How do you use the instruction register and manage memory in a computer system?

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vead
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Hello ,
I need help , rom memory is memory that store program. we can only read rom memory.

https://www.physicsforums.com/attachments/https://www.physicsforums.com/attachments/77060
look this diagram why does we connect both memory. I have read that when we start to read the program memory. we get data from program memory and store into instruction register. In my first diagram.

https://www.physicsforums.com/attachments/https://www.physicsforums.com/attachments/77061

how and where to use Instruction register ?
 
Attachments
  • Schematic of a PROGRAM MEMORY chip with address pins A0 to A15 on the left, data pins D0 to D7 on the right, and control pins labelled ROM select, ROM read, and Vcc on the lower left.
    PC2.png
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  • Schematic of an EPROM chip labelled 27256 beside a static RAM chip labelled 62256, with address lines A0 to A14 and data lines D0 to D7 wired in parallel to both, plus CE, OE, VPP and WE control pins.
    Untitled.jpg
    37.5 KB · Views: 494
vead said:
Hello ,
I need help , rom memory is memory that store program. we can only read rom memory.

https://www.physicsforums.com/attachments/https://www.physicsforums.com/attachments/77060
look this diagram why does we connect both memory. I have read that when we start to read the program memory. we get data from program memory and store into instruction register. In my first diagram.

https://www.physicsforums.com/attachments/https://www.physicsforums.com/attachments/77061

how and where to use Instruction register ?

You have different CE~ and OE~ signals for the EPROM and RAM memory chips. They are not both enabled at the same time. You would start with the program reading from ROM address 0x0000 or 0x0001 (depending on the processor architecture), and executing the instructions that you read from there. As the program progresses, it can use part of the RAM memory chip. You would typically have a "memory map" that defines what part of that 64KB ROM and what part of the 64KB RAM you access. You would not generally use all of either chip, since they generally share the same "memory map". Make sense?
 
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