Determining Number of Bits in MAR

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SUMMARY

The discussion focuses on calculating the required bits for the Memory Address Register (MAR) and Memory Buffer Register (MBR) in a memory system configured as 64K x 32. For a word-addressable memory, 16 bits are necessary for the MAR, while for byte-addressable memory, 18 bits are required. The MBR must always match the size of the memory unit, which is 32 bits in this case. The calculations confirm that the unit size is relevant when determining the MAR for byte-addressable systems.

PREREQUISITES
  • Understanding of memory architecture concepts
  • Familiarity with word-addressable and byte-addressable memory systems
  • Knowledge of binary number representation
  • Basic arithmetic operations involving powers of two
NEXT STEPS
  • Study the differences between word-addressable and byte-addressable memory systems
  • Learn about the implications of memory unit size on MAR and MBR
  • Explore memory addressing techniques in computer architecture
  • Investigate the role of registers in CPU architecture
USEFUL FOR

Computer architecture students, hardware engineers, and anyone involved in designing or understanding memory systems and their registers.

SpiffWilkie
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I'm trying to determine how many bits are required in the memory buffer register and in the memory address register given certain memory systems.

For example, given the below system, how many bits are needed in MAR, and MBR if the memory is word addressable and how many bits if the memory is byte-addressable.

64K x 32


From what I understand the bits in the MBR are equal to the number of bits in the memory unit, or 32 for this example. I'm having a hard time grasping the MAR concept, though.
If I'm using word address memory, would it just be 16 bits (26 x 210 = 64K)? I guess, I'm asking if the unit size is irrelevant in that case.
If the memory is byte-addressable, would it be 18 bits? 216 x 22 (since each unit is 4 bytes?)

Thanks for any insight.
 
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SpiffWilkie said:
I'm trying to determine how many bits are required in the memory buffer register and in the memory address register given certain memory systems.

For example, given the below system, how many bits are needed in MAR, and MBR if the memory is word addressable and how many bits if the memory is byte-addressable.

64K x 32


From what I understand the bits in the MBR are equal to the number of bits in the memory unit, or 32 for this example. I'm having a hard time grasping the MAR concept, though.
If I'm using word address memory, would it just be 16 bits (26 x 210 = 64K)? I guess, I'm asking if the unit size is irrelevant in that case.
If the memory is byte-addressable, would it be 18 bits? 216 x 22 (since each unit is 4 bytes?)

Thanks for any insight.

All of your numbers look correct to me.
 

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