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
, PART ONE
Solutions to End-of-Chapter Problems
,Chapter 1
Section 1-1
1. G = R1 = 22kQ
1
=45.5Il S
2. 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)
4. 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
1 1
T = f = 31.8 Hz = 31.4 rns
7. f = T1 = 271Il s = 37.0 Hz
1
, Vrms 3.5 V
8. Vpp = 0.354 = 0.354 = 9.90 V
Vrms 0.707 Vp
9. Vavg = 0.636 Vp = 1.11
10. The 5th hannonic is (5)(500Hz) = 2.5 kHz
11. Odd harmonics
Section 1-3
12. For the circuit shown in Figure 1-25 (text): Rm = 3.58 kCl
8.9 kQ)
Vrn= Vol = (12 V) ( 5.6 kQ = 7.55 V Vm=7.S5V~
Rrn= 1.5 kQ + (5.6 kQ 113.3 kQ) = 3.58 kQ
The equivalent circuit is shown in Figure 1-2.
T
FIGURE 1-2
13. 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:
15
VL = (7.55 V)( 08kk~) = 1.65 V
For RL = 2.7 kQ:
V L = (7.55 V)(i.i8 kk~) = 3.25 V
For RL = 1.0 kQ:
VL = (7.55 V)( lt8~~) = 3.79 V
14. 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.
2
for
Analog Fundamentals
A Systems Approach
1st Edition
Thomas L. Floyd
David M. Buchla
Part 1: Page 1-89
Part 2: Page 90-152
, PART ONE
Solutions to End-of-Chapter Problems
,Chapter 1
Section 1-1
1. G = R1 = 22kQ
1
=45.5Il S
2. 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)
4. 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
1 1
T = f = 31.8 Hz = 31.4 rns
7. f = T1 = 271Il s = 37.0 Hz
1
, Vrms 3.5 V
8. Vpp = 0.354 = 0.354 = 9.90 V
Vrms 0.707 Vp
9. Vavg = 0.636 Vp = 1.11
10. The 5th hannonic is (5)(500Hz) = 2.5 kHz
11. Odd harmonics
Section 1-3
12. For the circuit shown in Figure 1-25 (text): Rm = 3.58 kCl
8.9 kQ)
Vrn= Vol = (12 V) ( 5.6 kQ = 7.55 V Vm=7.S5V~
Rrn= 1.5 kQ + (5.6 kQ 113.3 kQ) = 3.58 kQ
The equivalent circuit is shown in Figure 1-2.
T
FIGURE 1-2
13. 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:
15
VL = (7.55 V)( 08kk~) = 1.65 V
For RL = 2.7 kQ:
V L = (7.55 V)(i.i8 kk~) = 3.25 V
For RL = 1.0 kQ:
VL = (7.55 V)( lt8~~) = 3.79 V
14. 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.
2
