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# 0132130084_ism12 - 458 CHAPTER 12 Exercises E12.1 (a) v GS...

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Unformatted text preview: 458 CHAPTER 12 Exercises E12.1 (a) v GS = 1 V and v DS = 5 V: Because we have v GS < V to , the FET is in cutoff. (b) v GS = 3 V and v DS = 0.5 V: Because v GS > V to and v GD = v GS − v DS = 2.5 > V to , the FET is in the triode region. (c) v GS = 3 V and v DS = 6 V: Because v GS > V to and v GD = v GS − v DS = − 3 V < V to , the FET is in the saturation region. (d) v GS = 5 V and v DS = 6 V: Because v GS > V to and v GD = v GS − v DS = 1 V which is less than V to , the FET is in the saturation region. E12.2 First we notice that for V, 1 or = GS v the transistor is in cutoff, and the drain current is zero. Next we compute the drain current in the saturation region for each value of v GS : 2 6 2 1 2 1 mA/V 1 ) 2 / 80 )( 10 50 ( ) / ( = × = = − L W KP K 2 ) ( to GS D V v K i − = The boundary between the triode and saturation regions occurs at to GS DS V v v − = GS v (V) D i (mA) DS v at boundary 2 1 1 3 4 2 4 9 3 In saturation, i D is constant, and in the triode region the characteristics are parabolas passing through the origin. The apex of the parabolas are on the boundary between the triode and saturation regions. The plots are shown in Figure 12.7 in the book. E12.3 First we notice that for V, 1 or − = GS v the transistor is in cutoff, and the drain current is zero. Next we compute the drain current in the saturation region for each value of v GS : 459 2 6 2 1 2 1 mA/V 25 . 1 ) 2 / 200 )( 10 25 ( ) / ( = × = = − L W KP K 2 ) ( to GS D V v K i − = The boundary between the triode and saturation regions occurs at to GS DS V v v − = . GS v (V) D i (mA) DS v at boundary − 2 1.25 − 1 − 3 5 − 2 − 4 11.25 − 3 In saturation, i D is constant, and in the triode region the characteristics are parabolas passing through the origin. The apex of the parabolas are on the boundary between the triode and saturation regions. The plots are shown in Figure 12.9 in the book. E12.4 We have 3 ) 2000 sin( ) ( ) ( + = + = t V t v t v GG in GS π Thus we have 4 max = GS V V, 3 = GSQ V V, and 2 min = GS V V. The characteristics and the load line are: 460 For v in = +1 we have v GS = 4 and the instantaneous operating point is A . Similarly for v in = − 1 we have v GS = 2 V and the instantaneous operating point is at B . We find V DSQ ≅ 11 V, V DS min ≅ 6 V, V DS max ≅ 14 V. E12.5 First, we compute V 7 2 1 2 = + = R R R V V DD G and 2 6 2 1 2 1 mA/V 5 . ) 10 / 200 )( 10 50 ( ) / ( = × = = − L W KP K As in Example 12.2, we need to solve: ( ) 2 1 2 2 = − +         − + K R V V V V K R V S G to GSQ to S GSQ Substituting values, we have 6 2 = − − GSQ GSQ V V The roots are V GSQ = − 2 V and 3 V. The correct root is V GSQ = 3 V which yields I DQ = K ( V GSQ − V to ) 2 = 2 mA. Finally, we have V DSQ = V DD − R S I DQ = 16 V....
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## This note was uploaded on 11/20/2011 for the course ELE 220 taught by Professor Anas during the Spring '11 term at American Dubai.

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0132130084_ism12 - 458 CHAPTER 12 Exercises E12.1 (a) v GS...

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