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EE 240 - HOMEWORK - Solution 1

# EE 240 - HOMEWORK - Solution 1 - OUT SOLUTION V OUT =...

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EE240-A2-2010 Assignment #1 SOLUTION Power-Limited Voltage Divider The problem is to design a voltage divider (as in the circuit diagram below) in which the output voltage is a predetermined fraction, f, of the supply voltage, 12V, and the power drain on the voltage source is close to, but no more, than 100 mW. Therefore, you need to determine R 1 and R 2 that satisfy these conditions. Find the pair of R 1 and R 2 that gives the correct V OUT and has a total power drain from the source of 100 mW. Now for the real-life complication. Resistors don’t come in all values so after you determine the exact R 1 and R 2 to get the correct V OUT , select the closest available ones from the standard values for 5% resistors. Look for these values in the posted file, StandardR&C.pdf. Recalculate the power to make sure you are still within the limit and determine the percentage error in your V OUT . Percentage Error = 100 x (Obtained V OUT – Desired V OUT )/ (Desired V

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Unformatted text preview: OUT ) SOLUTION V OUT = f*12V *= 12V * R 2 /(R 1 +R 2 ) (R 1 and R 2 make a voltage divider) So R 2 /(R 1 +R 2 ) = f (1) To calculate the power drain, The source current is I = 12V/(R 1 +R 2 ) so the source power supplied is P S = I*12V = 144V 2 /(R 1 +R 2 ) = 0.1 Watt Therefore (R 1 +R 2 ) = 1440 Ω (2) to get exactly 100 mW. Combine (1) and (2) to get R 2 = f*1440 R 1 = 1440 - R 2 . As an example, suppose f = 0.65. R 1 R 2 V OUT Then we want an output voltage of 7.80V with R 2 = 936 and R 1 = 504 Ω. To stay within the power limit of 100 mW, we must pick the resistors that have (R 1 +R 2 ) > 1440. (This makes sure the current is restricted below the max allowed.) The closest values satisfying this are R 2 = 910 Ω and R 1 = 560 Ω; R 1 +R 2 = 1470. Then the output voltage is V OUT = 12V* 910/(1470) = 7.429V instead of the desired 7.80V. Percentage Error = 100 x (7.429 - 7.80)/7.80 = - 4.75%....
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