ee577_sp07_midterm_solutions_morning

ee577_sp07_midterm_solutions_morning - 1 EE577a Midterm...

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Unformatted text preview: 1 EE577a Midterm Instructor: Prof. Pedram March 8, 2007 Morning Session and DEN Students Name: _________TA____________________ Student ID #: ________________________________ Problem 1: _____________ / 10 Problem 2: _____________ / 10 Problem 3: _____________ / 10 Problem 4: _____________ / 10 Problem 5: _____________ / 10 Problem 6: _____________ / 10 Problem 7: _____________ / 10 Problem 8: _____________ / 10 Problem 9: _____________ / 10 Problem 10: _____________ / 10 Total: _____________ / 100 2 1) Calculate the noise margin low (NM L ) for a resistive-load inverter circuit with the following parameters: V DD = 5V V T0,n = 1.0V, R=2.5K Ω k n = (W/L)k’ n =120 μ A/V 2 R V DD V in V out Solution: ( ) ( ) 2 , 2 2 OL OL n t dd n OL dd V V V V k R V V − − = − (2 point) V OL =2.81V (1 point) ) 2 ( ... ) ( 2 ) 1 ( ... 2 I V V K I R V V th in out dd = − = − Solving (1) and (2), out th in dd th in out dd V V V R K V V V K R V V = − ⋅ − ⇔ − = − 2 2 ) ( 2 ) ( 2 (2 points) Now, differentiating the above equation, we get: ) ( th in in out V V R K V V − ⋅ − = ∂ ∂ ⇔ (2 points) Therefore, 1 ) ( − = − ⋅ − = ∂ ∂ th in in out V V R K V V (1 points) 33 . 4 3 / 13 1 = = + ⋅ = ⇔ th IL V R K V (1 point) 52 . 1 81 . 2 33 . 4 = − = NM (1 point) 3 2) Use the average capacitance current equation to calculate the 80% to 20% output fall time for a CMOS inverter operating on a V DD of 3V and driving an output capacitance of 100fF. NMOS PMOS k 115 μ A/V 2 30 μ A/V 2 V T,0 0.45V -0.40V λ 0.05 V-1-0.10 V-1 Solution: Both 80% and 20% are in linear regions for NMOS. (2 points) Therefore, } ] 1 [ ] ) ( 2 [ 2 ] 1 [ ] ) ( 2 [ 2 { 2 1 % 20 2 % 80 2 ds ds ds th gs ds ds ds th gs avg V V V V V k V V V V V k I ⋅ + ⋅ − − + ⋅ + ⋅ − − = λ λ } ] 6 . 05 . 1 [ ] 6 . 6 . ) 45 . 3 ( 2 [ 2 10 115 ] 4 . 2 05 . 1 [ ] 4 . 2 4 . 2 ) 45 . 3 ( 2 [ 2 10 115 { 2 1 % 20 2 6 % 80 2 6 ⋅ + ⋅ − − × + ⋅ + ⋅ − − × = − − A μ 6 . 288 = (4 points) Now, Δ V =0.8 × V DD-0.2 × V DD =2.4 − 0.6=1.8V 0....
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This note was uploaded on 03/19/2008 for the course EE 577B taught by Professor Bhatti during the Spring '08 term at USC.

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ee577_sp07_midterm_solutions_morning - 1 EE577a Midterm...

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