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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 opamp. 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.
 Summer '07
 Boser
 Volt

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