EE 240 - HOMEWORK - Solution 2

EE 240 - HOMEWORK - Solution 2 - (B Calculate the...

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EE240-A2-2010 Assignment #2 SOLUTION Variable Current Motor Controller The problem is to design a variable current divider (as in the circuit diagram below) which varies the current in a 10- Ω motor between 0A and a maximum of f*200mA where f is the same fraction that you used in Assignment #1 (described at the bottom of the page in case you have forgotten). The supply voltage V S is 12V. This is accomplished by using a 0-10 Ω "pot" (short for potentiometer, a variable resistor), R P , indicated as a resistor with an arrow through it in the diagram. When the pot is set to 0 Ω , the motor is shorted out and stops because the current from the source goes directly to ground. When R P is its maximum value of 10 Ω , the motor gets its maximum current. (A) Analyze this circuit and determine R that gives I MOTOR = f*200mA when R P = 10 Ω. The efficiency of such a system is the power consumed by the motor divided by the power supplied by the source.
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Unformatted text preview: (B) Calculate the efficiency of your system at maximum motor current. SOLUTION When R P = R M = 10 Ω , the two parallel currents are equal and their sum equals the total current from the 12V source. The effective resistance of R P and R M in parallel is 5 Ω and the total current from the source is 2*f*200mA. The source current is I = 12V/(R + 5 Ω ) = 2*f*200mA Therefore, R = 12V /(2* f*200mA ) - 5 Ω R 1 0 - V OUT MOTOR R M = 10 Ω Max = f*200mA R = ??? Ω R P = 0-10 Ω I MOTOR The efficiency is the motor power over the total source power supplied (P S = I*12V) Therefore, Eff. = [(f*0.200A) 2 R M ] / [12V * (2 * f*0.200A)] Eff. = [(f*0.200A) R M ] / (24V) = [I M *R M ] / (24V) EXAMPLE: Say f = 0.650. Then I M = 130 mA = 0.130 A R = 12/(0.4f) - 5 = 41.15 Ω . Eff = (0.130A*10 Ω )/24V = 0.054 = 5.4% Higher efficiency would be obtained with a higher motor current limit....
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EE 240 - HOMEWORK - Solution 2 - (B Calculate the...

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