How do you draw timing diagram of jnz in 8085?

Answer 1

3 machine cycles...

1.Opcode fetch

2.read cycle

3.read cycle

Answer 2

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.

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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.

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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.

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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.

Answer 3

So now

6 T state Opcode fetch

+ Memory write * 2 (PC pushed to stack) ( 3+3 )

+ Memory read * 2 (New value of PC fetched from memory) ( 3+3 )

ie 6 + 3 + 3 + 3 + 3 = 18 T-states.

" It is so simple to be happy,

Answer 4

it has 5 machine cycles,

1)opcode fetch-after decoding the meaning of this opcode,

2)MEMORY READ--the content of memory location shown by stack pointer(sp) is transferred to up,sp->sp+1 .

3)MEMORY READ-- the content of memory location shown by stack pointer(sp) is transferred to up.

4)MEM WRITE--content of H reg goes to sp & sp->sp-1.

5)MEM WRITE--content of L reg goes to sp.

[by Vishal Barun]

Answer 5

Refer http://wiki.answers.com/Q/How_to_draw_timing_diagram_for_8085_microprocessor_instruction

Answer 6

It will consists of a FETCH cycle followed by 1 or more wait cycles. Once HLT is executed it inserts wait states in every clock period until it is retested or an interrupt occurs.

Answer 7

sorry