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10W-quiz-3-sol

# 10W-quiz-3-sol - ECE102(Winter 2010 Quiz 3 Problem 1 In the...

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ECE102 (Winter 2010), Quiz 3 Problem 1. In the circuit below, due to manufacturing errors transistors are not identi- cal. They both have k = 100 μ A/V 2 , λ = 0, and V t = 1 V, but ( W/L ) 1 = 1 . 5( W/L ) 2 = 20. Find V 1 , V 2 , and V 3 . (10pts). V 3 Q 1 Q 2 V 2 V 1 5 V 40k μ 200 A 40k Assume both transistors in saturation. Since V GS 1 = V GS 2 = 0 V 3 = V 3 : I D 1 = 0 . 5 k ( W/L ) 1 ( V GS 1 V t ) 2 I D 2 = 0 . 5 k ( W/L ) 2 ( V GS 2 V t ) 2 I D 1 I D 2 = ( W/L ) 1 ( W/L ) 2 = 1 . 5 I D 1 = 1 . 5 I D 2 KCL I D 1 + I D 2 = 200 μ A The last two equations can be solved to ±nd I D 1 = 120 μ A and I D 2 = 80 μ A. Then: I D 1 = 0 . 5 k ( W/L ) 1 ( V GS 1 V t ) 2 120 × 10 6 = 0 . 5 × 100 × 10 6 × 20 × ( V GS 1 1) 2 ( V GS 1 1) 2 = 0 . 12 The two roots are V GS 1 = 0 . 65 V (not correct as NMOS would be in cut-o²) and V GS 1 = 1 . 35 V. Then, V 3 = V GS 1 = 1 . 35 V. KVL 5 = 40 × 10 3 I D 1 + V 1 V 1 = 0 . 2 V KVL 5 = 40 × 10 3 I D 2 + V 2 V 2 = 1 . 8 V We need to proves that both transistors are in saturation: V DS 1 = V 1 V 3 = 0 . 2 ( 1 . 35) = 1 . 55 > V GS 1 V t = 1 . 35 0 . 35 = 1 V DS 2 = V 2 V 3 = 1 . 8 ( 1 . 35) = 3 . 15 > V GS 2

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10W-quiz-3-sol - ECE102(Winter 2010 Quiz 3 Problem 1 In the...

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