Icwi-microprocessor, Assembly Language

Assignment Help:

The definitions of the bits in ICWI are following:

Always set to the value 1. It directs the received byte to ICWI as oppose to OCW2 or OCW3.

Which also utilize the even address (A0 = 0).

Bit 3 (LTIM) - Determines whether the level-triggered mode (LTIM = 1) or the edge-triggered mode (LTIM = 0) is to be utilized. The edge-triggered mode causes the IRR bit to be cleared while the corresponding ISR bit is set.

Bit 2 (ADD) - not utilized in an 8086/8088 system only used in an 8080 or 8085 system.

Bit 1 (SNGL) - denoted whether or not the 8259A is cascaded with other 8259As. SNGL = 1 when just one 8259A is in the interrupt system.

Bit 0 (IC4) -  this is set to value 1 if an ICW4 is to be output to during the initialization sequence.

 For an 8086/8088 system this bit ought to be always be set to 1 because bit 0 in JCW4 ought be set to 1.Bits 7-3 of ICW2 are tilled from bits 7-3 of the second byte output by the CPU during the initialization of the 8259A, and bits 2-0 are set accordingly the level of interrupt request, for instance a request on IR6 would cause them to be set to 110. ICW3 is important just in systems including more than one 8259A and is output to only if SNGL value is equal to 0. ICW4 is output to only if IC4 (ICWI) is set to value 1; or else, the contents of ICW4 are cleared.  The bits in ICW4 are described as follows:

Bits 7-5 - it is always set to 0.

Bit 4 (SFNM) - If it is set to 1, the special fully nested mode is utilized. This mode is utilized in systems having more than one 8259A.

Bit 3 (BUF) - if BUF = 1 indicates that the SP/EN is to be utilized as an output to disable the system's8286 transceivers whereas the CPU inputs data from the 8259A. If no transceivers are present, then BUF should be set to value 0 and, in systems involving just one 8259A, a 1 should be applied to the SP/EN pin.

Bit 2 (M/S) - this bit is ignored when BUF value is zero. For a system that have only one 8259A, this bit should be1; or else, it should be the value1 for the master and value0 for the slaves.

Bit 1 (AEOI) - when AEOI = 1, then the ISR bit that caused the interrupt is cleared at the end of the second INTA pulse.

Bit 0 (µPM) -  when µPM = 1 denote the 8259A is in an 8086/8088 system. This bit being 0 implies an 8085 or 8080 system.  A usual program sequence for setting the contents of

ICWs, which suppose that the even address of the 8259A is 0080, is: MOV AL, 13H

OUT     80H, AL MOV AL, 18H

OUT     81H, AL MOV AL, ODH OUT 81H, AL

The first 2 instructions cause the requests to be edge triggered, show that only one 8259A which is used, and inform the 8259A that an ICW4 will be output. The next 2 instructions cause the 5 most important bits of the interrupt type to be set to value 00011. ICWS is not output to because SNGL = 1; so the final two instructions set ICW4 to OD, which informs the 8259A about the special wholly nested mode is not to be utilized, the SP/EN is utilized to disable transceivers, the 8259A is a master, EOI commands ought to be used to clear the ISR bit, and the 8259A is a part of the 8086 or 8088 system.

 


Related Discussions:- Icwi-microprocessor

Internal hardware-interrupts-microprocessor, Internal Hardware-Interrupts ...

Internal Hardware-Interrupts Internal hardware-interrupts are the outcome of sure situations that occur during the execution of a program, for example. Divide by 0. The interru

Using straight line method for depreciation, Request a depreciation of the...

Request a depreciation of the item, year of purchase, cost of item, number of years to be depreciated (estimated life ) and,the method of depreciation . Method of depreciation sh

Generating random number using 8086, I need to generate a random number bby...

I need to generate a random number bby using 8086 assembly language

Assembly language, Assembly Language: Inside the 8085, instructions ar...

Assembly Language: Inside the 8085, instructions are really stored like binary numbers, not a very good manner to look at them and very difficult to decipher. An assembler is

Nonrecursive Factorial, Write a nonrecursive version of the Factorial proce...

Write a nonrecursive version of the Factorial procedure (Section 8.3.2) that uses a loop. (A VideoNote for this exercise is posted on the Web site.) Write a short program that inte

Relocate program and data, ) What is the difference between re-locatable pr...

) What is the difference between re-locatable program and re-locatable data?

Overview of intel pro-pentium, Overview of Intel Pro-Pentium : The 2 c...

Overview of Intel Pro-Pentium : The 2 chief players in the PC CPU market are Motorola and Intel.  Intel has enjoyed incredible success with its processors since the early 1980

Test-logical instruction-microprocessor, TEST : Logical Compare Instructio...

TEST : Logical Compare Instruction: The TEST instruction performs bit by bit logical AND operation on the 2 operands. Each bit of the result is then set to value I, if the equival

Assembly assignment help, Problems: 1. Write a single program. Each of th...

Problems: 1. Write a single program. Each of the problems (2-4) should be written within a procedure. Your “main” procedure should call each procedure. Before calling each proc

Assembly Language Program, which uses BIOS interrupt INT 21 to read current...

which uses BIOS interrupt INT 21 to read current system time and displays it on the top-left corner of screen.

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd