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The ports connected to the LEDs, buttons and buzzer are then initialized.

Finally, the interrupts are activated, and the application waits for the execution of one of two interrupts.

The Basic Timer1 interrupt executes at a frequency of once every second. When this interrupt is occurs, it begins by switching the state of LED1 and LED2. Afterwards, it accesses the memory to fetch the next musical note to be performed. The routine ends with memory pointer management.

The Port 1 ISR begins by evaluating the source of the interrupt. The sound volume is reduced if the button SW1 is pressed. The sound volume is increased if button SW2 is pressed.

System configuration

Timer_b

It is the responsibility of Timer_B to produce the PWM signal that activates the Buzzer. Timer_B counts until the value contained in the TBCCR0 register is reached. It does not generate an interrupt, and must be sourced by SMCLK clock signal:

TBCTL = TBSSEL_2 | CNTL_0 | TBCLGRP_0 |MC_1 | ID_0;

Each PWM signal produced by Timer_B corresponds to a musical note. The relationship between the frequency and the musical note is given in Table 1.

Note SI0 DO RE MI FA SOL LA SI DO2
Freq [Hz] 503 524 587 662 701 787 878 1004 1048

Timer_B has a frequency clock input equal to 7.995392 MHz.

The value to write in the TBCCR0 register in order to generate the desired frequency is:

// TBCCR0 value of the musical notes #define SI0 15895#define DO 15258 #define RE 13620#define MI 12077 #define FA 11405#define SOL 10159 #define LA 9106#define SI 7963 #define DO2 7629TBCCTL4 = OUTMOD_3; // CCR4 interrupt enabledTBCCR4 = space[0]/2;

Timer_a configuration

TACTL = TASSEL_2 |MC_2 | ID_0 | TAIE; // SMCLK, continuous mode up to 0xffff TACCTL1 = CM1 | CCIS_0 | CAP | CCIE;// Capture on rising edge, Cap mode,// Cap/Com int. enable, TACCR1 input signal selected//********************************************************* // Timer A ISR//********************************************************* #pragma vector=TIMERA1_VECTOR__interrupt void TimerA1_ISR (void) {switch (TAIV) {case TAIV_TACCR1: if (capture == 0){T1 = TACCR1; flag = 1;capture = 1; }else { if (flag == 1) {T2 = TACCR1; if (T2>T1) T = T2-T1;} else{TAR = 0; }capture = 0; flag = 0;} break; case TAIV_TACCR2: break;case TAIV_TAIFG:tick++; if (tick == 60){LCD_freq(); tick = 0;} if (flag == 1)flag = 0; break; default: break;} }

Basic timer1

The Basic Timer1 generates an interrupt once every second. It uses two counters in series, where the BTCNT2 counter input uses the BTCNT1 counter output divided by 256. The BTCNT1 counter input is the ACLK clock signal with a frequency of 32.768 kHz.

If BTCNT2 counter selected output is divided by 128, what is the time period associated with the Basic Timer1 interrupt? _________

What are the values to write to the configuration registers?

BTCTL = BTDIV | BT_fCLK2_DIV128; // (ACLK/256)/128 IE2 |= BTIE; // enable BT interrupt//*********************************************************// Basic Timer ISR. Run with 1 sec period //*********************************************************#pragma vector=BASICTIMER_VECTOR __interrupt void basic_timer_ISR(void){ unsigned int read_data; // read data from file , frequency in kHzP2OUT^=0x06; // toogle LED1 and LED2counter++;if (counter == 5){ counter = 0;read_data = 200; TBCCR0 = 7995392/read_data;TBCCR4 = TBCCR0/2; }}

I/o ports configuration

// SW1 and SW2 configuration (Port1) P1SEL&= 0x00; // P1.0 and P1.2 I/O P1DIR&= 0x00; // P1.0 and P1.2 as inputs P1IFG = 0x00;P1IES&= 0xFF // high-to-low transition interrupt P1IE |= 0xFF; // enable port interrupts// LED1 and LED2 configuration (Port2):P2DIR |= 0x06; // P2.2 and P2.1 as outputs P2OUT = 0x04; // LED1 on and LED2 off// Buzzer port configuration (Port3)P3SEL |= 0x20; // P3.5 as special function P3DIR |= 0x20; // P3.5 as digital output

Fll+ configuration

FLL_CTL0 |= DCOPLUS + XCAP18PF; //DCO+ set,freq=xtal*D*N+1 SCFI0 |= FN_4; // x2 DCO freq, 8MHz nominal DCOSCFQCTL = 121; // (121+1) x 32768 x 2 = 7.99 MHz

Analysis of operation

System clocks inspection

The MCLK, SMCLK and ACLK system clocks are available at ports P1.1, P1.4 and P1.5 respectively. These ports are located on the SW2, RESET_CC and VREG_EN lines, which are available on the H2 Header pins 2, 5 and 6. All these resources are available because the Chipcon RF module is not installed and SW2 is not used.

Using the Registers view, set bits 1, 4 and 5 of P1SEL and P1DIR registers to choose the secondary function of their ports, that is, configured as outputs. Connect an oscilloscope probe at these positions to monitor the clock signals.

What are the values measured for each of the system clocks?

ACLK: _____________________

SMCLK: ____________________

MCLK: _____________________

Tbccr4 unit output frequency

With the help of an oscilloscope, it is possible to evaluate the operation of the application. Alternatively, it is possible to listen to the sound produced. By removing jumper JP1 and connecting the oscilloscope to this pin, it is possible to view the PWM signal produced by the microcontroller. The duty-cycle can be reduced or increased by pressing the push buttons SW1 and SW2.

Port p1 interrupt source decoding

All Port P1 interrupt lines share the same interrupt vector. The decoding is done through the P1IFG register.

This process can be observed by entering a breakpoint at the first line of the ISR code.

Execute the application.

The application’s execution is suspended at the breakpoint by pressing either button SW1 or SW2. From this point onwards, run the lines of code step-by-step and observe changes in the register values.

Measurement of electrical current drawn

The power consumption was discussed in the previous point. The electrical power required by the system during operation is measured by replacing the jumper on the Header PWR1 by an ammeter, which indicates the electric current taken by device during operation.

What is the value read? __________

This example and many others are available on the MSP430 Teaching ROM.

Request this ROM, and our other Teaching Materials here (External Link)

Questions & Answers

what is math number
Tric Reply
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
Alfred Reply
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
Kala Reply
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
Moses Reply
12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Shirleen Reply
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Adu
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
kinnecy Reply
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
Kimberly Reply
Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
August Reply
What is the expressiin for seven less than four times the number of nickels
Leonardo Reply
How do i figure this problem out.
how do you translate this in Algebraic Expressions
linda Reply
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
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Source:  OpenStax, Teaching and classroom laboratories based on the “ez430” and "experimenter's board" msp430 microcontroller platforms and code composer essentials. OpenStax CNX. May 19, 2009 Download for free at http://cnx.org/content/col10706/1.3
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