Fig. 1 Node/Mesh example: Bridge ckt with a pair of LEDs Each LED conducting in
ID: 1831222 • Letter: F
Question
Fig. 1 Node/Mesh example: Bridge ckt with a pair of LEDs Each LED conducting in the forward direction drops ~ 2V. Part 1: Node/Mesh example (with LH3330 Light Emitting Diodes) The ckt in Fig. 1 has component values: Is = 20mA, R1 = 240 Ohm , R2 =150 Ohm , R3 is to be found, R4 =330 Ohm , and R5 =110 Ohm . Each forward-biased LH3330 (or equivalent part no.) red LED can be modeled as an ideal voltage source of value 2V. Use mesh and/or node analysis to find the value of R3 that makes the LED's equally bright (i.e. each LED passes the same amount of current). Show your work.Explanation / Answer
Left loop: 0.02 Right loop: i3current through the bottom LED = (0.02 - i3) We want the top LED to pass the same amount of current.
Top loop: R3(0.02 - i3) + 2 + 150((0.02 - i3) - i3) +240((0.02 - i3) - 0.02) = 0 0.02R3 - i3R3 + 2 + 3 -150i3 - 150i3 + 4.8 - 240i3 - 4.8 = 0 (0.02 - i3)R3 = -5 + 540i3 R3 = (-5 + 540i3) / (0.02 -i3) (1)
Loop i3: 110i3 -2 + 330(i3 - 0.02) + 150(i3- (0.02 - i3)) = 0 110i3 - 2 + 330i3 - 6.6 + 150i3 - 3 -150i3 = 0 440i3 = 11.6 i3 =(29/1100) (2)
From (1) & 2):
R3 = (-5 + 540i3) / (0.02 - i3) = (-10160/7) = -1451.42
i R3 = i top Led = (0.02- i3) = (-7/1100) = -6.36 mA
i R1 = ((0.02 - i3) - 0.02) = (-29/1100) = -26.36mA
i R2 = ((0.02 - i3) - i3) = (-9/275) = -32.73mA
i R4 = i bottomLED = (0.02 - i3) = (-7/1100) = -6.36 mA
i R5 = i3 = 26.36mA
C = 110i3 = 2.9 V
KVL: -B + 150(i3 - (0.02 - i3)) + C = 0 B = (859/110) B = 7.81 V
KVL: -A + 240(0.02 - (0.02 - i3)) + B = 0 A = (163/110) = 1.48V
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