Draw and explain timing diagram for read operation in minimum mode of 8086.
1 Answer

The timing diagram for read operation in minimum mode is shown in fig below:

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These are explained in steps.

  1. When processor is ready to initiate the bus cycle, it applies a pulse to ALE during T1. Before the falling edge of ALE, the address, BHE, M/IO, DEN and DT/R must be stable i.e. DEN = high and DT/R = 0 for input or DT/R = 1 for output.
  2. At the trailing edge of ALE, ICs 74LS373 or 8282 latches the address.
  3. During T2 the address signals are disabled and S3-S7 ale available on AD16/S3-AD19/S6 and BHE/S7. Also DEN is lowered to enable transceiver.
  4. In case of input operation, RD is activated during T2 and AD° to AD15 go in high impedance preparing for input.
  5. If memory or I/O interface can perform the transfer immediately; there are no wait states and data is output on the bus during T3.

  6. After the data is accepted by the processor, RD is raised high at the beginning of T4.

  7. Upon detecting this transition during T.4, the memory or I/O device will disable its data signals.
  8. For an output operation, processor applies WR = 0 and then the data on the data bus during T2.
  9. In T4, WR is raised high and data signals are disabled.
  10. For either input or output operation, DEN is raised during 14 to disable the transceiver. Also M/I0 is set according to the next transfer at this time or during next T1 state. Thus length of bus cycle in 8086 is four clock cycle. If the bus is to be inactive after completion of bus cycle, then the gap between the successive cycles is filled by ideal state clock cycles.

When the memory or I/O device is not able to respond quickly during transfer, wait states (Tw) are inserted between T3 and T4 by disabling the READY input of the 8086. The bus activity during wait state is same as during T3.

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