the bus through which the data are transmitted or received within microprocessor and with peripherals is called as data bus.when used internally to microprocessor called internal data 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.
8 bit
8
Because its a microcontroller with an 8 bit data bus width.
No, there were no 4-bit consoles. The 2600 uses the 8-bit 6507 CPU and an 8-bit bus.
based on the size of the data bus they determine whether it is a 8 bit or 16 bit . here in 8086 it has 16 bit data bus hence it is called as 16 bit microprocessor
Absolutely not. A 2MHz 8-bit bus can transfer 2MB per second; a 4MHz 8-bit bus can transfer 4MB per second.
a. 8 bit b. 16 bit c.32 bit d.64 bit
a. 8 bit b. 16 bit c.32 bit d.64 bit
The difference between the 8086 and the 8088 is that the 8086 has a 16 bit data bus and that the 8088 has an 8 bit data bus. Both processors are the same 16 bit processor, and both have a 20 bit address bus. The 8086 is twice as fast as the 8088 in terms of data transfer rate on the bus for the same bus clock speed.
Assuming the raw bus speed is the same, you should be able to move 4 times as much data across a 32-bit bus as across an 8-bit bus, because there are 4 times as many bits. It's never that simple, though. The 8-bit bus on the original IBM PC was a 1.3MHz bus, and so could move about 1,300,000 bytes per second, flat out. The 32-bit bus specified by the VESA consortium for the 486 was a 33MHz bus and so could move 133,000,000 bytes per second, flat out. The multiplier in this case as closer to 100x.
Not necessarily. In the 8085, for instance, this is true. In the 8088, however, the processor is a 16 bit processor with an 8 bit data bus. The same is true for the 80386sx - it is a 32 bit processor on a 16 bit bus.