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The correct sequence of execution of Read/Write instruction in 8086 Min mode of operation is :

  • A.Set \(MN/\overline{MX}\) high, Output high on \(M/\overline{IO}\), Set ALE, Transfer data
  • B.Set \(MN/\overline{MX}\) low, Output high on \(M/\overline{IO}\), Transfer data, Set ALE
  • C.Output high on \(M/\overline{IO}\), Transfer data, Set ALE, Set \(MN/\overline{MX}\) high
  • D.Transfer data, Set ALE, Set \(MN/\overline{MX}\) high, Output high on \(M/\overline{IO}\)

Answer: A

The sequence is set MN/MX high, output high on M/IO, set ALE, transfer data.

Minimum mode is selected by tying MN/MX high. In a bus cycle, the 8086 first drives M/IO to show whether a memory or I/O location is addressed (high for memory), puts the address on the multiplexed bus and pulses ALE so that the address is latched, and only then, with the address held in the latches, transfers data on the same lines with RD or WR.

The other options transfer data before the address is latched, set ALE last, or use MN/MX low, which selects maximum mode.

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