timing

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