Intel 8085

The memory capacity of the 8085 microprocessor is 64 kb because the address bus is 16 bits, and you can address 216, or 64kb, with a 16 bit address bus.

group of wires or lines that are used to transfer adress between microprocesssor and i/o devices one way and of 20 bits

while data bus uses bothway transfer of data between microprocessor and i/o devices of 16 bits

pta nhi hai ?

44

Threads

Instruction decoder takes bits stored in the instruction register and decodes it and tells to CPU what it need to do for it and enable the components for the operation.

Simply, instruction decoder is like a dictionary. It tells the meaning of the instruction.

[object Object]

In the 8085 class machine, machine cycles are measured in microseconds to nanoseconds. For instance, at a clock frequency of 3 MHz, one clock is 333 nanoseconds, and one memory read or write, without wait states, is one microsecond. One CALL instruction, again without wait states, is 6 microseconds.

There are seven bits (one unused) in the accumulator prior to executing the SIM instruction that are of value. Numbering is 7 through 0, where 7 is the most significant...

7-Serial Out data - This sets the SOD pin

6-Enable serial data - if 0, ignore bit 7

5-Unused

4-Reset RST 7.5 flip-flop

3-Mask set enable - if 0, ignore bits 0, 1, and 2

2-Mask 7.5 - if 0, enable RST 7.5

1-Mask 6.5 - if 0, enable RST 6.5

0-Mask 5.5 - if 0, enable RST 5.5

the name of it

lol

CALL, requiring 18 clock cycles.

Direct addressing mode

The process of transferring instruction codes from memory location to instruction queue register is called opcode fetch.

To clear only Accumulator & not the Flag Register, you can use:

1) ANI 00 H

2) MVI A, 00 H

To clear both Accumulator as well as Flag Register, use:

LXI H, 0000 H

PUSH H

POP A.

A bi-directional buffer includes first and second unidirectional buffers connected for retransmitting signals in opposite directions between first and second buses. When an external bus driver pulls the first bus low, the first unidirectional buffer pulls the second bus low and generates a signal inhibiting the second unidirectional buffer from actively driving the first bus. When the external bus driver allows the first bus to return to the high logic level, the first unidirectional buffer temporarily supplies a high charging current to the second bus to quickly pull it up. Similarly, when an external bus driver pulls the second bus low, the second unidirectional buffer pulls the first bus low and generates a signal inhibiting the first unidirectional buffer from actively driving the second bus. When the external bus driver allows the second bus to return to the high logic level, the second buffer temporarily supplies a high charging current to the first bus to quickly pull it up. The bi-directional buffer includes a register for storing and reading out data representing successive logic states of a signal on the first bus, thereby providing a history of data appearing on the bus

Two ground pins are used in the 8086 microprocessor to increase the bus pull-down current capacity.

The disadvantage of the bit slice microprocessor is that it is slow that it why it was slow with earlier processors.

alp for lcm of a no

in the case of edge trigger, it may generate unwanted interrupt when input signal has glitch and so on.

on the other hand if edge trigger not seen in some special situation (eg. when process in the service routin) level trigger preffered!

display

A program which is used to count the number of numbers in an array using a 8085 microprocessor is known as a assembly language program.

I am not sure about 8086, but I can tell you the whole procedure in 8085.

PUSH instruction always pushes two bytes of data i.e. total 16 bits.

Example:

Assume that Stack is already initialized and SP is at 2008 address location. Then PUSH B instruction will have following steps:

1) The stack pointer (SP) will be pointing to the uppermost position of the stack (actually stack works in opposite order in terms of Addresses. e.g. if SP is now at address 2008, then PUSH instruction will store the contents on 2007 & 2006).

2) The contents of register B & C will be saved on to the stack such that contents of register B will be at 2007 & that of C will be at 2006 address location.

3) The SP is now modified to 2006.

While to some it might seem like this will increase pollution, I will show that it doesn't.

There exists 4 general purpose registers in 8086 namely Ax(Accumulator),Bx(Base), Cx(Count), Dx(Data).Further these 4 registers are divided into 8 based on higher and lower bits i.e.,. AX- AH &AL, BX-BH & BL, CX- CH & CL, DX- DH& DL.

In 8086, there are 4 segment registers:

1. Stack segment

2. Code segment

3. Data segment

4. Extra segment

Data transfer in the 8085 microprocessor can be done with the help of several instructions like the MOV,MVI,OUT and IN . Lets say, we want to move the contents of a register R1 to register R2, then we could use the instruction MOV R2,R1 ; and so on...