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Visitor II
March 7, 2006
Question

ST10F276 Read setup time

  • March 7, 2006
  • 5 replies
  • 1076 views
Posted on March 07, 2006 at 08:45

ST10F276 Read setup time

    This topic has been closed for replies.

    5 replies

    urikAuthor
    Visitor II
    March 7, 2006
    Posted on March 07, 2006 at 03:15

    Hello,

    what is the data set-up time when i use the demuxed bus, with no read delay?

    is that T15 ?

    if so does it means that the setup time is 18.5 ns? or if i use a 55 ns SRAM i can latch the data after 55ns and not 55 + 18.5 ns?

    thanks,

    Uri.

    Visitor II
    March 7, 2006
    Posted on March 07, 2006 at 05:12

    Hello,

    Yes, the data set-up time when using the demuxed bus, with no read delay is t15.

    If Fcpu = 40Mhz: t15= 18.5 + tc (ns) & tc = 2TCL x (15 - MCTC) ; with 15 - MCTC is the number of wait states.

    --> so t15 value is Fcpu and external memory access time dependent.

    Let's take an example:

    ST10 Frequency is 40 Mhz. Thus the CPU cycle is 25ns. Assuming that the memory has an access time of 55ns= 25ns + 30ns, in order to access this memory, 2 wait states should be

    inserted in the EBC cycle.

    In this case:

    t15= 18.5 + 25x2 = 68.5ns.

    If CPU Clock is Variable (from 1 to 64MHz):

    t15 = 3TCL – 19 + tC.

    I hope i was clear.

    Regards,

    Najoua.

    [ This message was edited by: Najoua on 07-03-2006 09:44 ]

    urikAuthor
    Visitor II
    March 7, 2006
    Posted on March 07, 2006 at 05:51

    Najoua Hi,

    you were not so clear , does the T15 <= 18.5 + Ext mem acc time

    or T15 <= Ext mem acc time - 18.5 ?

    if i am not worng for 55ns i need 3 wait states?

    can you please clear this point?

    thanks,

    Uri.

    Visitor II
    March 7, 2006
    Posted on March 07, 2006 at 07:04

    Hi Ukatry,

    I will clarify,

    t15 <= 18.5 + tc;

    tc = 2TCL x (15 - MCTC) which depends on Fcpu and the number of wait states in order to be able to acceed the memory.

    For a memory having an access time of 55ns:

    - at least 2 wait states must be added when Fcpu = 40Mhz because: 55 ns = 25 ns + 30 ns and 30/25 = 1.2 so 2 waitstates are needed.

    - at least 3 wait states must be added when Fcpu = 64 Mhz

    because: 55 ns = 15.6 ns + 39.4 ns and 39.4/15.6 = 2.5 so 3 waitstates are needed.

    I hope it is clear now.

    Regards,

    Najoua.

    [ This message was edited by: Najoua on 07-03-2006 11:39 ]

    urikAuthor
    Visitor II
    March 7, 2006
    Posted on March 07, 2006 at 08:45

    now it is clear , thank you very much.