Exam 2 Solutions 11

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Unformatted text preview: R2 -- > infinity, 0 (changes of Rin has no effect on Vout) Therefore, we should have as LARGE R2 as possible. Problem 2(a) 0 1) KCL @Vb: 2) KCL @ Vc: 0 3) KCL @ Vout: 0 Vb 1k Vout Vb 1k 0 0 0 Three unknowns: Vb, Vd, and Vout; three equations. After solving for the system, we get Vout 50Vin ------------------------ Problem 2(b) 0 1) KCL @Vb: 2) KCL @ Vc: 0 0 Vb 1k Vout Vb 1k 0 0 0 3) KCL @ Vout: Three unknowns: Vb, Vd, and Vout; three equations. After solving for the system, we get Vout Vin ∙ Then, IL = Vout / RL = Vin ∙ -------------------------------------------------- Problem 3(a) We want to be able to change the gain without changing the internal design of the op-amp. In addition, negative feedback provides stable gain as well. Problem 3(b) Amp*100 = 15V. Amp, max = 0.15V. The maximum amplitude without clipping is 0.15V. Problem 3(c) When clipping occurs, a pure sinusoidal wave is no longer sinusoidal. The distorted wave now contains waves of some high frequencies (since such clipped waves can be modeled as a combination of numerous sinusoids of multiple frequencies. Because of these additional frequencies, the sound is no longer the same. Problem 3(d) dB = 20 log (Gain) = 20 log (100) = 40 dB. Problem 4 0) Label all the nodes as shown above. 1) From Golden Rule #1, Vc = Vb. 2) KCL @ Vb: 0 3) KCL @ Vc: 0 0 0 From the two KCL equations, We can solve for Vout. Vb = aVin Vout 2a 1 Vin Vout 2a 1 Vin ---------------------...
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This note was uploaded on 01/24/2014 for the course EE 100 taught by Professor Boser during the Summer '07 term at University of California, Berkeley.

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