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Altera_Forum
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12 years ago

Error (170040): Can't place all RAM cells in design

Dear All....

I have an issue here.... quartus gives this error while compiling... my compilation report is saying 74% of memory used. But i am getting this error still !!!! Is there anything i can do to avoid this and use the whole memory ???? or is this unavoidable !!? Kindly help asap, its a bit urgent. Thanks in advance.

The design contains NIOS II with two processors using 203776 bits... and around 14 ROMs

error:

Error (170040): Can't place all RAM cells in design Info (170034): Selected device has 66 memory locations of type M9K. The current design requires 70 memory locations of type M9K to successfully fit.

Info (170033): Memory usage required for the design in the current device: 106% M9K memory block locations required

compilation report:

Flow Status Flow Failed - Tue Jul 23 10:27:53 2013

Quartus II 32-bit Version 12.0 Build 232 07/05/2012 SP 1 SJ Web Edition

Revision Name qcon_v0

Top-level Entity Name qcon_v0

Family Cyclone III

Device EP3C25Q240C8

Timing Models Final

Total logic elements 17,352 / 24,624 ( 70 % )

Total combinational functions 14,339 / 24,624 ( 58 % )

Dedicated logic registers 8,059 / 24,624 ( 33 % )

Total registers 8059

Total pins 130 / 149 ( 87 % )

Total virtual pins 0

total memory bits 450,288 / 608,256 ( 74 % )

Embedded Multiplier 9-bit elements 132 / 132 ( 100 % )

Total PLLs 1 / 4 ( 25 % )

3 Replies

  • Altera_Forum's avatar
    Altera_Forum
    Icon for Honored Contributor rankHonored Contributor
    12 years ago

    its mroe to do with how efficiently you use the memory. Unused parts of an M9K cannot be reused. an M9K has 1bit x 9K locations (or other conjugations of that, like 9bits x 1k)

  • Altera_Forum's avatar
    Altera_Forum
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    12 years ago

    --- Quote Start ---

    its mroe to do with how efficiently you use the memory. Unused parts of an M9K cannot be reused. an M9K has 1bit x 9K locations (or other conjugations of that, like 9bits x 1k)

    --- Quote End ---

    Ok.. so i must check the individual ROM's and try to make use of whole M9K blocks and try to avoid overflowing into a new M9K blocks. am i right ? i shall try it, thanks Tricky...
  • Altera_Forum's avatar
    Altera_Forum
    Icon for Honored Contributor rankHonored Contributor
    12 years ago

    --- Quote Start ---

    its mroe to do with how efficiently you use the memory. Unused parts of an M9K cannot be reused. an M9K has 1bit x 9K locations (or other conjugations of that, like 9bits x 1k)

    --- Quote End ---

    Hello,in my design I use M4K with the depth of 256 and the width of 13, but after fitter processing I saw in the "fitter ram summary" the so-called implentation depth of 256 and the implentation width of 2, where is the other 11 bits, so strange? Waiting the answer.Thank you!