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    how many t-states are required for execution of sta 2000h instruction?

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    Instruction type STA a16 in 8085 Microprocessor

    Instruction type STA a16 in 8085 Microprocessor - In 8085 Instruction set, STA is a mnemonic that stands for STore Accumulator contents in memory. In this instr ...

    Instruction type STA a16 in 8085 Microprocessor

    Microprocessor8085

    In 8085 Instruction set, STA is a mnemonic that stands for STore Accumulator contents in memory. In this instruction,Accumulator8-bit content will be stored to a memory location whose 16-bit address is indicated in the instruction as a16. This instruction uses absolute addressing for specifying the destination. This instruction occupies 3-Bytes of memory. First Byte is required for the opcode, and next successive 2-Bytes provide the 16-bit address divided into 8-bits each consecutively.

    Mnemonics, Operand Opcode(in HEX) Bytes

    STA Address 32 3

    Let us consider STA 4050H as an example instruction of this type. It is a 3-Byte instruction. The first Byte will contain the opcode hex value 32H. As in 8085 assembly language coding supports low order Byte of the address should be mentioned at first then the high order Byte of the address should be mentioned next. So next Byte in memory will hold 50H and after that 40H will be kept in the last third Byte. Let us suppose the initial content of Accumulator is ABH and initial content of memory location 4050H is CDH. So after execution, Accumulator content will remain as ABH and 4050H location’s content will become ABH replacing its previous content CDH. The content tracing of this instruction has been shown below −

    Before After (A) ABH ABH (4050H) CDH ABH

    Address Hex Codes Mnemonic Comment

    2008 2A STA 4050H Content of the memory location 4050H <- A

    2009 50 Low order Byte of the address

    200A 40 High order Byte of the address

    The timing diagram of this instruction STA 4050H is as follows −

    Summary − So this instruction SDA 4050H requires 3-Bytes, 4-Machine Cycles (Opcode Fetch, Memory Read, Memory Read, Memory Write) and 13 T-States for execution as shown in the timing diagram.

    Samual Sam

    Published on 30-Nov-2018 12:08:46

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    [Solved] How many machine cycles are required by STA instruction?

    Machine Cycle: Time taken to execute one OPERATION is known as a machine cycle.  One instruction will contain 1 to 5 machine cycles. T-State: The portion

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    How many machine cycles are required by STA instruction?

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    Machine Cycle: Time taken to execute one OPERATION is known as a machine cycle.  One instruction will contain 1 to 5 machine cycles.T-State: The portion of a machine cycle executed in one internal clock pulse is known as T-state.

    T states starts at the falling edge of a clock pulse.

    STA address

    It requires 4 Machine cycles.

    1 Opcode fetch (4T)

    2 Memory read (3T + 3T)

    1 Memory write (3T)

    Total number T-states = 13T

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    More Instruction Set of 8085 Questions

    Q1. If a program with 5 instructions is executed in 7 clock cycles, then CPI is ______.Q2. Determine the contents of accumulator if the instruction RAL is executed twice. Assume the contents of accumulator is AAH and CY = 0.Q3. Calculate the time required to execute the entire instruction cycle if two machine codes, 0011 1110 and 0011 0010, are stored in memory locations 2000H and 2001H, respectively. If the clock frequency is 2 MHz, the first machine code represents opcode to load data byte in the accumulator and the second code represents data to be loaded in the accumulator.Q4. The number of cycles required to execute the given instruction is: MOV DPTR,#data16Q5. How is the status of the carry, auxiliary carry and parity flag affected if write instruction? MOV A#9C ADD A, #64HQ6. The 8085 instruction that doubles the value in accumulator isQ7. The 8085 assembly language instruction that stores the contents of H and L registers into the memory locations 2050H and 2051H, respectively, is:Q8. Directions: It consists of two statements, one labelled as the ‘Statement (I)’ and the other as ‘Statement (II). Examine these two statements carefully and select the answer using the codes given below: Statement (I): Branch instructions in a microprocessor are used to change the sequence of program. Statement (II): All logical instructions are branch instructions.Q9. How many machine cycles are required by STA instruction?

    More Microprocessors Questions

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    : : LXID,DISP LP:     CALL         SUB      : :

    It is desired that control be returned to LP + DISP + 3 when the RET instruction is executed in the subroutine. The set of instructions that precede the RET instruction in the subroutine are:

    Q8. An 8085 assembly language program is given below. Assume that the carry flag is initially unset. The content of the accumulator after the execution of the program is MVI A, 07H RLC MOV B, A RLC RLC ADD B RRCQ9. 8086 एक ___बिट इंटीगर प्रोसेसर है।Q10. RAM stands for ________ in computer terminology

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    Timing diagram of MOV Instruction in Microprocessor

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    Timing diagram of MOV Instruction in Microprocessor

    Difficulty Level : Medium

    Last Updated : 31 May, 2022

    Problem – Draw the timing diagram of the given instruction in 8085,

    MOV B, C

    Given instruction copy the contents of the source register into the destination register and the contents of the source register are not altered.

    Example:

    MOV B, C Opcode: MOV Operand: B and C

    Bis is the destination register and C is the source register whose contents need to be transferred to the destination register. Algorithm – The instruction MOV B, C is of 1 byte; therefore the complete instruction will be stored in a single memory address. For example:

    2000: MOV B, C

    Only opcode fetching is required for this instruction and thus we need 4 T states for the timing diagram. For the opcode fetch the IO/M (low active) = 0, S1 = 1 and S0 = 1. The timing diagram of MOV instruction is shown below:

    In Opcode fetch ( t1-t4 T states):

    00 – lower bit of address where the opcode is stored, i.e., 00.20 – higher bit of address where the opcode is stored, i.e., 20.ALE – provides signal for multiplexed address and data bus. Only in t1 is it used as an address bus to fetch a lower bit of address otherwise it will be used as the data bus.RD (low active) – signal is 1 in t1 & t4 as no data is read by the microprocessor. Signal is 0 in t2 & t3 because here the data is read by a microprocessor.WR (low active) – signal is 1 throughout, no data is written by a microprocessor.IO/M (low active) – signal is 1 throughout because the operation is performing on memory.S0 and S1 – both are 1 in case of opcode fetching.

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    Improved By : aayushiagarwal1106 Article Tags : microprocessor system-programming

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