The transistor in the circuit in Figure P3.27 has parameters V T N = 0.8 V and K n = 0.25 mA/V 2 . Sketch the load line and plot the Q−point for (a) V D D = 4 V, R D = 1 k Ω and (b) V D D = 5 V, R D = 3 k Ω . What is the operating bias region for each condition? Figure P3.27

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Answer 1

Textbook Question

Chapter 3, Problem 3.27P

The transistor in the circuit in Figure P3.27 has parameters V T N = 0.8 V and K n = 0.25 mA/V 2 . Sketch the load line and plot the Q−point for (a) V D D = 4 V, R D = 1 k Ω and (b) V D D = 5 V, R D = 3 k Ω . What is the operating bias region for each condition?

Chapter 3, Problem 3.27P, The transistor in the circuit in Figure P3.27 has parameters VTN=0.8V and Kn=0.25mA/V2 . Sketch the
Figure P3.27

(a)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.27P

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 3.27P , additional homework tip 1

VTN=0.8 V, Kn=0.25 mAV2VDD=4 V, RD=1 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=4 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=4−0.8VDS(sat)=3.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=4−1000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(4−0.8)(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[6.4(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[(4−1000 I D)(6.4−4+1000 I D)]ID=0.25×10−3[(4−1000 I D)(2.4+1000 I D)]ID=0.25×10−3[9.6+4000ID−2400ID−106ID2]ID=2.4×10−3+0.4ID−250ID2250ID2+0.6ID−2.4×10−3=0ID=2.12 mA

From equation (1):

VDS=4−1000IDVDS=4−1000(2.12× 10 −3)VDS=1.88 V

From above calculations:

VDS<VDS(sat)

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

The value of Q -point is:

(VDSQ,IDQ)=(1.88 V, 2.12 mA)

The diagram of load line is sketched as follows:

From equation (1):

VDS=4−1000IDID=−1×10−3VDS+4×10−3

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

ID=−1×10−3VDS+4×10−30=−1×10−3VDS+4×10−3VDS=4 V

At VDS=0 ,

ID=−1×10−3×0+4×10−3ID=4 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.27P , 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 3.27P

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 3.27P , additional homework tip 3

VTN=0.8 V, Kn=0.25 mAV2VDD=5 V, RD=3 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=5 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=5−0.8VDS(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=5−3000ID ...(2)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(5−0.8)(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[8.4(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[(5−3000 I D)(8.4−5+3000 I D)]ID=0.25×10−3[(5−3000 I D)(3.4+3000 I D)]ID=0.25×10−3[17+15000ID−10,200ID−9×106ID2]ID=4.25×10−3+1.2ID−2250ID22250ID2−0.2ID−4.25×10−3=0ID=1.42 mA

From equation (2):

VDS=5−3000IDVDS=5−3000(1.42× 10 −3)VDS=0.74 V

From above calculations:

VDS<VDS(sat)

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

The value of Q-point is:

(VDSQ,IDQ)=(0.74 V, 1.42 mA)

The diagram of load line is sketched as follows:

From equation (1):

VDS=5−3000IDID=−(13)×10−3VDS+(53)×10−3

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VDS=5 V

At VDS=0 ,

ID=1.67 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.27P , additional homework tip 4

Figure 2

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Answer 2

Textbook Question

Chapter 3, Problem 3.27P

The transistor in the circuit in Figure P3.27 has parameters V T N = 0.8 V and K n = 0.25 mA/V 2 . Sketch the load line and plot the Q−point for (a) V D D = 4 V, R D = 1 k Ω and (b) V D D = 5 V, R D = 3 k Ω . What is the operating bias region for each condition?

Chapter 3, Problem 3.27P, The transistor in the circuit in Figure P3.27 has parameters VTN=0.8V and Kn=0.25mA/V2 . Sketch the
Figure P3.27

(a)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.27P

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 3.27P , additional homework tip 1

VTN=0.8 V, Kn=0.25 mAV2VDD=4 V, RD=1 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=4 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=4−0.8VDS(sat)=3.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=4−1000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(4−0.8)(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[6.4(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[(4−1000 I D)(6.4−4+1000 I D)]ID=0.25×10−3[(4−1000 I D)(2.4+1000 I D)]ID=0.25×10−3[9.6+4000ID−2400ID−106ID2]ID=2.4×10−3+0.4ID−250ID2250ID2+0.6ID−2.4×10−3=0ID=2.12 mA

From equation (1):

VDS=4−1000IDVDS=4−1000(2.12× 10 −3)VDS=1.88 V

From above calculations:

VDS<VDS(sat)

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

The value of Q -point is:

(VDSQ,IDQ)=(1.88 V, 2.12 mA)

The diagram of load line is sketched as follows:

From equation (1):

VDS=4−1000IDID=−1×10−3VDS+4×10−3

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

ID=−1×10−3VDS+4×10−30=−1×10−3VDS+4×10−3VDS=4 V

At VDS=0 ,

ID=−1×10−3×0+4×10−3ID=4 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.27P , 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 3.27P

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 3.27P , additional homework tip 3

VTN=0.8 V, Kn=0.25 mAV2VDD=5 V, RD=3 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=5 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=5−0.8VDS(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=5−3000ID ...(2)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(5−0.8)(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[8.4(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[(5−3000 I D)(8.4−5+3000 I D)]ID=0.25×10−3[(5−3000 I D)(3.4+3000 I D)]ID=0.25×10−3[17+15000ID−10,200ID−9×106ID2]ID=4.25×10−3+1.2ID−2250ID22250ID2−0.2ID−4.25×10−3=0ID=1.42 mA

From equation (2):

VDS=5−3000IDVDS=5−3000(1.42× 10 −3)VDS=0.74 V

From above calculations:

VDS<VDS(sat)

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

The value of Q-point is:

(VDSQ,IDQ)=(0.74 V, 1.42 mA)

The diagram of load line is sketched as follows:

From equation (1):

VDS=5−3000IDID=−(13)×10−3VDS+(53)×10−3

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VDS=5 V

At VDS=0 ,

ID=1.67 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.27P , additional homework tip 4

Figure 2

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Answer 3

Textbook Question

Chapter 3, Problem 3.27P

The transistor in the circuit in Figure P3.27 has parameters V T N = 0.8 V and K n = 0.25 mA/V 2 . Sketch the load line and plot the Q−point for (a) V D D = 4 V, R D = 1 k Ω and (b) V D D = 5 V, R D = 3 k Ω . What is the operating bias region for each condition?

Chapter 3, Problem 3.27P, The transistor in the circuit in Figure P3.27 has parameters VTN=0.8V and Kn=0.25mA/V2 . Sketch the
Figure P3.27

(a)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.27P

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 3.27P , additional homework tip 1

VTN=0.8 V, Kn=0.25 mAV2VDD=4 V, RD=1 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=4 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=4−0.8VDS(sat)=3.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=4−1000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(4−0.8)(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[6.4(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[(4−1000 I D)(6.4−4+1000 I D)]ID=0.25×10−3[(4−1000 I D)(2.4+1000 I D)]ID=0.25×10−3[9.6+4000ID−2400ID−106ID2]ID=2.4×10−3+0.4ID−250ID2250ID2+0.6ID−2.4×10−3=0ID=2.12 mA

From equation (1):

VDS=4−1000IDVDS=4−1000(2.12× 10 −3)VDS=1.88 V

From above calculations:

VDS<VDS(sat)

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

The value of Q -point is:

(VDSQ,IDQ)=(1.88 V, 2.12 mA)

The diagram of load line is sketched as follows:

From equation (1):

VDS=4−1000IDID=−1×10−3VDS+4×10−3

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

ID=−1×10−3VDS+4×10−30=−1×10−3VDS+4×10−3VDS=4 V

At VDS=0 ,

ID=−1×10−3×0+4×10−3ID=4 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.27P , 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 3.27P

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 3.27P , additional homework tip 3

VTN=0.8 V, Kn=0.25 mAV2VDD=5 V, RD=3 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=5 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=5−0.8VDS(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=5−3000ID ...(2)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(5−0.8)(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[8.4(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[(5−3000 I D)(8.4−5+3000 I D)]ID=0.25×10−3[(5−3000 I D)(3.4+3000 I D)]ID=0.25×10−3[17+15000ID−10,200ID−9×106ID2]ID=4.25×10−3+1.2ID−2250ID22250ID2−0.2ID−4.25×10−3=0ID=1.42 mA

From equation (2):

VDS=5−3000IDVDS=5−3000(1.42× 10 −3)VDS=0.74 V

From above calculations:

VDS<VDS(sat)

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

The value of Q-point is:

(VDSQ,IDQ)=(0.74 V, 1.42 mA)

The diagram of load line is sketched as follows:

From equation (1):

VDS=5−3000IDID=−(13)×10−3VDS+(53)×10−3

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VDS=5 V

At VDS=0 ,

ID=1.67 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.27P , additional homework tip 4

Figure 2

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Answer 4

Textbook Question

Chapter 3, Problem 3.27P

The transistor in the circuit in Figure P3.27 has parameters V T N = 0.8 V and K n = 0.25 mA/V 2 . Sketch the load line and plot the Q−point for (a) V D D = 4 V, R D = 1 k Ω and (b) V D D = 5 V, R D = 3 k Ω . What is the operating bias region for each condition?

Chapter 3, Problem 3.27P, The transistor in the circuit in Figure P3.27 has parameters VTN=0.8V and Kn=0.25mA/V2 . Sketch the
Figure P3.27

(a)

Expert Solution

To determine

To sketch: A load line and labelQ -point.

To find: The region of operation of transistor.

Answer to Problem 3.27P

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 3.27P , additional homework tip 1

VTN=0.8 V, Kn=0.25 mAV2VDD=4 V, RD=1 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=4 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=4−0.8VDS(sat)=3.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=4−1000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(4−0.8)(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[6.4(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[(4−1000 I D)(6.4−4+1000 I D)]ID=0.25×10−3[(4−1000 I D)(2.4+1000 I D)]ID=0.25×10−3[9.6+4000ID−2400ID−106ID2]ID=2.4×10−3+0.4ID−250ID2250ID2+0.6ID−2.4×10−3=0ID=2.12 mA

From equation (1):

VDS=4−1000IDVDS=4−1000(2.12× 10 −3)VDS=1.88 V

From above calculations:

VDS<VDS(sat)

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

The value of Q -point is:

(VDSQ,IDQ)=(1.88 V, 2.12 mA)

The diagram of load line is sketched as follows:

From equation (1):

VDS=4−1000IDID=−1×10−3VDS+4×10−3

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

ID=−1×10−3VDS+4×10−30=−1×10−3VDS+4×10−3VDS=4 V

At VDS=0 ,

ID=−1×10−3×0+4×10−3ID=4 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.27P , 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 3.27P

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 3.27P , additional homework tip 3

VTN=0.8 V, Kn=0.25 mAV2VDD=5 V, RD=3 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=5 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=5−0.8VDS(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=5−3000ID ...(2)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(5−0.8)(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[8.4(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[(5−3000 I D)(8.4−5+3000 I D)]ID=0.25×10−3[(5−3000 I D)(3.4+3000 I D)]ID=0.25×10−3[17+15000ID−10,200ID−9×106ID2]ID=4.25×10−3+1.2ID−2250ID22250ID2−0.2ID−4.25×10−3=0ID=1.42 mA

From equation (2):

VDS=5−3000IDVDS=5−3000(1.42× 10 −3)VDS=0.74 V

From above calculations:

VDS<VDS(sat)

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

The value of Q-point is:

(VDSQ,IDQ)=(0.74 V, 1.42 mA)

The diagram of load line is sketched as follows:

From equation (1):

VDS=5−3000IDID=−(13)×10−3VDS+(53)×10−3

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VDS=5 V

At VDS=0 ,

ID=1.67 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.27P , additional homework tip 4

Figure 2

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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 3.27P

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 3.27P , additional homework tip 1

VTN=0.8 V, Kn=0.25 mAV2VDD=4 V, RD=1 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=4 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=4−0.8VDS(sat)=3.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=4−1000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(4−0.8)(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[6.4(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[(4−1000 I D)(6.4−4+1000 I D)]ID=0.25×10−3[(4−1000 I D)(2.4+1000 I D)]ID=0.25×10−3[9.6+4000ID−2400ID−106ID2]ID=2.4×10−3+0.4ID−250ID2250ID2+0.6ID−2.4×10−3=0ID=2.12 mA

From equation (1):

VDS=4−1000IDVDS=4−1000(2.12× 10 −3)VDS=1.88 V

From above calculations:

VDS<VDS(sat)

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

The value of Q -point is:

(VDSQ,IDQ)=(1.88 V, 2.12 mA)

The diagram of load line is sketched as follows:

From equation (1):

VDS=4−1000IDID=−1×10−3VDS+4×10−3

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

ID=−1×10−3VDS+4×10−30=−1×10−3VDS+4×10−3VDS=4 V

At VDS=0 ,

ID=−1×10−3×0+4×10−3ID=4 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.27P , 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 3.27P

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 3.27P , additional homework tip 3

VTN=0.8 V, Kn=0.25 mAV2VDD=5 V, RD=3 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=5 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=5−0.8VDS(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=5−3000ID ...(2)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(5−0.8)(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[8.4(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[(5−3000 I D)(8.4−5+3000 I D)]ID=0.25×10−3[(5−3000 I D)(3.4+3000 I D)]ID=0.25×10−3[17+15000ID−10,200ID−9×106ID2]ID=4.25×10−3+1.2ID−2250ID22250ID2−0.2ID−4.25×10−3=0ID=1.42 mA

From equation (2):

VDS=5−3000IDVDS=5−3000(1.42× 10 −3)VDS=0.74 V

From above calculations:

VDS<VDS(sat)

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

The value of Q-point is:

(VDSQ,IDQ)=(0.74 V, 1.42 mA)

The diagram of load line is sketched as follows:

From equation (1):

VDS=5−3000IDID=−(13)×10−3VDS+(53)×10−3

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VDS=5 V

At VDS=0 ,

ID=1.67 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.27P , 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 3.27P

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 3.27P , additional homework tip 1

VTN=0.8 V, Kn=0.25 mAV2VDD=4 V, RD=1 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=4 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=4−0.8VDS(sat)=3.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=4−1000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(4−0.8)(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[6.4(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[(4−1000 I D)(6.4−4+1000 I D)]ID=0.25×10−3[(4−1000 I D)(2.4+1000 I D)]ID=0.25×10−3[9.6+4000ID−2400ID−106ID2]ID=2.4×10−3+0.4ID−250ID2250ID2+0.6ID−2.4×10−3=0ID=2.12 mA

From equation (1):

VDS=4−1000IDVDS=4−1000(2.12× 10 −3)VDS=1.88 V

From above calculations:

VDS<VDS(sat)

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

The value of Q -point is:

(VDSQ,IDQ)=(1.88 V, 2.12 mA)

The diagram of load line is sketched as follows:

From equation (1):

VDS=4−1000IDID=−1×10−3VDS+4×10−3

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

ID=−1×10−3VDS+4×10−30=−1×10−3VDS+4×10−3VDS=4 V

At VDS=0 ,

ID=−1×10−3×0+4×10−3ID=4 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.27P , 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 3.27P

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 3.27P , additional homework tip 1

VTN=0.8 V, Kn=0.25 mAV2VDD=4 V, RD=1 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=4 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=4−0.8VDS(sat)=3.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=4−1000ID ...(1)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(4−0.8)(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[6.4(4−1000 I D)−( 4−1000 I D )2]ID=0.25×10−3[(4−1000 I D)(6.4−4+1000 I D)]ID=0.25×10−3[(4−1000 I D)(2.4+1000 I D)]ID=0.25×10−3[9.6+4000ID−2400ID−106ID2]ID=2.4×10−3+0.4ID−250ID2250ID2+0.6ID−2.4×10−3=0ID=2.12 mA

From equation (1):

VDS=4−1000IDVDS=4−1000(2.12× 10 −3)VDS=1.88 V

From above calculations:

VDS<VDS(sat)

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

The value of Q -point is:

(VDSQ,IDQ)=(1.88 V, 2.12 mA)

The diagram of load line is sketched as follows:

From equation (1):

VDS=4−1000IDID=−1×10−3VDS+4×10−3

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

ID=−1×10−3VDS+4×10−30=−1×10−3VDS+4×10−3VDS=4 V

At VDS=0 ,

ID=−1×10−3×0+4×10−3ID=4 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.27P , 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 3.27P

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 3.27P , additional homework tip 3

VTN=0.8 V, Kn=0.25 mAV2VDD=5 V, RD=3 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=5 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=5−0.8VDS(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=5−3000ID ...(2)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(5−0.8)(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[8.4(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[(5−3000 I D)(8.4−5+3000 I D)]ID=0.25×10−3[(5−3000 I D)(3.4+3000 I D)]ID=0.25×10−3[17+15000ID−10,200ID−9×106ID2]ID=4.25×10−3+1.2ID−2250ID22250ID2−0.2ID−4.25×10−3=0ID=1.42 mA

From equation (2):

VDS=5−3000IDVDS=5−3000(1.42× 10 −3)VDS=0.74 V

From above calculations:

VDS<VDS(sat)

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

The value of Q-point is:

(VDSQ,IDQ)=(0.74 V, 1.42 mA)

The diagram of load line is sketched as follows:

From equation (1):

VDS=5−3000IDID=−(13)×10−3VDS+(53)×10−3

It is equation of straight line with negative slope.

y=−mx+c

At ID=0 ,

VDS=5 V

At VDS=0 ,

ID=1.67 mA

Microelectronics: Circuit Analysis and Design, Chapter 3, Problem 3.27P , 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 3.27P

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 3.27P , additional homework tip 3

VTN=0.8 V, Kn=0.25 mAV2VDD=5 V, RD=3 kΩ

Calculation:

The value of VGS is:

VGS=VG−VSVGS=VDD−0VGS=5 V

The value of VDS(sat) is:

VDS(sat)=VGS−VTNVDS(sat)=5−0.8VDS(sat)=4.2 V

Applying Kirchhoff’s voltage in drain-source terminal:

VDS=VDD−IDRDVDS=5−3000ID ...(2)

Assuming the Mosfet operates in triode region:

The expression of drain current in triode region:

ID=Kn[2( V GS− V TN)VDS−VDS2]ID=0.25×10−3[2(5−0.8)(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[8.4(5−3000 I D)−( 5−3000 I D )2]ID=0.25×10−3[(5−3000 I D)(8.4−5+3000 I D)]ID=0.25×10−3[(5−3000 I D)(3.4+3000 I D)]ID=0.25×10−3[17+15000ID−10,200ID−9×106ID2]ID=4.25×10−3+1.2ID−2250ID22250ID2−0.2ID−4.25×10−3=0ID=1.42 mA