Exercise 4 Problem 1, 2 & 3

Exercise 4 Problem 1, 2 & 3

par Alexandre Nicolas Lechartier,
Number of replies: 1

Dear Prof. Enz, dear TAs,

I have some questions about exercise 4.

1) In exercise 4 problem 1, I don't understand how from eq.2, the element \frac{A_m^{5/3}}{W^1.75} transforms to  \frac{l_{tot}^{5/3}}{W^0.083} in equation 4. May I have some hints about that?

2) Why do we use L \approx 35 nH in problem 2?

3) In problem 3, isn't 2WC_{ox} = 0.18fF instead of 17fF? If not, may I have some explanations over the reason?

Thank you,

Alexandre

In reply to Alexandre Nicolas Lechartier

Re: Exercise 4 Problem 1, 2 & 3

par Arthur Eugène Dietrich,
Hello Alexandre,

1) If I'm not mistaken, A_m = W*l_tot, so A_m^(5/3)/W^1.75 = (W*l_tot)^(5/3)/W^1.75 =(l_tot)^(5/3)/W^0.083 as 1.75-5/3 = 0.083.

2) I'm not sure but as specified in the question, we need to design the same inductor as in the 2nd point of the problem. In the correction of problem 1, the inductance is 35nH so we re-use this value.

3) For this question, I have no idea so if a TA can answer it, it would be great!