CHAPTER 9
Solutions for Exercises
E9.1 The equivalent circuit for the sensor and the input resistance of the amplifier is shown in Figure 9.2 in the book. Thus the input voltage is Rin v in = v sensor Rsensor + Rin We want the input voltage with an i
CHAPTER 10
Solutions for Exercises
E10.1 Solving Equation 10.1 for the saturation current and substituting values, we have iD Is = exp(vD / nVT ) 1 10 4 = exp(0.600 / 0.026) 1 = 9.502 10 15 A Then for vD = 0.650 V, we have iD = I s exp(vD / nVT )
CHAPTER 12
Solutions for Exercises
E12.1 (a) vGS = 1 V and vDS = 5 V: Because we have vGS < Vto, the FET is in cutoff. (b) vGS = 3 V and vDS = 0.5 V: Because vGS > Vto and vGD = vGS - vDS = 2.5 > Vto, the FET is in the triode region. (c) vGS = 3 V an
CHAPTER 14
Solutions for Exercises
E14.1
(a) iA =
vA RA
iB =
vB RB
iF = iA + iB =
v A vB + RA RB
v v v o = RF iF = RF A + B R A RB vA = RA . (b) For the vA source, RinA = iA (c) Similarly RinB = RB . (d) In part (a) we found that the outp
CHAPTER 15
Solutions for Exercises
E15.1 If one grasps the wire with the right hand and with the thumb pointing north, the fingers point west under the wire and curl around to point east above the wire. If one places the fingers of the right hand on
CHAPTER 16
Solutions for Exercises
E16.1 The input power to the dc motor is Pin = Vsource I source = Pout + Ploss Substituting values and solving for the source current we have 220I source = 50 746 + 3350 I source = 184.8 A Also we have P 50 746 =
CHAPTER 17
Solutions for Exercises
E17.1 From Equation 17.5, we have
Bgap = Kia (t ) cos( ) + Kib (t ) cos( 120 ) + Kic (t ) cos( 240 )
Using the expressions given in the Exercise statement for the currents, we have
Bgap = KI m cos(t ) cos( ) +