Why it is necessary to demultiplex AD0 to AD7 address lines?

How ale signal demultiplex ad0 ad7 bus?

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.

Explain the need to demultiplex the bus AD7-AD0 in 8085 microprocessor?

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.

How many multiplexed lines in 8086?

The address lines A0..A15 are multiplexed with the data lines D0..D15 on the pins AD0..AD15

Draw a schemetic to demultiplex bus ad0-ad7 using any octal latch in 8085 microprocessor?

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.

How does ALE signal demultiplex explain with diagram?

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.

How many data lines in 8086?

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.

Why address and data lines are demultiplexed in 8085?

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

How address decoding is done in Intel 8085 microprocessor?

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.

What is ALE in 8086 in microprocessor?

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.

How is an instruction fetched from memory into CPU in the 8085 microprocessor?

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.

Difference between 8085 and 8086?

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).

How do you draw timing diagram of jnz in 8085?

If this is a homework assignment, you really should consider doing it yourself The MVI instruction in the 8085 microprocessor contains 7 or 10 T-Cycles, each one clock cycle, not including wait states. Each cycle starts on the falling edge of CLK. <> <> <> T1a - ALE goes high for one half clock. During this time, S0, S1, IO/M-, A15-A8, and AD7-AD0 become valid, and are guaranteed valid at the falling edge of ALE. (AD7-AD0 represent A7-A0, and must be strobed by external hardware.) A15-A0 will be the address of the MVI instruction. Somewhat after ALE, AD7-AD0 will float. T2a - RD- goes low for one clock cycle. While RD- is low, the external hardware has permission to drive AD7-AD0. It must supply the opcode for MVI. READY is sampled at the beginning of T2 - If it is low, T2 will be repeated, until READY is sampled high. T3a - RD- remains low for one more half clock cycle. The external hardware must guarantee AD7-AD0 valid by the beginning of T3a. The 8085 samples AD7-AD0 at the beginning of T3a. This will give it the MVI opcode. T4a - Nothing happens externally. All lines persist their prior state. The 8085 processes the MVI opcode and sets itself up for the required actions. <> <> <> T1b - This is the same timing as T1a, except that the address is one greater. T2b - This is the same timing as T2a. During this time, the external hardware must drive the immediate value of the MVI instruction onto AD7-AD0. T3b - This is the same timing as T3a. At the conclusion of T3b the 8085 knows the value to store in the destination. If the destination was an internal register, the instruction is complete. If the destination was M, the cycles continue. <> <> <> T1c - This is the same timing as T1a, except that the address is the contents of the HL register, H sent on A15-A8, and L sent on AD7-AD0. T2c - This is the same timing as T1a, except that WR- is used instead of RD-, and the AD7-AD0 lines do not float - they emit the immediate value retrieved in T3b. The AD7-AD0 line will change sometime between ALE and WR-. T3c - This is the same timing as T3a, except that WR- goes high at the beginning instead of at the halfway point. The external hardware is expected to save the AD7-AD0 lines into the address specified during T1c on the rising edge of WR-. The 8085 will persist the AD7-AD0 lines for one half clock cycle to guarantee the AD7-AD0 lines.