Demultiplexing the AD0 to AD7 address lines is necessary to separate the address and data signals in microprocessor systems, particularly in those where a single set of lines is used for both functions. This allows for more efficient use of the bus, as the same lines can carry address information during one phase of operation and data during another. By demultiplexing, the system can clearly distinguish between when it needs to read or write data versus when it is accessing specific memory addresses, ensuring accurate communication and reducing the risk of errors. Additionally, it helps in simplifying the design of the bus architecture and enhances overall system performance.
As AD7-AD0 lines serve a dual purpose they have to be demultiplexed to get all the information. The address's high order bits remain on the bus for 3 clock periods. ... An external latch is used to save the value of AD7-AD0 when it is carrying the address bits so that the entire address remains for the 3 clock cycles.
The AD0-AD7 lines in an 8085 are multiplexed to reduce the pin count of the IC. Several added features were added to the 8085 from the 8080 design, and Intel did not want to require a larger package.
The address lines A0..A15 are multiplexed with the data lines D0..D15 on the pins AD0..AD15
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.
Demultiplexing the bus AD7-AD0 The Intel 8085 is an 8-bit microprocessor. Its data bus is 8-bit wide and hence, 8bits of data can be transmitted in parallel form or to the microprocessor. The Intel8085 requires a 16-bit wide address bus as the memory addresses are of 16 bits. The 8 most significant bits of the address are transmitted by the address bus(A8-A15). The 8 least significant bits of the address are transmitted byaddress/data bus (AD7-AD0). The address/data bus transmits data and addressinformation at different times. This is the basic need for demultiplexing the busAD7-AD0.
The AD0-AD7 lines are multiplexed to optimize the use of limited I/O pins in microcontrollers and microprocessors. By combining address and data functions on the same lines, the system can reduce the number of physical connections needed, allowing for a more compact design. This multiplexing requires additional control signals to differentiate between address and data phases, but it ultimately enhances efficiency in data transfer and system performance.
There are eight datalines, D0 through D7, in the 8085 microprocessor. They are shared, or multiplexed with the eight low order address lines, A0 through A7, and are called AD0 through AD7 on the pinout drawing.
Basically , Demultiplexing is breaking of multiplexed signal .Recall that A/D0 -A/D15 and A16/S3-A19/S6 are the multiplexed signals in 8086.To do so, three demultiplexing latches are used .ALE (Address Enable Latch) is used for strobe Demultiplexing.8086 is 16bit data lines and 20 bit address line microprocessor.BY the Demultiplexing ,we Get A0-A19 separate Address lines and D0-D15 Data lines . Ajmal Shahbaz
Address decoding in the Intel 8085 microprocessor is done by latching the 8 bits of the AD0-AD7 bus during the ALE pulse, holding on the falling edge of ALE. After ALE, the latched results become A0-A7, and the AD0-AD7 bus becomes D0-D7.
ALE is a signal that means that the data bus contains the lower order address bus values. External hardware should strobe the data bus during ALE time, and lock it on the falling edge of ALE.
An instruction fetch in the 8085 is similar to an operand fetch... During T1, ALE pulses high for one half cycle. On the falling edge, external logic is expected to strobe the AD0-AD7 lines to form the A0-A7 lines. A8-A15, IO/M-, S0, and S1 are also presented, but they stay valid after ALE. S0 is high for opcode fetch, and low for operand fetch. RD- goes true (low) at the end of T1. If READY is false at the end of T1, TWAIT is entered, and all lines are persisted, with TWAIT repeated as necessary until READY is true. At the end of T2, the CPU strobes the data presented on AD0-AD7 by external logic. At the midpoint of T3, RD- goes false (high) and the external logic must stop driving AD0-AD7. T4 is used to decode and process the opcode. External logic does nothing, since there is no ALE. If the opcode requires extra data, such as immediate data or an address, T1, TWAIT, T2, and T3 are repeated to fetch the additional bytes, although S0 is low during these cycles.
Differentiate Between 8085 and 8086Solution:SN808580861It is a 8-bit micro processorIt is a 16-bit processor2It contains 16-bit address bus and 8-bit data busIt contains 20-bit address bus and 16-bit data bus..3It doesn't have memory segmentation feature8086 has a special concept called as memory segmentation. It allows parallel processing4There is no overflow flag8086 their exists a overflow flag along with condition code flags5In 8085 the clock speed is 3MHZwhere as in 8086 the clock speed is 5MHZ.6In 8085 minimum maximum mode is not present8086 is designed to operate in two modes, Minimum and Maximum.7The first 8 lines of address bus and 8 lines of data bus are multiplexed AD0- AD7It has multiplexed address and data bus AD0- AD15 and A16 - A19.8Lower SpeedHigher Throughput (Speed) (This is achieved by a concept called pipelining).