IngramGaryEE3350Homework2

IngramGaryEE3350Homework2 - Gary Ingram Homework #2...

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Gary Ingram Homework #2 07/26/10 1) 3.4 In Fig. 3.22a, calculate the load current, load voltage, load power, diode power, and total power. Since this is an ideal diode, and it is in forward bias, it is equivalent to a closed switch, and there is no voltage loss across the diode. Load Current ± ² ³´µ¶· ¸µµµ¹ ² º»¼½ Load Voltage ¾ ± ² º»¾ Load Power ¿ ± ² ³º»¼½·³º»¾· ² »ÀÁ Diode Power ¿ à ² ³»¾·³º»¼½· ² »Â Total Power ¿ Ä ² »ÀÁ 2) 3.29 What value should R 2 be in Fig. 3-23b to set up a diode current of 0.25mA? Using loop analysis and the 2D approximation, I will replace the diode with a voltage source and a closed gate, knowing the current Å Æ is equal to .25mA, I will solve for Å Æ and plug in .25 mA and solve for R 2 . ³ÇÈÉ Ê Ë Æ ·Å Ì Ê ³ÍË Æ ·Å Æ ² ÌÆ ³ÍË Æ ·Å Ì Ê ³ÎÉ Ê Ë Æ ·Å Æ ² ÍÈÀ Ï Å Æ ² ³ÇÀ ÆÆÐÑ Í Ò·³Ë Æ Í ÌÐÎÐÀ ÒÈÏ· ³Ë Æ Ê ÒÆÐÎÀ ÏÌÒ· ² ÈÀ ÆÎÓÔ Ë Æ
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IngramGaryEE3350Homework2 - Gary Ingram Homework #2...

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