exam_pastc

Exam_pastc - 2 3.5V V D V S R D V DD I DS 8V Given V T =1V K = 0.25mA/V 2 2k Ω Find V DS and I DS for V G = 3.5V MOS is in saturation mode mA m V

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Unformatted text preview: 2 3.5V V D V S R D V DD I DS 8V Given V T =1V K = 0.25mA/V 2 2k Ω Find V DS and I DS for V G = 3.5V MOS is in saturation mode mA m V V K I T GS DS 5625 . 1 ) 1 5 . 3 ( 25 . ) ( 2 2 = − = − = V k m R I V V D DS DD DS 875 . 4 2 ) 5625 . 1 ( 8 ≅ − = − = mA k I DS 5625 . 1 2 875 . 4 8 = − = ∴ 4 (18) Given V T = 1V K = 0.25mA/V 2 1 = ∴ I R 1 V 1 I1 8V V 2 V 3 6k Ω 12k Ω R 2 4. (a) If V 2 = 0.5V, find I 1 and V 3 . (b) If V 2 = 2V, find I 1 and V 3 . State clearly the reasons for your answer. Given V T = 1V, K = 0.25mA/V 2 . Given that at triode region, I DS = 2K[(V GS - V T )V DS –V DS 2 / 2] at saturation region, I DS = K[(V GS - V T ) 2 ] (19). (19) If V 2 = 0.5V a V V 8 3 = ∴ If V 2 = 3V b mA m V V K I T GS 1 ) 1 3 ( 25 . ) ( 2 2 1 = − = − = ∴ V k m R I V 2 ) 6 ( 1 8 8 1 1 3 = − = − = ∴ NMOS is saturated since 1. 3 > 1 2. 2 = 3 - 1 Assume NMOS is in saturation NMOS is saturated since 1. V GS > V T 2. V DS = V GS- V T V GS < V T NMOS cut off 0.5 < 1 R 1 V 1 I1 8V V 2 V 3 6k Ω 12k Ω R 2 (26) hence V O = 3 + 1 = 4V V DS = 12 V = V DD is impossible since MOS is not cut off (V DGS > V T ) Vi V O R D V DD I D 13V 2k Ω 9V 3V 2k Ω 5 (26) Given V T = 1V K = 0.5mA/V 2 NMOS is in triode since 1. V GS > V T 2. V DS < V GS- V T NMOS is triode since 1. 6 > 1 2. 1 < 6 - 1 V or V V V V V V V V V m k V V V m V V V V K k V R V V I DS DS DS DS DS DS DS DS DS DS DS DS DS T GS DS D DS DD D 1 10 10 11 ] 2 5 [ 2 10 ] 2 5 [ 1 2 10 )] 2 ) 1 2 8 [( 1 ] 2 ) [( 2 2 10 2 2 2 2 2 = ∴ = + − ∴ − = − ∴ − = − ∴ − − − = − − = −...
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This note was uploaded on 10/28/2011 for the course ELEC 101 taught by Professor Chan during the Spring '09 term at HKUST.

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Exam_pastc - 2 3.5V V D V S R D V DD I DS 8V Given V T =1V K = 0.25mA/V 2 2k Ω Find V DS and I DS for V G = 3.5V MOS is in saturation mode mA m V

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