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# /diy/ - Do-It-Yourself

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 >> Anonymous Wed Mar 3 14:34:18 2021 No.2046073   [View] File: 18 KB, 550x357, file.png [View same] [iqdb] [saucenao] [google] Doing a practice test for some diode analysis using CVD. I have some calculation errors that I'm not sure how I got them. The answers are given below, my Va doesn't match theirs by a little so I'm not sure if it's just my paranoia or I genuinely I got it wrong (they said Vx but I think its because my professor forgot it was Va so Vx = Va). Here's how I did it:1) For diode analysis, we're told to make some assumptions on diodes' state first. My assumption for this analysis is that D1 = on, D2 = off. 2) I changed D1 to a voltage source with 0.7V (this is what I'm using for turn-on voltage), and my D2 to open circuit.3) Because the current doesn't travel through R3 and R4, I just ignore those 2 resistors and know that the Va is probe is equal to the voltage across that parallel line with D1 and R2.4) I did mesh analysis, drawing a clockwise loop:-5v (given voltage source) + (1500)(I) [voltage across resistor 1] + 0.7v (diode) + (1500)(I) [v across resistor 2] = 0Solving for the current, I get 0.00143A. Since the current going through R1 and R2 is the same, they're both 1.43 mA which matches with the answer given below.-------Here is probably where I did something wrong.---5) I assume that I need to know Vb because at the end of the branch/line(?), it is ground, so if I know Vb then that'll equal to Va. I_r1 = (5v - Vb)/(1500)Vb = 5v - 0.00143A * 1500Vb = 2.855VVa=VbVa (answer key at bottom) = 2.850VVb (mine) = 2.855VWas my comprehension/method of doing the question wrong or is it just some rounding that led to my paranoia for missing that 0.005V?And to verify my work, since Vb = 2.855V is greater than 0.7V, the diode should turn on which should agree to my assumption.

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