Intel 8086 and 8088

The number of data lines required in a 16-bit microprocessor is 16, in the internal view, but could be something else, such as 8, in the external view. In the 8086/8088, the processor is 16 bits, and its internal data bus is 16 bits. The 8086 provides 16 bits on the external bus, but the 8088 provides 8 bits, even though they are the same internal processor - the 8088 simply uses two memory cycles to move one word.

The largest number that can be represented on a 16 bit bus depends on the interpretation of that number. An unsigned 16 bit number ranges from 0 to 65535, so 65535 is the largest value when unsigned. A 16 bit two's complement signed number ranges from -32768 to +32767, so +32767 is the largest value when signed.

8

Mov ax,1234

mov bx,2345

add ax,bx

mov @(some memory location ) say 2200,ax

hlt

64

Stack/explicit.

pata nai

No. They have a different instruction set. However, the processors are sufficiently similar that an 8086 assembler could assemble an 8085 program, given appropriate constants and macros. Some things, however, such as RIM, SIM, RST, etc. do not have equivalents in the 8086.

The original 8088 processor had a maximum clock frequency of 5 MHz. As implemented in the original IBM PC, it ran at 4.77 MHz. There were variations of the 8088 that could run at 8 MHz.

An offset spatula is sometimes thought of as a commercial tool. It is similar to the standard kitchen device we all know and use except that it has an extra "kick" or extension up from the working surface. The flat blade extends from the leading edge back to a vertical right angle which is extended up a distance of an inch or two and from there it slants back up to the handle. If one googles "offset spatula" and clicks on a link (particularly one that offers the tools for sale), it is probable that a picture will be available.

The debug flag in the 8086/8088 microprocessor causes one instruction to be executed, followed immediately by a debug interrupt. The flag then gets turned off, so you get get nested debug interrupts. This is used by a debugger program to single step a process that it is debugging (the debugee, so to speak).

Segmented memory in the 8086/8088 means that the effective address is added to 16 times the segment address. In the question, an effective address of 14A3H with DS of 7000H, means the physical address is 714A3H. However, the BIU in the 8086 only addresses words, not bytes, so the address generated is 714A2H. If the operation required more than one byte of data, subsequent even addresses of 714A4H, 714A6H, etc. would be generated.

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yes it can be used

If the FAN-IN exceeds FAN-OUT, then the VIH and VIL levels would not be met, and bus operation would be erratic at best and non-functional at worst. If the capacitance load were exceeded, then slow bus transactions might work, while fast bus transactions might fail. All of this would be dependent on the preceeding state of the bus, so "erratic" would be the least of your troubles. Its also possible that, with inadequate buffering, the bus would ring and, on a scope, would appear to work, but, at strobe time, the logic value would be unreliable.

DVST stands for Direct View Storage Tube. It is one of the display devices in which an electron flood gun and writing gun is present. The flood gun floods electrons to a wire grid on which already the writing gun has written some image. The electrons from the flood gun will be repelled back by the negatively charged wire grid which has been charged so by the writing electron beam. The part of the wire grid which has not been charged -ve will allow the electrons to pass through and the electrons will collide on the screen and produce the image.

Adv.-1)Refreshing CRT is not required.

2)Complex picture can be displayed in high resolution without flicker

3)It has flat screen

The 8085 is an 8 bit microprocessor. It cannot directly handle 32 bit data. That said, it is possible to write a routine that can handle 32 bit data, just 8 bits at a time.

In order to demultiplex the address and data bus, you provide latches that sample the multiplexed bus. At ALE=true, they follow the bus. At ALE=falling edge, they lock onto the last value of the bus. The latches will then become the address bus, while the original bus becomes the data bus.

A 14 bit address can specify 214 or 16,384 different locations.

To ensure that any chemicals, hazardous substances and dangerous goods used as part of the companies operations are identified and controlled in a manner to minimise the risk of adverse health effects to employees, customers, contractors and the public.

(frame no * page size) + offset value = physical add

where

frame value is the value present in the corresponding page number

offset value is the last n bits of the logical address

page no is the first m-n bits of logical address

2^m is the logical address

2^n is the page size

The difference between the 8086 and the 8088 is that the 8086 has a 16 bit data bus and that the 8088 has an 8 bit data bus.

Both processors are the same 16 bit processor, and both have a 20 bit address bus. The 8086 is twice as fast as the 8088 in terms of data transfer rate on the bus for the same bus clock speed.

Clear direction flag (CLD) means that index registers will be incremented when looping through a string. Opposite of STD (set direction flag), which means the index registers of the string will be decremented when looping.

0FH is a literal constant for the integer value 15. In the 8086/8088, as an opcode, it is the lock prefix, which means that no other CPU or bus master can take the bus during the execution of the instruction.

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