The transistor in the circuit in Figure P3.29 has parameters V T P = − 0.8 V and K p = 0.20 mA/V 2 . Sketch the load line and plot the Q−point for (a) V D D = 3.5 V, R D = 1.2 k Ω and (b) V D D = 5 V, R D = 4 k Ω . What is the operating bias region for each condition? Figure P3.29

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 3, Problem 3.29P

The transistor in the circuit in Figure P3.29 has parameters V T P = − 0.8 V and K p = 0.20 mA/V 2 . Sketch the load line and plot the Q−point for (a) V D D = 3.5 V, R D = 1.2 k Ω and (b) V D D = 5 V, R D = 4 k Ω . What is the operating bias region for each condition?

Chapter 3, Problem 3.29P, The transistor in the circuit in Figure P3.29 has parameters VTP=0.8V and Kp=0.20mA/V2 . Sketch the
Figure P3.29

(a)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 1.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 1

VTP=−0.8 V, Kp=0.2 mAV2VDD=3.5 V, RD=1.2 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=3.5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=3.5−0.8VSD(sat)=2.7 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=3.5−1200ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(3.5−0.8)(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[5.4(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[(3.5−1200 I D)(5.4−3.5+1200 I D)]ID=0.2×10−3[(3.5−1200 I D)(1.9+1200 I D)]ID=0.2×10−3[6.65+4200ID−2280ID−1.44×106ID2]ID=1.33×10−3+0.384ID−288ID2288ID2+0.616ID−1.33×10−3=0ID=1.33 mA

From equation (1):

VSD=3.5−1200IDVSD=3.5−1200(1.33× 10 −3)VSD=1.9 V

From above calculations:

VSD<VSD(sat)

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

(VSDQ,IDQ)=(1.9 V, 1.33 mA)

The diagram of load line is sketched as follows:

From equation (1):

VSD=3.5−1200IDID=−(1 1200)VSD+( 3.5 1200)

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VSD=3.5 V

At VSD=0 ,

ID=3.51200ID=2.92 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 2

Figure 1

(b)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 2.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 3

VTP=−0.8 V, Kp=0.2 mAV2VDD=5 V, RD=4 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=5−0.8VSD(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=5−4000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(5−0.8)(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[8.4(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[(5−4000 I D)(8.4−5+4000 I D)]ID=0.2×10−3[(5−4000 I D)(3.4+4000 I D)]ID=0.2×10−3[17+20,000ID−13,600ID−16×106ID2]ID=3.4×10−3+1.28ID−3200ID23200ID2−0.28ID−3.4×10−3=0ID=1.075 mA

From equation (1):

VSD=5−4000IDVSD=5−4000(1.075× 10 −3)VSD=0.7 V

From above calculations:

VSD<VSD(sat)

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

(VSDQ,IDQ)=(0.7 V, 1.075 mA)

The diagram of load line is sketched as follows:

From equation (1):

VSD=5−4000IDID=−(1 4000)VSD+(5 4000)

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VSD=5 V

At VSD=0 ,

ID=54000ID=1.25 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 4

Figure 2

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

A transistor has the CB input characteristics shown in Figure 3–11 (p. 106 ). If alpha = 0.95, find IC when VBE = 0.72 V and VCB = 10 V.

Vec +12V Re 1kS2 Re 470k2 BC107 Vc Ve Figure 3: Practical Fixed Bias Transistor Circuit 2. Measure the DC voltages Vc and VB using digital multi-meters. Determine the quiescent base current, collector current, and collector- emitter voltage.

Determine VB, VE, VC, VCE, IB, IE, and IC in Figure. The 2N3904 is a general purpose transistor with a typical BDC 200 Vcc +30 V WWII VCE VB R₁ • 22 ΚΩ IC(mA) Chọn... * Chọn... * IB(UA) Chọn... * IE(MA) Chọn... ◆ Chọn... * Chọn... * Chọn... * VE VC R₂ ´ 10 ΚΩ www Rc 1.0 ΚΩ 2N3904 PDC=200 RE 1.0 ΚΩ

Answer 2

Textbook Question

Chapter 3, Problem 3.29P

The transistor in the circuit in Figure P3.29 has parameters V T P = − 0.8 V and K p = 0.20 mA/V 2 . Sketch the load line and plot the Q−point for (a) V D D = 3.5 V, R D = 1.2 k Ω and (b) V D D = 5 V, R D = 4 k Ω . What is the operating bias region for each condition?

Chapter 3, Problem 3.29P, The transistor in the circuit in Figure P3.29 has parameters VTP=0.8V and Kp=0.20mA/V2 . Sketch the
Figure P3.29

(a)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 1.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 1

VTP=−0.8 V, Kp=0.2 mAV2VDD=3.5 V, RD=1.2 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=3.5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=3.5−0.8VSD(sat)=2.7 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=3.5−1200ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(3.5−0.8)(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[5.4(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[(3.5−1200 I D)(5.4−3.5+1200 I D)]ID=0.2×10−3[(3.5−1200 I D)(1.9+1200 I D)]ID=0.2×10−3[6.65+4200ID−2280ID−1.44×106ID2]ID=1.33×10−3+0.384ID−288ID2288ID2+0.616ID−1.33×10−3=0ID=1.33 mA

From equation (1):

VSD=3.5−1200IDVSD=3.5−1200(1.33× 10 −3)VSD=1.9 V

From above calculations:

VSD<VSD(sat)

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

(VSDQ,IDQ)=(1.9 V, 1.33 mA)

The diagram of load line is sketched as follows:

From equation (1):

VSD=3.5−1200IDID=−(1 1200)VSD+( 3.5 1200)

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VSD=3.5 V

At VSD=0 ,

ID=3.51200ID=2.92 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 2

Figure 1

(b)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 2.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 3

VTP=−0.8 V, Kp=0.2 mAV2VDD=5 V, RD=4 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=5−0.8VSD(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=5−4000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(5−0.8)(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[8.4(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[(5−4000 I D)(8.4−5+4000 I D)]ID=0.2×10−3[(5−4000 I D)(3.4+4000 I D)]ID=0.2×10−3[17+20,000ID−13,600ID−16×106ID2]ID=3.4×10−3+1.28ID−3200ID23200ID2−0.28ID−3.4×10−3=0ID=1.075 mA

From equation (1):

VSD=5−4000IDVSD=5−4000(1.075× 10 −3)VSD=0.7 V

From above calculations:

VSD<VSD(sat)

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

(VSDQ,IDQ)=(0.7 V, 1.075 mA)

The diagram of load line is sketched as follows:

From equation (1):

VSD=5−4000IDID=−(1 4000)VSD+(5 4000)

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VSD=5 V

At VSD=0 ,

ID=54000ID=1.25 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 4

Figure 2

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 3, Problem 3.29P

The transistor in the circuit in Figure P3.29 has parameters V T P = − 0.8 V and K p = 0.20 mA/V 2 . Sketch the load line and plot the Q−point for (a) V D D = 3.5 V, R D = 1.2 k Ω and (b) V D D = 5 V, R D = 4 k Ω . What is the operating bias region for each condition?

Chapter 3, Problem 3.29P, The transistor in the circuit in Figure P3.29 has parameters VTP=0.8V and Kp=0.20mA/V2 . Sketch the
Figure P3.29

(a)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 1.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 1

VTP=−0.8 V, Kp=0.2 mAV2VDD=3.5 V, RD=1.2 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=3.5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=3.5−0.8VSD(sat)=2.7 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=3.5−1200ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(3.5−0.8)(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[5.4(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[(3.5−1200 I D)(5.4−3.5+1200 I D)]ID=0.2×10−3[(3.5−1200 I D)(1.9+1200 I D)]ID=0.2×10−3[6.65+4200ID−2280ID−1.44×106ID2]ID=1.33×10−3+0.384ID−288ID2288ID2+0.616ID−1.33×10−3=0ID=1.33 mA

From equation (1):

VSD=3.5−1200IDVSD=3.5−1200(1.33× 10 −3)VSD=1.9 V

From above calculations:

VSD<VSD(sat)

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

(VSDQ,IDQ)=(1.9 V, 1.33 mA)

The diagram of load line is sketched as follows:

From equation (1):

VSD=3.5−1200IDID=−(1 1200)VSD+( 3.5 1200)

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VSD=3.5 V

At VSD=0 ,

ID=3.51200ID=2.92 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 2

Figure 1

(b)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 2.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 3

VTP=−0.8 V, Kp=0.2 mAV2VDD=5 V, RD=4 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=5−0.8VSD(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=5−4000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(5−0.8)(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[8.4(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[(5−4000 I D)(8.4−5+4000 I D)]ID=0.2×10−3[(5−4000 I D)(3.4+4000 I D)]ID=0.2×10−3[17+20,000ID−13,600ID−16×106ID2]ID=3.4×10−3+1.28ID−3200ID23200ID2−0.28ID−3.4×10−3=0ID=1.075 mA

From equation (1):

VSD=5−4000IDVSD=5−4000(1.075× 10 −3)VSD=0.7 V

From above calculations:

VSD<VSD(sat)

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

(VSDQ,IDQ)=(0.7 V, 1.075 mA)

The diagram of load line is sketched as follows:

From equation (1):

VSD=5−4000IDID=−(1 4000)VSD+(5 4000)

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VSD=5 V

At VSD=0 ,

ID=54000ID=1.25 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 4

Figure 2

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 3, Problem 3.29P

The transistor in the circuit in Figure P3.29 has parameters V T P = − 0.8 V and K p = 0.20 mA/V 2 . Sketch the load line and plot the Q−point for (a) V D D = 3.5 V, R D = 1.2 k Ω and (b) V D D = 5 V, R D = 4 k Ω . What is the operating bias region for each condition?

Chapter 3, Problem 3.29P, The transistor in the circuit in Figure P3.29 has parameters VTP=0.8V and Kp=0.20mA/V2 . Sketch the
Figure P3.29

(a)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 1.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 1

VTP=−0.8 V, Kp=0.2 mAV2VDD=3.5 V, RD=1.2 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=3.5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=3.5−0.8VSD(sat)=2.7 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=3.5−1200ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(3.5−0.8)(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[5.4(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[(3.5−1200 I D)(5.4−3.5+1200 I D)]ID=0.2×10−3[(3.5−1200 I D)(1.9+1200 I D)]ID=0.2×10−3[6.65+4200ID−2280ID−1.44×106ID2]ID=1.33×10−3+0.384ID−288ID2288ID2+0.616ID−1.33×10−3=0ID=1.33 mA

From equation (1):

VSD=3.5−1200IDVSD=3.5−1200(1.33× 10 −3)VSD=1.9 V

From above calculations:

VSD<VSD(sat)

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

(VSDQ,IDQ)=(1.9 V, 1.33 mA)

The diagram of load line is sketched as follows:

From equation (1):

VSD=3.5−1200IDID=−(1 1200)VSD+( 3.5 1200)

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VSD=3.5 V

At VSD=0 ,

ID=3.51200ID=2.92 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 2

Figure 1

(b)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 2.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 3

VTP=−0.8 V, Kp=0.2 mAV2VDD=5 V, RD=4 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=5−0.8VSD(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=5−4000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(5−0.8)(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[8.4(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[(5−4000 I D)(8.4−5+4000 I D)]ID=0.2×10−3[(5−4000 I D)(3.4+4000 I D)]ID=0.2×10−3[17+20,000ID−13,600ID−16×106ID2]ID=3.4×10−3+1.28ID−3200ID23200ID2−0.28ID−3.4×10−3=0ID=1.075 mA

From equation (1):

VSD=5−4000IDVSD=5−4000(1.075× 10 −3)VSD=0.7 V

From above calculations:

VSD<VSD(sat)

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

(VSDQ,IDQ)=(0.7 V, 1.075 mA)

The diagram of load line is sketched as follows:

From equation (1):

VSD=5−4000IDID=−(1 4000)VSD+(5 4000)

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VSD=5 V

At VSD=0 ,

ID=54000ID=1.25 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 4

Figure 2

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 1.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 1

VTP=−0.8 V, Kp=0.2 mAV2VDD=3.5 V, RD=1.2 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=3.5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=3.5−0.8VSD(sat)=2.7 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=3.5−1200ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(3.5−0.8)(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[5.4(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[(3.5−1200 I D)(5.4−3.5+1200 I D)]ID=0.2×10−3[(3.5−1200 I D)(1.9+1200 I D)]ID=0.2×10−3[6.65+4200ID−2280ID−1.44×106ID2]ID=1.33×10−3+0.384ID−288ID2288ID2+0.616ID−1.33×10−3=0ID=1.33 mA

From equation (1):

VSD=3.5−1200IDVSD=3.5−1200(1.33× 10 −3)VSD=1.9 V

From above calculations:

VSD<VSD(sat)

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

(VSDQ,IDQ)=(1.9 V, 1.33 mA)

The diagram of load line is sketched as follows:

From equation (1):

VSD=3.5−1200IDID=−(1 1200)VSD+( 3.5 1200)

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VSD=3.5 V

At VSD=0 ,

ID=3.51200ID=2.92 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 2

Figure 1

(b)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 2.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 3

VTP=−0.8 V, Kp=0.2 mAV2VDD=5 V, RD=4 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=5−0.8VSD(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=5−4000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(5−0.8)(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[8.4(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[(5−4000 I D)(8.4−5+4000 I D)]ID=0.2×10−3[(5−4000 I D)(3.4+4000 I D)]ID=0.2×10−3[17+20,000ID−13,600ID−16×106ID2]ID=3.4×10−3+1.28ID−3200ID23200ID2−0.28ID−3.4×10−3=0ID=1.075 mA

From equation (1):

VSD=5−4000IDVSD=5−4000(1.075× 10 −3)VSD=0.7 V

From above calculations:

VSD<VSD(sat)

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

(VSDQ,IDQ)=(0.7 V, 1.075 mA)

The diagram of load line is sketched as follows:

From equation (1):

VSD=5−4000IDID=−(1 4000)VSD+(5 4000)

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VSD=5 V

At VSD=0 ,

ID=54000ID=1.25 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 4

Figure 2

Answer 8

(a)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 1.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 1

VTP=−0.8 V, Kp=0.2 mAV2VDD=3.5 V, RD=1.2 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=3.5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=3.5−0.8VSD(sat)=2.7 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=3.5−1200ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(3.5−0.8)(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[5.4(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[(3.5−1200 I D)(5.4−3.5+1200 I D)]ID=0.2×10−3[(3.5−1200 I D)(1.9+1200 I D)]ID=0.2×10−3[6.65+4200ID−2280ID−1.44×106ID2]ID=1.33×10−3+0.384ID−288ID2288ID2+0.616ID−1.33×10−3=0ID=1.33 mA

From equation (1):

VSD=3.5−1200IDVSD=3.5−1200(1.33× 10 −3)VSD=1.9 V

From above calculations:

VSD<VSD(sat)

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

(VSDQ,IDQ)=(1.9 V, 1.33 mA)

The diagram of load line is sketched as follows:

From equation (1):

VSD=3.5−1200IDID=−(1 1200)VSD+( 3.5 1200)

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VSD=3.5 V

At VSD=0 ,

ID=3.51200ID=2.92 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 2

Figure 1

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer 13

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 1.

The transistor operates in triode region.

Answer 14

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 1

VTP=−0.8 V, Kp=0.2 mAV2VDD=3.5 V, RD=1.2 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=3.5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=3.5−0.8VSD(sat)=2.7 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=3.5−1200ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(3.5−0.8)(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[5.4(3.5−1200 I D)−( 3.5−1200 I D )2]ID=0.2×10−3[(3.5−1200 I D)(5.4−3.5+1200 I D)]ID=0.2×10−3[(3.5−1200 I D)(1.9+1200 I D)]ID=0.2×10−3[6.65+4200ID−2280ID−1.44×106ID2]ID=1.33×10−3+0.384ID−288ID2288ID2+0.616ID−1.33×10−3=0ID=1.33 mA

From equation (1):

VSD=3.5−1200IDVSD=3.5−1200(1.33× 10 −3)VSD=1.9 V

From above calculations:

VSD<VSD(sat)

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

(VSDQ,IDQ)=(1.9 V, 1.33 mA)

The diagram of load line is sketched as follows:

From equation (1):

VSD=3.5−1200IDID=−(1 1200)VSD+( 3.5 1200)

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VSD=3.5 V

At VSD=0 ,

ID=3.51200ID=2.92 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 2

Figure 1

Answer 15

(b)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 2.

The transistor operates in triode region.

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 3

VTP=−0.8 V, Kp=0.2 mAV2VDD=5 V, RD=4 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=5−0.8VSD(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=5−4000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(5−0.8)(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[8.4(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[(5−4000 I D)(8.4−5+4000 I D)]ID=0.2×10−3[(5−4000 I D)(3.4+4000 I D)]ID=0.2×10−3[17+20,000ID−13,600ID−16×106ID2]ID=3.4×10−3+1.28ID−3200ID23200ID2−0.28ID−3.4×10−3=0ID=1.075 mA

From equation (1):

VSD=5−4000IDVSD=5−4000(1.075× 10 −3)VSD=0.7 V

From above calculations:

VSD<VSD(sat)

Hence, the assumption is correct and the transistor is biased in triode region.

The value of Q -point is:

(VSDQ,IDQ)=(0.7 V, 1.075 mA)

The diagram of load line is sketched as follows:

From equation (1):

VSD=5−4000IDID=−(1 4000)VSD+(5 4000)

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VSD=5 V

At VSD=0 ,

ID=54000ID=1.25 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 4

Figure 2

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer 20

Answer to Problem 3.29P

A load line along with Q -point is shown in Figure 2.

The transistor operates in triode region.

Answer 21

Explanation of Solution

Given Information:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.29P , additional homework tip 3

VTP=−0.8 V, Kp=0.2 mAV2VDD=5 V, RD=4 kΩ

Calculation:

The value of VSG is:

VSG=VS−VGVSG=VDD−0VSG=5 V

The value of VSD(sat) is:

VSD(sat)=VGS−|VTP|VSD(sat)=5−0.8VSD(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VSD=VDD−IDRDVSD=5−4000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V SG−| V TP |)VSD−VSD2]ID=0.2×10−3[2(5−0.8)(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[8.4(5−4000 I D)−( 5−4000 I D )2]ID=0.2×10−3[(5−4000 I D)(8.4−5+4000 I D)]ID=0.2×10−3[(5−4000 I D)(3.4+4000 I D)]ID=0.2×10−3[17+20,000ID−13,600ID−16×106ID2]ID=3.4×10−3+1.28ID−3200ID23200ID2−0.28ID−3.4×10−3=0ID=1.075 mA