Solutions Manual - Chapter 1 Section 1-1 1 1 1. G = R = 22kQ =45.5IlS

Solution’s Manual for Analog Fundamentals A Systems Approach 1st Edition Thomas L. Floyd David M. Buchla

Part 1: Page 1-89

Part 2: Page 90-152 1 / 4

PART ONE

Solutions to End-of-Chapter Problems 2 / 4

Chapter 1 Section 1-1

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• G = R = 22kQ =45.5IlS
• The resistance of a diode decreases as the voltage increases.

3. The ac resistance at V = 0.7 V and] = 5.0 rnA is approximately:

(0.72 V -0.675 V) = 333 Q

(5.5 rnA -4.0 rnA)

• An W curve with decreasing ac resistance as voltage increases is shown in Figure 1-1.

/ / / I»(mA) VM

FIGURE 1-1

Section 1-2

5. For the sine wave given by the equation vet) = 100 sin(200 t +0.52):

(a) V = 100 V

V:Vg = (0.636)(100 V) = 63.6 V

(0 = 200 rad/s (b) v(2.0 ms) = 100 sin(200 rad/s (200 ms) +0.52 rad) = 100 sin (40.52) = 31.5 V

6. For the voltage v(t) = 100 sin (200 t + 0.52):

f = 200 rad!s = 31.8 Hz 21t rad! cycle

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T = f = 31.8 Hz = 31.4 rns

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• f = T = 27 Ils = 37.0 Hz

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V rms 3.5 V

• V pp = 0.354 = 0.354 = 9.90 V

V rms 0.707 Vp

• Vavg = 0.636 Vp = 1.11

10.The 5 th hannonic is (5)(500Hz) = 2.5 kHz

• Odd harmonics

Section 1-3

12. For the circuit shown in Figure 1-25 (text):

( 5.6 kQ) Vrn= Vol = (12 V) 8.9 kQ = 7.55 V Rrn= 1.5 kQ + (5.6 kQ 113.3 kQ) = 3.58 kQ The equivalent circuit is shown in Figure 1-2.Rm = 3.58 kCl Vm=7.S5V~ T

FIGURE 1-2

• For the circuit in Figure 1-25 (text), Rrn = 3.58 kQ and Vrn = 7.55 V (see problem

12).For RL =

1.0 kQ:

V L = (7.55 V)( 15

• kk~) = 1.65 V

For RL = 2.7 kQ:

VL = (7.55 V)(i.i8 kk~) = 3.25 V

For RL = 1.0 kQ:

VL = (7.55 V)( lt8 ~~) = 3.79 V

• From problem 12, Rrn = 3.58 kQ and Vrn = 7.55 V

RN = Rrn = 3.58 kQ Vrn 7.55 V IN= Rrn = 3.58 kQ = 2.11 rnA

15. The load line (for Figure 1-26 of the text) crosses the y-axis at:

Vrn 15 V I SAT = R rn = 200 kQ = 75 JlA

and crosses the x-axis at:

Vco = Vrn= 15 V The load line drawn between these two points as shown. The IV curve for a 150 kQ resistor is shown in Figure 1-3.

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