EE2005-Tutorial5-Solution - NATIONAL UNIVERSITY of...

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NATIONAL UNIVERSITY of SINGAPORE Department of Electrical and Computer Engineering EE2005 – Electronics Tutorial 5 Solution Q1. (a) Equate i DN = i DP , we have K[(v I – V th )v o – 0.5v o 2 ]= 0.5K(V DD – v I -V th ) 2 Therefore, (v I – V th )v o – 0.5v o 2 = 0.5(V DD – v I -V th ) 2 --- Eqn. (1) (1 mark) Differentiating with respect to v I , (v I – V th ) I o dv dv + 1.v o – 0.5(2)v o I o dv dv = 0.5(2)(V DD – v I -V th )(-1) Thus, (v I – V th ) I o dv dv +v o – v o I o dv dv = -(V DD – v I -V th ) . --- Eqn. (2) (1 mark) (b) Substitute v I = V IH and I o dv dv = -1 into Eqn. (2): (V IH – V th )(-1)+v o – v o (-1)= -(V DD – V IH -V th ) -V IH + V th +2v o = -V DD + V IH +V th Thus, v o = V IH - 0.5V DD ( 2 m a r k s ) (c) Substitute v I = V IH and v o = V IH – 0.5 V DD into Eqn. (1): (V IH – V th )( V IH – 0.5 V DD ) – 0.5(V IH – 0.5 V DD ) 2 = 0.5(V DD – V IH -V th ) 2 Rearrange, V IH (V DD – 2V th )= (5/8) V DD 2 – 1.5V DD V th +0.5V th 2 V IH
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EE2005-Tutorial5-Solution - NATIONAL UNIVERSITY of...

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