Minimum number of RW required

SystemVerilog
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1

Hi,

Can anyone answer the below questions with solution.
Scenarion: RAM with 8 bit Address bus, 8 bit Wdata bus, and 8 bit Rdata bus.

  1. Minimum number of WR operation required if 2 bits of Wdata is swapped?
  2. Minimum number of WR operation required if 2 bits of Wdata is interconnected?
  3. Minimum number of WR operation required if 2 bits of Rdata is swapped?
  4. Minimum number of WR operation required if 2 bits of Rdata is interconnected?
  5. Minimum number of WR operation required if 2 bits of Addr is swapped?
  6. Minimum number of WR operation required if 2 bits of Addr is interconnected?

Please provide the solution with reasoning

2

In reply to rexjohn4u:

Hi,

The scenario could be of different, As per my understanding i could describe one scenario:

  1. we have 256 address locations so let us say that 2 data bits are swapped or interconnected, In this case if you write in location 0x01 it may write in 0x03 or it may be swapped or written to 0x02. So over here we have to write the walking locations of the RAM with a unique data Let us say as data = address
    So the write operation is to be performed on locations 0x01, 0x02, 0x04, 0x08, 0x10, 0x20,0x40,0x80, As the address are swapped or interconnected the data is going to be stored in the different locations and read back all the address locations.

With this we could cover all the scenarios that you have discussed.

Thanks,
Nikhil

3

Hi Nikhil,

Thanks for the reply. If i brief on your solution, “,minimum” number of WR required to identify the faulty 2 bits in case of address bits swapped is 256?

Lets say if we have 2 more faulty bit in address bus(total 4), what will be the minimum WR required.

How it differs when Wdata and Rdata bus got this faulty 2 bits?

Regards,
Rex

4

In reply to rexjohn4u:

Hi Rex,

What ever the number of faults are the writes are not 256 the number of write for a memory space of 8 bits are the writes performed on the following locations of address 8’b00000001, 8’b00000010, 8’b00000100 … so the number of writes are always 8 but the number of reads are 256.

As per your question even two more bits are also false then the writes could be to some different location so the writes are the number of writes and reads are same in terms of my understanding.

Thanks,
Nikhil

5

Hi Nikhil,

Thank You for the detailed explanation.

Regards,
Rex