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G. Rizzoni, Principles and Applications of Electrical Engineering, 6 Edition Problem solutions, Chapter 10####### 10. 1PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and Chapter 10: Bipolar Junction Transistors: Operation, CircuitModels, and ...

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 10

 Chapter 10: Bipolar Junction Transistors: Operation, Circuit Models, and Applications – Instructor Notes Chapter 10 introduces bipolar junction transistors. The material on transistors is divided into two independent chapters, one on bipolar devices, and one on field-effect devices. The two chapters are functionally independent, except for the fact that Section 10.1, introducing the concept of transistors as amplifiers and switches, can be covered prior to starting Chapter 11 if the instructor decides to only teach field-effect devices, or to cover them before bipolar devices. Section 10.2 introduces the fundamental ideas behind the operation of bipolar transistors, and illustrates the calculation of the state and operating point of basic transistor circuits. The discussion of the properties of the BJT in Section 10.2 is centered around a description of the base and collector characteristics, and purposely avoids a detailed description of the physics of the device, with the intent of providing an intuitive understanding of the transistor as an amplifier and electronic switch. The second part of the chapter has been reorganized for clarity. Section 10.3 introduces large-signal models of the BJT, and also includes the box Focus on Methodology: Using device data sheets (pp. 559-561). Example 10.4 (LED Driver) and the box Focus on Measurements: Large Signal Amplifier for Diode Thermometer (pp. 566-568) provide two application examples. New to the 5th Edition are examples 10.5 and 10.6, that present simple but practically useful battery charger and DC motor drive BJT circuits. These examples are accompanied by related homework problems (10.25-10.27). Section 10.4 defines the concept of operating point and illustrates the selection of a bias point, introducing the idea of a small-signal amplifier in the most basic way. Finally, Section 10.5 introduces the analysis of BJT switches and presents TTL gates. The end-of-chapter problems are straightforward applications of the concepts illustrated in the chapter. The 5th Edition of this book includes 17 new problems; some of the 4th Edition problems were removed, increasing the end-of-chapter problem count from 40 to 51.

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

Understand the basic principles of amplification and switching. Section 10.1. Understand the physical operation of bipolar transistors, and identify their state. Section 10.2 Understand the large-signal model of the bipolar transistor, and apply it to simple amplifier circuits. Section 10.3. Determine and select the operating point of a bipolar transistor circuit; understand the principle of small signal amplifiers. Section 10.4. Understand the operation of bipolar transistor as a switch and analyze basic analog and digital gate circuits. Section 10.5.

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 10

 Section 10.2: Operation of the Bipolar Junction Transistor  Problem 10.1 For each transistor shown in Figure P10.1, determine whether the BE and BC junctions are forward- or reverse-biased, and determine the operating mode.

Solution: Known quantities: Transistor diagrams, as shown in Figure P10.1: (a) pnp, VEB = 0.6 V and VEC = 4.0 V (b) npn, VCB = 0.7 V and VCE = 0.2 V (c) npn, VBE = 0.7 V and VCE = 0.3 V (d) pnp, VBC = 0.6 V and VEC = 5.4 V

Find: For each transistor shown in Figure P10.1, determine whether the BE and BC junctions are forward or reverse biased, and determine the operating region.

Analysis: (a) VBE = - 0.6 V for a pnp transistor implies that the BE junction is forward-biased. VBC = VEC - VEB = 3.4 V. The CB junction is reverse-biased. Therefore, the transistor is in the active region. (b) VBC = - 0.7 V for a npn transistor implies that the CB junction is reverse-biased. VBE = VBC - VEC = -0.5 V. The BE junction is reverse-biased. Therefore, the transistor is in the cutoff region. (c) VBE = 0.7 V for a npn transistor implies that the BE junction is forward-biased. VBC = VEC - VEB = 0.4 V. The CB junction is forward-biased. Therefore, the transistor is in the saturation region. (d) VBC = 0.6 V for a pnp transistor implies that the CB junction is reverse-biased. VBE = VBC – VEC = - 4.8 V. The BE junction is forward-biased. Therefore, the transistor is in the active region.

 Problem 10.2 Determine the mode of operation for the following transistors:

a) b) c) d)

npn, VBE = 0.8 V and VCE = 0.4 V npn, VCB = 1.4 V and VCE = 2.1 V pnp, VCB = 0.9 V and VCE = 0.4 V npn, VBE = - 1.2 V and VCB = 0.6 V

Solution: Known quantities: Transistor type and operating characteristics: e) npn, VBE = 0.8 V and VCE = 0.4 V 10.2 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 10 f) npn, VCB = 1.4 V and VCE = 2.1 V g) pnp, VCB = 0.9 V and VCE = 0.4 V h) npn, VBE = - 1.2 V and VCB = 0.6 V

Find: The region of operation for each transistor

Analysis: a)

Since VBE = 0.8 V, the BE junction is forward-biased. VCB = VCE + VEB = - 0.4 V. Thus, the CB junction is forward-biased. Therefore, the transistor is in the saturation region.

b) VBE = VBC + VCE = 0.7 V. The BE junction is forwardbiased. VCB = 1.4 V. The CB junction is reverse-biased. Therefore, the transistor is in the active region. c)

VCB = 0.9 V for a pnp transistor implies that the CB junction is forward-biased.

VBE = VBC – VCE = - 1.3 V. The BE junction is forward-biased. Therefore, the transistor is in the saturation region. d) With VBE = - 1.2 V, the BE junction is reverse-biased. VCB = - 0.6 V. The CB junction is reverse-biased. Therefore, the transistor is in the cutoff region.

 Problem 10.3 Given the circuit of Figure P10.3, determine the operating point of the transistor. Let β = 100 ÷ 200, RB = 100 kΩ, Rc = 200ohm VCC = 7V.

Solution: Known quantities: The circuit of Figure P10.3: RB , RC ,VCC , β = 100 ÷ 200

Find: The operating point of the transistor

Analysis: The base current is:

I BQ =

VCC − VBEQ RB

=

VCC − VBEon 7 − 0.7 = = 63 µA RB 10 5

The collector current varies between 10.3 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 10

I C1 = I CQ min = β min I BQ = 100 ⋅ 63 × 10− 6 = 6.3mA 100.63.(10^-6) = 6.3 mA which corresponds

VCE1 = VCC − RC I C1 = 7 − 200⋅ 6.3 ⋅ 10− 3 = 5.74V and the value

I C 2 = I CQ max = β maxI BQ = 200 ⋅63 ⋅10 −6 =12.6mA This corresponds

VCE 2 = VCC − RC I C 2 = 7 − 200⋅ 12.6 ⋅ 10− 3 = 4.48V The operating point of the transistor is between P and Q.

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 10

 Problem 10.4 Refer to Figure 10.4 and assume that for a pnp transistor the emitter and base currents are IE = 5ma and IB = 0.2 mA, respectively. The voltage drops across the emitter-base and collector-base junctions are VEB = 0.67V and VCB = 7.8V. Find: a. VCE; b. the collector current; c. The total power dissipated in the transistor, defined here as P = VCE IC + VBE IB.

Solution: Known quantities: The magnitude of a pnp transistor's emitter and base current, and the magnitudes of the voltages across the emitterbase and collector-base junctions: IE = 5 mA, IB = 0.2 mA and VEB = 0.67 V, VCB = 7.8 V.

Find: a) VCE. b) IC. c) The total power dissipated in the transistor, defined as P = VCE I C + VBE I B .

Analysis: a) VCE = VCB - VEB = 7.8 - 0.67 = 7.13 V. b) IC = IE - IB = 5 - 0.2 = 4.8mA. c)

The total power dissipated in the transistor can be found to be: P ≈ VCE I C = 7.13× 4.8 × 10− = 34.22 mW 3

 Problem 10.5

For the circuit shown in Figure P10.5, determine the emitter current ฀฀฀฀ and the collector-emitter voltage ฀฀฀฀฀฀ , as defined in Figure P10.9. Assume ฀฀฀ ฀ = 0.62 V.

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 10

Known quantities: ฀฀฀ ฀ = 0.62 V and the information in Figure P10.5

Find: The emitter current, ฀฀฀฀ , and the collector-emitter voltage, ฀฀฀฀฀฀

Analysis: Emitter current: First, determine the operating mode of the transistor. Assume that the BJT is in cutoff mode. Then: ฀฀฀ ฀ = 12 V

฀฀฀ ฀ = 19 V

฀฀฀ ฀ = 0 V

The voltage across the EBJ is: ฀฀฀฀฀฀ = ฀฀฀ ฀ − ฀฀฀ ฀ = 12 − 0 = 12 ฀฀ Therefore, the EBJ is forward-biased and not in cutoff mode. Because the voltage at the collector is higher than the voltage at the base (i.e., the CBJ is reverse-biased), the transistor is in active mode. Therefore: ฀฀฀฀฀฀ = ฀฀฀ ฀ = 0.62 ฀฀ ฀฀฀ ฀ = 12 ฀฀ ฀฀฀ ฀ = 12 − 0.62 = 11.38 ฀฀ Calculate the emitter current using Ohm’s law: ฀฀฀฀ 11.38 = .46 ฀฀฀฀ ฀฀฀ ฀ = = ฀฀฀ ฀ 25000 Collector-emitter voltage: Because the constant of proportionality (฀฀ ) is not given, assume that it is very large. That is: ฀฀฀ ฀ ≈ ฀฀฀฀ With this assumption in place, use KVL to solve for the collector-emitter voltage: ฀฀฀฀฀฀ = ฀฀฀฀฀฀ − ฀฀฀฀ ฀฀฀ ฀ − ฀฀฀฀ ฀฀฀ ฀ = 19 − 0.00046 ∗ 12000 − 0.00046 ∗ 25000 = 1.98 ฀฀

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 10 The collector-emitter voltage is well above 0.2 V. Therefore, we can be assured that the transistor is, indeed, in active mode.

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

 Problem 10.6 Given the circuit of Figure P10.6, determine VCE and IC . Assume β = 80, R1 = 15 kΩ, R2 = 25 kΩ, Rc = 2 kΩ, VBB = 5V, VCC = 10V, and VAA = −4V.

Solution: Known quantities: The circuit of Figure P10.6, assuming the BJT has R1 , R2 , R C , V CC , V BB ,V AA , β .

Find: The operating point and the region in which the transistor operates

Analysis: Using Thèvenin theorem, we obtain the following circuit:

where

V eq =

R2 R1 V − V = 1.625V R1 + R 2 BB R1 + R2 AA

Req =

R1 R2 = 9.375kΩ R1 + R2

The sign and the value of Veq allow to be in conduction; the base current is:

I BQ =

Veq − VBEQ Veq − VBEon = = 98.7 µ A Req R eq

The collector current is:

I CQ = βI BQ = 7.896 mA

V CEQ = V CC − RC I CQ = −5.792V 10.7 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 10

This negative value indicates that BJT doesn`t work in active zone. We have to compute the collector and base currents, considering the saturation: VBEsat

I Bsat =

Veq −V BEsat = 93.33µA R eq

ICsat =

VCC −V CEsat = 4.9mA RC

= 0.75V ,VCEsat = 0.2V

Now, we check the condition that guarantees BJT works in saturation zone:

I Bsat >

ICsat hFE

93.33⋅ 10− 6 > 61.25⋅ 10− 6

 Problem 10.7 Given the circuit of Figure P10.7, determine the emitter current IE and the collector-base voltage VCB. Assume the offset voltage is Vγ = 0.6V.

Solution: Known quantities: The circuit of Figure P10.7, assuming the BJT has V γ = 0.6 V

Find: The emitter current and the collector-base voltage

Analysis: Applying KVL to the right-hand side of the circuit,

−VCC + I E R E + VEB = 0 V − V EB 20 − 0.6 = 497.4 “A . Since β >> 1 , I C ≈ I E = 497.4 “A = I E = CC RE 39 ⋅ 103 Applying KVL to the left-hand side: VCB + I C RC − V DD = 0

VCB = VDD − I C RC = 20 − 497.4 ⋅ 20 ⋅10−3 = 10.05V

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 10

 Problem 10.8

Given the circuit of Figure P10.8, determine ฀฀฀฀฀฀ and ฀฀฀฀ . Assume฀฀1 = 50 ฀฀Ω, ฀฀2 = 10 ฀฀Ω, ฀฀฀ ฀ = 600 Ω, ฀฀฀ ฀ = 400 Ω, ฀฀฀฀฀฀ = 0.9 V, ฀฀฀ ฀ = 25 ฀฀A, ฀฀2 = 200 ฀฀A, and ฀฀฀฀฀฀ = 18 V.

Known quantities: ฀฀1 = 50 ฀฀Ω; ฀฀2 = 10 ฀฀; ฀฀฀ ฀ = 600 Ω; ฀฀฀ ฀ = 400 Ω; ฀฀฀฀฀฀ = 0.9 V; ฀฀฀ ฀ = 25 ฀฀A; ฀฀2 = 200 ฀฀A; ฀฀฀฀฀฀ = 18 V

Find: The collector-emitter voltage, ฀฀฀฀฀฀ , and the collector current, ฀฀฀฀

Analysis: Collector current: Using KCL, the current through ฀฀1 , ฀฀1 , can be written: ฀฀1 = ฀฀2 + ฀฀฀ ฀ = 200 ฀฀−6 + 25 ฀฀−6 = 225 ฀฀฀฀ Use KVL and Ohm’s law to determine the base voltage, ฀฀฀฀ : ฀฀฀ ฀ = ฀฀฀฀฀฀ − ฀฀1 ฀฀1 = 18 − 225 ฀฀−6 ∗ 50 ฀฀3 = 6.75 ฀฀ Therefore, the emitter voltage, ฀฀฀฀ is: ฀฀฀ ฀ = ฀฀฀ ฀ − ฀฀฀฀฀฀ = 6.75 − 0.9 = 5.85 ฀฀ Now, the emitter current may be determined using Ohm’s law: ฀฀฀฀ 5.85 ฀฀฀ ฀ = = = 14.63 ฀฀฀฀ ฀฀฀ ฀ 400 Ω 10.9 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 10 Using equation 10.8: ฀฀฀ ฀ = ฀฀฀ ฀ + ฀฀฀฀ The collector current can be determined by the equation: ฀฀฀ ฀ = ฀฀฀ ฀ − ฀฀฀ ฀ = 14.63 ฀฀−3 − 25 ฀฀−6 = 14.61 ฀฀฀฀ Collector-emitter voltage: ฀฀฀฀฀฀ can be determined by using Ohm’s law to determine ฀฀฀฀ : ฀฀฀ ฀ = ฀฀฀฀฀฀ − ฀฀฀฀ ฀฀฀ ฀ = 14.61 ฀฀−3 ∗ 600 Ω = 9.23 ฀฀ Therefore, the collector-emitter voltage is: ฀฀฀฀฀฀ = ฀฀฀ ฀ − ฀฀฀ ฀ = 9.23 − 5.85 = 3.38 ฀฀ Final answer: ฀฀฀฀฀฀ = ฀฀. ฀฀฀฀ V, ฀฀฀ ฀ = ฀฀฀฀. ฀฀฀฀ mA

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 10

 Problem 10.9 The collector characteristics for a certain transistor are shown in Figure P10.9. a. Find the ratio IC/IB for VCE = 10V and IB = 100, 200, and 600 μA. b. If the maximum allowable collector power dissipation is P = iCvCE = 0.5W for IB = 500 μA, find VCE .

Solution: Known quantities: The collector characteristics for a certain transistor, as shown in Figure P10.9.

Find: a) The ratio IC/IB for VCE = 10 V and I B = 100µ A, 200µ A, and 600 µ A b) VCE, assuming the maximum allowable collector power dissipation is 0.5 W for I B = 500 µA .

Analysis: a)

For IB = 100 µA and VCE = 10 V, from the characteristics, we have IC = 17 mA. The ratio IC / IB is 170. For IB = 200 µA and VCE = 10 V, from the characteristics, we have IC = 33 mA. The ratio IC / IB is 165. For IB = 600 µA and VCE = 10 V, from the characteristics, we have IC = 86 mA. The ratio IC / IB is 143.

b) For IB = 500 µA, and if we consider an average β from a., we have IC = 159· 500 10-3= 79.5 mA. The power P 0.5 = = 6.29 V . dissipated by the transistor is P = V CE I C + VBE I B ≈ VCE IC , therefore: VCE ≈ I C 79.5 ⋅ 10−3

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 10

 Problem 10.10 Given the circuit of Figure P10.10, determine the current IR. Let RB = 30 kΩ, RC1 = 1kΩ, RC2 = 3kΩ, R = 7kΩ, VBB1 = 4V , VBB2 = 3V, VCC = 10V, β1 = 40, and β2 = 60.

Solution: Known quantities: Figure P10.10, assuming RB , RC 1 , RC 2 , R,V BB1 ,V BB 2 ,VCC , β1 , β 2 .

Find: IR

Analysis: The base currents in 2 BJT are:

I B1 =

V BB 1 − V BE = 110µA RB

I B2 =

V BB2 − V BE = 76.66µA RB

The collector currents are:

I C1 = β1 I B1 = 4 .4 mA I C 2 = β 2 I B 2 = 4.59mA From the collector circuits:

VCC = RC 1 I1 + VCE 1 = RC 1 (IC 1 + I ) + VCE 1 VCC = RC 2 I 2 + VCE 2 = RC 2 (I C 2 − I ) + VCE 21 subtracting member to member

RC1 IC1 − RC 2 I C 2 + ( RC1 + RC 2 ) I + VCE1 − VCE 2 = 0 Also

VCE 1 − VCE 2 = RI ⇒ I =

RC 2IC 2 − RC 1IC 1 ≅ 851µA RC1 + RC 2 + R



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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 10

  Problem 10.11 For the circuit shown in Figure P10.11, determine

IR. Let RB = 50 kΩ, RC = 1 kΩ, R = 2 kΩ, VBB = 2V, VCC = 12V, and β = 120.

Solution: Known quantities: Figure P10.11, assuming RB , RC , R ,VBB ,VCC ,β .

Find: The current IR

Analysis: The base and collector currents are:

IB =

VBB − VBE = 26 µA RB

IC = hFE IB = β I B = 120.26.10 −6 = 3.1mA Considering that

VCE = RI R , from collector circuit we get:

VCC = RC (I C + I R ) + VCE = RC (I C + I R ) + RI R = RC I C + I R (R + RC ) ⇓ V − RC I C I R = CC = 3mA R + RC

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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 10

 Problem 10.12 For the circuit shown in Figure P10.12, verify that the transistor is in saturation. Let RB = 8kΩ, RE = 260ohm, RC = 1.1kΩ, VCC = 13 V, VBB = 7 V,

and β = 100 ÷ 300.

Solution: Known quantities: The circuit shown in Figure P10.12: let

RB , RE , RC ,VCC ,V BB ,β = 100 ÷ 300

...