Rizzoni 6e SM CH11 - solution PDF

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####### 11. 1PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andChapter 11: Field effect Transistors: Operation, Circuit, Models, andApplications – Instructor NotesChapter 11 introduces field-effect transistors. Should the instructor choose to ...

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11

Chapter 11: Field effect Transistors: Operation, Circuit, Models, and Applications – Instructor Notes Chapter 11 introduces field-effect transistors. Should the instructor choose to only teach field-effect devices (or to cover FETs before BJTs), Section 10.1 can be used as an introductory section prior to starting Chapter 11. Section 11.1 briefly reviews the classification and symbols for the major families of field-effect devices. Section 11.2 introduces the fundamental ideas behind the operation of N-channel field-effect enhancement-mode transistors. A brief explanation of P-channel devices is also presented in this section. Section 11.3 , illustrates the calculation of the state and operating point of basic field-effect transistor circuits. Section 11.4 outlines the operation of MOSFET large-signal amplifiers, and presents two practical examples (11.6 and 11.7), rfelated to a battery charging circuit and a DC motor drive circuit. These examples are analogous to those presented in Chapter 10 for BJT large-signal amplifiers, giving the instructor the opportunity to make a comparison of the two technologies, if so desired. Finally, Section 11.5 introduces the analysis of MOSFET switches and presents CMOS gates. The box Focus on Measurements: MOSFET bidirectional analog gate (pp. 572-573) presents ananalog application of CMOS technology. The end-of-chapter problems are straightforward applications of the concepts illustrated in the chapter. The 5th Edition of this book includes 13 new problems; some of the 4th Edition problems were removed, increasing the end-of-chapter problem count from 23 to 35.

Learning Objectives 1. 2. 3. 4. 5. 6.

Understand the classification of field-effect transistors. Section 11.1. Learn the basic operation of enhancement-mode MOSFETs by understanding their i-v curves and defining equations. Section 11.2. Learn how enhancement-mode MOSFET circuits are biased. Section 11.3. Understand the concept and operation of FET large-signal amplifiers. Section 11.4 Understand the concept and operation of FET switches. Section 11.5. Analyze FET switches and digital gates. Section 11.5.

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11

Section 11.2: n-channel MOSFET Operation Problem 11.1 The transistors shown in Figure P11.1 have |Vt| = 3 V. Determine the operating region.

Solution: Known quantities: For the transistors shown in Figure P11.1, VT = 3 V .

Find: The operating state of each transistor.

Analysis: a)

This is an n-channel enhancement MOSFET, with VT = 3 V. To operate in the triode region, the condition is:

v DS < v GS − VT . To operate in the saturation region, the condition is: v DS ≥ vGS − VT . To turn the transistor on, the condition is: vGS > VT . We can compute: vGS = −2.5 V Since vGS is lesser than VT, the transistor is in the cut-off region. b) This is a p-channel enhancement MOSFET, with VT = – 3 V. To operate in the triode region, the

vSD < vSG − VT . To operate in the

condition is: saturation

region,

the

condition

is:

v SD ≥ vSG − VT . To turn the transistor on, the condition is:

vSG > − VT .

We can compute:

v SG = −2 V v SD = 1 V

vSG − VT = −2 − 3 = −5 V vSD = 1 V > v SG − VT = −5 V . Therefore, the transistor is in the saturation region. c)

This is a p-channel enhancement MOSFET, with VT = – 3 V. To operate in the triode region, the condition is:

vSD < vSG − VT . To operate in the saturation region, the condition is: v SD ≥ vSG − VT . To turn the transistor on, the condition is:

vSG > − VT .

We can compute:

v SG = 5 V v SD = 1 V 11.2 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11

v SG − VT = 5 − 3 = 2 V vSD = 1 V < v SG − VT = 2 V Therefore, the transistor is in the triode region. d) This is an n-channel enhancement MOSFET, with VT = 3 V. To operate in the triode region, the condition is: v DS < vGS − VT . To operate in the saturation region, the condition is: v DS ≥ vGS − VT . To turn the transistor on, the condition is: vGS > VT . We have:

vGS = −2 V Since vGS is lesser than VT, the transistor is in the cut-off region.

Problem 11.2 The three terminals of an n-channel enhancement-mode MOSFET are at potentials of 4, 5, and 10 V with respect to ground. Draw the circuit symbol, with the appropriate voltages at each terminal, if the device is operating a. In the ohmic region. b. In the saturation region.

Solution: Known quantities: The potentials of an n-channel enhancement-mode MOSFET (4, 5, and 10 V respectively).

Find: The circuit symbol, if the device is operating: a) In the ohmic state. b) In the saturation region.

Analysis: a)

To operate in the ohmic region, the condition is: v DS 0, v DS > 0 . +

vGD = 10V

vDS = 4V +

+

G

vGS = 5V -

b) To operate in the saturation region, the condition is: v DS

S

-

≥ v GS − VT and VT > 0, v DS > 0 .

The circuit for operation in the saturation region is shown below. 11.3 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11

D

-

+

vGD = 4V

vDS = 10V +

+

G

vGS = 5V -

S

-

Problem 11.3 An enhancement-type NMOS transistor with Vt = 2 V has its source grounded and a 3-V DC source connected to the gate. Determine the operating state if a. vD = 0.5 V b. vD = 1 V c. vD = 5 V

Solution: Known quantities: The threshold voltage, VT = 2 V, of an enhancement-type NMOS that has its source grounded and a 3 V DC source connected to the gate.

Find: The operating state if: a) vD = 0.5 V . b) vD = 1 V . c)

vD = 5 V

Analysis: vDS = vD = 0.5 V a) vGS − VT = 3 − 2 = 1 V v DS < v GS −VT The transistor is in the triode region. vDS = vD = 1 V b) vGS − VT = 3 − 2 = 1 V v DS = v GS −VT The transistor is either in the triode or in the saturation region. vDS = vD = 5 V c) vGS − VT = 3 − 2 = 1 V v DS > v GS −VT 11.4 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11

The transistor is in the saturation region.

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11

Problem 11.4 In the circuit shown in Figure P11.4, the p-channel transistor has |Vt| = 2 V and K = 10 mA/V2. Find R and vS for iD = 0.4 mA.

Solution: Known quantities: The threshold voltage, |Vt| = 2 V, of the p-channel transistor 2 shown in Figure P11.4. k = 10 mA/V .

Find: R and vD for id = 0.4 mA .

Analysis: The device shown is a p-channel enhancement mode MOSFET, with , |Vt| = 2 V and , VDG = 0 V. To operate in the saturation region we require: Since

v SD ≥ v SG − VT .

vDG = vSD − vSG = 0 > − VT = −2 V , the transistor is in the saturation region. Knowing

we can write: 0.4 = 10(v SG − 2) and determine v DG 2

k = 10 mA/V2 ,

= v SD = v SG = 2.2 V . R can be found as follows:

20 − vD 20 − 2.2 = 44.5 kΩ = R= − iD 0.4 ⋅ 10 3

Problem 11.5 An enhancement-type NMOS transistor has VT = 2.5 V and iD = 0.8mA when vGS = vDS = 4 V. Find the value of iD for vGS = 5 V.

Solution: Known quantities: The threshold voltage, VT = 2.5 V, of an enhancement-type NMOS transistor. iD = 0.8 mA when vGS = vDS = 4 V.

Find: The value of iD for vGS = 5 V.

Analysis: Because

v DS > vGS − VT , the transistor is in the saturation region:

i D = k ⋅ ( vGS − VT ) 2 = k ⋅ (4 − 2.5) 2 = 0.0008 A ⇒ k = 3.55⋅ 10 −4 . For vGS = 5 V we have:

iD = 3.55⋅ 10 −4 ⋅ (5 − 2.5) 2 = 0.0022A .

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11

Problem 11.6 The NMOS transistor shown in Figure P11.6 has ฀฀฀ ฀ = 1.5 V and ฀ ฀ = 0.4 mA/฀฀ 2 . If ฀฀฀ ฀ is a pulse with 0 to 5 V, find the voltage levels of the pulse signal at the drain.

Known quantities: Amplitude of ฀฀฀ ฀ , ฀฀฀ ฀ = 1.5 V, and ฀ ฀ = 0.4 mA/฀฀ 2

Find: The voltage levels of the pulse signal at the drain, ฀฀฀฀

Analysis: To find ฀฀฀฀ , determine the drain voltage at each level of the pulse.

฀฀฀ ฀ = ฀฀ V:

First, calculate the overdrive voltage using the equation: ฀฀฀฀฀฀ = ฀฀฀฀฀฀ − ฀฀฀฀ = 0 − 1.5 = −1.5 ฀฀

The drain to source voltage ฀฀฀฀฀฀ is most likely greater than ฀฀฀฀฀฀ . So, by equation 11.3, the transistor is operating in saturation mode. Next, calculate the drain current, ฀฀฀฀ , using equation 11.8: 2 ฀฀฀ ฀ ≈ ฀฀฀฀ ฀฀฀฀

= 0.0004 ∗ (−1.52 ) = 0.9 ฀฀฀฀ 11.7

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11

Find the drain voltage using KVL:

฀฀฀ ฀ = −฀฀฀฀ ฀ ฀ + ฀฀฀฀฀฀

= −0.0009 ∗ 1000 + 5 = ฀฀. ฀฀ ฀฀

Check the assumption by calculating ฀฀฀฀฀฀ = 4.1 V. Then, ฀฀฀฀฀฀ > ฀฀฀฀฀฀ , which, by equation 11.8, means the transistor is in the saturation region. Thus, the assumption is confirmed to be correct. ฀฀฀ ฀ = ฀฀ V:

Again, calculate the overdrive voltage:

฀฀฀฀฀฀ = ฀฀฀฀฀฀ − ฀฀฀฀ = 5 − 1.5 = 3.5 ฀฀

This time, assume that the transistor is in the triode region. Use KVL at the drain to write the following equation: ฀฀฀฀฀฀ = ฀฀฀฀฀฀ − ฀฀฀฀ ฀฀฀฀

Next, use equation 11.6 to obtain:

= 5 − ฀฀฀฀ 1000

฀฀(2฀฀฀฀฀฀ − ฀฀฀฀฀฀)฀฀฀฀฀฀ − ฀฀฀ ฀ = 0

Substitute the result from KVL into equation 11.6:

0 = ฀฀(2฀฀฀฀฀฀ − 5 + ฀฀฀฀ 1000) ∗ (5 − ฀฀฀฀ 1000) − ฀฀฀฀ = 0.0004(2 + ฀฀฀฀ 1000) ∗ (5 − ฀฀฀฀ 1000) − ฀฀฀฀

= −400฀฀2฀฀+ 0.2฀฀฀ ฀ + 0.004

Solve this result using the quadratic equation to obtain the solutions: ฀฀฀ ฀ = 3.43 ฀฀฀฀

฀฀฀ ฀ = −2.93 ฀฀฀฀

Select ฀฀฀ ฀ = 3.43 mA and plug it into the drain-source voltage equation: ฀฀฀฀฀฀ = 5 − ฀฀฀฀ 1000

= 5 − 0.00343 ∗ 1000 = 1.57 ฀฀

Final answer: ฀฀฀ ฀ = ฀฀. ฀฀ V, ฀฀฀ ฀ = ฀฀ V; ฀฀฀ ฀ = ฀฀. ฀฀฀฀ V, ฀฀฀ ฀ = ฀฀ V

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11

Problem 11.7 In the circuit shown in Figure P11.7, a drain voltage of 0.1 V is established. Find the current ฀฀฀฀ for ฀฀฀ ฀ = 1 V, and ฀ ฀ = 0.5 mA/฀฀ 2 .

Known quantities: ฀฀฀ ฀ = 0.1 V, ฀฀฀ ฀ = 1 V, and ฀ ฀ = 0.5 mA/฀฀ 2

Find: The drain current, ฀฀฀฀

Analysis: Note that the transistor is an NMOS and determine its operating mode. Due to ฀฀฀฀฀฀ , the source voltage and gate voltage are: ฀฀฀ ฀ = 0 ฀฀

฀฀฀ ฀ = 15 ฀฀

Next, find the overdrive voltage: ฀฀฀฀฀฀ = ฀฀฀฀฀฀ − ฀฀฀฀ = 15 − 1

= 14 ฀฀

Because ฀฀฀฀฀฀ < ฀฀฀฀฀฀ , the transistor is in the triode region (฀฀฀฀฀฀ = 0.1 V). Note that ฀฀฀฀฀฀ > ฀฀฀฀ , so the transistor is not in the cutoff region. Calculate the drain current using equation 11.6: ฀฀฀ ฀ = ฀฀(2฀฀฀฀฀฀ − ฀฀฀฀฀฀ )฀฀฀฀฀฀

= 0.0005 ∗ (2 ∗ 14 − 0.1) ∗ 0.1

Final answer: ฀฀฀ ฀ = ฀฀. ฀฀ mA

= 1.4 ฀฀฀฀

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11

Problem 11.8 An n-channel enhancement-mode MOSFET, shown in Figure P11.8, is operated in the ohmic region. Size the circuit so that quiescent drain current iDQ = 4mA. Let VDD = 15V , K = 0.3mA/V 2, VT = 3.3V.

Solution: Known quantities: The circuit shown in Figure P11.8: V DD , K , VT , R1, R 2 , R D

Find: Find the operating point.

Analysis: The equation that links V GSQ and the drain current is:

I DQ = K (VGSQ − VT )

2

The voltage V GSQ coincides with the voltage across R2:

R2 2 VDD = 8.57V ⇒ I DQ = K (VGSQ − VT ) = 0.914mA R1 + R2 The voltage V DSQ is: VGSQ =

V DSQ =V DD − RD I DQ = 17.25V The Mosfet works in active region if VDSQ > VGSQ −VT = 4.57V

Problem 11.9 Compute the power dissipated by the circuit in Figure P11.9. Let VDD = VSS = 15V , R1 = R2 = 90kohm, RD = 0.1kohm, VT = 3.5V, K = 0.816mA/V 2.

Solution: Known quantities: The circuit shown in Figure P11.9: VDD , K , VT , R1 , R 2 , RD , RS

Find: Find the operating region of the MOSFET.

Analysis: Using Thevenin theorem the circuit becomes: 11.10 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11

VGG = where

RG = From the gate mesh: VGG

R 1R 2 90 × 10 3 × 90 × 10 3 = = 45 kΩ R1 + R2 90 × 10 3 + 90 × 10 3 = VGSQ + RS I DQ

I DQ = K (VGSQ − VT

Using

⇒I

2 DQ

R2 90 × 10 3 VDD = 15 = 7.5V R1 + R2 90 × 10 3 + 90 × 10 3

)

2

= K (VGG − RS I DQ − VT ) ⇒ 2

2(VGG − VT )   1  V GG − V T − +  I DQ +  2 RS  KRS  RS 

2

  = 0 

Solving the equation:

0.0112 A( Unacceptable) I DQ =   0.0023A The voltage between gate and source is:

VGSQ = VGG − RS I DQ = 5.8V While the voltage between drain and source is:

VDSQ = VDD − ( RS + RD ) I DQ = 11.1 It is easy to check that the condition of operation in the active zone is verified

VDSQ > VGSQ − VT 11.1 > 2.8

Problem 11.10 Find the operating region of the enhancement-type NMOS transistor shown in Figure P11.8. Let ฀฀฀฀฀฀ = 20 V, ฀ ฀ = 0.2 mA/฀฀ 2 , ฀฀฀ ฀ = 4 V, ฀฀1 = 4 ฀฀Ω, ฀฀2 = 3 ฀฀Ω, and ฀฀฀ ฀ = 3 ฀฀Ω.

Known quantities: ฀฀฀฀฀฀ = 20 V, ฀ ฀ = 0.2 mA/฀฀ 2 , ฀฀฀ ฀ = 4 V, ฀฀1 = 4 ฀฀Ω, ฀฀2 = 3 ฀฀Ω, ฀฀฀ ฀ = 3 ฀฀Ω 11.11 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11

Find: The operating region of the enhancement-type NMOS transistor

Analysis: Determine the gate voltage, ฀฀฀ ฀ , of the transistor using the voltage division between ฀฀1 and ฀฀2 : ฀฀2 ฀฀฀ ฀ = ฀฀฀฀฀฀ ฀฀1 + ฀฀2 = 20 ∗

3

4+3

= 8.57 ฀฀

Determine the gate-source voltage:

฀฀฀฀฀฀ = ฀฀฀ ฀ − ฀฀฀฀

= 8.57 − 0

= 8.57 ฀฀

Determine the overdrive voltage:

฀฀฀฀฀฀ = ฀฀฀฀฀฀ − ฀฀฀฀ = 8.57 − 4

= 4.57 ฀฀

Assume that the transistor is in the saturation region and check the assumption. Determine the drain current using equation 11.8: 2 ฀฀฀ ฀ ≈ ฀฀฀฀ ฀฀฀฀ = 0.0002 ∗ (4.572 )

Next, use KVL to determine the drain voltage:

= 4.18 ฀฀฀฀

฀฀฀ ฀ = ฀฀฀฀฀฀ − ฀฀฀฀ ฀฀฀฀

= 20 − 0.00418 ∗ 3000

= 7.46 ฀฀

The drain-source voltage is ฀฀฀฀฀฀ = 7.46; therefore, the transistor is in the operating region (i.e., ฀฀฀฀฀฀ > ฀฀฀฀฀฀ ) by equation 11.8. Final answer: saturation region

Problem 11.11

Find the operating region of the enhancement-type NMOS transistor shown in Figure P11.11. Let ฀฀฀฀฀฀ = 18 V, ฀ ฀ = 0.3 mA/฀฀ 2 , ฀฀฀ ฀ = 3 V, ฀฀1 = 5.5 ฀฀Ω, ฀฀2 = 4.5 ฀฀Ω, ฀฀฀ ฀ = 2 ฀฀Ω, and ฀฀฀ ฀ = 1 ฀฀Ω. 11.12

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11

Known quantities: ฀฀฀฀฀฀ = 18 V, ฀ ฀ = 0.3 mA/฀฀ 2 , ฀฀฀ ฀ = 3 V, ฀฀1 = 5.5 ฀฀Ω, ฀฀2 = 4.5 ฀฀Ω, ฀฀฀ ฀ = 2 ฀฀Ω, ฀฀฀ ฀ = 1 ฀฀Ω

Find: The operating region of the enhancement-type NMOS transistor

Analysis: Assume that the transistor is in the saturation region. Calculate the gate voltage from the voltage division between ฀฀1 and ฀฀2 : ฀฀2 ฀฀฀ ฀ = ฀฀฀฀฀฀ (฀฀1 + ฀฀2 ) = 18 ∗

4.5 5.5 + 4.5

= 8.1 ฀฀

Apply KVL at the gate and note that the gate current is 0 A, due to the infinite input resistance of the MOSFET:

Use equation 11.8 to obtain:

฀฀฀฀฀฀ = 8.1 − ฀฀฀฀ ฀฀฀฀

฀฀฀ ฀ = ฀฀(฀฀฀฀฀฀ − ฀฀฀฀ )2 Substitute the result from KVL into this equation: ฀฀฀ ฀ = ฀฀(8.1 − ฀฀฀฀ ฀฀฀ ฀ − ฀฀฀฀ )2

= 0.0003 ∗ (5.1 − ฀฀฀฀ 1000)2

= 300฀฀2฀฀− 3.06฀฀฀ ฀ + 0.0078

Solve this result using the quadratic equation to receive the following solutions: 11.13 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 11 ฀฀฀ ฀ = 5.2 ฀฀฀฀ ฀฀฀ ฀ = 5.0 ฀฀฀฀ Plugging thi...