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If the microprocessor bus has a 32-bit local address bus and a 16-bit local data bus what is the impact on the system speed?
Wiki User
∙ 14y ago
•If the local address bus is 32 bits, the whole address can be transferred at once and decoded in memory. However, because the data bus is only 16 bits, it will require 2 cycles to fetch a 32-bit instruction or operand.
Wiki User
∙ 14y ago
Wiki User
∙ 11y agoit will require two steps to decode address and it will also require two steps to fetch the data (assumption that processor is 32bit).
Wiki User
∙ 13y agoI assume your question is begin with "in a 32 bit microprocessor", then these 16 bit bus will slow down the whole process by their limited address.
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What type of microprocessor is 8086?
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If you assume that it has a 16-bit data bus, then it would be 128k so the microprocessor can access 2^16 points, which is 64k (from it being a 16bit address) 16bits = 2 bytes (memory) so through a 16 bit memory, it can access 2*64k, which is 128k alternatively, if its 8bit memory, 8bits=1byte 1*64k = 64k I'm no expert, and i was searching for the answer myself, hope this helped
Consider two microprocessors having 8 and 16bit wide external data buses respectivelyThe two processors are identical other wise and their bus cycles take just as longsuppose all instructions?
During a single bus cycle, the 8-bit microprocessor transfers one byte while the 16-bit microprocessor transfers two bytes. The 16-bit microprocessor has twice the data transfer rate.
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An instruction is a command to the microprocessor to perform a given task on specified data. Each instruction has two parts: One is task to be performed,called the operation code (opcode). Second is the data to be operated on, called the operand. It can be specified in various ways,it may include 8bit/16bit data, an internal register, a memory location , or 8bit/16bit address. In some instructions, the operand is implicit. The 8085 instruction set is classified into three groups according to Word size. They are- 1. One word / 1 byte instructions 2. Ttwo word / 2byte instructions 3. Three word / 3byte instructions
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Standard VGA is 16bit.
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In microprocessors, from 8086 there is 20bit address bus. This address bus is so organised that reading a word(16bit string from 8086 to 80286 and 32bit string from 80386 to PIV) starting at even address is different from reading a word starting from odd address. suppose a word starts at 20000h address so it will placed in 20000h and 20001h respectively,then, this word is read by the microprocessor in 1 clock cycle i.e. the byte at 20000h and byte at 20001h are simultaneously read. Hence the word is read in 1 clock cycle. but had a word started at 20001h-- 1 byte in 20001h and the other byte in 20002h then the microprocessor is unable to read the word in 1 clock cycle. It takes 1 clock cycle to read the word from 20001h and another clock cycle to read the word from 20002h. Hence a word stored a odd location slows down the process of reading it. this entire process is referred as Even banking and Odd banking.
Which processor had a 32-bit internal bus and a 16-bit external bus?
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The Java compiler itself (javac) is a 32-bit application.
One memory technology uses only a 16bit path what acronym is used for this type of memory?
Rimm
Why are P8 and P9 connectors keyed a certain way?
so they fit correctly into an 8bit or 16bit slot
