Solution MT-1 April 2022

ETEE 310  Microprocessors and Microcontrollers

Time 1.5 Hours                                                               MM: 30

 

Q1. Differentiate between PC and SP

Sl	PC	SP
1	Program counter (PC) points to the next instruction in memory to be executed	Stack Ponter(SP) points to the top of the stack memory from where an item could be accessed
2	It is auto incremented for normal sequential program	Incremented for PUSH and for POP operation an item accessed and the SP decremented for stack-up configuration

 

Q1. (b) Briefly describe different memory technologies used in microprocessor system.

A microprocessor system does not contain any memory internally other than the set of registers. Memory can be interfaced externally with the microprocessor. They can be in the form of primary memories which are random access memory (RAM ) and read only memory ( ROM)

RAM comes as SRAM, SDRAM, DRAM, DDRAM

ROM comes in various types as Masked ROM, PROM, EPROM, EEPROM

Q.1 (c) What are one byte, two byte, and three byte instructions?

Microprocessor instruction sets can be classified depending on the size of the instruction.

One byte instructions – These instructions have implied operands and take one byte space in memory.  Examples of such instructions are CMC, STC, CLC,

Two byte instructions – These instructions are of two byte in size. The first byte is the OPCODE and the second byte is the operand. Examples are MVI B, 55h, ANI 35h

Three byte instructions –  These instructions are three byte in size. First byte is the OPCODE and the subsequent two byte are the operand or address of the operand. Examples – LXI H, 2050h; LDA 2000h

Q.1 (d) Explain the concept of memory segmentation in 8086.

8086 microprocessor supports 1MB of memory. To address 1MB it requires a 20 bit address, but it does not have a 20 bit address register. Instead, all address registers (segment/pointer registers) are 16 bit.

Concept of memory segmentation allows 1 MB memory to be accessed using only the 16-bit registers. Thus the  1 MB memory of 8086 is logically divided into four segments, each segment is of 64KB in size which could be easily accessed using any 8086 address registers. These segments are :

Code Segment

Data Segment

Stack Segment

Extra Segment

To address the memory we then require two addresses these are the segment value of the address and the effective address. The physical address of 20 bit will be formed using thses two addresses:

Physical address = segment value x 10h + EA

Q.1 (e) Discuss the function of INTR, NMI, ALE, READY.

INTR: It is an interrupt pin on the microprocessor indicating an interrupt signal. It is a vectored and maskable interrupt, and the lowest priority.

NMI : NMI is a non-maskable interrupt pin on a microprocessor. Signal of very critical in nature are applied to this pin. Once an interrupt signal arrives on this pin, the ISR for that interrupt will be processed. This has the highest priority in 8086 system.

ALE: Full form of the ALE is address latch enable. In a microprocessor system, this pin is connected to the enable pin of the address latch. When ALE signal is active it enables the latch to latch the address at the output. Thus dividing the AD lines into two separate lines called as address lines ot address bus and the data bus.

READY : This signal is used to synchronize the slow speed peripherals with the microprocessor. This signal is used by the peripherals and memory devices in order to show the readiness for the next operation

Q2.(a) What is an assembly language program? Explain it with the help of an example.

An assembly language is a symbolic machine language having one-to-one correspondence with the OPCODE of an instruction. This language was evolved to make the human task of remembering and writing the programs for a given processor.

As the program or instructions are written using a human readable mnemonic or symbols, an assemble software is required for a system to convert the symbolic program into the machine program which is to be executed by the microprocessor.

Example of an assembly program:

The following code exemplifies an assembly program to add the values of two registers.

MVI A, 55h                 ; Mov an immediate value Hex 55 into accumulator

MVI B, 35h                 ; Mov an immediate value Hex 35 into register B

MVI D, 0h                   ; Clear to save carry

ADD B                         ; Adds the conten of accumulator and register B; A=A+B

STA 2050h                 ; saves the sum at address 2050

JNC      save_carry

INR D                          ; increment D if carry arises

Save_carry:    MOV A, D                    ; mov carry to accumulator

STA 2051h                 ; save carry at address 2051h

HLT

Q.2(b) What is a subroutine? Write a software routing for 8085

A subroutine is a subprogram written to perform a specific task or routine for a main program. Once written, a subroutine can be called any number of times from within the main program to perform that assigned task. The concept of the subroutine allows the code reusability and relieves the coder from the effort of writing and understanding the program.

Software delay in assembly can be very easily implemented using the subroutine using the concept of loop statements.

We only need to understand the concept of T states an instruction take to execute. Sum all the T states in a loop inside the subroutine and multiply with the time for each T state.

Here is a simple example of writing a 1 ms delay for 8085 having the clock frequency of 2MHz

1 ms = (no. of T States * N)* time/Tstate

15 Count *0.5us = 1ms + 3*0.5us

Count = 134

Delay:

MVI C, Count

again : NOP                            1T

DCR C                         4T

JNZ again                    10/7T

1ms= (15* Count – 3)*0.5us

Q3.(a) Connect 128K word ( 128×16) RAM with the system lines of 8086 microprocessor. Assume suitable address

Total address space 128K word = 258 K byte.

No of address lines = 18 (A17 – A0)

A19	A18	A17	A16	A15	A14	A13	A12	A11	A10	A9	A8	A7	A6	A5	A4	A3	A2	A1	A0
0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0	0	1	1	1	1	1	1	1	1	1	1	1	1	1	1	1	1	1	1

The use of BHE’ and A0 is evident from the following table

BHE’    AO       usage

0          0          Both bytes accessed

0          1          High Byte (Odd memory bank)

1          0          Low byte (Even memory bank)

1          1          None selected

128 K word= 256 byte of memory interfacing with 8086 system.

Q3.(b) Draw the timing diagram of minimum mode memory read cycle

Q4. (a) Write a program for 8086 microprocessor (using assemble directive ) to add two numbers of 10 bytes each.

Data segment

    A          DB       26h.24h,32h, 34h, 45h,65h,67h,76h,66h, 55h

            B          DB       27h.74h,32h, 34h, 15h,05h,67h,76h,56h, 25h

SUM    DB 11 dup(0)

Data ends

Code segment

Assume CS: code, DS: data

                  MOV AX, data

                           MOV DS, AX

LEA SI, A

LEA BX, B

LEA DI, SUM

MOV CL, 0Ah

MOV CH, 00h

nextByte :       MOV AL, [SI]

ADC AL, [BX]

MOV [DI], AL

INC SI

INC BX

DEC DI

DEC CL

JNZ nextByte

JNC save_carry

INC CH

save_carry:     DEC DI

                        MOV [DI], CH

            Code ends