Counter And Time Delay In 8085 Microprocessor Pdf
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Programs of 8085 on counters and delay.pdf FREE PDF DOWNLOAD NOW!!! Learn more Info for Support. What are counters and time delays in 8085 microprocessor.
1.A counter is designed simply by loading an appropriate number into one of the registers and using INR(increment by 1) & DCR(decrement by 1) instructions. 2.A loop is established to update the count,and each count is checked to determine whether it has reached the final number or not.if not then the loop is again repeated. 3.These counters have 1 drawback.i.e.counting is performed at such high speed that only the last count can be observed.to observe counting there must be a proper time delay between counts. 1.A counter is designed simply by loading an appropriate number into one of the registers and using INR(increment by 1) & DCR(decrement by 1) instructions. 2.A loop is established to update the count,and each count is checked to determine whether it has reached the final number or not.if not then the loop is again repeated. 3.These counters have 1 drawback.i.e.counting is performed at such high speed that only the last count can be observed.to observe counting there must be a proper time delay between counts.
A minimum mode of 8086 configuration depicts a stand alone system of computer where no other processor is connected. This is similar to 8085 block diagram with the following d ifference. The Data transceiver block which helps the signals traveling a longer distance to get boosted up. Two control signals data transmit/ receive are connected to the direction input of transceiver (Transmitter/Receiver) and DEN.
signal works as enable for this block. This is the same as Read cycle Timing Diagram except that the DT/R. line goes high indicating it is a Data Transmission operation for the processor to memory / peripheral. Again DEN. line goes low to validate data and WR.
line goes low, indicating a Write operation In the maximum mode of operation of 8086, wherein either a numeric coprocessor of the type 8087 or another processor is interfaced with 8086. The Memory, Address Bus, Data Buses are shared resources between the two processors. The control signals for Maximum mode of operation are generated by the Bus Controller chip 8788.
The three status outputs S0., S1., S2. from the processor are input to 8788. The outputs of the bus controller are the Control Signals, namely DEN, DT/R., IORC., IOWTC., MWTC., MRDC., ALE etc. These control signals perform the same task as the minimum mode operation. However the DEN is an active HIGH signal which has to be converted to active LOW by means of an inverter. How to draw timing diagram?discuss the various steps You first need to understand the machine cycles of 8085 The status signals are as follows IO/M(bar):- 1 IO 0 Memor y S1 S0 Process - 0 0 Halt 0 1 Write 1 0 Read 1 1 Opcode fetch 1)Opcode fetch ( Compulsory Machine cycle) This cycle requires 4 T-states.
Counter And Time Delay In 8085 Microprocessor
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=1 2)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 = 0 3) 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 = 1 4) & 5) IO write and read Simlar to the above two, only IO/M = 1 These 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,B Draw only opcode fetch as no further memory acces is required as operands specified in registers only B) MVI A,32H Draw opcode fetch and memory read as operand(1 byte) has to be fetched from memory C) LXI H, 2000H Draw Opcode Fetch and two memory Reads as two bytes, 00H and 20H, (lower byte fetched first) have to be read from memory. D) STA 2000H This instruction stores the value of accumulator(8 bit) at the location specified. 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. 1.A counter is designed simply by loading an appropriate number into one of the registers and using INR(increment by 1) & DCR(decrement by 1) instructions.
2.A loo p is established to update the count,and each count is checked to determine whether it has reached the final number or not.if not then the loop is again repeated. 3.These counters have 1 drawback.i.e.counting is performed at such high speed that only the last count can be observed.to observe counting there must be a proper time delay between counts.