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What is ALE in 8085 in microprocessor?
ALE=Address Latch Enabled.(pin number 30 in 8085)8085 has a special pin referred as ALE, which indicates whether multiplex bus functions as an address bus or a data bus. Whenever 8085 starts any new operation, ALE signal goes to logic 1 for about 1/2 clock cycle, at about the falling edge of CLK. If ALE=1 then multiplex bus functions as address bus. After that half clock cycle, it goes to logic 0 for nearly 3 or 4 clock cycles. If ALE=0 then multiplex bus acts as a data bus.The ALE pin helps to enable the latching of lower order ADDR bus. The AD0-AD7 pins, as well as other control pins such as S0, S1, IO/M-, and the other address pins A8-A15, are setup to be correct at the falling edge of ALE.
How do you draw timing diagram for 8085 microprocessor CMP instruction?
how to draw timing diagram?discuss the various stepsYou first need to understand the machine cycles of 8085The status signals are as followsIO/M(bar) :--- 1 IO 0 MemoryS1 | S0 | Process-----------------------------------------------------------0 | 0 | Halt0 | 1 | Write1 | 0 | Read1 | 1 | Opcode fetch1)Opcode fetch ( Compulsory Machine cycle)This cycle requires 4 T-states.1st T state ALE is high and lower byte of address from PC(Program Counter) is placed on the multiplexed data/address bus.In the second T-state, after checking the status of READY pin, RD(bar) goes low the opcode is placed on the data bus, This state continues in the 3rd T-State.The fourth T-state is used by the uP to decode the instruction and to generate the relevant control signals. The state of the address bus is unspecified( This T-state is used by some DMA controllers to transfer data in hidden/transperant mode)IO/M_ = 0 S1=1 S0=12)Memory read(for 1 byte)Three T states, similar to the first 3 T states of opcode fetch( as first 3 states of opcode fetch is effectively memory read)IO/M_ 0 S1 = 1 S0 = 03) Memory Write(for 1 byte)Similar to Write but instead of RD bar WR bar is used. Also the data stays on the bus a little longer than READ*.IO/M_ 0 S1 = 0 S0 = 14) & 5) IO write and readSimlar to the above two, only IO/M_ = 1These are the basic machine cycles you will require to draw timing diagrams for most instructions. There are additional cycles such as INTA bar and Bus idle. If anyone requires diagrams for these cycles, message me and i will explain them later.Also some instructions like CALL require 6 T-state Opcode fetch. For this you can draw the 4 T state Opcode fetch but 4th T state extended to the fifth and sixth T state.------------------------------------------------------------------------------------------Now, to draw the timing diagram for any instruction you need to understand what exactly the instruction does. I will explain a few. If you need a specific instruction, msg me.A) MOV A,BDraw only opcode fetch as no further memory acces is required as operands specified in registers onlyB) MVI A,32HDraw opcode fetch and memory read as operand(1 byte) has to be fetched from memoryC) LXI H, 2000HDraw Opcode Fetch and two memory Reads as two bytes, 00H and 20H, (lower byte fetched first) have to be read from memory.D) STA 2000HThis instruction stores the value of accumulator(8 bit) at the location specified.Opcode fetch + Memory read * 2 (byte address) + Memory write * 1(1 byte)i.e 13 T-states 4+3+3+3During the memory write the address bus contains the address fetched by the memory read cycle earlierE) CALL addresss(can be specifed in terms of a label)During a call instruction the uP pushes the current value of program counter(16 bit ie 2 byte) to the stack and then copies the new value from the memory(specified in the instruction)6 T state Opcode fetch+ Memory write * 2 (PC pushed to stack)+ Memory read * 2 (New value of PC fetched from memory)ie 6 + 3 + 3 + 3 + 3 = 18 T-statesNote that during the memory write cycle the address bus contains the address of the top of the stack(Stack Pointer)F)JMP 16-bit address3 Cycles as Follows4 T-State Opcode Fetch+ 2 * Memory Read ( 16 bit = 2 bytes)ie 4 + 3 + 3 = 10 T-states.Note that separate cycle is not required for loading the address into the PC as PC is a register.
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.
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.
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 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.
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 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.
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.
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.
State the functions for ALE and TRAP pins in 8085?
In an 8085, the ALE pin pulses high for about one half clock cycle at the beginning of each machine cycle, approximately near the falling edge of CLK. At the falling edge of ALE, external hardware is expected to strobe the values of AD0-AD7 and hold them as the low order address values A0-A7. Other lines, such as IO/M-, S0, S1, and A8-A15 are also guaranteed valid at the falling edge of ALE, but they do not need to be strobed as they are not multiplexed. (All these lines do change during ALE, at about the 1/3 point depending on clock frequency, so they are considered invalid until the falling edge of ALE.) Following ALE, the AD0-AD7 bus lines become the data bus, D0-D7. This multiplexed scheme saves 8 pins on the chip design. In an 8085, the TRAP pin is a non maskable interrupt with highest priority. It must go high and stay high to be recognized, and it will not be recognized again until it goes low and then high. (Edge and level triggered.)The recognition point is on the falling edge of CLK, one clock cycle before the ALE that follows the last machine cycle of the instruction. When recognized, an internal RST instruction ocurrs, with a vector address of 0024H.
What is ALE in 8085 in microprocessor?
ALE=Address Latch Enabled.(pin number 30 in 8085)8085 has a special pin referred as ALE, which indicates whether multiplex bus functions as an address bus or a data bus. Whenever 8085 starts any new operation, ALE signal goes to logic 1 for about 1/2 clock cycle, at about the falling edge of CLK. If ALE=1 then multiplex bus functions as address bus. After that half clock cycle, it goes to logic 0 for nearly 3 or 4 clock cycles. If ALE=0 then multiplex bus acts as a data bus.The ALE pin helps to enable the latching of lower order ADDR bus. The AD0-AD7 pins, as well as other control pins such as S0, S1, IO/M-, and the other address pins A8-A15, are setup to be correct at the falling edge of ALE.
What is meant by low order and high order address bus?
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.
