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Unformatted text preview: CHAPTER 12 Solutions for 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....
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- Spring '08
- Electrical Engineering