Solutions_3 - EE 310 Problem Set No. 3 Solutions 2.50 (a)...

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EE 310 Problem Set No. 3 Solutions 2.50 (a) (i) 5 = I υ V, 1 D and 2 D on ( ) 5 . 0 6 . 0 5 . 0 5 5 5 6 . 0 5 + = + + O O O O ( ) 7 . 0 0 . 2 0 . 2 20 . 0 20 . 0 2 . 1 0 . 1 88 . 0 = + + + = + + O O V (ii) 5 = I V 455 . 0 5 5 . 0 5 . 0 = + = I O V (b) (i) 5 = I V, 4 . 4 = O V (ii) 5 = I V, 6 . 0 = O V 2.59 (a) For 5 . 0 = I V, 0 3 2 1 = = = D D D I I I , 5 . 0 = O V (b) For 5 . 1 = I V, 1 D on; 0 3 2 = = D D I I 0667 . 0 8 4 7 . 0 5 . 1 1 = + = D I mA ( )( ) 23 . 1 8 0667 . 0 7 . 0 = + = O V (c) For 3 = I V, 1 D and 2 D conducting, 0 3 = D I 6 7 . 1 8 7 . 0 4 3 + = O O O ( ) 069 . 2 1667 . 0 125 . 0 25 . 0 2833 . 0 0875 . 0 75 . 0 = + + = + + O O V Then 171 . 0 8 7 . 0 069 . 2 1 = = D I mA 0615 . 0 6 7 . 1 069 . 2 2 = = D I mA (d) For 5 = I V, all diodes conducting 4 7 . 2 6 7 . 1 8 7 . 0 4 5 + + = O O O O ( ) 25 . 0 1667 . 0 125 . 0 25 . 0 675 . 0 2833 . 0 0875 . 0 25 . 1 + + + = + + + O So 90 . 2 = O V Then 275 . 0 8 7 . 0 90 . 2 1 = = D I mA 20 . 0 6 7 . 1 90 . 2 2 = = D I mA 05 . 0 4 7 . 2 90 . 2 3 = = D I mA
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EE 310 Problem Set No. 3 Solutions 3.3 (a) Enhancement-mode (b) From Graph V T Now = 1.5 V ( ) ( ) ( ) ( ) 2 2 2 2 0.03 2 1.5 0.25 0.12 0.15 3 1.5 2.25 0.0666 0.39 4 1.5 6.25 0.0624 0.77 5 1.5 12.25 0.0629 n nn n n n K KK K K K = −= ⇒= = = = = = = From last three, 2 (Avg) 0.0640 mA/V n K = (c) 2 2 (sat) 0.0640(3.5 1.5) (sat) 0.256 mA for 3.5 V (sat) 0.0640(4.5 1.5) (sat) 0.576 mA for 4.5 V D D GS D D GS i iV i = −⇒ = = = = = 3.5 (a) ( ) 8 . 1 4 . 0 2 . 2 = = = TN GS DS V V sat V V ( ) ⇒ = > = 8 . 1
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This note was uploaded on 03/11/2012 for the course E E 310 taught by Professor Wharton,markjedwards,perrysto during the Spring '10 term at Pennsylvania State University, University Park.

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Solutions_3 - EE 310 Problem Set No. 3 Solutions 2.50 (a)...

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