Multiplexed bus transfers different information at different well defined times through the same line. It advantages include, reduce physical/manufacturing complexity since fewer pins are used, and it prevent multiple device for competing for the same bus at the same time thus enhancing efficiency.
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The question is unclear, but it looks like it meant to ask when the first microprocessor was invented. That was the Intel 4004, invented in 1971. It was the first single chip microprocessor, and the first microprocessor available commercially. The 4004 was a 4 bit processor, with a 12 bit address bus and an 8 bit opcode. Everything was multiplexed, 4 bits at a time, in 8 clock cycles per instruction.
hi, as per my knowledge it has got CAN bus wiring System,
Multiplexing of the data and address buses is done to reduce the pin count on the microprocessor chip. The address information is emitted at the beginning of a memory cycle, and external logic is expected to latch that address. Then the bus becomes the data bus and the required data is transferred to or from memory using the latched address.In the 8085, this saves 8 pins at the cost of 1 pin, ALE. In the 8086/8088, this saves 16 pins at the cost of 1 pin, ALE. In some architectures or modes, there is no ALE, but the external logic is still required to know when to latch the address based on some other criteria.As an example of that, in the original Intel 4004, the microprocessor's bus was 4 bits, while the address bus was 12 bits. There were 8 clock cycles. In the first three, external logic was expected to latch the three 4 bit parts of the 12 bit address. Similarly, in the next two, the resultant opcode, which as 8 bits, was multiplexed by the external logic into two 4 bit parts. (The 4004 was only a 16 pin chip, but it packed a lot of complexity in its day, being the world's first microprocessor.)
The address bus is a section of the bus that emits the address of the desired instruction or operand.
The higher order address bus is not multiplexed with data bus of 8085 because that is the way Intel designed the processor. Besides, the data bus is only 8 bits and the address bus is 16 bits. If you were to multiplex the whole address bus on the data bus, you would need two T1 (ALE) states, and that would be excess logic. Back to the original answer - that is simply the way Intel designed the processor.
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.
ad is multiplex address data line bus
The 8085 was the next generation of the 8080, providing operation on a single +5V power supply, a multiplexed address/data bus, integration of the system controller and clock generator, new automatically vectored interrupts, a few 16 bit instructions, and serial I/O.
since data can be read /write from/to the microprocessor, hence data bus is bidirectional. if data is required read from microprocessor then it will be pointing to a memory location by the address bus, by indicating which location data its required to read. similarly to write a data to a location, again the microprocessor will be to that particular location by holding that address in address bus. hence it will be unidirectional.
The 8085 microprocessor is used IC 74LS373 to latch the address of 8085. Basically latch is consists of 8 flip flops. Generally we used D-flip flops (Delay).The clock of these flip flops are connected together and available as a output pin called enable.Working : The address will appear on AD0 AD7 lines. The ALE will go high and forcingEnable = 1. This will make latch enable and ready to work. Before address disappears ALE = 0. This will make latch disable. AD0 - AD7 will now be used as data bus.Hence, AD0 - AD7 (low order) address bus of the 8085 microprocessor is multiplexed (time-shared) with the data bus. The buses need to be demultiplexed.
Bus cycle is a single transaction between the main memory and the CPU.