Max Memory Space Addressed by 18 Address Bits

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Discussion Overview

The discussion revolves around the maximum memory space that can be addressed using 18 address bits, exploring the relationship between address bits and memory capacity. Participants examine the mathematical implications and practical considerations of memory addressing in computing systems.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant inquires about the formula for determining the largest memory space addressable by 18 bits.
  • Another participant calculates the number of combinations of 18 bits as 2^18, resulting in 262,144.
  • A subsequent post clarifies that if each memory location holds one byte, then the total addressable space is indeed 262,144 bytes.
  • One participant notes that while 2^18 addresses are possible, not all chips utilize all available addresses, citing specific examples like the PIC16F8X range.
  • A participant references a Digital Circuits book stating that 18 address leads correspond to 256K, questioning the relationship to the previously calculated 262,144.
  • Another participant explains that 256K is equivalent to 256 * 1,024, clarifying that "k" in this context refers to 1,024 rather than 1,000.

Areas of Agreement / Disagreement

Participants generally agree on the calculation of addressable memory space as 262,144 bytes, but there is some contention regarding the practical application of this addressing in specific chip architectures, as well as the interpretation of "256K."

Contextual Notes

There are assumptions regarding the size of memory locations and the practical limitations of certain hardware that are not fully explored, leading to potential discrepancies in understanding the total addressable memory space.

needhelp83
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The largest memory space that can be addressed by 18 addressing bits is...


Is there a formula for this perhaps?
 
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How many different combinations of 18 ones and zeros are there?

- Warren
 
2^n
n=18

2^18= 262,144
 
Would that be 262,144 bytes then
 
Assuming that each "memory location" holds one byte, yes.
 
Thanks for the help?
 
Something to note - while you might be able to have addresses going all the way up to 2^18 a lot of chips do not have memory locations for all of the addresses. Such as variants in the PIC16F8X range.
 
I noticed in a Digital Circuits book that with 18 address leads it should be 256K. So where does my 262,144 come from with 2^18. Just a thought?
 
256 * 1,024 = 262,144. The "k" in this context means 1,024, not 1,000.

- Warren
 

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