Plot log 10 I D versus V D over the range 0.1 ≤ V D ≤ 0.7 V for (a) I S = 10 − 12 and (b) I S = 10 − 14 A .

Question

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Answer 1

Textbook Question

Chapter 1, Problem 1.30P

Plot log 10 I D versus V D over the range 0.1 ≤ V D ≤ 0.7 V for (a) I S = 10 − 12 and (b) I S = 10 − 14 A .

(a)

Expert Solution

To determine

To plot: The graph of log10ID versus VD .

Answer to Problem 1.30P

The graph of log10ID versus VD for the IS=10−12 is shown in Figure 1.

Explanation of Solution

Given:

Reverse saturation current IS=10−12 A

Diode Voltage range, 0.1≤VD≤0.7

Concept Used:

ID=IS(exp( V D V T ))

Calculation:

For IS=10−12

For VD=0.1V

ID=10−12(exp( 0.1 0.026 ))=4.68127×10−11 Alog10(ID)=−10.3296

For VD=0.2V

ID=10−12(exp( 0.2 0.026 ))=2.1914×10−9 Alog10(ID)=−8.6593

For VD=0.3 V

ID=10−12(exp( 0.3 0.026 ))=1.02591×10−7 Alog10(ID)=−6.9889

For VD=0.4 V

ID=10−12(exp( 0.4 0.026 ))=4.8023×10−6 Alog10(ID)=−5.3185

For VD=0.5 V

ID=10−12(exp( 0.5 0.026 ))=2.248×10−4 Alog10(ID)=−3.6482

For VD=0.6 V

ID=10−12(exp( 0.6 0.026 ))=1.05240×10−2 Alog10(ID)=−1.9778

For VD=0.7 V

ID=10−12(exp( 0.7 0.026 ))=4.92656×10−1 Alog10(ID)=−0.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 1

Figure 1

(b)

Expert Solution

To determine

To plot: The graph of log10ID versus VD .

Answer to Problem 1.30P

The graph of log10ID Versus VD for the IS=10−14 is shown in Figure 2.

Explanation of Solution

Given:

Reverse saturation current IS=10−14

Concept Used:

Formulae;

ID=IS(exp( V D V T ))

Calculation:

For IS=10−14

For VD=0.1V

ID=10−14(exp( 0.1 0.026 ))=4.68127×10−13 Alog10(ID)=−12.3296

For VD=0.2V

ID=10−14(exp( 0.2 0.026 ))=2.1914×10−11 Alog10(ID)=−10.6593

For VD=0.3 V

ID=10−14(exp( 0.3 0.026 ))=1.02591×10−9 Alog10(ID)=−8.9889

For VD=0.4 V

ID=10−14(exp( 0.4 0.026 ))=4.8023×10−8 Alog10(ID)=−7.3185

For VD=0.5 V

ID=10−14(exp( 0.5 0.026 ))=2.248×10−6 Alog10(ID)=−5.6482

For VD=0.6 V

ID=10−14(exp( 0.6 0.026 ))=1.05240×10−4 Alog10(ID)=−3.9778

For VD=0.7 V

ID=10−14(exp( 0.7 0.026 ))=4.92656×10−3 Alog10(ID)=−2.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 2

Figure 2

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5. Answer the following questions. Take help from ChatGPT to answer these questions (if you need). Write the answers briefly using your own words with no more than two sentences, and check whether ChatGPT is giving you the appropriate answers in the context of our class. a) What is the Bode plot? What kind of input do we consider for the frequency-response- based method? b) What is the advantage of design using the frequency-response method? c) Define gain margin, phase margin, gain crossover frequency, and phase crossover frequency.

Phase (deg) 3. The Bode diagram of a system is shown below. Magnitude (dB) System: sys -10 Frequency (rad/s): 0.141 Magnitude (dB): -15.6 -20 -30 40 -50 -60 0 -45 -90 -135 101 10° Bode Diagram System: sys Frequency (radis): 10 Magnitude (dB): -18.9 System: sys Frequency (rad/s): 10 Phase (deg):-52.2 101 Frequency (rad/s) 102 103 Find the steady-state output of the system for each of the following inputs. a) u(t) = 100 b) u(t) 100 cos(10 t + 10°) = c) u(t) = 500 + 200 cos(10 t + 10°)

Phase (deg) 270 4. Consider a closed-loop system with unity (negative) feedback. The Bode diagram of the open-loop transfer function is given below. Magnitude (dB) -500 -150 -50 10 dB System Frequency (eds): 6.63 Magnitude (B) 0.0778 Буку Frequency(): 10.1 Magnitude ()-705 Frequency(6.63 Phase (deg): -144 Frequency (rad): 10.1 Phase (deg): -180 101 Frequency (rad) a) Find the gain margin, phase margin, gain crossover frequency, and phase crossover frequency. b) Is the closed-loop system stable? What is the steady-state error for step-input?

Answer 2

Textbook Question

Chapter 1, Problem 1.30P

Plot log 10 I D versus V D over the range 0.1 ≤ V D ≤ 0.7 V for (a) I S = 10 − 12 and (b) I S = 10 − 14 A .

(a)

Expert Solution

To determine

To plot: The graph of log10ID versus VD .

Answer to Problem 1.30P

The graph of log10ID versus VD for the IS=10−12 is shown in Figure 1.

Explanation of Solution

Given:

Reverse saturation current IS=10−12 A

Diode Voltage range, 0.1≤VD≤0.7

Concept Used:

ID=IS(exp( V D V T ))

Calculation:

For IS=10−12

For VD=0.1V

ID=10−12(exp( 0.1 0.026 ))=4.68127×10−11 Alog10(ID)=−10.3296

For VD=0.2V

ID=10−12(exp( 0.2 0.026 ))=2.1914×10−9 Alog10(ID)=−8.6593

For VD=0.3 V

ID=10−12(exp( 0.3 0.026 ))=1.02591×10−7 Alog10(ID)=−6.9889

For VD=0.4 V

ID=10−12(exp( 0.4 0.026 ))=4.8023×10−6 Alog10(ID)=−5.3185

For VD=0.5 V

ID=10−12(exp( 0.5 0.026 ))=2.248×10−4 Alog10(ID)=−3.6482

For VD=0.6 V

ID=10−12(exp( 0.6 0.026 ))=1.05240×10−2 Alog10(ID)=−1.9778

For VD=0.7 V

ID=10−12(exp( 0.7 0.026 ))=4.92656×10−1 Alog10(ID)=−0.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 1

Figure 1

(b)

Expert Solution

To determine

To plot: The graph of log10ID versus VD .

Answer to Problem 1.30P

The graph of log10ID Versus VD for the IS=10−14 is shown in Figure 2.

Explanation of Solution

Given:

Reverse saturation current IS=10−14

Concept Used:

Formulae;

ID=IS(exp( V D V T ))

Calculation:

For IS=10−14

For VD=0.1V

ID=10−14(exp( 0.1 0.026 ))=4.68127×10−13 Alog10(ID)=−12.3296

For VD=0.2V

ID=10−14(exp( 0.2 0.026 ))=2.1914×10−11 Alog10(ID)=−10.6593

For VD=0.3 V

ID=10−14(exp( 0.3 0.026 ))=1.02591×10−9 Alog10(ID)=−8.9889

For VD=0.4 V

ID=10−14(exp( 0.4 0.026 ))=4.8023×10−8 Alog10(ID)=−7.3185

For VD=0.5 V

ID=10−14(exp( 0.5 0.026 ))=2.248×10−6 Alog10(ID)=−5.6482

For VD=0.6 V

ID=10−14(exp( 0.6 0.026 ))=1.05240×10−4 Alog10(ID)=−3.9778

For VD=0.7 V

ID=10−14(exp( 0.7 0.026 ))=4.92656×10−3 Alog10(ID)=−2.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 2

Figure 2

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Answer 3

Textbook Question

Chapter 1, Problem 1.30P

Plot log 10 I D versus V D over the range 0.1 ≤ V D ≤ 0.7 V for (a) I S = 10 − 12 and (b) I S = 10 − 14 A .

(a)

Expert Solution

To determine

To plot: The graph of log10ID versus VD .

Answer to Problem 1.30P

The graph of log10ID versus VD for the IS=10−12 is shown in Figure 1.

Explanation of Solution

Given:

Reverse saturation current IS=10−12 A

Diode Voltage range, 0.1≤VD≤0.7

Concept Used:

ID=IS(exp( V D V T ))

Calculation:

For IS=10−12

For VD=0.1V

ID=10−12(exp( 0.1 0.026 ))=4.68127×10−11 Alog10(ID)=−10.3296

For VD=0.2V

ID=10−12(exp( 0.2 0.026 ))=2.1914×10−9 Alog10(ID)=−8.6593

For VD=0.3 V

ID=10−12(exp( 0.3 0.026 ))=1.02591×10−7 Alog10(ID)=−6.9889

For VD=0.4 V

ID=10−12(exp( 0.4 0.026 ))=4.8023×10−6 Alog10(ID)=−5.3185

For VD=0.5 V

ID=10−12(exp( 0.5 0.026 ))=2.248×10−4 Alog10(ID)=−3.6482

For VD=0.6 V

ID=10−12(exp( 0.6 0.026 ))=1.05240×10−2 Alog10(ID)=−1.9778

For VD=0.7 V

ID=10−12(exp( 0.7 0.026 ))=4.92656×10−1 Alog10(ID)=−0.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 1

Figure 1

(b)

Expert Solution

To determine

To plot: The graph of log10ID versus VD .

Answer to Problem 1.30P

The graph of log10ID Versus VD for the IS=10−14 is shown in Figure 2.

Explanation of Solution

Given:

Reverse saturation current IS=10−14

Concept Used:

Formulae;

ID=IS(exp( V D V T ))

Calculation:

For IS=10−14

For VD=0.1V

ID=10−14(exp( 0.1 0.026 ))=4.68127×10−13 Alog10(ID)=−12.3296

For VD=0.2V

ID=10−14(exp( 0.2 0.026 ))=2.1914×10−11 Alog10(ID)=−10.6593

For VD=0.3 V

ID=10−14(exp( 0.3 0.026 ))=1.02591×10−9 Alog10(ID)=−8.9889

For VD=0.4 V

ID=10−14(exp( 0.4 0.026 ))=4.8023×10−8 Alog10(ID)=−7.3185

For VD=0.5 V

ID=10−14(exp( 0.5 0.026 ))=2.248×10−6 Alog10(ID)=−5.6482

For VD=0.6 V

ID=10−14(exp( 0.6 0.026 ))=1.05240×10−4 Alog10(ID)=−3.9778

For VD=0.7 V

ID=10−14(exp( 0.7 0.026 ))=4.92656×10−3 Alog10(ID)=−2.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 2

Figure 2

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Answer 4

Textbook Question

Chapter 1, Problem 1.30P

Plot log 10 I D versus V D over the range 0.1 ≤ V D ≤ 0.7 V for (a) I S = 10 − 12 and (b) I S = 10 − 14 A .

(a)

Expert Solution

To determine

To plot: The graph of log10ID versus VD .

Answer to Problem 1.30P

The graph of log10ID versus VD for the IS=10−12 is shown in Figure 1.

Explanation of Solution

Given:

Reverse saturation current IS=10−12 A

Diode Voltage range, 0.1≤VD≤0.7

Concept Used:

ID=IS(exp( V D V T ))

Calculation:

For IS=10−12

For VD=0.1V

ID=10−12(exp( 0.1 0.026 ))=4.68127×10−11 Alog10(ID)=−10.3296

For VD=0.2V

ID=10−12(exp( 0.2 0.026 ))=2.1914×10−9 Alog10(ID)=−8.6593

For VD=0.3 V

ID=10−12(exp( 0.3 0.026 ))=1.02591×10−7 Alog10(ID)=−6.9889

For VD=0.4 V

ID=10−12(exp( 0.4 0.026 ))=4.8023×10−6 Alog10(ID)=−5.3185

For VD=0.5 V

ID=10−12(exp( 0.5 0.026 ))=2.248×10−4 Alog10(ID)=−3.6482

For VD=0.6 V

ID=10−12(exp( 0.6 0.026 ))=1.05240×10−2 Alog10(ID)=−1.9778

For VD=0.7 V

ID=10−12(exp( 0.7 0.026 ))=4.92656×10−1 Alog10(ID)=−0.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 1

Figure 1

(b)

Expert Solution

To determine

To plot: The graph of log10ID versus VD .

Answer to Problem 1.30P

The graph of log10ID Versus VD for the IS=10−14 is shown in Figure 2.

Explanation of Solution

Given:

Reverse saturation current IS=10−14

Concept Used:

Formulae;

ID=IS(exp( V D V T ))

Calculation:

For IS=10−14

For VD=0.1V

ID=10−14(exp( 0.1 0.026 ))=4.68127×10−13 Alog10(ID)=−12.3296

For VD=0.2V

ID=10−14(exp( 0.2 0.026 ))=2.1914×10−11 Alog10(ID)=−10.6593

For VD=0.3 V

ID=10−14(exp( 0.3 0.026 ))=1.02591×10−9 Alog10(ID)=−8.9889

For VD=0.4 V

ID=10−14(exp( 0.4 0.026 ))=4.8023×10−8 Alog10(ID)=−7.3185

For VD=0.5 V

ID=10−14(exp( 0.5 0.026 ))=2.248×10−6 Alog10(ID)=−5.6482

For VD=0.6 V

ID=10−14(exp( 0.6 0.026 ))=1.05240×10−4 Alog10(ID)=−3.9778

For VD=0.7 V

ID=10−14(exp( 0.7 0.026 ))=4.92656×10−3 Alog10(ID)=−2.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 2

Figure 2

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

To plot: The graph of log10ID versus VD .

Answer to Problem 1.30P

The graph of log10ID versus VD for the IS=10−12 is shown in Figure 1.

Explanation of Solution

Given:

Reverse saturation current IS=10−12 A

Diode Voltage range, 0.1≤VD≤0.7

Concept Used:

ID=IS(exp( V D V T ))

Calculation:

For IS=10−12

For VD=0.1V

ID=10−12(exp( 0.1 0.026 ))=4.68127×10−11 Alog10(ID)=−10.3296

For VD=0.2V

ID=10−12(exp( 0.2 0.026 ))=2.1914×10−9 Alog10(ID)=−8.6593

For VD=0.3 V

ID=10−12(exp( 0.3 0.026 ))=1.02591×10−7 Alog10(ID)=−6.9889

For VD=0.4 V

ID=10−12(exp( 0.4 0.026 ))=4.8023×10−6 Alog10(ID)=−5.3185

For VD=0.5 V

ID=10−12(exp( 0.5 0.026 ))=2.248×10−4 Alog10(ID)=−3.6482

For VD=0.6 V

ID=10−12(exp( 0.6 0.026 ))=1.05240×10−2 Alog10(ID)=−1.9778

For VD=0.7 V

ID=10−12(exp( 0.7 0.026 ))=4.92656×10−1 Alog10(ID)=−0.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 1

Figure 1

(b)

Expert Solution

To determine

To plot: The graph of log10ID versus VD .

Answer to Problem 1.30P

The graph of log10ID Versus VD for the IS=10−14 is shown in Figure 2.

Explanation of Solution

Given:

Reverse saturation current IS=10−14

Concept Used:

Formulae;

ID=IS(exp( V D V T ))

Calculation:

For IS=10−14

For VD=0.1V

ID=10−14(exp( 0.1 0.026 ))=4.68127×10−13 Alog10(ID)=−12.3296

For VD=0.2V

ID=10−14(exp( 0.2 0.026 ))=2.1914×10−11 Alog10(ID)=−10.6593

For VD=0.3 V

ID=10−14(exp( 0.3 0.026 ))=1.02591×10−9 Alog10(ID)=−8.9889

For VD=0.4 V

ID=10−14(exp( 0.4 0.026 ))=4.8023×10−8 Alog10(ID)=−7.3185

For VD=0.5 V

ID=10−14(exp( 0.5 0.026 ))=2.248×10−6 Alog10(ID)=−5.6482

For VD=0.6 V

ID=10−14(exp( 0.6 0.026 ))=1.05240×10−4 Alog10(ID)=−3.9778

For VD=0.7 V

ID=10−14(exp( 0.7 0.026 ))=4.92656×10−3 Alog10(ID)=−2.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 2

Figure 2

Answer 8

(a)

Expert Solution

To determine

To plot: The graph of log10ID versus VD .

Answer to Problem 1.30P

The graph of log10ID versus VD for the IS=10−12 is shown in Figure 1.

Explanation of Solution

Given:

Reverse saturation current IS=10−12 A

Diode Voltage range, 0.1≤VD≤0.7

Concept Used:

ID=IS(exp( V D V T ))

Calculation:

For IS=10−12

For VD=0.1V

ID=10−12(exp( 0.1 0.026 ))=4.68127×10−11 Alog10(ID)=−10.3296

For VD=0.2V

ID=10−12(exp( 0.2 0.026 ))=2.1914×10−9 Alog10(ID)=−8.6593

For VD=0.3 V

ID=10−12(exp( 0.3 0.026 ))=1.02591×10−7 Alog10(ID)=−6.9889

For VD=0.4 V

ID=10−12(exp( 0.4 0.026 ))=4.8023×10−6 Alog10(ID)=−5.3185

For VD=0.5 V

ID=10−12(exp( 0.5 0.026 ))=2.248×10−4 Alog10(ID)=−3.6482

For VD=0.6 V

ID=10−12(exp( 0.6 0.026 ))=1.05240×10−2 Alog10(ID)=−1.9778

For VD=0.7 V

ID=10−12(exp( 0.7 0.026 ))=4.92656×10−1 Alog10(ID)=−0.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 1

Figure 1

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To plot: The graph of log10ID versus VD .

Answer 13

Answer to Problem 1.30P

The graph of log10ID versus VD for the IS=10−12 is shown in Figure 1.

Answer 14

Explanation of Solution

Given:

Reverse saturation current IS=10−12 A

Diode Voltage range, 0.1≤VD≤0.7

Concept Used:

ID=IS(exp( V D V T ))

Calculation:

For IS=10−12

For VD=0.1V

ID=10−12(exp( 0.1 0.026 ))=4.68127×10−11 Alog10(ID)=−10.3296

For VD=0.2V

ID=10−12(exp( 0.2 0.026 ))=2.1914×10−9 Alog10(ID)=−8.6593

For VD=0.3 V

ID=10−12(exp( 0.3 0.026 ))=1.02591×10−7 Alog10(ID)=−6.9889

For VD=0.4 V

ID=10−12(exp( 0.4 0.026 ))=4.8023×10−6 Alog10(ID)=−5.3185

For VD=0.5 V

ID=10−12(exp( 0.5 0.026 ))=2.248×10−4 Alog10(ID)=−3.6482

For VD=0.6 V

ID=10−12(exp( 0.6 0.026 ))=1.05240×10−2 Alog10(ID)=−1.9778

For VD=0.7 V

ID=10−12(exp( 0.7 0.026 ))=4.92656×10−1 Alog10(ID)=−0.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 1

Figure 1

Answer 15

(b)

Expert Solution

To determine

To plot: The graph of log10ID versus VD .

Answer to Problem 1.30P

The graph of log10ID Versus VD for the IS=10−14 is shown in Figure 2.

Explanation of Solution

Given:

Reverse saturation current IS=10−14

Concept Used:

Formulae;

ID=IS(exp( V D V T ))

Calculation:

For IS=10−14

For VD=0.1V

ID=10−14(exp( 0.1 0.026 ))=4.68127×10−13 Alog10(ID)=−12.3296

For VD=0.2V

ID=10−14(exp( 0.2 0.026 ))=2.1914×10−11 Alog10(ID)=−10.6593

For VD=0.3 V

ID=10−14(exp( 0.3 0.026 ))=1.02591×10−9 Alog10(ID)=−8.9889

For VD=0.4 V

ID=10−14(exp( 0.4 0.026 ))=4.8023×10−8 Alog10(ID)=−7.3185

For VD=0.5 V

ID=10−14(exp( 0.5 0.026 ))=2.248×10−6 Alog10(ID)=−5.6482

For VD=0.6 V

ID=10−14(exp( 0.6 0.026 ))=1.05240×10−4 Alog10(ID)=−3.9778

For VD=0.7 V

ID=10−14(exp( 0.7 0.026 ))=4.92656×10−3 Alog10(ID)=−2.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 2

Figure 2

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To plot: The graph of log10ID versus VD .

Answer 20

Answer to Problem 1.30P

The graph of log10ID Versus VD for the IS=10−14 is shown in Figure 2.

Answer 21

Explanation of Solution

Given:

Reverse saturation current IS=10−14

Concept Used:

Formulae;

ID=IS(exp( V D V T ))

Calculation:

For IS=10−14

For VD=0.1V

ID=10−14(exp( 0.1 0.026 ))=4.68127×10−13 Alog10(ID)=−12.3296

For VD=0.2V

ID=10−14(exp( 0.2 0.026 ))=2.1914×10−11 Alog10(ID)=−10.6593

For VD=0.3 V

ID=10−14(exp( 0.3 0.026 ))=1.02591×10−9 Alog10(ID)=−8.9889

For VD=0.4 V

ID=10−14(exp( 0.4 0.026 ))=4.8023×10−8 Alog10(ID)=−7.3185

For VD=0.5 V

ID=10−14(exp( 0.5 0.026 ))=2.248×10−6 Alog10(ID)=−5.6482

For VD=0.6 V

ID=10−14(exp( 0.6 0.026 ))=1.05240×10−4 Alog10(ID)=−3.9778

For VD=0.7 V

ID=10−14(exp( 0.7 0.026 ))=4.92656×10−3 Alog10(ID)=−2.30745

Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.30P , additional homework tip 2

Figure 2

Answer 22

Ch. 1 - Calculate the intrinsic carrier concentration in...Ch. 1 - (a) Calculate the majority and minority carrier...Ch. 1 - Consider ntype GaAs at T=300K doped to a...Ch. 1 - Consider silicon at T=300K . Assume the hole...Ch. 1 - Determine the intrinsic carrier concentration in...Ch. 1 - (a) Consider silicon at T=300K . Assume that...Ch. 1 - Using the results of TYU1.2, determine the drift...Ch. 1 - The electron and hole diffusion coefficients in...Ch. 1 - A sample of silicon at T=300K is doped to...Ch. 1 - (a) Calculate Vbi for a GaAs pn junction at T=300K...

Plot log 10 I D versus V D over the range 0.1 ≤ V D ≤ 0.7 V for (a) I S = 10 − 12 and (b) I S = 10 − 14 A .

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