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

Cache memory-microprocessor, Cache Memory Caching is a technology based...

Cache Memory Caching is a technology based on the memory subsystem of any computer. The majoraim of a cache is to accelerate the computer while keeping the cost of the computer

Basic microprocessor architecture and interface, Basic Microprocessor Archi...

Basic Microprocessor Architecture and Interface : Introduction: Intel launches its first 4-bit microprocessor 4004 in the year 1971 and 8-bit microprocessor 8008 in the y

Architecture of 8088-microprocessor, Architecture Of 8088 The register ...

Architecture Of 8088 The register set of 8088 is accurately the same as in to 8086. The architecture of 8088 is also same to 8086 except for 2 changes; a) 8088 has 4-byte instr

Logical instruction-microprocessor, Logical Instruction : This type of...

Logical Instruction : This type of instructions is utilized for carrying out the bit by bit shift, basic logical operations or rotate. All of the condition code flags are affe

Program to perform one byte bcd addition-assembly program, Program : Write...

Program : Write a program to perform a one byte BCD addition. Solution : It is consider that the operands are in BCD form, but the CPU considers it as hexadecimal and acco

Zero flag, Zero flag: The next line compares the value in register. A ...

Zero flag: The next line compares the value in register. A with the value 1. If they are equivalent, the Zero flag is set (to 1). The next line then jumps to start: only if th

C#, * * * * **** ...

* * * * **** * * * * * How can i print this help me pls

Multiplication using shift and add instruction, Multiply two numbers by usi...

Multiply two numbers by using shift and rotate instruction

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 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