Low order means the low end, and high order means the high end. Think of a decimal constant - lets pick 1234 - 12 would be considered the high order and 34 would be considered the low order.
In the 8085, the address bus is A15 (high order bit) to A0 (low order bit). Since the processor is byte organized, you can say this as A15-A8 is the high order byte and A7-A0 is the low order byte.
The low order address bus on the 8085 is the strobed contents of the data bus during ALE time. AD7-AD0 --> ALE --> A7-A0. Setup and hold time is such that the data bus is valid for address at the trailing edge of ALE.
The 8086 has a 20 bit address bus and a 16 bit data bus. The low order 16 bits of the address bus share the same 16 pins as the data bus. The low order 16 bits of the address are emitted in the first clock cycle of a memory access cycle. External logic is expected to latch that address. Then the bus becomes a data bus. The high order 4 bits of the address bus are handled separately.The determination of operand size (8 bit vs 16 bit) is made by BHE and A0. If BHE is high, it is a 16 bit operand at an even address. If BHE is low and A0 is low, it is an 8 bit operand at an even address. If BHE is low and A0 is high, it is an 8 bit operand at an odd address.
The control and high order address buses are unidirectional.
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In order to demultiplex the address and data bus, you provide latches that sample the multiplexed bus. At ALE=true, they follow the bus. At ALE=falling edge, they lock onto the last value of the bus. The latches will then become the address bus, while the original bus becomes the data bus.
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.
If this is a homework assignment, please consider trying it yourself first, otherwise the value of the reinforcement to the lesson offered by the homework will be lost on you.The low order address bus and the data bus on an 8085 are separated using an 8-bit latch that is strobed using ALE during the first T-cycle of a bus cycle. The bus represents valid address information on the falling edge of ALE. Soon thereafter, somewhat less than one half clock cycle, the bus changes meaning to become the data bus, and will float for a read or become output data for a write.The high order address bus is already separated, and does not need to be latched.
In computing, a physical address, also real address or binary address, is the memory address that is electronically presented on a computer address bus circuitry in order to enable the data bus to access a particular storage cell of main memory.
The address bus is a section of the bus that emits the address of the desired instruction or operand.
There is no such thing as a CPU Drive. If you meant "What does a CPU communicate with", then the answer is everything inside your computer.
Yes, the address bus is uni-directional.
The address bus in the 8085 is 16 bits wide.
The data and address buses are multiplexed in order to save pin count on the chip. In the first clock cycle of a read or write cycle, the address is emitted on the address/data bus. The ALE signal is used to strobe the address, after which the address/data bus becomes the data bus. External logic is expected to strobe the address at the trailing edge of ALE. ALE is generated directly by the 8085, and by the 8086/8088 in minimum mode. In maximum mode in the 8086/8088, ALE is generated by the 8288 Bus Controller.