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Size of 8086 address bus?

The 8086/8088 has an internal 20-bit address bus and 16-bit data bus. Externally, the address bus is 20-bits, and the data bus is 16-bits for the 8086 and 8-bits for the 8088.The data bus in the 8086 is 16 bits in size, while the address bus is 20.


How do you measure data bus in width?

The width of a data bus is measured in bits, indicating how many bits of data can be transmitted simultaneously. For example, a 32-bit data bus can carry 32 bits of data in parallel at once, while a 64-bit data bus can carry 64 bits. This measurement is crucial as it directly affects the data transfer rate and overall performance of a computer system. In essence, a wider data bus allows for more data to be processed concurrently, enhancing throughput.


How many bits is the PSP?

Technically, it's 256-bit. That's a rumor, but whatever.


How many times can a 16 bit bus can transfer ata time?

16-bit bus means the bus has 16 parallel wires for data transfer. That allows it to transfer 16 bits at a time.


Size of data bus in 8086?

It is my understanding that the 80286 has a 16-bit data bus. This was a doubling of the original 8086/8088 processors.


How wide can the external data bus be?

The external data bus can be as wide as desired, given the necessary compromises between performance, complexity, and cost. The wider the bus, the faster the theoretical aggregate data transfer rate. In the 8085 and 8088, the external data bus is 8 bits wide; in the 8086, it is 16 bits wide; in the 80386, it is 32 bits wide; and in the modern incarnations of 64 bit processors, it is 64 bits wide.


In a Pentium the front side data bus is how many bits wide?

For the Pentium the front side data bus is 64 bits wide. The back side is 32 bits wide.


How many bits wide is the AGP bus?

32 bit


The width of a data bus is called?

The width of a data bus is referred to as the data path size. An example would be a 16 bit bus can transmit 16 bits of information


How many bits are there in data and address inputs of memory?

The number of bits in data and address inputs of memory varies depending on the architecture of the system. Commonly, modern systems use 32-bit or 64-bit architectures, meaning they can handle 32 or 64 bits of data at a time, respectively. The address input, which determines how much memory can be directly accessed, is often the same as the data bus width; for instance, a 64-bit architecture typically has a 64-bit address bus, allowing access to a larger memory space.


Consider a hypothetical 32-bit microprocessor having 32-bit instructions composed of two fields the first byte contains the opcode and the remainder the immediate operand or an operand address What?

• What is the maximum directly addressable memory capacity (in bytes)? 2(32-8) = 16,777,216 bytes = 16 MB • Discuss the impact on the system speed if the microprocessor data bus has • a 32-bit local address bus and a 16-bit local data bus. Instruction and data transfers would take three bus cycles each - one for the address and two for the data. • a 16-bit local address bus and a 16-bit local data bus. Instruction and data transfers would take four bus cycles each - two for the address and two for the data. • How many bits are needed for the program counter and the instruction register? 24 bits for the PC (24-bit addresses), 32 bits for the IR (32-bit addresses)


How many address bus and deta bus in 8085?

The 8085 microprocessor has a 16-bit address bus and an 8-bit data bus. This means it can address up to 2^16 (or 65,536) memory locations, while it can transfer 8 bits of data at a time. The combination of these buses allows the 8085 to efficiently access and process data from memory.