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Week 1 Homework Ecet 365

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Submitted By MarkJ127
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Pages 454–455, Problems 8.9, 8.10, D8.13, D8.14, D8.15.

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?
The shaftwill rotate 360°/(4 · n) per step.
360 / (4 * 24) = 3.75 degrees of change occurs on each step.
If a stepper motor with 50 teeth North and South has 200 spr then 24 teeth should be about 100 spr.
Rps = 4f/N = 4 * 1MHz / 100 = 40000 rps.

8.10 Draw a figure similar to Figure 8.80 showing how half stepping works.
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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%.

D8.14 Design an interface for a ±12 V, 500 mA bipolar stepper motor. There are 200 steps/revolution. Write software to spin the motor at 1 rps.

D8.15 An alarm is powered by 120 VAC, and the on/off state of the alarm is controlled by an EM relay. Design the interface between a computer output port and the EM relay. The dropout voltage is 3.5 V, and the coil current needs at least 100 mA. Limit the coil voltage to 6 V. Write three software functions: Alarm_Init, Alarm_On, and

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Ecet 3650

...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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