How does 8086 ΜP fetch 16 bits of data from 8 bits of memory location in one cycle?

The 8086 processor provides a 16 bit data bus. So It is capable of transferring 16 bits in one cycle but each memory location is only of a byte(8 bits), therefore we need two cycles to access 16 bits(8 bit each) from two different memory locations. The solution to this problem is Memory Banking.

What is the problem in accessing a word with an odd address compared to that with an even address?

When a word is read from an odd address, one byte is in one word, the other byte in another. You cannot address both bytes at the same time. Even if the memory had a separate address bus for each byte, the processor does not have the pins to address both address busses at the same time.

Why in 8086 External RAM is divided into even and odd banks justify with example?

8086 has 20-bit addressing model for memory access. Each address represents a single byte – however, the natural word size of 8086 is 2 bytes, so you need a way to read two bytes at the same time – hence, two banks.

How the odd and even memory banks got selected in 8086?

For the programmer, the 8086 memory address space is a sequence of one mega-byte in which one location stores an 8-bit binary code/data and two consecutive locations store 16-bit binary code/data. The even memory bank is selected by the address line Ao and the odd memory bank is selected by the control signal BHE.

Why is 8086 a 16-bit microprocessor?

Originally Answered: Why is the Intel 8086 CPU called a 16-bit CPU? The registers and memory word size of the 8086 are 16 bit – meaning it can operate on 16 bit values and address 65536 bytes of memory at a time.

Why is the size of a 16-bit data register?

The size of these registers is 16 bits because the memory addresses are 16 bits. They are: Program Counter This register is used to sequence the execution of the instructions. The function of the program counter is to point to the memory address from which the next byte is to be fetched.

How can even and odd addresses be achieved for memory organization in 8086?

Since each even and odd bank constitutes 2^19 address, only 19 bits can be used for addressing , a1 to 19 are used for addressing , and a0 is used for enabling / disabling the even bank , another signal, BHE is used for enabling / disabling odd bank. When signal is low a bank is enabled and high bank is disabled.

What does addressing problem mean?

To address a problem normally means to tackle a problem or to apply oneself to a problem (the first definition you gave). To deal with suggests to me that you solve the problem, which in my opinion goes too far.

Why is memory divided into different banks?

Cache memory is divided in banks to evade the effects of the bank cycle time (see above) [=> missing “bank cycle” definition, above]. When data is stored or retrieved consecutively each bank has enough time to recover before the next request for that bank arrives.

Why memory is divided into segments in 8086?

Segmentation is the process in which the main memory of the computer is divided into different segments and each segment has its own base address. It is basically used to enhance the speed of execution of the computer system, so that processor is able to fetch and execute the data from the memory easily and fast.

Why is 8086 a 16 bit microprocessor?

How many 16 bit registers are there in 8086?

The 8086 has eight more or less general 16-bit registers (including the stack pointer but excluding the instruction pointer, flag register and segment registers). Four of them, AX, BX, CX, DX, can also be accessed as twice as many 8-bit registers (see figure) while the other four, SI, DI, BP, SP, are 16-bit only.

How many bits can a 8086 processor transfer in one cycle?

The 8086 processor provides a 16 bit data bus. So It is capable of transferring 16 bits in one cycle but each memory location is only of a byte (8 bits), therefore we need two cycles to access 16 bits (8 bit each) from two different memory locations.

How many addresses does an 8086 microprocessor have?

The 8086 has 20-bit address bus, so it can address 2^20 or 1,048,576 addresses. Each address represents a stored byte. To make it possible to read or write a word with one machine cycle, the memory for an 8086 is set up in to 2 banks of up to 524,288 bytes each.

What is the read bus timing of an 8088 microprocessor?

15. The Intel 8088 microprocessor has a read bus timing similar to that of Figure 3.18, but requires four processor clock cycles. The valid data is on the bus for an amount of time that extends into the fourth processor clock cycle. Assume a processor clock rate of 8 MHz.

How to get 20-bit physical address from 8086 address line?

It cannot get the 20-bit Physical adress using the 8086 Address Line or 16-bit Segment Registers alone. In order to access memory location, you cannot pass 20-bit address directly to the processor.