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

Seg-segment-assemblers directive-microprocessor, SEG : Segment of a Label:...

SEG : Segment of a Label:- The SEG operator is which is used to decide the segment address of the, variable, label or procedure and substitutes the segment base address in plac

Microprocessor, from pin description it seems that 8086 has 16 address/data...

from pin description it seems that 8086 has 16 address/data lines i.e.AD0_AD15.The physical address is however is larger than 2^16.How this condition can be handled

Implemented with an arm assembly language program, The problem to be solved...

The problem to be solved and implemented with an ARM assembly language program You are asked to do some image processing on an image composed of characters shaped in For exa

Program, 2. Write a program to separate out positive and negative numbers f...

2. Write a program to separate out positive and negative numbers from a given series of 16-bit hexadecimal numbers.

General bus operation-microprocessor, General Bus Operation The 8086 ha...

General Bus Operation The 8086 has a joined data and address bus commonly referred to as a time multiplexed address and data bus. The major reason behind  multiplexing address

Write an assembly language program to fi nd the maximum, Write an assembly ...

Write an assembly language program to find the maximum of: y = x 6 - 14x 2 + 56x for the range -2 ≤ x ≤ 4, by stepping one by one through the range. The program should in

Input output interface-microprocessor, I/O interface I/O  devices such ...

I/O interface I/O  devices such as displays and keyboards  establish  communication of computer with outside world. Devices may be interfaced in 2 ways Memory mapped I/O and I/

Interrupt table-how interrupt table processed-microprocessor, Interrupt Tab...

Interrupt Table Each interrupt level has a booked memory location, called an interrupt vector.  All these vectors (or pointers) are stored in the interrupt table. Table lies at

Solotuon, using 8086 assembly language that interchange upper four bits to ...

using 8086 assembly language that interchange upper four bits to lower four bits. assume that data store in byte memory and it written back to same location. and assume the data as

Program, wap proram for bthe addition of two 3*3 matrix

wap proram for bthe addition of two 3*3 matrix

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