0132130084_ism12

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 29

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online