In the Art of Assembly ebook, I have read that 8088 processor require four clock cycle to access memory (in whatever clock speed & memory speed is). How is it calculated? Is it time for processor to prepare for calculate address and put data on the bus?

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    $\begingroup$ I'm not sure that a question about memory access times on a specific processor is really computer science. Community opinions? $\endgroup$ – David Richerby Jul 29 '16 at 13:36
  • $\begingroup$ Also, could you please give a link to the ebook? $\endgroup$ – David Richerby Jul 29 '16 at 13:37
  • $\begingroup$ sorry for that, i am really beginner and self-taught. Art of Assembly $\endgroup$ – user55424 Aug 1 '16 at 0:51
  • $\begingroup$ This is the same ebook I have included link to via edit - this is a mirror. $\endgroup$ – Evil Aug 1 '16 at 1:13
  • $\begingroup$ can you explain me what is "mirror" mean ? i am not native english $\endgroup$ – user55424 Aug 1 '16 at 1:34

The 8088 is $16 bit$ microprocessor with $8 bit$ data bus, it requires $2$ cycles to access word. During the second read counter is automatticaly incremented. In 8088 $1$ cycle is needed to apply valid physical memory address (and read or write). The missing $1$ cycle is for preparation. The cycles are not calculated, simply read from specification steps performed.


  1. acquire valid address to address bus.
  2. RD (or WR) signal is issued, DEN is asserted (if WR then data is put onto the data bus). DEN connects data bus buffers to external data.
  3. allow memory to access data, if RD the data bus is sampled ot the end of cycle.
  4. bus signals are deactivated, data sampling is finished (RD). If WR data is transfered to memory.

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