BRAM utilization - how to calculate

dear all

I have a question about calculating a BRAM utilzation.

I am using XC2VP30 and data sheet says

-----------------------------
    Block SelectRAM+
-----------------------------
18kb Blocks | Max. BRAM (kb)
-----------------------------
136         | 2448
-----------------------------

and in my MHS file, BRAM is specified as

-----------------------------
...
BEGIN lmb_bram_if_cntlr  # 64 kb
 PARAMETER C_BASEADDR = 0x00000000
 PARAMETER C_BASEADDR = 0x0000ffff
 BUS_INTERFACE SLMB = ilmb
...
END

BEGIN lmb_bram_if_cntlr # 64 kb
 PARAMETER C_BASEADDR = 0x00000000
 PARAMETER C_BASEADDR = 0x0000ffff
 BUS_INTERFACE SLMB = dlmb
...
END
-----------------------------

as far as I understand, the BRAM utilization should be

(64 + 64 ) kb / 2448 kb = 5.3 %

But log file says

-----------------------------
Device utilization summary

Number of Block RAMs  : 32 out of 136   23%
...
-----------------------------

I understand this as a NUMBER of BLOCKS out of total NUMBER of BLOCKS.
Anyway, the memory utilization is very low.

Questions are
- Is my understanding correct ?
- How can we utilize more efficiently ?

Thankyou and thankyou for previous Q & A too.

"Jack" <JEmoderatz@yahoo.com> schrieb im Newsbeitrag
news:1107883607.877297.20670@z14g2000cwz.googlegroups.com...
> dear all
>
> I have a question about calculating a BRAM utilzation.
>
> I am using XC2VP30 and data sheet says
>
> -----------------------------
>     Block SelectRAM+
> -----------------------------
> 18kb Blocks | Max. BRAM (kb)
> -----------------------------
> 136         | 2448
> -----------------------------

Kbits !!

>
> and in my MHS file, BRAM is specified as
>
> -----------------------------
> ...
> BEGIN lmb_bram_if_cntlr  # 64 kb
>  PARAMETER C_BASEADDR = 0x00000000
>  PARAMETER C_BASEADDR = 0x0000ffff
>  BUS_INTERFACE SLMB = ilmb
> ...
> END
>
> BEGIN lmb_bram_if_cntlr # 64 kb
>  PARAMETER C_BASEADDR = 0x00000000
>  PARAMETER C_BASEADDR = 0x0000ffff
>  BUS_INTERFACE SLMB = dlmb
> ...
> END
> -----------------------------
>
> as far as I understand, the BRAM utilization should be
>
> (64 + 64 ) kb / 2448 kb = 5.3 %

NO
64 KByte ! the ilmb and dlmb use A and B ports of the same 64KByte block


> Number of Block RAMs  : 32 out of 136   23%

correct 32 brams, each 2KB = 64Kb

when counting bits each bram has 18bits so 2 bits are wasted,
so the bit percent will differ a little

Antti

Each BRAM has 18kbit of memory
136 BRAM has 136*18kbit = 2448 kbit (not kbyte)

64 kbyte of memory is requiring 32 BRAMs (each BRAM has 2 kbyte of memory)

So 64 kbyte of memory is requiring 32 BRAMs which is 32/136 = 23% of the 
available BRAMs

G�ran


Jack wrote:
> dear all
> 
> I have a question about calculating a BRAM utilzation.
> 
> I am using XC2VP30 and data sheet says
> 
> -----------------------------
>     Block SelectRAM+
> -----------------------------
> 18kb Blocks | Max. BRAM (kb)
> -----------------------------
> 136         | 2448
> -----------------------------
> 
> and in my MHS file, BRAM is specified as
> 
> -----------------------------
> ...
> BEGIN lmb_bram_if_cntlr  # 64 kb
>  PARAMETER C_BASEADDR = 0x00000000
>  PARAMETER C_BASEADDR = 0x0000ffff
>  BUS_INTERFACE SLMB = ilmb
> ...
> END
> 
> BEGIN lmb_bram_if_cntlr # 64 kb
>  PARAMETER C_BASEADDR = 0x00000000
>  PARAMETER C_BASEADDR = 0x0000ffff
>  BUS_INTERFACE SLMB = dlmb
> ...
> END
> -----------------------------
> 
> as far as I understand, the BRAM utilization should be
> 
> (64 + 64 ) kb / 2448 kb = 5.3 %
> 
> But log file says
> 
> -----------------------------
> Device utilization summary
> 
> Number of Block RAMs  : 32 out of 136   23%
> ...
> -----------------------------
> 
> I understand this as a NUMBER of BLOCKS out of total NUMBER of BLOCKS.
> Anyway, the memory utilization is very low.
> 
> Questions are
> - Is my understanding correct ?
> - How can we utilize more efficiently ?
> 
> Thankyou and thankyou for previous Q & A too.
>