Rizzoni5eSM_CH10
-
Upload
raffaele-longo -
Category
Documents
-
view
512 -
download
3
Transcript of Rizzoni5eSM_CH10
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.1 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.
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. Understand the basic principles of amplification and switching. Section 10.1. 2. Understand the physical operation of bipolar transistors, and identify their state. Section
10.2 3. Understand the large-signal model of the bipolar transistor, and apply it to simple
amplifier circuits. Section 10.3. 4. Determine and select the operating point of a bipolar transistor circuit; understand the
principle of small signal amplifiers. Section 10.4. 5. Understand the operation of bipolar transistor as a switch and analyze basic analog and
digital gate circuits. Section 10.5.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
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.
Section 10.2: Operation of the Bipolar Junction Transistor
Problem 10.1
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
Solution:
Known quantities:
Transistor type and operating characteristics:
a) npn, VBE = 0.8 V and VCE = 0.4 V
b) npn, VCB = 1.4 V and VCE = 2.1 V
c) pnp, VCB = 0.9 V and VCE = 0.4 V
d) 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.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
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.
b) VBE = VBC + VCE = 0.7 V. The BE junction is forward-biased.
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
Solution:
Known quantities:
The circuit of Figure P10.3: 100==B
C
I
Iβ .
Find:
The operating point and the state of the transistor.
Analysis:
V 6.0=BEV and the BE junction is forward biased.
A 9.13820
6.012
1
µ=−
=−
=R
VVI BECCB
mA39.1=⋅= BC II β
Writing KVL around the right-hand side of the circuit:
0=+++− EECECCCC RIVRIV
VCE = VCC − IC RC − IC + IB( )RE = 12− (1.39)(2.2)− (1.39+ 0.0139)(0.910) = 7.664 V
VBC = VBE +VCE = 0.6 _ 7.664 = 8.264 V
⇒> BECE VV
The transistor is in the active region.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.4 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.
Problem 10.4
Solution:
Known quantities:
The magnitude of a pnp transistor's emitter and base current, and the magnitudes of the voltages across the emitter-
base and collector-base junctions:
IE = 6 mA, I
B = 0.1 mA and V
EB = 0.65 V, V
CB = 7.3 V.
Find:
a) VCE.
b) IC.
c) The total power dissipated in the transistor, defined as BBECCE IVIVP += .
Analysis:
a) VCE = V
CB - V
EB = 7.3 - 0.65 = 6.65 V.
b) IC = I
E - I
B = 6 - 0.1 = 5.9 mA.
c) The total power dissipated in the transistor can be found to be: mW 39109.565.6 3 =××=≈ −CCE IVP
Problem 10.5
Solution:
Known quantities:
The circuit of Figure P10.5, assuming the BJT has Vγ = 0.6 V.
Find:
The emitter current and the collector-base voltage.
Analysis:
Applying KVL to the right-hand side of the circuit, A 52030000
156.0
30000
15µ−=
+−=
+−= BE
EV
I
Then, on the left-hand side, assuming β >> 1:
( ) V 8.171010105201010
010
36 =×××−−=−=
⇒=++−−
CCCB
CBCC
RIV
VRI
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.5 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.
Problem 10.6
Solution:
Known quantities:
The circuit of Figure P10.6, assuming the BJT has
V 6.0=BEV and β =150.
Find:
The operating point and the region in which the transistor
operates.
Analysis:
Define RC = 3.3 kΩ, RE = 1.2 kΩ, R1 = 62 kΩ, R2 = 15 kΩ, VCC = 18 V
By applying Thevenin’s theorem from base and mass, we have
V857.7101515112001025.2330018
mA25.2
µA15)1(
V5.3
kΩ078.12||
63
21
2
21
=⋅⋅⋅−⋅⋅−=−−=
==
≅++
−=
≅+
=
==
−−EECCCCCE
BC
EB
BEBBB
CCBB
B
IRIRVV
II
RR
VVI
VRR
RV
RRR
β
β
From the value of VCE it is clear that the BJT is in the active region.
Problem 10.7
Solution:
Known quantities:
The circuit of Figure P10.7, assuming the BJT has
V6.0=γV .
Find:
The emitter current and the collector-base voltage.
Analysis:
Applying KVL to the right-hand side of the circuit,
0=++− EBEECC VRIV
µA4.4971039
6.0203
=⋅
−=
−=
E
EBCCE
R
VVI . Since 1>>β , µA4.497=≈ EC II
Applying KVL to the left-hand side: 0=−+ DDCCCB VRIV
VCB = VDD − IC RC = 20− 497.4 ⋅20 ⋅10−3 = 10.05V
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.6 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.
Problem 10.8
Solution:
Known quantities:
The circuit of Figure P10.7, assuming the emitter resistor is changed to Ωk 22 and the BJT has V6.0=γV .
Find:
The operating point of the transistor.
Analysis:
Aµ8.8811022
6.0203
=⋅
−=
−=
E
EBCCE
R
VVI , µA8.881=≈ EC II
V364.2310208.88120 =−⋅⋅−=−= CCDDCB RIVV
Problem 10.9
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
A 600 and A,200 A, 100 µµµ=BI
b) VCE, assuming the maximum allowable collector power
dissipation is 0.5 W for A 500 µ=BI .
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
dissipated by the transistor is CCEBBECCE IVIVIVP ≈+= , therefore: VCE ≈P
IC
=0.5
79.5 ⋅10−3= 6.29 V.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
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.
Problem 10.10
Solution:
Known quantities:
Figure P10.10, assuming both transistors are silicon-based with 100=β .
Find:
a) IC1, VC1, VCE1. b) IC2, VC2, VCE2.
Analysis:
a) From KVL: ⇒=++− 030 111 BEBB VRI
µA07.3910750
7.03031 =
⋅
−=BI
⇒=⋅= mA907.311 BC II β
V779.52.6907.33030 111 =⋅−=−= CCC IRV
V779.511 == CCE VV .
b) Again, from KVL: ⇒=++− 0779.5 222 EEBE RIV mA081.1107.4
7.0779.532 =
⋅
−=EI
and mA07.1101
100081.1
122 =
⋅=
+=
ββ
EC II .
Also, ⇒=+++− 0)(30 2222 CEECC VRRI V574.3)7.420()07.1(302 =+⋅−=CEV .
Finally, V603.8)20()07.1(3030
22
22 =⋅−=⇒
−= C
C
CC V
R
VI .
Problem 10.11
Solution:
Known quantities:
Collector characteristics of the 2N3904 npn transistor, see
data sheet pg. 560.
Find:
The operating point of the transistor in Figure P10.11, and the value of β at this point.
Analysis:
Construct a load line. Writing KVL, we have: 0500050 =+⋅+− CEC VI .
Then, if 0=CI , V50=CEV ; and if 0=CEV , mA10=CI . The load line is shown superimposed on the collector
characteristic below:
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.8 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.
The operating point is at the intersection of the load line
and the A 20 µ=BI line of the characteristic. Therefore,
mA 5≈CQI and V 20≈CEQV .
Under these conditions, an A 5 µ increase in BI yields an
increase in CI of approximately mA 156 =− .
Therefore,
200105
1016
3
=⋅
⋅=
∆∆
≈−
−
B
C
I
Iβ
The same result can be obtained by checking the hFE gain
from the data-sheets corresponding to 5 mA.
Problem 10.12
Solution:
Known quantities:
The circuit shown in Figure P10.12. With reference to Figure 10.20, assume V6.0=γV , V2.0=satV .
Find:
The operating point of the transistor, by computing the ratio of collector current to base current.
Analysis:
V2.0== satCE VV , therefore mA8.92.010
=−
=C
CR
I V6.0== γVVBE , therefore µA1026.07.5
=−
=B
BR
I
β<<=⋅
⋅=
−
−08.96
10102
108.96
3
B
C
I
I
Load line
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
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.
Problem 10.13
Solution:
Known quantities:
The circuit in Figure 10.28 in the text. VCC=20 V, RC=5kΩ, RE=1kΩ.
Find:
The region of operation of the transistor.
Analysis:
(a) IE = IC + IB = 1 mA + 20µA = 1.02 mA
VE = 1000 IE = 1.02 V
VRC = 5000 IC = 5 V
VCB = 20 - VRC -VBE -VE
= 20 - 5 - 0.7 - 1.02 = 13.28 V
The CB junction is reverse-biased. Therefore, the transistor is operating in the active region.
(b) IE = 3.2 mA + 0.3 mA = 3.5 mA
VE = 3.5 V
VRC = 16 V
VCB = 20 -16 - 0.8 - 3.5 = -30 mV
The CB junction is forward-biased. Therefore, the transistor is operating in the saturation region.
(c) IE = 3 mA + 1.5 mA = 4.5 mA
VE = 4.5 V
VRC = 15 V
VCB = 20 - 15 - 0.85 - 4.5 = -0.35 V
The CB junction is forward-biased. Therefore, the transistor is operating in the saturation region.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.10 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.
Problem 10.14
Solution:
Known quantities:
The circuit of Figure P10.14, VCEsat=0.1V, VBEsat=0.6V, and β=50.
Find:
The base voltage required to saturate the transistor.
Analysis:
The collector current is
mA 9.111
1.012=
−=CI
The base current is
IB =IC
β=
11.9
50= 0.238 mA = 238µA
And since
mA 10
BEsatBBB
VVI
−=
Therefore,
V V 98.26.0k10mA 238.0 =+Ω⋅=BBV
Problem 10.15
Solution:
Known quantities:
β=60; VBE=0.6V; VCB=7.2V; |IE|=4mA.
Find:
a) IB; b) VCE;
Analysis:
(a) Since
BCBE IIII )1( +=+= β
we can compute
IB =IE
β +1=
4
61= 65.6 mA
(b) VCE = VCB - VBE
= 7.2 - 0.6 = 6.6 V
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.11 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.
Problem 10.16
Solution:
Known quantities:
Collector characteristics of 2N3904 npn transistor; Transistor circuits;
Find:
The operating point;
Analysis:
From KVL,
or
If 0=CEV , mAk
IC 99.410
9.49== , and if 0=CI , VVCE 9.49= . The load line
is shown superimposed on the collector characteristic below:
The operating point is at the intersection of the load line and the AIB µ20=
line of the characteristic. Therefore, mAICQ 3≈ and VVCEQ 8≈ .
Under these conditions, a Aµ10 increase in BI yields an increase in CI of
approximately mAmAmA 235 =− . Therefore,
20010
2==
∆∆
≈A
mA
I
I
B
C
µβ
Addition of the emitter resistor effectively increased the current gain by
decreasing the magnitude of the slope of the load line.
0)20(5550 =++++− AIkVkI CCEC µ
9.491.05010 =−=+ CCE kIV
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.12 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.
Section 10.3: BJT Large-Signal Model
Problem 10.17
Solution:
Known quantities:
For the circuit shown in Figure 10.14 in the text:
mW100,mA10V,4.1
95,V,2.0V,7.0,V5,kΩ1,mA5,V5,V0
max =≥=
========
PIV
VVVRIVV
LEDLED
CEsatCCBBonoff
γ
γ β
Find:
Range of RC.
Analysis:
Ω=−−
≤−−
= 34001.0
2.04.15
LED
CEsatLEDCCC
I
VVVR
γ
From the maximum power
Ω=−−
>
===
47
mA714.1
1.0
max
maxmax
LED
CEsatLEDCCC
LEDLED
I
VVVR
V
PI
γ
γ
Therefore, RC ∈[47, 340] Ω
Problem 10.18
Solution:
Known quantities:
For the circuit shown in Figure 10.18 in the text: Ω======= 500188.5,V,6V,75.0,V12,kΩ33,V1.1 SCEQBECCBD RVVVRV β
Find:
The resistance RC.
Analysis:
The current through the resistance RB is given by
µA6.1033000
75.01.1=
−=
−=
B
BEQDB
R
VVI
The current through RS is: mA8.9500
1.16=
−=
−=
S
DCEQS
R
VVI
It follows that the current through the resistance RC is mA8.11=+= SBCQ III β
Finally, Ω=−
=−
= 5.5080118.0
612
CQ
CEQCCC
I
VVR
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.13 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.
Problem 10.19
Solution:
Known quantities:
For the circuit shown in Figure 10.14 in the text:
mW100,mA10V,4.1
95,V,2.0V,7.0,V5,Ω340,mA5,V5,V0
max
max
=≥=
========
PIV
VVVRIVV
LEDLED
CEsatCCCBonoff
γ
γ β
Find: Range of RB.
Analysis:
If the BJT is in saturation
mA10=−−
=C
CEsatLEDCCC
R
VVVI
γ
In order to guarantee that the BJT is in saturation
Ω=−
≥
Ω=−
=−
≤
860
k85.40
95
01.0
7.05
/
maxB
onB
C
onB
I
VVR
I
VVR
γ
γ
β
Problem 10.20
Solution:
Known quantities:
For the circuit shown in Figure 10.14 in the text:
mW100,mA10V,4.1
V,2.0V,7.0,V5,Ω340,kΩ10,mA5,V5,V0
max
max
=≥=
========
PIV
VVVRRIVV
LEDLED
CEsatCCCBBonoff
γ
γ
Find: Minimum value of β that will ensure the correct operation of the LED.
Analysis:
mA43.010000
3.4==
−=
B
onB
R
VVI
γ
25.231043.0
01.03
minmin =
⋅==
−B
LED
I
Iβ
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.14 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.
Problem 10.21
Solution:
Known quantities:
For the circuit shown in Figure 10.14 in the text:
mW100,mA10V,4.1
V,2.0V,7.0,V5,Ω340,kΩ10,mA5,V3.3,V0
max
max
=≥=
========
PIV
VVVRRIVV
LEDLED
CEsatCCCBBonoff
γ
γ
Find:
Minimum value of β that will ensure the correct operation of the LED.
Analysis:
mA26.010000
7.03.3=
−=
−=
B
onB
R
VVI
γ
5.381026.0
01.03
minmin =
⋅==
−B
LED
I
Iβ
Problem 10.22
Solution:
Known quantities:
For the circuit shown in Figure 10.14 in the text:
A1
V,1V,7.0,V13,Ω12,kΩ1,mA1,V5,V0 max
≥
========
C
CEsatCCBBonoff
I
VVVRRIVV γ
Find: Minimum value of β that will ensure the correct operation of the fuel injector.
Analysis:
A112
113=
−=
−=
R
VVI CEsatCCC
1000101
13
maxmin =
⋅==
−B
C
I
Iβ
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.15 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.
Problem 10.23
Solution:
Known quantities:
For the circuit shown in Figure 10.14 in the text:
A1
V,1V,7.0,V13,Ω12,2000,mA1,V5,V0 max
≥
========
C
CEsatCCBonoff
I
VVVRIVV γβ
Find: The range of RB that will ensure the correct operation of the fuel injector.
Analysis:
If the BJT is in saturation
A1=−
=R
VVI CEsatCCC
Because this is the minimum value allowed for the current to drive the fuel injector, it is necessary to guarantee that the BJT is in saturation. In order to guarantee that the BJT is in saturation
Ω=−
≥
Ω=−
=−
≤
k3.4
k6.8
2000
1
7.05
/
maxB
onB
C
onB
I
VVR
I
VVR
γ
γ
β
Problem 10.24
Solution:
Known quantities:
For the circuit shown in Figure 10.14 in the text:
A1
V,1V,7.0,V13,Ω12,2000,mA1,V3.3,V0 max
≥
========
C
CEsatCCBonoff
I
VVVRIVV γβ
Find: The range of RB that will ensure the correct operation of the fuel injector.
Analysis:
If the BJT is in saturation
A1=−
=R
VVI CEsatCCC
Because this is the minimum value allowed for the current to drive the fuel injector, it is necessary to guarantee that the BJT is in saturation. In order to guarantee that the BJT is in saturation
Ω=−
≥
Ω=−
=−
≤
k6.2
k2.5
2000
1
7.03.3
/
maxB
onB
C
onB
I
VVR
I
VVR
γ
γ
β
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.16 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.
Problem 10.25
Solution:
Known quantities:
The circuit of Figure P10.25: IC = 40 mA; Transistor large signal parameters. Find: Design a constant-current battery charging circuit, that is, find the values of VCC, R1, R2 that will cause the transistor Q1 to act as a 40-mA constant current source.
Assumptions:
Assume that the transistor is forward biased. Use the large-signal model with β = 100.
Analysis:
The battery charging current is 40 mA, IC = 40 mA.
Thus, the emitter current must be mA4.401
=+
= EE IIβ
β.
Since the base-emitter junction voltage is assumed to be 0.6 V, then resistor R2 has a voltage: V 56.06.52 =−=−= γVVV z , so the required value of R2 to be:
Ω=== 8.1230404.0
52
EI
VR
Since the only purpose of R1 is to bias the Zener diode, we can select a value that will supply enough current fro the Zener to operate, for example R1 > 100 Ω, so that there will be as little current flow through this resistance as possible. Finally, we need to select an appropriate supply voltage. VCC must be greater than or equal to the sum of the battery voltage, the CE junction voltage and the voltage across R2. That is, 59 ++≥ CECC VV . A collector supply of 24 V
will be more than adequate for this task.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.17 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.
Problem 10.26
Solution:
Known quantities:
The circuit of Figure of P10.26. Find: Analyze the operation of the circuit and explain how EI is decreasing
until the battery is full. Find the values of VCC, R1 that will result in a practical design.
Assumptions:
Assume that the transistor is forward biased.
Analysis:
When the Zener Diode works in its reverse breakdown area, it provides a constant voltage: V 11=zV . That means:
V 11== ZB VV . When the transistor is forward biased, according to KVL,
batteryBEBEZ VVRIV ++⋅= γ , where BER is the base resistance.
As the battery gets charged, the actual battery charging voltage batteryV will increase from 9.6 V to 10.4 V.
As batteryV increases gradually, ZV and γV stay unchanged, then we can see that BEI will decrease gradually.
So ( ) BEE II 1+= β will also decrease at the same time. Since the only purpose of R1 is to bias the Zener diode, we can select a value that will supply enough current fro the Zener to operate, for example R1 > 100 Ω, so that there will be as little current flow through this resistance as possible. Finally, we need to select an appropriate supply voltage. VCC must be greater than or equal to the sum of the battery voltage, the CE junction voltage. That is, CECC VV +≥11 . A collector supply of 12 V should be adequate for this
task.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.18 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.
Problem 10.27
Solution:
Known quantities:
The circuit of Figure P10.27: Vin= 5 V Find: Values of Rb.
Assumptions:
Assume that the transistors are in the active region. Use the large-signal model with β = 40 for each transistor.
Analysis:
The emitter current from Q1, iE1 = (β+1) iB1 becomes the base current for Q2, and therefore, iC2 = β iE1 = β (β+1) iB1. The Q1 base current is given by the expression
b
inB
R
VVVi
γγ −−=1
Therefore the motor current will reach maximum when Vin= 5 V:
( ) A 34.01max =
−−+=
b
inC
R
VVVi
γγββ
So, ( ) ( ) Ω=−
⋅=−
+= 329,18)2.15(
34.0
41402
34.0
1γ
ββVVR inb
Since 18.33 kΩ is a standard resistor value, we should select Rb = 18.33 kΩ, which will result in a slightly lower maximum current.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.19 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.
Section 10.4: Selecting an Operating Point for a BJT
Problem 10.28
Solution:
Known quantities:
The circuit of Figure of 10.22 in the text, V.10 ,50 V,5 , K1 min ===Ω= CCBBC VVR β
Find: The range of BR to make the transistor in the saturation state.
Analysis:
Assuming V 2.0=CEsatV , the current CI is:
mA 8.9=−
=C
CEsatCCC
R
VVI
Therefore, mA 196.0==b
II CB
Assuming V 6.0== BEsatVVγ , we have
Ω=−
= k 45.22B
BEBBB
I
VVR
That is Ω<< k 45.220 BR
Problem 10.29
Solution:
Known quantities:
The circuit of Figure of 10.22 in the text, V.5 ,50 ,10K , K1 min ==Ω=Ω= CCBC VRR β
Find: The range of BBV to make the transistor in the saturation state.
Analysis:
Assume V 2.0=CEsatV , the current CI can be found as
mA 8.4=−
=C
CEsatCCC
R
VVI
Therefore, A 96mA 096.0 µ===b
II CB
Assuming V 6.0== BEsatVVγ , we have
V 56.1=+= BEsatBBBB VRIV
That is V 56.1>BBV
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.20 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.
Problem 10.30
Solution:
Known quantities:
The circuit of Figure 10.20 in the text, V.10 ,100 A,20 , k2 ===Ω= CCBBC VIR βµ
Find: .,,, CBCEEC VVII
Analysis:
V 4.56.06 Then
V 6.0 Assume
V 62210
mA 02.21020101
)1(
mA 21020100
6
6
=−=−=
=
=×−=−=
=××=
+=+=
=××==
−
−
BECECB
BE
CCCCCE
BCBE
BC
VVV
V
RIVV
IIII
II
β
β
Problem 10.31
Solution:
Known quantities:
For the circuit shown in Figure P10.31: V20=CCV 130=β MΩ8.11 =R Ωk3002 =R
Ωk3=CR Ωk1=ER
Ωk1=LR Ωk6.0=SR mV )1028.6cos( 1 3tvS ×= .
Find:
The Thèvenin equivalent of the part of the circuit containing 1R , 2R , and
CCV with respect to the terminals of 2R . Redraw the schematic using the
Thèvenin equivalent.
Analysis:
Extracting the part of the circuit specified, the Thèvenin equivalent voltage is the open circuit voltage. The equivalent resistance is obtained by suppressing the ideal independent voltage source:
Note that CCV must remain in the circuit because it supplies current to
other parts of the circuit:
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.21 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.
Problem 10.32
Solution:
Known quantities:
For the circuit shown in Figure P10.32: V 12=CCV 130=β Ωk821 =R Ωk222 =R Ωk5.0=ER Ω16=LR .
Find:
CEQV at the DC operating point.
Analysis:
Simplify the circuit by obtaining the Thèvenin equivalent of the biasing network (R1,, R2, VCC) in the base circuit:
Redraw the circuit using the Thèvenin equivalent. The "DC blocking" or "AC coupling" capacitors act as open
circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions
of current and polarities of voltages.
Assume the transistor is operating in its active region. Then, the base-emitter junction is forward biased.
( ) ( ) µA18.22500113017350
7.0538.2
1 = =
R + + R
V - V = I
EB
BEQBBBQ ⋅++
−⋅β
( ) ( )
V55.105.0906.212
:KVL
mA906.21018.2211301 6
= = R I - V = V
0 = V + V - R I -
= + = I + = I
EEQCCCEQ
CCCEQEEQ
BQEQ
⋅−
⋅⋅ −β
The collector-emitter voltage is greater than its saturation value (0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid.
Ωk35.172282
2282Suppress
V538.22282
2212:VD
= = R + R
R R = R = R:V
= = R + R
R V = V = V = V
21
21eqBCC
21
2CCOCTHBB
+⋅
+⋅
0 = R I] + [ + V + R I + V -
0 = R I + V + R I + V -
I] + [ = I [Si] 700 V
EBQBEQBBQBB
EEQBEQBBQBB
BQEQBEQ
1
:KVL
1mV
β
β≈
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.22 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.
Problem 10.33
Solution:
Known quantities:
For the circuit shown in Figure P10.33: V12=CCV 100=β V 4=EEV Ωk100=BR
Ωk3=CR Ωk3=ER
Ωk6=LR Ωk6.0=SR mV )1028.6cos( 1 3tvS ×= .
Find:
CEQV and the region of operation.
Analysis:
The "DC blocking" or "AC coupling" capacitors act as open circuits for
DC; therefore, the signal source and load can be neglected since this is a
DC problem. Specify directions of current and polarities of voltages.
Assume the transistor is operating in its active region; then, the base-
emitter junction is forward biased and:
[ ]
V 06.11
300010100.8273000109.818124
0 :KVL
A 0.82710189.8)1100()1(
A9.81810189.8)100(
A189.8)3000)(1100(100000
7.04
1
0 :KVL
)1(
][ mV 700
66
6
6
=
⋅⋅⋅−⋅⋅−+=−−+=
⇒=+−−−+
=⋅⋅+=⋅+=
=⋅⋅=⋅=
=++
−=
++
−=
⇒=+++−
+=⋅=
≈
−−
−
−
EEQCCQCCEECEQ
CCCCQCEQEEQEE
BQEQ
BQCQ
EB
BEQEEBQ
EEQBEQBBQEE
BQEQBQCQ
BEQ
RIRIVVV
VRIVRIV
II
II
RR
VVI
RIVRIV
IIII
SiV
µβ
µβ
µβ
ββ
The collector-emitter voltage is greater (more positive) than its saturation value (+ 0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid. Notes: 1. DC power may be supplied to an npn BJT circuit by connecting the positive terminal
of a DC source to the collector circuit, or, by connecting the negative terminal of a DC source to the emitter circuit, or, as was done here, both.
2. In a pnp BJT circuit the polarities of the sources must be reversed. Negative to collector and positive to emitter.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.23 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.
Problem 10.34
Solution:
Known quantities:
For the circuit shown in Figure P10.34: V12=CCV 130=β Ωk325=BR Ωk9.1=CR
Ωk3.2=ER
Ωk10=LR Ωk5.0=SR mV )1028.6cos( 1 3tvS ×= .
Find:
CEQV and the region of operation.
Analysis:
The "DC blocking" or "AC coupling" capacitors act as open circuits for DC;
therefore, the signal source and load can be neglected since this is a DC
problem. Specify directions of current and polarities of voltages. Assume the
transistor is operating in its active region; then, the base-emitter junction is
forward biased. The base and collector currents both flow through the collector
resistor in this circuit.
The collector-emitter voltage is greater than its saturation value (0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid.
0)(1)(-
0 :KVL
V 896.4
3.2691.19.1691.112
0 :KVL
mA 691.11096.12)1130()1(
A91.1210))9.13.2()1130(325(
7.012
))(1(
)1(
0 :KCL
)1(
][ mV 700
6
3
=+−−++
⇒=+−−−−
=
⋅−⋅−=−−=
⇒=+−−−
=⋅⋅+=⋅+==
=⋅+⋅++
−=
+++
−=
+=+=
⇒=−+
+=⋅=
≈
−
CCBBQBEQCEBQ
CCCRCBBQBEQEEQ
EEQCRCCCCEQ
CCCRCCEQEEQ
BQEQRC
CEB
BEQCCBQ
BQBQCQRC
RCCQBQ
BQEQBQCQ
BEQ
VRIVRRI
VRIRIVRI
RIRIVV
VRIVRI
III
RRR
VVI
IIII
III
IIII
SiV
β
β
µβ
β
ββ
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.24 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.
Problem 10.35
Solution:
Known quantities:
For the circuit shown in Figure P10.35: V3=Sv 100=β Ωk60=BR
Find:
a) The value of ER so that EI is 1 mA.
b) CR so that CV is 5 V.
c) The small-signal equivalent circuit of the amplifier for Ω= k5LR
d) The voltage gain.
Analysis:
(a) With Ω= k 60BR and V 3=BV , applying KVL, we have
EBBB RIRI )1(6.03 β+++=
EB
RI
101k60
4.2
+Ω=
mARk
IE
E 110160
4.2101 =
+Ω=
Therefore,
Ω=−⋅
= k 81.1101
604.2101ER
(b) EECCCE RIRIV −−=15
From (a), we have mA 99.01=
+=
ββ
EC II
Therefore, Ω=−−
= k 27.899.0
81.1515CR
(c) The small signal equivalent circuit is shown below
(d) iwB
SB
hR
VI
+=∆
∆−=
oeLCout
hRIv
1 Bfe
oe
outC Ih
h
VI ∆+=∆
1
Since hoe is not given, we can reasonably assume that 1/hoe is very large. Therefore,
15.4100
−=+⋅
−==ieB
L
s
outV
hR
R
v
vA
Ω=×
=∂
=−
kI
Vh
BQIB
BEie 6.60
100099.0
6.03
∂
OUTvvS
+
-
∆ IB
C
E-
h ie
oeh1
I∆ C
R C
R L
+
BR B
∆ I Bh fe
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.25 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.
Problem 10.36
Solution:
Known quantities:
For the circuit shown in Figure P10.36: Ωk200=CR
Find:
e) The operating point of the transistor. f) Voltage gain inout vv ; current gain inout ii
g) Input resistance ir
h) Output resistance or
Analysis:
(a) V 1.621
2 =+
=RR
RVV CCB
Ω== 87.3749|| 21 RRRB
Assuming V 6.0=BEV , we have
V 5.5=−= BEBEV VVV
mA 22==E
EE
R
VI
mA 088.01=
+=b
II EB
and
( )V 12.55.51021.912200-15
5.5
3-
.
=−⋅⋅=
−−=−= CCCCECCE IRVVVV
(b) The AC equivalent circuit is shown on the right:
Ω=×
≈=−
kI
Vh
BQIB
BEie 82.6
10088.0
6.03∂
∂
BCBEout IIIRv )1250(250)( +=+=
BieBoutieBin IhIvhIv ⋅⋅+=+= 251250
Therefore, the voltage gain is
902.0==in
outV
v
vA and
BCBout IIIi ⋅+++= )1(β
BBieBBB
inBin RIhII
R
vIi )251250( ⋅⋅++=+=
and the current gain is
84.12)251250(
)1(=
⋅⋅+++
=BBieBB
B
in
out
RIhII
I
i
i β
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.26 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.
(c) To find the input resistance we compute:
BieBin IhIv ⋅⋅+= 251250
BBieBBin RIhIIi )251250( ⋅⋅++=
Therefore. the input resistance is
Ω== 3558in
ini
i
vr
(d) To find the output resistance we compute
BCBEout IIIRv )1250(250)( +=+=
BCBout IIIi ⋅+++= )1(β
Therefore, the output resistance is
Ω== 250out
outo
i
vr
Problem 10.37
Solution:
Known quantities:
The circuit shown in Figure P10.37(a), P10.37(b):
Find:
The duration of the fuel injector pulse.
Analysis:
(a) With VCE = 0.3 V, VBE = 0.9 V and VBATT = 13
V, TC = 100°, from Figure P9.6(d), we have KC = 0,
VCIT = 16/13= 1.23 ms. The signal duration is:
τ = 1×10-3×0 + 1.23×10-3 = 1.23 ms
When Vsignal is applied, the base current is
IB = VBATT /80= 0.1625 A
Thus, the transistor will be in the saturation region.
Therefore,
Vinj = VBATT - VCE = 13 - 0.3 = 12.7 V
The time constant of the injector circuit is:
τ' = L/R= 0.1 ms << τ = 1.23 ms
As Vsignal = 0, the transistor is in the cut-off region. The differential equation governing the injector current is:
1×10-3 dIinj/dt+ 10Iinj = Vinj,
and Iinj(0) = 0. Thus,
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.27 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.
Iinj = Vinj/10- Vinj/10∙e-10000t
The time when Iinj ≥ 0.1 is then found to be
tinj = -(ln(1.17/1.27))/104= 8.2 µs
That is, 8.2 ms after the Vsignal is applied, the fuel will be
injected into the intake manifold.
(b) From Figure P10.37(d), we have
KC = - 1/60∙TC + 5/3; at TC = 20, KC = 4/3. Also, VCIT =
16/8.6= 1.86 ms
The signal duration therefore is
τ = 1×10-3×4/3+ 1.86×10-3 = 3.19 ms
When Vsignal is applied, the base current is
IB = VBATT /80= 0.1075 A
Thus, the transistor will be in the saturation region. Therefore,
Vinj = VBATT - VCE = 8.6 - 0.3 = 8.3 V
The time constant of the injection circuit is:
τ' = L/R= 0.1 ms << τ = 3.19 ms
When Vsignal is 0, the transistor is in the cut-off region. Using the same differential equation and initial condition
as in part (a),
Iinj = Vinj/10- Vinj/10∙e-10000t
The time when Iinj ≥ 0.1 can be found as
t = 12.84 µs
That is, 12.84 µs after the Vsignal is applied, the fuel will be injected into the intake manifold.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.28 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.
Problem 10.38
Solution:
Known quantities:
For the circuit shown in Figure P10.38: V 8.0=γV
Find:
The maximum of frequency with which the light can be switched.
Analysis:
From the power dissipation of the relay,
W5.02
WR
VP == ,
we can compute
RW = 50 Ω
When vS is 5 V, the transistor is in the saturation region and the relay current at steady state is:
mA 10650
2.055=
−=RI
When vS is 0 V, the transistor will be in cut off, and we have the following equation
050005.0 =+ RR I
dt
dI
That is
010103 =⋅+ R
R Idt
dI
Solving the above equation with
IR(0) = 0.106 A,
IR(t) = 0.106e-10000t
vR(t) = - 25 IR = - 2.65e-10000t
The relay will be cut-off as soon as vS is 0.
Therefore, the switching frequency will be the frequency of the input signal vS.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.29 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.
Problem 10.39
Solution:
Known quantities:
For the circuit shown in Figure P10.39: 70: ;130: 21 == ββ QQ .
Find:
The overall current gain.
Analysis:
The AC circuit is shown on the right:
The current gain is
9300)1(hh
h
1
112fe1
1
12fe1
1
22fe1
2
2
1
1
1
21
1
=++=+=
+=
+=
+==
B
Bfefe
B
Efe
B
Bfe
B
C
B
C
B
CC
Bi
I
Ihh
I
Ih
I
Ih
I
I
I
I
I
II
I
IA
Problem 10.40
Solution:
Known quantities:
For the circuit shown in Figure P10.40: V 6.0=γV .
Find:
1R and 2R .
Analysis:
a) It’s given by the problem.
b)
For 5=CEQV , VVVEC RR 20=+ .
Assume EC II ≈ .
VV
VVmAk
I
B
EC
7.87.08
885.2
20
=+=
=⇒==
For 20=β , mAIB 4.020
8== .
hie1
IB1
I B2h ie2
IC1
IC2h fe1IB1
h fe2I B2
B'C'
E'
I
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.30 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.
For 50=β , and mAIC 8.8≤ ,
VV
mAI
B
B
5.97.08.8
176.050
8.8
=+=
=≤
5.910176.0
7.8104.0
3
3
=×−
=×−−
−
BB
BB
RV
RV
Solving, Ω= kRB 35.5 .
VVB 84.9=
Since )25(21
1
21
1
RR
RV
RR
RV CCB +
=+
=
and 21
21
RR
RRRB +
= , we can solve for
Ω= kR 82.81 and Ω= kR 59.132 .
c)
Assume VVCE 23max
≈ and VVCE 3min
≈ .
Then CEQV should be set to the middle of this range, or VVCEQ 132
323=
+= .
From KVL,
2515001000 =++ CQCEQEQ IVI
or,
25)(1500)1(1000 =+++ BQCEQBQ IVI ββ
Solving,
121325251000 =−=BQI
AIBQ µ81.47=∴
VIV BQEQ 829.4)1(1000 =+= β
VVVV BEEQBQ 529.5=+=
Again from KVL,
( ) 529.5)81.47(22 11 =−⇒=− AIRVIIR RBQBQR µ
And 471.19529.5252522 22 =−=⇒=+ RBQR IRVIR
Choose a typical value for 2R , say Ω= kR 102 . Then,
mAR
IR 947.1471.19
22
==
And Ω=−
= 291181.47947.1
529.51
AmAR
µ
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.31 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.
Section 10.4: BJT Switches and Gates
Problem 10.41
Solution:
Known quantities:
The circuit given in Figure P10.41.
Find:
Show that the given circuit functions as an OR gate if the
output is taken at v01.
Analysis:
Construct a state table. This table clearly describes an AND
gate when the output is taken at 1ov .
v1 v2 Q1 Q2 Q3 vo1 vo2
0 0 off off on 0 5V
0 5V off on off 5V 0
5V 0 on off off 5V 0
5V 5V on on off 5V 0
Problem 10.42
Solution:
Known quantities:
The circuit given in Figure P10.41.
Find:
Show that the given circuit functions as a NOR gate if the output is
taken at v02.
Analysis:
See the state table constructed for Problem 10.41. This table clearly describes a NOR gate when the output is taken at 2ov .
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.32 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.
Problem 10.43
Solution:
Known quantities:
The circuit given in Figure P10.43.
Find:
Show that the given circuit functions as an AND gate if the output is taken at v01.
Analysis:
Construct a state table. This table clearly describes an AND gate when
the output is taken at 1ov .
v1 v2 Q1 Q2 Q3 vo1 vo2
0 0 off off on 0 5V
0 5V off on on 0 5V
5V 0 on off on 0 5V
5V 5V on on off 5V 0
Problem 10.44
Solution:
Known quantities:
The circuit given in Figure P10.43.
Find:
Show that the given circuit functions as a NAND gate if the output is taken at v02.
Analysis:
See the state table constructed for Problem 10.32. This table clearly describes a NAND gate when the output is taken at 2ov .
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.33 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.
Problem 10.45
Solution:
Known quantities:
In the circuit given in Figure P10.45 the minimum value of vin for a high input is 2.0 V. Assume that the transistor Q1 has a β of at least 10.
Find:
The range for resistor RB that can guarantee that the transistor is on.
Analysis:
mA4.22000
2.05=
−=ci , therefore, i
B = i
C/β = 0.24 mA.
(vin
)min = 2.0 V and (vin
)max = 5.0 V, therefore, applying KVL: -vin +RB iB + 0.6 = 0
or B
inB
i
vR
6.0−= . Substituting for (v
in)min and (vin)max , we find the following range for RB
:
Ω333.18Ω833.5 kRk B ≤≤
Problem 10.46
Solution:
Known quantities:
For the circuit given in Figure P10.46: Ωk27,kΩ10 2121 ==== BBCC RRRR .
Find:
a) vB, vout, and the state of the transistor Q1 when vin is low. b) vB, vout, and the state of the transistor Q1 when vin is high.
Analysis:
a) vin is low ⇒ Q1 is cutoff ⇒ vB = 5 V ⇒ Q2 is in saturation ⇒ vout = low = 0.2 V.
b) vin is high ⇒ Q1 is in saturation ⇒ vB = 0.2 V ⇒ Q2 is cutoff ⇒ vout = high = 5 V.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.34 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.
Problem 10.47
Solution:
Known quantities:
For the inverter given in Figure P10.47: Ωk5,Ωk221 === BCC RRR .
Find:
The minimum values of β1 and β2 to ensure that Q1 and Q2 saturate when vin is high.
Analysis:
mA4.22000
2.05=
−=ci , therefore, mA
5.2
β=ci . Applying
KVL: 06.06.06.05 1 =++++− BBiR
Therefore, iB1
= 0.64 mA. 2111 500600
BBE iii +=⋅= β or
21
5.22.164.0 ββ +=⋅
Choose β2 = 10 ⇒ β
1 = 2.27.
Problem 10.48
Solution:
Known quantities:
For the inverter given in Figure P10.47: 4,Ωk2,Ωk5.2 2121 ==== ββCC RR .
Find:
Show that Q1 saturates when vin is high. Find a condition for
2CR to ensure that Q2 also saturates.
Analysis:
mA2.3mA8.04000
2.311 =⇒== CB ii
Applying KCL:
mA8;mA22.3500
6002222 =⋅==⇒=+ BCBB iiii β
Applying KVL: Ω600008.02.05 22 =⇒⋅=− CC RR
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.35 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.
Problem 10.49
Solution:
Known quantities:
The basic circuit of a TTL gate, shown in Figure P10.49.
Find:
The logic function performed by this circuit.
Analysis:
The circuit performs the function of a 2-input NAND gate. The analysis is similar to Example 10.10.
Problem 10.50
Solution:
Known quantities:
The circuit diagram of a three-input TTL NAND gate, given in Figure P10.50.
Find:
vB1, vB2, vB3, vC2, and vout, assuming that all the input voltages are high.
Analysis:
Q2 and Q3 conduct, while Q4 is cutoff. vB1 = 1.8 V, vB2 = 1.2 V, vB3 = 0.6 V, and vC2 = vout = 0.2 V.
G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 10
10.36 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.
Problem 10.51
Solution:
Known quantities:
Figure P10.51.
Find:
Show that when two or more emitter-follower outputs are connected to a common load, as shown in Figure P10.51, the OR operation results; that is, v0 = v1 OR v2.
Analysis:
v2 v
1 Q
1 Q
2 v
0
L L L L L
L H H L H
H L L H H
H H H H H L : Low; H : High.