- Asked 2/14
- 2073 Magh · 6+2 marks
Timing diagram of LXI D, 2465H
Short answer
LXI D,2465H has three cycles: opcode fetch (11H), read 65H to E, read 24H to D, total 10 T-states. With 6 MHz crystal, T=0.333 µs, time = 10×0.333 = 3.33 µs.
Draw and explain the timing diagram of LXI D, 2465H. Calculate the time required to execute this instruction if the crystal frequency is 6 MHz.
LXI D,2465H (opcode 11H) loads the register pair DE with 2465H, that is D = 24H and E = 65H. It is a 3-byte instruction (11H, 65H, 24H) with three machine cycles: opcode fetch (4T), memory read (3T), memory read (3T) = 10 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 | 11H (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 | 65H (data in) | 0 | RD low, rises at end | 0 1 0 | |
| M3 (memory read, 3T) | T1 | 20H | 02H | 1 | - | 0 1 0 |
| T2 | 20H | Hi-Z (bus floats) | 0 | RD low | 0 1 0 | |
| T3 | 20H | 24H (data in) | 0 | RD low, rises at end | 0 1 0 |
| M1 (OF) | M2 (MR) | M3 (MR) |
|T1 T2 T3 T4|T1 T2 T3 |T1 T2 T3 |
ALE |HI lo lo lo|HI lo lo |HI lo lo |
RD |hi LO LO hi|hi LO LO |hi LO LO |
WR |hi hi hi hi|hi hi hi |hi hi hi |
Explanation
- M1: opcode 11H is fetched from 2000H and decoded in T4.
- M2: address 2001H; data byte 65H is read and stored in E (low-order register of the pair).
- M3: address 2002H; data byte 24H is read and stored in D.
No write operation takes place, so WR stays high throughout.
Time
Crystal = 6 MHz, so the internal clock = 6 / 2 = 3 MHz and T = 0.333 microsecond.
Time = 10 x 0.333 = 3.33 microseconds.
✓ Answer checked by Claude Opus 5.5 (high effort)
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