Question

Suppose a microprocessor has a 16-bit address bus and a 16-bit data bus, and the addressable unit is a byte.

A. If the memory of this microprocessor is 16-bit memory (memory that can input and output data at a time of 16 bits), what is the maximum amount of memory space that the processor can directly access?
B. If the memory of this microprocessor is 8-bit memory (memory that can input and output data 8 bits at a time), what is the maximum amount of memory space that the processor can directly access?
C. If this processor has separate I/O space, what features should it have?
D. If the number of I/O ports is given in 8 bits, how many 8-bit I/O ports (I/O ports that can input and output 8-bit data at one time) can you have?
How many 16-bit I/O ports (I/O ports that can input and output 16-bit data at a time) can you have?

Answer 1

• The instruction consists of opcode and operands. Given the instruction set of size 12, 4 bits are required for opcode (2^4 = 16).
  
• As there are total 64 registers, 6 bits are required for identifying a register.
  
• As the instruction contains 3 registers (2 source + 1 designation), 3 * 6 = 18 bit are required for register identifiers.
  
• 12 bits are required for immediate value as given.
  
• Total bits for an instruction = 4 + 18 + 12 = 34 bits
  
• The instructions are required to be stored in a byte-aligned fashion. The nearest byte boundary after 34 bits is at 40 bits (5 bytes).
  
• Hence, for 100 instructions, the memory required is 5 * 100 = 500 bytes
• The instruction bus would be generally as wide as the instructions. Or smaller, as the IC (Integrated Circuit or processor) will then wait for the next instruction and combine two to form the bigger instruction. Thus, you can send a 16-bit instruction over an 8-bit bus.

• 65,536 bytes

  16 address bits, 16 address pins
  
  The memory on these CPUs is addressable at the byte level. This leads to a memory addressable limit of 2¹⁶ × 1 byte = 65,536 bytes or 64 kilobytes.

• A 16-bit integer can store 2¹⁶ (or 65,536) distinct values. In an unsigned representation, these values are the integers between 0 and 65,535; using two's complement, possible values range from −32,768 to 32,767. Hence, a processor with 16-bit memory addresses can directly access 64 KB of byte-addressable memory.

• Each 1 or 0 in a binary number iscalled a bit. From there, a group of 4bits is called a nibble, and 8-bitsmakes a byte. ... It could be 16-bits, 32, 64, or even more.

• If n=2, you can address 2 locations (0, 1, 2, and 3). As you can see, number of addressable locations = n^2. This means that n=log(1024) to the base 2. Thus, n=10.

• For example, an 8-bit-byte-addressable machine with a 20-bit address bus (e.g. Intel 8086) can address 2²⁰ (1,048,576) memory locations, or one MiB of memory, while a 32-bit bus (e.g. Intel 80386) addresses 2³² (4,294,967,296) locations, or a 4 GiB address space.