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Unformatted text preview: EECS 40, Fall 2008 Prof. ChangHasnain Homework #11 + Solutions This is just a practice homework, you dont have to turn it in! Transistor Basics 1. P12.9 LoadLine Analysis 2. P12.24 3. Consider the following circuit with V DD = 5V, R 1 = 20k , R 2 = 20k , R S = 3k , R D = 10k R D R 1 R 2 R S V G V DD M 1 V O a. Find the value of V G . b. From the given I D vs. V GS curve, find the values of I DQ , V GSQ . c. From the graph, what is the value of V To ? d. What region of operation is the transistor in? e. Find V o . f. Repeat for R D = 15k (Assume the I D vs. V GS curve is the same)? Bias Circuits 4. P12.31 5. P12.28 0.5 1 1.5 2.5 3 2 3.5 0.6 0.4 0.2 0.8 1.0 I D (mA) V GS (V) 0 0 SmallSignal Model 6. P12.44 7. P12.50 8. P12.52 9. P12.57 Solve a) and b) and redo the problem assuming r d = 100k (instead of infinity). How does this affect the gain? How does it affect the input/output resistance? CMOS Logic 10. P12.58 Instead of drawing simple switches draw the equivalent circuits assuming the a transistor has an equivalent resistance of R when it is on and can be represented by an open circuit when it is off. Solutions: Problem 1 With V GS = V DS = 5V, the transistor operate in the saturation region with we have I d = K*(V GS V TO ) 2 . Solving for K and substituting values we obtain K = 125uA/V 2 . However, we have K = (W/L) (KP/2). Solving for W/L and substituting values we obtain W/L = 5. If L = 2um, W = 10um. Problem 2 Using KVL, we find that V GS (t) = sin (2000 t) + 7 10 = 3 + sin (2000 t) Therefore, the maximum V GS = 2, the quiescent V GS = 3, and the minimum V GS =  4 V....
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 Fall '07
 ChangHasnain
 Transistor

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