Intel 8085

LHLD 2600h

is use to store data of 2600h in L register and data of 2601h in H register

While

SHLD 2600h

is use to store data of L in 2600h memory location and data of H in 2601h memory location data of H register

Log2 260 is 8.022, so it would seem that 9 bits are required to handle 260 instructions. In practice, however, the opcode is a multiple of 8 bits, so most instructions are 8 bits, with a few being 16 bits.

In order to add 32 bit numbers in the 8085, you need to add them 8 bits at a time, tracking the carrys between each add.

LXI B,first_number

LXI H,second_number

LXI D,result

LDAX B ;first byte - no carry in

ADD M

STAX D

INX B; point to next byte

INX D

INX H

LDAX B ;second byte - carry in

ADC M ;note the ADC instead of ADD

STAX D

INX B; point to next byte

INX D

INX H

LDAX B ;third byte - carry in

ADC M

STAX D

INX B; point to next byte

INX D

INX H

LDAX B ;fourth - carry in

ADC M

STAX D

There are 74 instructions in the 8085 microprocessor.

I am not quite sure how significant this is. The reason, however, is that accessing a register, which is part of the CPU, is faster than accessing RAM memory, which is located on a separate chip.

I am not quite sure how significant this is. The reason, however, is that accessing a register, which is part of the CPU, is faster than accessing RAM memory, which is located on a separate chip.

I am not quite sure how significant this is. The reason, however, is that accessing a register, which is part of the CPU, is faster than accessing RAM memory, which is located on a separate chip.

I am not quite sure how significant this is. The reason, however, is that accessing a register, which is part of the CPU, is faster than accessing RAM memory, which is located on a separate chip.

The return address.

Optical fiber

When you first get onto theslap.com, if you go to the top right of the page, you will see the word Register next to the word Login. If you do not have a slap account, you just click on Register, then you can then create a username and password.

Interrupt acknowledge is the process of acknowledging a hardware interrupt, obtaining an interrupt vector address, and initiating the interrupt service routine in software. The INTA- (Interrupt Acknowledge) pin has the same timing as RD-, and external hardware is expected to provide an opcode in response to it. In the case of TRAP, RST7.5, RST6.5, and RST5.5, there is no specific interrupt acknowledge cycle like there is for INTR, but everything else is the same.

these essential elements are

1,operation code (op code) and

2.address field

both compose the computer instruction

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Im pretty sure it is located on the mother board

But, there cna be objectives, they say the microprocessor is he CPU.

So this is based on my knowledge.

The 8085 does not have 32 bit capabilities, nor does it have a multiply, so to do 32 bit multiplication you would need to write a routine. There are several possibilities. One would be to setup a 64 bit result area, and a loop that scans each bit in one multiplicand, adding the second multiplicand to the result, and then shifting the result.

This is simply 32 adds with 32 multiplies by two.

Even though the 8085 is only an 8 bit processor, some things can be done 16 bits at a time, particularly involving the HL register.

stack register is a special purpose register.

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.

8 bits

There are six classes of interrupt:

*Supervisor Call Interrupts (SVC)

*I/O Interrupts

*External Interrupts

*Restart Interupts

*Program Check Interrupt

*Machine Check Interrupts

The address bus is unidirectional(only in one direction) in the processor. So, the flow of information on this bus is from the microprocessor to the attached device(memory module).

1st address for operand.

2nd address for another operand.

3rd address for store the result.

4th address for next instruction.

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.

According to Win Ballada ,CPA, MBA that the repetitious procedure can be avoided with the introduction of an alternative device called cash priority program. this program which is prepared at the start of the liquidation processwill help the partner's project when they can expert to be included in the cash distribution.

when interrupts requests arrive from two or more devices simultaneously , the process has to decide which request should be serviced first and which one should delayed. the processor takes the decision with the help of interrupt priorities.