hw2_ext_sol

# hw2_ext_sol - p = 4W 3 Since we are assuming the ideal...

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EE 101 Handout # 6-a Prof. A. El Gamal Jan 24, 2003 Homework #2 Extra Problem Solutions 1. By cascading the ampliFers we get that R in = R in , R out = R out and a = a 2 R in R in + R out 2. Let’s assume that the diode is o±. So, i = 0 and the current controlled voltage source has value 0, i.e., is a short circuit. The circuit reduces to the current source connected to a 2Ω resistor. We Fnd the diode is reverse biased (has - 2V across it), so our assumption is correct. The power is supplied by the current source and is
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Unformatted text preview: p = 4W 3. Since we are assuming the ideal op-amp model, v + = v-= 0, i + = i-= 0 and the current through the resistor R 1 is i s . So, the potential of node 1 is v 1 =-i s R 1 . The current through R 2 is i 1 =-i s R 1 R 2 . By KCL we get i = i 1-i s =-i s R 1 R 2-i s =-i s ( R 1 R 2 +1). i s i R 1 R L R 2 i 1 1 1...
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## This note was uploaded on 04/06/2010 for the course EE 102 taught by Professor Pingli during the Fall '08 term at Shanghai Jiao Tong University.

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