Lecture 17

# Bermel solution 3 k 12 v 100 f 300 3 09 ln 2202012 200

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Unformatted text preview: ln = ECE 201-4, Prof. Bermel Solution 3 kΩ 12 V + - 100 µF (−300 )3 Ω − 0.9 ln − = 2/20/2012 200 µF − 0.0036 − 0.0036 = −0.0036 = 1− / / ECE 201-4, Prof. Bermel = Here: Qf=3.6 mC T=0.9 s Io=4 mA Solution 12 V + - 3 kΩ 200 µF 100 µF = = 1 = 2 2/20/2012 / 1− / (t&gt;0) 1− / ECE 201-4, Prof. Bermel Here: Qf=3.6 mC T=0.9 s Io=4 mA Solution Current (mA) 4 3.5 3 2.5 2 1.5 1 0.5 0 -2 2/20/2012 -1 0 1 ECE 201-4, Prof. Bermel 2 3 4 Solution Energy Stored (mJ) 22 20 18 16 14 12 10 8 6 4 2 0 -2 2/20/2012 -1 0 1 ECE 201-4, Prof. Bermel 2 3 4 Example 3 • How does this circuit modify a standard AC input voltage? What is its likely purpose? 2/20/2012 ECE 201-4, Prof. Bermel Solution • Diodes are a “one-way street” for electrons • Ideal Diode: , &gt;0 = 0, otherwise • Realistic (Shockley...
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