Intel 8085

The response to another interrupt request during servicing of an interrupt depends on the setting of the interrupt enable flag and/or the interrupt mask. If the interrupt service routine is thread safe, it will process correctly. More probably, problems will arise so the proper procedure is to either set the interrupt mask (8085) to not allow this or lower priority interrupts before reenabling interrupt, or to leave interrupts disabled until this iteration is complete.

If interrupts are disabled or masked, a subsequent interrupt will be posted but not processed until the first is completed.

During a single bus cycle, the 8-bit microprocessor transfers one byte while the 16-bit microprocessor transfers two bytes. The 16-bit microprocessor has twice the data transfer rate.

it has high stability large q factor and the frequency that doesn't drift with aging. It is used as a clock source most of the times.

It is employed to hold temporarily the right hand instruction from a word in memory..

For example, The IAS machine's basic unit of information was a 40-bit, so that two instructions could be stored in each 40-bit memory location. Each instruction consisted of an 8-bit {operation code} and a 12-bit address.

Hence the IBR (Instruction Buffer Register) is used to temporarily hold Right hand instruction for the next use.

The troubleshooting manual will be shorter and more concise; the booklet will be more descriptive and complex.

flag manipulation instructions are which changes the flag conditions

for example STC - set the carry flag

STI - set the interrupt flag so interrupts will be enabled

SATISH MN

Use following instructions:

MOV A,R1

MOV R4,A

There is insufficient information in the question to properly answer it. You need to specify which FPGA you are interested in. Please restate the question.

As far as the bus interface is concerned, there is no real difference between data and instructions. Except for the S0 pin, an opcode fetch will look the same as a memory read. There is one extra clock cycle following an opcode fetch, which is used by the CPU to decode and process the opcode, but the bus does not care because there is no sequence initiation with ALE.

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RIM is Read Interrupt Mask. It is used to access the interrupt mask register, which contains the status of the three interrupt masks and the serial input data pin.

interrupt vector table

First you bend over and get a prostate exam. Then you shave your nuts and take the hair and put it on your head. Next walk to brad fishman and give him so hair.

evaluation of microprocesser 8085 upto this year

-128 to 127

I believe a nested interrupt, is where an interrupt is allowed to occur (and thus is handled) during an already occurring Interupt service ruotine.

I.E.

First interrupt occurs

ISR1 begins

second Interrupt occurs

ISR2 begins

ISR2 Finishes

ISR1 continues from where left off

ISR1 finishes

http://infocenter.arm.com/help/index.jsp?topic=/com.arm.doc.dvi0022a/ar01s02s09.html

No. Each processor core is 3.2gHz. The bottleneck is the bus interface, but the compromise is that each core tends to spend more time executing instructions than accessing memory. Even so, L1 and L2 cache improves the situation by caching memory, so you get the best of both worlds. If, however, all the cores randomly accessed wide areas of memory, performance would suffer.

Because that's how Intel designed the 8085.

In addition to the 8080 type interrupt on the INTR pin, Intel chose, for the 8085, to implement four new interrupts, RST 5.5, RST 6.5, RST 7.5, and TRAP, each of which would not require the interrupting device to provide a vector. The naming convention of x.5 was simply in recognition that Intel placed the implicit vector halfway between two other RST vectors. As an example, RST 6.5 is halfway between RST 6 and RST 7. Since RST 6 and RST 5 are eight bytes away from each other, placing RST 6.5 in between would place a limit of four bytes, and four bytes is enough to place a three byte JMP instruction.

The decimal and hex addresses of all of the vectors are...

RST 0 - 0 - 00H

RST 1 - 8 - 08H

RST 2 - 16 - 10H

RST 3 - 24 - 18H

RST 4 - 32 - 20H

TRAP - 36 - 24H

RST 5 - 40 - 28H

RST 5.5 - 44 - 2CH

RST 6 - 48 - 30H

RST 6.5 - 52 - 34H

RST 7 - 56 - 38H

RST 7.5 - 60 - 3CH

RAL:- Each binary bit of the accumulator is rotated left by one position through the Carry flag. Bit D7 is placed in the Carry flag, and the Carry flag is placed in the least significant position D0.CY is modified according to bit D7.S, Z, P, AC are not affected.RAR:- Each binary bit of the accumulator is rotated right by one position through the Carry flag.Bit D0 is placed in the Carry flag, and the Carry flag is placed in the most ignificant

position D7.CY is modified according to bit D0.S, Z, P, AC are not affected.....

....

Answer by Raj Choudhary.......

Pins 1 and 20 in the 8086 microprocessor are (both) power and signal ground (GND).

The stack register contains the address of the last item pushed on the stack. The stack is a region of memory used for temporary storage of instruction addresses and register values in a Last-In-First-Out (nested) structure. It is used for interrupts and subroutine calls.

Grove joined Intel in 1968

The first Pentium prosessor had 3.1 million transistors.For other pentiums, view the timeline here: http://www.Intel.com/pressroom/kits/quickreffam.htm

We are now in the neighborhood of three billion. The Power 7 and the Niagara have 2 billion and there are some with more. The amount changes from week to week so this answer is only good for a few days here in December six of 2011. LOL!