MICROPROCESSOR AND MICROCONTROLLER

MICROPROCESSOR AND MICROCONTROLLER

                                                              ASSIGNMENT-II

1a) What is the difference between Harvard Architecture and von Neumann Architecture?

The name Harvard Architecture comes from the Harvard Mark. The most obvious characteristic of the Harvard Architecture is that it has physically separate signals and storage for code and data memory. It is possible to access program memory and data memory simultaneously. Typically, code (or program) memory is read-only and data memory is read-write. Therefore, it is impossible for program contents to be modified by the program itself.

The von Neumann Architecture is named after the mathematician and early computer scientist John von Neumann. Von Neumann machines have shared signals and memory for code and data. Thus, the program can be easily modified by itself since it is stored in read-write memory.

1b) 8051 was developed using which technology?

Intel’s original MCS-51 family was developed using NMOS technology, but later versions, identified by a letter C in their name (e.g., 80C51) used CMOS technology and consume less power than their NMOS predecessors. This made them more suitable for battery-powered devices.

2a) With 12 MHz clock frequency how many instructions (of 1 machine cycle and 2 machine cycle) it can execute per second?

A cycle is, in reality, 12 pulses of the crystal. That is to say, if an instruction takes one machine cycle to execute, it will take 12 pulses of the crystal to execute. Since we know the crystal is pulsing 11,059,000 times per second and that one machine cycle is 12 pulses, we can calculate how many instruction cycles the 8051 can execute per second:

11,059,000 / 12 = 921,583

This means that the 8051 can execute 921,583 single-cycle instructions per second. Since a large number of 8051 instructions are single-cycle instructions it is often considered that the 8051 can execute roughly 1 million instructions per second, although in reality it is less–and, depending on the instructions being used, an estimate of about 600,000 instructions per second is more realistic.

2b).If the internal memory 20H contains AAH and 07H contains 55H. What is the content of register A and status of carry bit after executing the following code?

MOV C,07H

MOV A,#20H

ADDC A,07H

Answer: A=76H

             C=0 (RESET)

2c). If the XLAT frequency of 8051 is 8MHz, find the time taken to execute the following program.

              MOV R2,#04

              MOV R1,#06

 WAIT: DJNZ R2,WAIT

               MOV R2,#04  Machine cycle-1

               MOV R1,#6   Machine cycle-1

WAIT:     DJNZ R2,WAIT  Machine cycle-2 x 4

3a). Explain briefly the interrupts of 8051, indicate their vector addresses

 

3c). When any interrupt is enabled, then where the pointer moves immediately?

The pointer moves to the first location of the memory called the interrupt vector table.

Explanation: When any interrupt is enabled, then it goes to vector table where the address of the ISR is placed.

4a). Interface ADC0809 to 8051 and write ALP to convert the analog voltage connected to second channel. Display the digital value on LEDs connected to port-0.

4b). Which errors are more likely to get generated by conversion time and ADC resolution respectively in accordance to the digital signal processing?