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Message: <blockquote data-quote="Lalakajee" data-source="post: 9426104" data-attributes="member: 36636">512K means there are 512*1024 = 524288 (2^19) memory locations in memory unit, each 4 bytes (32 bits).Therefore databus width is equal to 32 bits.Address bus width 19bits (Note that the power of 2 is here!!<img src="/styles/default/xenforo/smilies/default/yes.gif"  class="smilie" loading="lazy" alt=":yes:" title="Yes    :yes:" data-shortname=":yes:" />)In other words we can say that register width 32 bits, memory address field width 19 bits and register adddress field width 7 bits. (120 approximately 2^7)Now we have,Total register width (IR) = 32 bitsAddress field width = 19bitsregister address field width?? = 32-(19+7)=6bitsVerify what we did<img src="/styles/default/xenforo/smilies/default/nerd.gif"  class="smilie" loading="lazy" alt=":nerd:" title="Nerd    :nerd:" data-shortname=":nerd:" />Address bits = 9 ---> 2^19 = 524288 (can access all memory locations)Register address bits = 7 ---> 2^7 =128 (can access all 120 register + additional 8)opcode bits ---> 6 2^6=64 (can define different 64 instrunctions for this CPU.)</blockquote>

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