Quick

1

• State the three main elements in any microprocessor system

• State the difference between a microprocessor and microcontroller • State the advantages of flash ROM,

compared with other memory types.

2

• The three main elements are: Processor, memory and I/O

• A microcontroller has processor, memory and I/O on one chip, while microprocessor needs separate memory and I/O chips to form a working system

• Flash ROM can be electrically re-written many times, but is non-volatile

Some Quick Exercises

• Microcontroller contains on chip memory, input/output (I/O) ports and the

processor CPU.

• State the main function of I/O ports • Name THREE (3) types of I/O

communication ports.

Answers

• communication between microcomputer and the outside world (external system) such as humans and machine

5

• How many bits does the

128k MCU program

memory contain?

Some Quick Exercises

• Why it’s a good practice to put all

hardware definitions into a common file? • State three PIC chip options, which are

determined by the configuration code. (#config … )

Answers

• Common File: Rewriting the hardware definitions for each and every programs to be used for the same board can be troublesome. All the

programs using the same hardware/circuit can just include the common hardware definitions file into the source code

• PIC Chip Options: Watchdog timer, High speed programming, Low voltage programming, In-circuit debugger, code protection, memory write, brown-out reset, power-up timer

9

• Why serial data communication is generally slower than parallel? • Why serial data communication can

achieve higher speed compared to parallel communication by being clocked at a greater rate?

10

• Serial vs Parallel: The data has to be converted to serial form in a shift register and transmitted one bit at a time on a single line, while parallel data is transferred 8 (or more) bits at a time.

• A number of factors allow serial to be clocked at a greater rate:

• Clock skewbetween different channels is not an issue (for unclocked asynchronous serial communicationlinks) • Serial connection requires fewer interconnecting cables,

occupies less space. The extra space allows for better isolation of the channel from surroundings

• Crosstalkis less of an issue compared to parallel, because fewer conductors in proximity.

Some Quick Exercises

• This function initializes the RS232… The oscillator frequency to be used is 10MHz. The baud rate required is 9600 bps. Calculate the value of X to obtain the required baud rate.

void init_rs232( ) {

BRGH = 1;

SPBRG = X; /* set baud rate */

TX9 = 0; /* ninebits?1:0,,,8- or 9-bit transmission */ SYNC = 0; /* asynchronous */

SPEN = 1; /* enable serial port pins */

TXEN = 1; /* enable the transmitter & (automatically)TXIF=1 */ CREN = 1; /* enable reception */

SREN = 0; /* no effect */ TXIE = 0; /* disable tx interrupts */ RCIE = 0; /* disable rx interrupts */ }

Answers

X = (Fosc / Baud Rate / 16) – 1 = (10MHz / 9600 / 16) - 1 = 65 – 1

13

• What are the advantages of in-circuit programming and debugging over the corresponding conventional development process?

14

• The program can be run, single

stepped and paused in the actual target hardware

• allowing hardware and timing faults to be identified as well a the usual syntax and logical errors

• the chip does not need to be removed from the application hardware once fitted, preventing possible damage

Some Quick Exercises

• What are the advantages of using circuit simulation software such as Proteus over the corresponding conventional

development process?

Answers

• The conventional process is to build prototype hardware, download the

program to the microcontroller and test it in circuit.

• Simulation allows the design to be tested and debugged before building hardware. • The schematic can then be converted into

a netlist and a layout to produce the final PCB without prototyping.

17

• What is a compiler (for C programming) and what does a compiler do?

18

• Compiler is a program that processes statements written in a programming language and turns them into machine language that a computer's processor uses.

• The compiler gets the syntax error in the written program before turning the C statements into machine instructions

Some Quick Exercises

• PIC is a family of Harvard architecture microcontrollers. Sketch the block diagram of a Harvard architecture microcontroller.

Answers

21

• Von Neumann is another type of microcontroller architecture besides Harvard architecture. Sketch the block diagram of a von Neumann architecture microcontroller.

22

• Block diagram of a von Neumann architecture :

Some Quick Exercises

• What are the advantages of Harvard architecture compared to von Neumann architecture?

Answers

• In a system using von Neumann architecture, the CPU can be either reading an instruction or reading/writing data from/to the memory. Both cannot occur at the same time since the

instructions and data use the same bus system. • In a system using the Harvard architecture, the CPU can both read an instruction and perform a data memory access at the same time, can thus be faster for a given circuit complexity because instruction fetches and data access do not contend for a single memory pathway.

25

• EEPROM is a type of memory widely used in computers and other electronic devices. State the type of EEPROM.

• EEPROM is realized by using arrays of floating gate transistors. State one advantage of floating gate transistors that is useful in EEPROM. • Flash memory is one type of EEPROM which is

able to achieve faster erase/write speed compared to traditional EEPROM. Why?

26

• Non-volatile memory which can be

electrically re-written many times.

extended periods of time even without a connection to a power supply.

• Flash memory works much faster than traditional EEPROMs because it writes data in chunks/blocks, usually 512 bytes in size, instead of 1 byte at a time.

Some Quick Exercises

#include <htc.h> #include <pic16f877a.h> #define _XTAL_FREQ 20000000 // Set the oscillator frequency to 20MHz void main( )

{ TRISC=0x00; //Port C as outputs T2CON = 0b01010000; // Timer 2 control register PR2 = 0b01111111;

CCP1CON = 0b00111100; //Set to PWM mode CCPR1L=63;

TMR2ON=1; //Turn on Timer 2 while(1)

{ } } //End of Program

Calculate the pulse width, duty cycle and frequency of the pulse generated in the program below. The circuit uses a 20MHz oscillator.

Answers

• Instruction clock frequency = 20/4 = 5MHz • Instruction clock period = 0.2 us

• Timer2 counts from 0 to 127 (0b01111111 ) • Pre-scaler set at 1

• Time for Timer 2 to reset = 127 x 0.2 us = 25.4 us • Post-scaler = 10, but unused in PWM mode • Pulse period = Time for Timer 2 to reset = 25.4 us

• Pulse frequency = 1/(pulse period) = 1/(25.4us) = 39.37kHz • Every time Timer2 counts until 63 (determined by CCPR1L) output

of CCP1 (Pin RC2 or C2) will turn from high to low, the time taken (Pulse width) is 63 x 0.2 us = 12.6us OR 1.26 x 10e-5 seconds • Duty cycle = 12.6us / 25.4us = 49.6%

29

• Describe TWO characteristics of an active-high switch.

30

• An active high switch pulls up the

logic level to HIGH when being

pressed/pushed.

• An active high switch requires a pull

down resistor to pull down the logic

level to LOW when the switch is

released.

The End

• That’s all for this time.

• Enjoy your time playing around with microcontroller and interfacing 