Write a program to monitor bit PD7 connected to a switch. If PD7 is HIGH create

Question

Write a program to monitor bit PD7 connected to a switch. If PD7 is HIGH create a square wave of 50% duty cycle on bit 0 of Port B and if PD7 is LOW create a square wave of 66% duty cycle on bit 3 of Port B. The 50% duty cycle means that the “on” and “off" states (or the high and low portions of the pulse) have the same length. Therefore, we toggle PBO with a time delay between each state. VCC | ATmega 4.7k 3 PD7 Switch 50% Duty Cycle On Time = Off Time PB0 A 66% duty cycle means that the “on” state is twice the "off" state. VCC 4.7k ATmega PD7 66% means On Time = 2 x Off time Switch PB3

Explanation / Answer

#include <avr/io.h>

#include <util/delay.h>

void pwm_init()

{

    // initialize TCCR0 as per requirement, say as follows

    TCCR0 |= (1<<WGM00)|(1<<COM01)|(1<<WGM01)|(1<<CS00);

  

    // make sure to make OC0 pin (pin PB3 for atmega32) as output pin

    DDRB |= (1<<PB3);

// PD7 as a input

DDRD |= (0<<PD7);

}

  

void main()

{

    uint8_t duty;

    duty = 128;       // duty cycle = 50% of 255 = 127.5 = 128

duty1=170; //dutycycle = 66,66% of 255 = 169.98=170

  

    // initialize timer in PWM mode

    pwm_init();

if (PIND & (1<<PD7)) //check condition of switch

{

    // run forever

    while(1)

    {

        OCR0 = duty;

    }

}

else

{

    // run forever

    while(1)

    {

        OCR0 = duty;

    }

}

}

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