...ECET 365 week 1 homework Pgs 454-455 Austin Salankey 5/3/2016 DeVry University 8.9 A stepper motor has 24 North teeth and 24 South teeth. What angle change occurs on each step? If a full step is output every 1 ms (and assuming it doesn’t slip), at what speed does the motor spin? ANSWER: 360°/(4n) = 360°/[4(24)] = 360°/96 = 3.75° Revolutions per Second (4f/N) = 4(1000s)/(96) = 4000/96 = 41.667rps -or- 41.667rps(60sec) = 2500rpm 8.10 Draw a figure similar to Figure 8.80 showing how half stepping works. D8.13) Design an interface for a +24 V, 500 mA geared DC motor. The time constant of the motor is 10 ms. Include software to adjust the delivered power from 0 to 100%. Answer: I chose the TIP120 because it can sink at least 3 X the current needed for this motor. The base current is: I_b=I_coil/h_fe =0.5A/1000= 0.5mA Disired interface resistor = R_(b )≤((V_OH- V_be ))/I_(b ) = 5KΩ Now to make PWM 10 times faster than the 10 mS we go to 1mS; or a frequency of 1kHz // MC912 C32 C // 1 ms PWM on PT0 void PWM_Init (void) { MODRR |= 0x01; // Port T, pin 0 is connected to PWM system PWME |= 0x01; // Enable PWM Channel 0 PWMPOL |= 0x01; // PT0 High then Low PWMCLK |= 0x01: // Clock SA PWMPRCLK = (PWMPRCLK & 0xF8) | 0x04; // A = E/16 PWMSCLA |= 5; //SA = A/10, 0.25*160 = 40 us PWMPER0 = 25; ...
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