Electronic Devices and Circuit Theory

Transcript

1 nstructor s Resource Manual to accompany Electronic Devices and Circuit Theory Tenth Edition Robert L. Boylestad Louis Nashelsky Upper Saddle River, New Jersey Columbus, Ohio

2 Copyright 009 by Pearson Education, nc., Upper Saddle River, New Jersey Pearson Prentice Hall. All rights reserved. Printed in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department. Pearson Prentice Hall is a trademark of Pearson Education, nc. Pearson is a registered trademark of Pearson plc Prentice Hall is a registered trademark of Pearson Education, nc. nstructors of classes using Boylestad/Nashelsky, Electronic Devices and Circuit Theory, 0 th edition, may reproduce material from the instructor s text solutions manual for classroom use SBN-3: SBN-0:

3 Contents Solutions to Problems in Text Solutions for Laboratory Manual 85 iii

4 Chapter. Copper has 0 orbiting electrons with only one electron in the outermost shell. The fact that the outermost shell with its 9 th electron is incomplete (subshell can contain electrons) and distant from the nucleus reveals that this electron is loosely bound to its parent atom. The application of an external electric field of the correct polarity can easily draw this loosely bound electron from its atomic structure for conduction. Both intrinsic silicon and germanium have complete outer shells due to the sharing (covalent bonding) of electrons between atoms. Electrons that are part of a complete shell structure require increased levels of applied attractive forces to be removed from their parent atom.. ntrinsic material: an intrinsic semiconductor is one that has been refined to be as pure as physically possible. That is, one with the fewest possible number of impurities. 3. Negative temperature coefficient: materials with negative temperature coefficients have decreasing resistance levels as the temperature increases. Covalent bonding: covalent bonding is the sharing of electrons between neighboring atoms to form complete outermost shells and a more stable lattice structure. 4. W Q (6 C)(3 ) 8 J e 48( J) J Q W J C C is the charge associated with 4 electrons. 6. GaP Gallium Phosphide E g.4 e ZnS Zinc Sulfide E g 3.67 e 7. An n-type semiconductor material has an excess of electrons for conduction established by doping an intrinsic material with donor atoms having more valence electrons than needed to establish the covalent bonding. The majority carrier is the electron while the minority carrier is the hole. A p-type semiconductor material is formed by doping an intrinsic material with acceptor atoms having an insufficient number of electrons in the valence shell to complete the covalent bonding thereby creating a hole in the covalent structure. The majority carrier is the hole while the minority carrier is the electron. 8. A donor atom has five electrons in its outermost valence shell while an acceptor atom has only 3 electrons in the valence shell. 9. Majority carriers are those carriers of a material that far exceed the number of any other carriers in the material. Minority carriers are those carriers of a material that are less in number than any other carrier of the material.

5 0. Same basic appearance as Fig..7 since arsenic also has 5 valence electrons (pentavalent).. Same basic appearance as Fig..9 since boron also has 3 valence electrons (trivalent) For forward bias, the positive potential is applied to the p-type material and the negative potential to the n-type material. 5. T K k,600/n,600/ (low value of D ) 5800 (5800)(0.6) kd TK D s e e (e.877 ) 7.97 ma 6. k,600/n,600/ 5800 (n for D 0.6 ) T K T C (5800)(0.6 ) k / T K e e e e 5 μa( ) ma / ( k T K s ) 7. (a) T K k,600/n,600/ 5800 kd (5800)( 0 ) T K 93 D s e 0.μA e (e ) ( ) μA D s 0. μa (b) The result is expected since the diode current under reverse-bias conditions should equal the saturation value. 8. (a) x y e x (b) y e 0 (c) For 0, e 0 and s ( ) 0 ma

6 9. T 0 C: s 0. μa T 30 C: s (0. μa) 0. μa (Doubles every 0 C rise in temperature) T 40 C: s (0. μa) 0.4 μa T 50 C: s (0.4 μa) 0.8 μa T 60 C: s (0.8 μa).6 μa.6 μa: 0. μa 6: increase due to rise in temperature of 40 C. 0. For most applications the silicon diode is the device of choice due to its higher temperature capability. Ge typically has a working limit of about 85 degrees centigrade while Si can be used at temperatures approaching 00 degrees centigrade. Silicon diodes also have a higher current handling capability. Germanium diodes are the better device for some RF small signal applications, where the smaller threshold voltage may prove advantageous.. From.9: 0 ma s 75 C 5 C 5 C pa pa.05 μa F decreased with increase in temperature. : : s increased with increase in temperature.05 μa: 0.0 pa :. An ideal device or system is one that has the characteristics we would prefer to have when using a device or system in a practical application. Usually, however, technology only permits a close replica of the desired characteristics. The ideal characteristics provide an excellent basis for comparison with the actual device characteristics permitting an estimate of how well the device or system will perform. On occasion, the ideal device or system can be assumed to obtain a good estimate of the overall response of the design. When assuming an ideal device or system there is no regard for component or manufacturing tolerances or any variation from device to device of a particular lot. 3. n the forward-bias region the 0 drop across the diode at any level of current results in a resistance level of zero ohms the on state conduction is established. n the reverse-bias region the zero current level at any reverse-bias voltage assures a very high resistance level the open circuit or off state conduction is interrupted. 4. The most important difference between the characteristics of a diode and a simple switch is that the switch, being mechanical, is capable of conducting current in either direction while the diode only allows charge to flow through the element in one direction (specifically the direction defined by the arrow of the symbol using conventional current flow). 5. D 0.66, D ma D 0.65 R DC 35 Ω ma D 3

7 6. At D 5 ma, D 0.8 D 0.8 R DC Ω D 5 ma As the forward diode current increases, the static resistance decreases. 7. D 0, D s 0. μa D 0 R DC D 0. μa 00 MΩ D 30, D s 0. μa D 30 R DC 300 MΩ 0.μA D As the reverse voltage increases, the reverse resistance increases directly (since the diode leakage current remains constant). 8. (a) r d (b) r d Δ Δ d d ma 5 ma 0 ma 6 m 6 m.6 Ω 0 ma D 3 Ω (c) quite close 9. D 0 ma, D 0.76 D 0.76 R DC 76 Ω 0 ma r d Δ Δ D d d R DC >> r d ma 5 ma 0 ma 3 Ω 30. D ma, r d D 5 ma, r d Δ Δ d d Δ Δ Ω ma 0 ma d d Ω 0 ma 0 ma 6 m 3. D ma, r d (6 Ω) 5 Ω vs 55 Ω (#30) D 6 m 6 m D 5 ma, r d.73 Ω vs Ω (#30) 5 ma D 3. r av Δ Δ d d Ω 3.5 ma. ma 4

8 33. r d Δ Δ d d Ω 7 ma 3 ma 4 ma (relatively close to average value of 4.4 Ω (#3)) 34. r av Δ Δ d d ma 0 ma 4 ma 4.9 Ω 35. Using the best approximation to the curve beyond D 0.7 : Δd r av 4 Ω Δ 5 ma 0 ma 5 ma d 36. (a) R 5 : C T 0.75 pf R 0 : C T.5 pf ΔC Δ T R.5 pf 0.75 pf 0.5 pf pf/ (b) R 0 : C T.5 pf R : C T 3 pf ΔC Δ T R.5 pf 3 pf.75 pf pf/ (c) 0.94 pf/: pf/ 5.88: 6: ncreased sensitivity near D From Fig..33 D 0, C D 3.3 pf D 0.5, C D 9 pf 38. The transition capacitance is due to the depletion region acting like a dielectric in the reversebias region, while the diffusion capacitance is determined by the rate of charge injection into the region just outside the depletion boundaries of a forward-biased device. Both capacitances are present in both the reverse- and forward-bias directions, but the transition capacitance is the dominant effect for reverse-biased diodes and the diffusion capacitance is the dominant effect for forward-biased conditions. 5

9 39. D 0., C D 7.3 pf X C 3.64 kω π fc π(6 MHz)(7.3 pf) D 0, C T 0.9 pf X C 9.47 kω π fc π(6 MHz)(0.9 pf) 40. f 0 0 kω ma t s + t t t rr 9 ns t s + t s 9 ns t s 3 ns t t t s 6 ns As the magnitude of the reverse-bias potential increases, the capacitance drops rapidly from a level of about 5 pf with no bias. For reverse-bias potentials in excess of 0 the capacitance levels off at about.5 pf. 43. At D 5, D 0. na and at D 00, D 0.45 na. Although the change in R is more than 00%, the level of R and the resulting change is relatively small for most applications. 44. Log scale: T A 5 C, R 0.5 na T A 00 C, R 60 na The change is significant. 60 na: 0.5 na 0: Yes, at 95 C R would increase to 64 na starting with 0.5 na (at 5 C) (and double the level every 0 C). 6

10 45. F 0. ma: r d 700 Ω F.5 ma: r d 70 Ω F 0 ma: r d 6 Ω The results support the fact that the dynamic or ac resistance decreases rapidly with increasing current levels. 46. T 5 C: P max 500 mw T 00 C: P max 60 mw P max F F Pmax 500 mw F 74.9 ma F 0.7 Pmax 60 mw F ma 0.7 F 74.9 ma: ma.9: : 47. Using the bottom right graph of Fig..37: F 500 T 5 C At F 50 ma, T 04 C 48. ΔZ 49. T C +0.07% 00% Z ( T T0) ( T 5) T T T T C ΔZ 00% ( T T ) Z 0 (5 4.8 ) 00% 0.053%/ C 5 (00 5 ) 7

11 5. (0 6.8 ) 00% 77% (4 6.8 ) The 0 Zener is therefore 77% of the distance between 6.8 and 4 measured from the 6.8 characteristic. At Z 0. ma, T C 0.06%/ C (5 3.6 ) 00% 44% ( ) The 5 Zener is therefore 44% of the distance between 3.6 and 6.8 measured from the 3.6 characteristic. At Z 0. ma, T C 0.05%/ C Zener: 0. ma: 400 Ω ma: 95 Ω 0 ma: 3 Ω The steeper the curve (higher d/d) the less the dynamic resistance. 54. T.0, which is considerably higher than germanium ( 0.3 ) or silicon ( 0.7 ). For germanium it is a 6.7: ratio, and for silicon a.86: ratio. 55. Fig..53 (f) F 3 ma Fig..53 (e) F (a) Relative 5 ma 0 ma % 4.4% increase 0.8 ratio: (b) Relative ma %.9% increase.38 ratio: (c) For currents greater than about 30 ma the percent increase is significantly less than for increasing currents of lesser magnitude. 8

12 57. (a) From Fig..53 (i) 75 (b) For the high-efficiency red unit of Fig..53: 0. ma 0 ma C x 0 ma x 00 C 0. ma/ C 9

13 Chapter. The load line will intersect at D E 8 R 330 Ω 4.4 ma and D 8. (a) (b) (c) D Q 0.9 D Q.5 ma R E D Q D Q 0.7 D Q. ma R E D Q D Q 0 D Q 4.4 ma R E D Q For (a) and (b), levels of D Q and D Q are quite close. Levels of part (c) are reasonably close but as expected due to level of applied voltage E. E 5. (a) D R.7 ma. kω The load line extends from D.7 ma to D , ma D Q D Q E 5 (b) D R 0.64 ma 0.47 kω The load line extends from D 0.64 ma to D , 9 ma D Q D Q E 5 (c) D R 7.78 ma 0.8 kω The load line extends from D 7.78 ma to D ,.5 ma D Q D Q The resulting values of 3. Load line through.5 ma. D.5 ma with R D Q are quite close, while D Q extends from ma to.5 ma. D Q 0 ma of characteristics and D 7 will intersect D axis as E 7 R R 7.5 ma 0.6 kω 0

14 E (a) D R D 3.3 ma R. kω D 0.7, R E D E 30 0 (b) D D R. kω D 0, R ma Yes, since E T the levels of D and R are quite close. 5. (a) 0 ma; diode reverse-biased. (b) 0Ω (Kirchhoff s voltage law) Ω A (c) 0 A; center branch open 0 Ω 6. (a) Diode forward-biased, Kirchhoff s voltage law (CW): o 0 o R D o R.955 ma. kω (b) Diode forward-biased, D.4 ma. kω+ 4.7 kω o 4.7 kω + D (.4 ma)(4.7 kω) (a) o k Ω( ) kω+ kω (0 ) (9 ) ).3 (b).95 ma. kω+ 4.7 kω 5.9 kω R (.95 ma)(4.7 kω) 9 o 9 7

15 8. (a) Determine the Thevenin equivalent circuit for the 0 ma source and. kω resistor. E Th R (0 ma)(. kω) R Th. kω (b) Diode forward-biased D ma 6.8 kω Kirchhoff s voltage law (CW): + o o 4.3 Diode forward-biased 0.7 D 6.6 ma. kω +. kω o D (. kω) (6.6 ma)(. kω) (a) (b) o o 0.3 o kω+ 3.3 kω 4.5 kω ma, o ( ma)(3.3 kω) (a) Both diodes forward-biased R ma 4.7 kω Assuming identical diodes: 4.06 ma D R.05 ma o (b) Right diode forward-biased: D ma. kω o (a) Ge diode on preventing Si diode from turning on : ma kω kω (b) ma 4.7 kω 4.7 kω o + (0.553 ma)(4.7 kω) 4.6

16 . Both diodes forward-biased: 0.7, 0.3 o o kω ma kω kω 0.47 kω ma 0.47 kω (Si diode) kω 0.47 kω 9.3 ma 0.85 ma 8.45 ma 3. For the parallel Si kω branches a Thevenin equivalent will result (for on diodes) in a single series branch of 0.7 and kω resistor as shown below: kω ma kω k 3. ma D Ω.55 ma 4. Both diodes off. The threshold voltage of 0.7 is unavailable for either diode. o 0 5. Both diodes on, o Both diodes on. o Both diodes off, o 0 8. The Si diode with 5 at the cathode is on while the other is off. The result is o at one terminal is more positive than 5 at the other input terminal. Therefore assume lower diode on and upper diode off. The result: o The result supports the above assumptions. 0. Since all the system terminals are at 0 the required difference of 0.7 across either diode cannot be established. Therefore, both diodes are off and o +0 as established by 0 supply connected to kω resistor. 3

17 . The Si diode requires more terminal voltage than the Ge diode to turn on. Therefore, with 5 at both input terminals, assume Si diode off and Ge diode on. The result: o The result supports the above assumptions. dc. dc 0.38 m m m m 6.8 R. kω.85 ma 3. Using dc 0.38( m T ) 0.38( m 0.7 ) Solving: m : for m : T dc 4. m L max kω 0.94 ma 4

18 max (. kω) ma. kω + max (. kω) 0.94 ma ma 3.78 ma Dmax Lmax 5. m (0 ) dc 0.38 m 0.38(55.56 ) Diode will conduct when v o 0.7 ; that is, 0 k Ω( v ) v o 0.7 i 0 kω+ kω Solving: v i 0.77 For v i 0.77 Si diode is on and v o 0.7. For v i < 0.77 Si diode is open and level of v o is determined by voltage divider rule: 0 k Ω( v ) v o i 0 kω+ kω v i For v i 0 : v o 0.909( 0 ) 9.09 When v o 0.7, vr v max i 0.7 max R 9.3 ma max kω 0 max (reverse) ma kω+ 0 kω 5

19 7. (a) P max 4 mw (0.7 ) D 4 mw D 0 ma 0.7 (b) 4.7 kω 56 kω 4.34 kω R max ma 4.34 kω max 36.7 ma (c) diode 8.36 ma (d) Yes, D 0 ma > 8.36 ma (e) diode 36.7 ma max 0 ma 8. (a) m (0 ) 69.7 L m i m D 69.7 (0.7 ) dc 0.636(68.3 ) (b) P m (load) + D (c) D (max) L m 68.3 R kω L 68.3 ma (d) P max D D (0.7 ) max (0.7 )(68.3 ma) 7.8 mw 9. 6

20 30. Positive half-cycle of v i : oltage-divider rule:. k Ω( i ) max o max. kω+. kω ( imax ) (00 ) Positive pulse of v i : Top left diode off, bottom left diode on. kω. kω. kω. k Ω(70 ) o peak. kω +. kω Negative pulse of v i : Top left diode on, bottom left diode off. k Ω(70 ) o peak. kω +. kω dc 0.636(56.67 ) dc m (50 ) (a) Si diode open for positive pulse of v i and v o 0 For 0 < v i 0.7 diode on and v o v i For v i 0, v o For v i 0.7, v o Polarity of v o across the. kω resistor acting as a load is the same. oltage-divider rule:. k Ω( i ) max o max. kω+. kω ( imax ) (00 ) 50 7

21 (b) For v i 5 the 5 battery will ensure the diode is forward-biased and v o v i 5. At v i 5 v o At v i 0 v o For v i > 5 the diode is reverse-biased and v o (a) Positive pulse of v i :. k Ω(0 0.7 ) o. kω+. kω Negative pulse of v i : diode open, v o (b) Positive pulse of v i : o Negative pulse of v i : diode open, v o (a) For v i 0 the diode is reverse-biased and v o 0. For v i 5, v i overpowers the battery and the diode is on. Applying Kirchhoff s voltage law in the clockwise direction: 5 + v o 0 v o 3 (b) For v i 0 the 0 level overpowers the 5 supply and the diode is on. Using the short-circuit equivalent for the diode we find v o v i 0. For v i 5, both v i and the 5 supply reverse-bias the diode and separate v i from v o. However, v o is connected directly through the. kω resistor to the 5 supply and v o 5. 8

22 35. (a) Diode on for v i 4.7 For v i > 4.7, o For v i < 4.7, diode off and v o v i (b) Again, diode on for v i 4.7 but v o now defined as the voltage across the diode For v i 4.7, v o 0.7 For v i < 4.7, diode off, D R 0 ma and. kω R (0 ma)r 0 Therefore, v o v i 4 At v i 0, v o 4 v i 8, v o For the positive region of v i : The right Si diode is reverse-biased. The left Si diode is on for levels of v i greater than n fact, v o 6 for v i 6. For v i < 6 both diodes are reverse-biased and v o v i. For the negative region of v i : The left Si diode is reverse-biased. The right Si diode is on for levels of v i more negative than n fact, v o 8 for v i 8. For v i > 8 both diodes are reverse-biased and v o v i. i R : For 8 < v i < 6 there is no conduction through the 0 kω resistor due to the lack of a complete circuit. Therefore, i R 0 ma. For v i 6 v R v i v o v i 6 For v i 0, v R and i R 4 0 kω 0.4 ma For v i 8 v R v i v o v i + 8 9

23 For v i 0 v R and i R 0. ma 0 kω 37. (a) Starting with v i 0, the diode is in the on state and the capacitor quickly charges to 0 +. During this interval of time v o is across the on diode (short-current equivalent) and v o 0. When v i switches to the +0 level the diode enters the off state (open-circuit equivalent) and v o v i + v C (b) Starting with v i 0, the diode is in the on state and the capacitor quickly charges up to 5 +. Note that v i +0 and the 5 supply are additive across the capacitor. During this time interval v o is across on diode and 5 supply and v o 5. When v i switches to the +0 level the diode enters the off state and v o v i + v C

24 38. (a) For negative half cycle capacitor charges to peak value of with polarity. The output v o is directly across the on diode resulting in v o 0.7 as a negative peak value. For next positive half cycle v o v i with peak value of v o (b) For positive half cycle capacitor charges to peak value of with polarity. The output v o For next negative half cycle v o v i 99.3 with negative peak value of v o Using the ideal diode approximation the vertical shift of part (a) would be 0 rather than 9.3 and 00 rather than 99.3 for part (b). Using the ideal diode approximation would certainly be appropriate in this case. 39. (a) τ RC (56 kω)(0. μf) 5.6 ms 5τ 8 ms (b) 5τ 8 ms T ms 0.5 ms, 56: (c) Positive pulse of v i : Diode on and v o Capacitor charges to Negative pulse of v i : Diode off and v o 0.3.3

25 40. Solution is network of Fig..76(b) using a 0 supply in place of the 5 source. 4. Network of Fig..78 with battery reversed. 4. (a) n the absence of the Zener diode 80 Ω(0 ) L 80 Ω+ 0 Ω 9 L 9 < Z 0 and diode non-conducting 0 Therefore, L R 0 Ω+ 80 Ω with Z 0 ma and L 9 50 ma (b) n the absence of the Zener diode 470 Ω(0 ) L 470 Ω+ 0 Ω 3.6 L 3.6 > Z 0 and Zener diode on Therefore, L 0 and R s 0 / R 0 /0 Ω ma Rs Rs s L L /R L 0 /470 Ω.8 ma and Z L ma.8 ma 4.7 ma R s (c) P Z max 400 mw Z Z (0 )( Z ) 400 mw Z 40 ma 0 L min R s Z ma 40 ma 5.45 ma max L 0 R L, Ω 5.45 ma Lmin Large R L reduces L and forces more of R s to pass through Zener diode. (d) n the absence of the Zener diode RL (0 ) L 0 R L + 0 Ω 0R L R L 0R L 00 R L 0 Ω

26 L 43. (a) Z, R L 60 Ω L 00 ma R L i 60 Ω(6 ) L Z RL + Rs 60 Ω+ Rs 70 + R s 960 R s 40 R s 0 Ω (b) P Z max Z Zmax ( )(00 ma).4 W L Z 44. Since L is fixed in magnitude the maximum value of RL RL maximum. The maximum level of of i. PZ 400 mw max Z 50 ma max Z 8 L Z 8 L ma RL RL 0 Ω R s Z + L 50 ma ma ma i Z R s Rs or i R R s s + Z (86.36 ma)(9 Ω) R s will occur when Z is a R s will in turn determine the maximum permissible level Any value of v i that exceeds 5.86 will result in a current Z that will exceed the maximum value. 45. At 30 we have to be sure Zener diode is on. R L i k Ω(30 ) L 0 RL + Rs kω+ Rs Solving, R s 0.5 kω At 50, ZM RS kω 60 ma, L 0 kω R S L 60 ma 0 ma 40 ma 0 ma 46. For v i +50 : Z forward-biased at 0.7 Z reverse-biased at the Zener potential and Z 0. Therefore, o Z Z 3

27 For v i 50 : Z reverse-biased at the Zener potential and Z 0. Z forward-biased at 0.7. Therefore, o Z Z For a 5 square wave neither Zener diode will reach its Zener potential. n fact, for either polarity of v i one Zener diode will be in an open-circuit state resulting in v o v i. 47. m.44(0 ) m (69.68 ) The P for each diode is m P (.44)( rms ) 4

28 Chapter 3.. A bipolar transistor utilizes holes and electrons in the injection or charge flow process, while unipolar devices utilize either electrons or holes, but not both, in the charge flow process. 3. Forward- and reverse-biased. 4. The leakage current CO is the minority carrier current in the collector E the largest B the smallest C E 9. B 00 C C 00 B E C + B 00 B + B 0 B 8 ma B E 79. μa 0 0 C 00 B 00(79. μa) 7.9 ma 0.. E 5 ma, CB : BE 800 m CB 0 : BE 770 m CB 0 : BE 750 m The change in CB is 0 : 0: The resulting change in BE is 800 m:750 m.07: (very slight) Δ (a) r av 5 Ω Δ 8 ma 0 (b) Yes, since 5 Ω is often negligible compared to the other resistance levels of the network. 3. (a) C E 4.5 ma (b) C E 4.5 ma (c) negligible: change cannot be detected on this set of characteristics. (d) C E 5

29 4. (a) Using Fig. 3.7 first, E 7 ma Then Fig. 3.8 results in C 7 ma (b) Using Fig. 3.8 first, E 5 ma Then Fig. 3.7 results in BE 0.78 (c) Using Fig. 3.0(b) E 5 ma results in BE 0.8 (d) Using Fig. 3.0(c) E 5 ma results in BE 0.7 (e) Yes, the difference in levels of BE can be ignored for most applications if voltages of several volts are present in the network. 5. (a) C α E (0.998)(4 ma) 3.99 ma 6. (b) E C + B C E B.8 ma 0.0 ma.78 ma C.78 ma α dc ma E α 0.98 (c) C β B B (40 μa).96 ma α ma E C α ma 7. i i /R i 500 m/0 Ω 5 ma L i 5 ma L L R L (5 ma)( kω) 5 o 5 A v i i 00 m 00 m 8. i Ri + R s 0 Ω+ 00 Ω 0 Ω L i.67 ma L L R (.67 ma)(5 kω) 8.35 o 8.35 A v i.67 ma (a) Fig. 3.4(b): B 35μA Fig. 3.4(a): C 3.6 ma (b) Fig. 3.4(a): CE.5 Fig. 3.4(b): BE 0.7 6

30 C ma. (a) β B 7 μa 7.65 β 7.65 (b) α β (c) CEO 0.3 ma (d) CBO ( α) CEO ( 0.99)(0.3 ma).4 μa. (a) Fig. 3.4(a): CEO 0.3 ma (b) Fig. 3.4(a): C.35 ma C.35 ma β dc 0 μa 35 B (c) α β 35 β CBO ( α) CEO ( 0.996)(0.3 ma). μa C 6.7 ma 3. (a) β dc B 80 μa C 0.85 ma (b) β dc 70 B 5 μa C 3.4 ma (c) β dc B 30 μa 3.33 (d) β dc does change from pt. to pt. on the characteristics. Low B, high CE higher betas High B, low CE lower betas 4. (a) β ac (b) β ac Δ Δ Δ Δ C B C B CE CE ma 6 ma.3 ma μa 70 μa 0 μa.4 ma 0.3 ma. ma μa 0 μa 0 μa ΔC 4.5 ma.35 ma.9 ma (c) β ac 95 Δ B CE 0 40 μa 0 μa 0 μa (d) β ac does change from point to point on the characteristics. The highest value was obtained at a higher level of CE and lower level of C. The separation between B curves is the greatest in this region. 7

31 (e) CE B β dc β ac C β dc /β ac 5 80 μa ma μa ma μa ma.55 As C decreased, the level of β dc and β ac increased. Note that the level of β dc and β ac in the center of the active region is close to the average value of the levels obtained. n each case β dc is larger than β ac, with the least difference occurring in the center of the active region. C.9 ma 5. β dc 6 B 5 μa β 6 α β E C /α.9 ma/ ma α (a) β α β 0 0 (b) α 0.99 β ma (c) B C. μa β 80 E C + B ma +. μa.0 ma e i be 0..9 o.9 A v i 0.95 E.9 e R.9 ma (rms) kω E 9. Output characteristics: Curves are essentially the same with new scales as shown. nput characteristics: Common-emitter input characteristics may be used directly for common-collector calculations. 8

32 30. P C max 30 mw CE C PC 30 mw max C C max, CE C 7 ma 4.9 max PC 30 mw max CE CE max, C.5 ma CE 0 max PC 30 mw max CE 0, C 3 ma CE 0 PC 30 mw max C 4 ma, CE C 4 ma 7.5 PC 30 mw max CE 5, C ma 5 CE 3. C PC max C max, CE Cmax PC max CB CB max, C CBmax 30 mw 5 6 ma 30 mw ma 5 PC 30 mw max C 4 ma, CB ma C PC 30 mw max CB 0, C 3 ma 0 CB 9

33 3. The operating temperature range is 55 C T J 50 C F 9 C ( 55 C) F 5 F 9 (50 C) F 5 67 F T J 30 F 33. C max 00 ma, CE max 30, P P D max 65 mw D 65 mw max C C max, CE C 00 ma 3.5 max PD 65 mw max CE CE max, C 0.83 ma CE 30 max PD 65 mw max C 00 ma, CE 6.5 C 00 ma PD 65 mw max CE 0, C 3.5 ma 0 CE 34. From Fig. 3.3 (a) CBO 50 na max βmin + βmax β avg CEO β CBO (00)(50 na) 5 μa 30

34 35. h FE (β dc ) with CE, T 5 C C 0. ma, h FE 0.43(00) 43 C 0 ma, h FE 0.98(00) 98 h fe (β ac ) with CE 0, T 5 C C 0. ma, h fe 7 C 0 ma, h fe 60 For both h FE and h fe the same increase in collector current resulted in a similar increase (relatively speaking) in the gain parameter. The levels are higher for h fe but note that CE is higher also. 36. As the reverse-bias potential increases in magnitude the input capacitance C ibo decreases (Fig. 3.3(b)). ncreasing reverse-bias potentials causes the width of the depletion region to A increase, thereby reducing the capacitance C d. 37. (a) At C ma, h fe 0 At C 0 ma, h fe 60 (b) The results confirm the conclusions of problems 3 and 4 that beta tends to increase with increasing collector current. 39. (a) β ac Δ Δ C B CE 3 6 ma. ma 3.8 ma 80 μa 60 μa 0 μa 90 (b) β dc C B ma μa (c) β ac 4 ma ma ma 8 μa 8 μa 0 μa 00 (d) β dc C B 3 ma μa (e) n both cases β dc is slightly higher than β ac ( 0%) (f)(g) n general β dc and β ac increase with increasing C for fixed CE and both decrease for decreasing levels of CE for a fixed E. However, if C increases while CE decreases when moving between two points on the characteristics, chances are the level of β dc or β ac may not change significantly. n other words, the expected increase due to an increase in collector current may be offset by a decrease in CE. The above data reveals that this is a strong possibility since the levels of β are relatively close. 3

35 Chapter 4. (a) BQ CC BE R 470 kω 470 kω B 3.55 μa (b) CQ β (90)(3.55 μa).93 ma BQ (c) R 6 (.93 ma)(.7 kω) 8.09 CEQ CC CQ C (d) C CE Q 8.09 (e) B BE 0.7 (f) E 0. (a) C β B 80(40 μa) 3. ma (b) R C (c) R B R 6 6 C CC C.875 kω 3. ma 3. ma C C R B kω 40 μa 40 μa B (d) CE C 6 3. (a) C E B 4 ma 0 μa 3.98 ma 4 ma (b) CC CE + C R C 7. + (3.98 ma)(. kω) (c) β C B 3.98 ma μa (d) R B R B CC BE 763 kω 0 μa B B 4. C sat CC 6 R.7 kω C 5.93 ma 5. (a) Load line intersects vertical axis at C 3 kω 7 ma and horizontal axis at CE. CC BE 0.7 (b) B 5 μa: R B 8 kω 5 μa B (c) C Q 3.4 ma, CE Q

36 (d) β C B 3.4 ma 36 5 μa (e) α β β (f) C sat CC R 3 kω 7 ma C (g) (h) P D (0.75 )(3.4 ma) mw CEQ CQ (i) (j) 6. (a) P s CC ( C + B ) (3.4 ma + 5 μa) 7.9 mw P R P s P D 7.9 mw mw mw BQ CC BE R + ( β + ) R 50 k Ω+ (0).5 kω kω B 9.8 μa E (b) CQ β (00)(9.8 μa).9 ma BQ (c) CE Q CC C (R C + R E ) 0 (.9 ma)(.4 kω +.5 kω) (d) C CC C R C 0 (.9 ma)(.4 kω) (e) B CC B R B 0 (9.8 μa)(50 kω) (f) E C CE (a) R C CC C C kω ma ma (b) E C : R E E E.4. kω ma (c) R B R B CC BE E 356 kω ma/80 5 μa B B (d) CE C E (e) B BE + E

37 E. 8. (a) C E R 3.09 ma E 0.68 kω C 3.09 ma β B 0 μa 54.5 (b) CC R C + CE + E (3.09 ma)(.7 kω) R B CC BE E (c) R B 0 μa B A μ B 747 kω 9. Csot CC 0 0 R + R.4 kω+.5 kω 3.9 kω C E 5.3 ma 0. (a) CC 4 C sat 6.8 ma RC + RE RC +. kω 4 R C +. kω 3.59 kω 6.8 ma R C.33 kω (b) β C B 4 ma μa R (4 ma)(. k ) B CC BE E Ω (c) R B 30 μa B A μ B kω (d) P D CE Q CQ (0 )(4 ma) 40 mw (e) P CR C (4 ma) (.33 kω) 37.8 mw. (a) Problem : C Q.93 ma, CE Q 8.09 (b) B Q 3.55 μa (the same) C β Q B (35)(3.55 μa) 4.39 ma Q R 6 (4.39 ma)(.7 kω) 4.5 CEQ CC CQ C 34

38 4.39 ma.93 ma (c) %Δ C 00% 49.83%.93 ma %Δ CE % 48.70% 8.09 Less than 50% due to level of accuracy carried through calculations. (d) Problem 6: C Q.9 ma, 8.6 ( 9.8 μa) CE Q (e) CC BE B Q RB + ( β + ) RE 50 k Ω+ (50 + )(.5 k Ω ) 6. μa C β Q B (50)(6. μa) 3.93 ma Q CE Q CC C (R C + R E ) 0 (3.93 ma)(.4 kω +.5 kω) ma.9 ma (f) %Δ C.9 ma 00% 34.59% %Δ CE % 46.76% (g) For both C and CE the % change is less for the emitter-stabilized.?. βr E 0R (80)(0.68 kω) 0(9. kω) 54.4 kω 9 kω (No!) (a) Use exact approach: R Th R R 6 kω 9. kω 7.94 kω R E Th CC (9. k Ω)(6 ) R + R 9. kω+ 6 kω.05 ETh BE B Q RTh + ( β + ) RE 7.94 k Ω+ (8)(0.68 k Ω).4 μa (b) β (80)(.4 μa).7 ma CQ BQ B Q (c) CE Q CC C Q (R C + R E ) 6 (.7 ma)(3.9 kω kω) 8.7 (d) C CC C R C 6 (.7 ma)(3.9 kω) 9.33 (e) E E R E C R E (.7 ma)(0.68 kω).6 (f) B E + BE

39 3. (a) C CC R C C kω.8 ma (b) E E R E C R E (.8 ma)(. kω).54 (c) B BE + E R (d) R : R R R R R CC B R R ma B ma R 5.6 kω 39.4 kω 4. (a) C β B (00)(0 μa) ma (b) E C + B ma + 0 μa.0 ma E E R E (.0 ma)(. kω).4 (c) CC C + C R C ( ma)(.7 kω) (d) CE C E (e) B E + BE (f) + R R B kω CC B R R + 0 μa μa + 0 μa μa kω μa 5. C sat CC 6 R + R 3.9 kω kω ma 4.58 kω C E 36

40 6. (a) βr E 0R (0)( kω) 0(8. kω) 0 kω 8 kω (checks) R B CC (8. k Ω)(8 ) R + R 39 kω+ 8. kω 3.3 E B BE E.43 C E R.43 ma kω E (b) CE CC C (R C + R E ) 8 (.43 ma)(3.3 kω + kω) 7.55 (c) B C.43 ma β μa (d) E E R E C R E (.43 ma)( kω).43 (e) B (a) R Th R R 39 kω 8. kω 6.78 kω R C CC 8. k Ω(8 ) E Th R + R 39 kω+ 8. kω 3.3 ETh BE B R + ( β + ) R 6.78 k Ω+ ()( k Ω) Th E μa 7.78 kω C β B (0)(9.0 μa).8 ma (vs..43 ma #6) (b) CE CC C (R C + R E ) 8 (.8 ma)(3.3 kω + kω) (vs #6) (c) 9.0 μa (vs. 0.5 μa #6) (d) E E R E C R E (.8 ma)( kω).8 (vs..43 #6) (e) B BE + E (vs. 3.3 #6) The results suggest that the approximate approach is valid if Eq is satisfied. R 9. k Ω(6 ) 8. (a) B CC R + R 6 kω+ 9. kω.05 E B BE E.35 E R.99 ma E 0.68 kω E.99 ma C Q 37

41 CE Q CC C (R C + R E ) 6 (.99 ma)(3.9 kω kω) B Q CQ.99 ma 4.88 μa β 80 (b) From Problem :.7 ma, 8.7, C Q CE Q B Q.4 μa (c) The differences of about 4% suggest that the exact approach should be employed when appropriate. 9. (a) Csat CC ma RC + RE 3RE + RE 4RE 4 4 R E 0.8 kω 4(7.5 ma) 30 ma R C 3R E 3(0.8 kω).4 kω (b) E E R E C R E (5 ma)(0.8 kω) 4 (c) B E + BE R (d) B CC R (4 ), 4.7 R + R R + 4 kω (e) β dc C B R 5.84 kω 5 ma 38.5 μa 9.8 (f) βr E 0R (9.8)(0.8 kω) 0(5.84 kω) kω 58.4 kω (checks) 0. (a) From problem b, C.7 ma From problem c, CE 8.7 (b) β changed to 0: From problem a, E Th.05, R Th 7.94 kω ETh BE B RTh + ( β + ) RE 7.94 k Ω + ()(0.68 k Ω) 4.96 μa C β B (0)(4.96 μa).8 ma CE CC C (R C + R E ) 6 (.8 ma)(3.9 kω kω)

42 (c).8 ma.7 ma % Δ C 00% 5.6%.7 ma % Δ CE 00% 5.0% 8.7 (d) c f 0c %Δ C 49.83% 34.59% 5.6% %Δ CE 48.70% 46.76% 5.0% Fixed-bias Emitter feedback oltagedivider (e) Quite obviously, the voltage-divider configuration is the least sensitive to changes in β...(a) Problem 6: Approximation approach: Problem 7: Exact analysis: C Q.8 ma, C Q.43 ma, CE Q 7.55 CE Q 8. The exact solution will be employed to demonstrate the effect of the change of β. Using the approximate approach would result in %Δ C 0% and %Δ CE 0%. (b) Problem 7: E Th 3.3, R Th 6.78 kω ETH BE B R + ( β + ) R 6.78 k Ω+ (80 + ) kω kω Th E.94 μa C β B (80)(.94 μa).33 ma CE CC C (R C + R E ) 8 (.33 ma)(3.3 kω + kω) ma.8 ma (c) %Δ C 00%.9%.8 ma %Δ CE %.68% 8. For situations where βr E > 0R the change in C and/or CE due to significant change in β will be relatively small. (d) %Δ C.9% vs % for problem. %Δ CE.68% vs % for problem. (e) oltage-divider configuration considerably less sensitive.. The resulting %Δ C and %Δ CE will be quite small. 39

43 CC BE (a) B RB + β ( RC + RE) 470 k Ω + (0)(3.6 kω+ 0.5 k Ω) 5.88 μa (b) C β B (0)(5.88 μa).9 ma (c) C CC C R C 6 (.9 ma)(3.6 kω) (a) B CC BE R + β ( R + R ) 6.90 kω+ 00(6. kω+.5 k Ω ) B C E 0.07 μa C β B (00)(0.07 μa).0 ma (b) C CC C R C 30 (.0 ma)(6. kω) (c) E E R E C R E (.0 ma)(.5 kω) 3.0 (d) CE CC C (R C + R E ) 30 (.0 ma)(6. kω +.5 kω) 4.5 CC BE (a) B RB + β ( RC + RE) 470 k Ω+ (90)(9. kω+ 9. k Ω) 0.09 μa C β B (90)(0.09 μa) 0.9 ma CE CC C (R C + R E ) (0.9 ma)(9. kω + 9. kω) 5.44 CC BE 0.7 (b) β 35, B RB + β ( RC + RE) 470 k Ω+ (35)(9. kω+ 9. k Ω) 7.8 μa C β B (35)(7.8 μa) ma CE CC C (R C + R E ) (0.983 ma)(9. kω + 9. kω) 4. (c) ma 0.9 ma % Δ C 00% 8.0% 0.9 ma % Δ CE 00% 4.45% 5.44 (d) The results for the collector feedback configuration are closer to the voltage-divider configuration than to the other two. However, the voltage-divider configuration continues to have the least sensitivities to change in β. 40

44 5. MΩ 0 Ω, R B 50 kω CC BE 0.7 B RB + β ( RC + RE) 50 k Ω+ (80)(4.7 kω+ 3.3 k Ω) 7. μa C β B (80)(7. μa).8 ma C CC C R C (.8 ma)(4.7 kω) 5.98 Full MΩ: R B,000 kω + 50 kω,50 kω.5 MΩ CC BE 0.7 B R + β ( R + R ).5 M Ω+ (80)(4.7 kω+ 3.3 k Ω) B C E 4.36 μa C β B (80)(4.36 μa) ma C CC C R C (0.785 ma)(4.7 kω) 8.3 C ranges from 5.98 to (a) E B BE E 3.3 (b) C E R.75 ma E. kω (c) C CC C R C 8 (.75 ma)(. kω).95 (d) CE C E R.95 4 B C B (e) B 4.09 μa RB RB 330 kω C.75 ma (f) β μa B CC + EE BE (a) B RB + ( β + ) RE 330 k Ω+ ()(. k Ω) 3.78 μa E (β + ) B ()(3.78 μa).88 ma EE + E R E E 0 E EE + E R E 6 + (.88 ma)(. kω).54 EE BE (a) B RB + ( β + ) RE 9. k Ω+ (0 + )5 kω 6. μa (b) C β B (0)(6. μa) ma (c) CE CC + EE C (R C + R E ) 6 + (0.744 ma)(7 kω) 7.9 (d) C CC C R C 6 (0.744 ma)( kω)

45 9. (a) E kω. kω 3.3 ma (b) C 0 (3.3 ma)(.8 kω) (c) CE (3.3 ma)(. kω +.8 kω) (a) βr E > 0R not satisfied Use exact approach: Network redrawn to determine the Thevenin equivalent: (b) C β B (30)(3.95 μa).8 ma R Th 50 k Ω 55 kω μa 50 kω + 50 kω E Th 8 + (35.9 μa)(50 kω) B 55 k Ω + (30 + )(7.5 k Ω) 3.95 μa (c) E 8 + (.8 ma)(7.5 kω) (d) CE (.8 ma)(9. kω kω) (a) B R B C BE R R 560 kω B B 3.04 μa CC (b) C R C C kω 3.9 kω.56 ma (c) β C B.56 ma 3.04 μa 96.3 (d) CE C 8 4

46 .5 ma 3. B C β μa R 0.7 B CC BE R B 36.6 kω B B 3.5 μa R 6 6 C CC CC C CEQ R C.5 ma.5 ma C C CQ.4 kω Standard values: R B 360 kω R C.4 kω 33. CC C sat R + R C 0 R + R 4 E E E 0 ma 0 ma 0 5R E 0 ma 5R E 0 0 ma kω R E k 5Ω 400 Ω R C 4R E.6 kω 5 ma B C β μa ma(0.4 k Ω) 9.3 R B RB / B 4.67 μa 4.67 μa 45.7 kω Standard values: R E 390 Ω, R C.6 kω, R B 430 kω E E R E 0.75 kω E C 4 ma ( ) R C CC CC C CE + Q E R C C C C 4 (8 + 3 ) kω 4 ma 4 ma 4 ma B E + BE R CC R (4 ) B 3.7 unknowns! R + R R + R use βr E 0R for increased stability (0)(0.75 kω) 0R R 8.5 kω Choose R 7.5 kω 43

47 Substituting in the above equation: 7.5 k Ω(4 ) kω+ R R 4.5 kω Standard values: R E 0.75 kω, R C 3.3 kω, R 7.5 kω, R 43 kω 35. E (8 ) CC 5 E 5.6 R E. kω (use. kω) E 5 ma CC 8 C + E R C CC C R 8.4 R C C.68 kω (use.6 kω) C 5 ma B BE + E R B CC R (8 ) 6.3 ( unknowns) R + R R + R C 5 ma β 35.4 B 37 μa βr E 0R (35.4)(. kω) 0(R ) R 5.4 kω (use 5 kω) (5.4 k Ω)(8 ) Substituting: kω+ R Solving, R 5.5 kω (use 5 kω) Standard values: R E. kω R C.6 kω R 5 kω R 5 kω 36. kω kω 8.65 ma 37. For current mirror: (3 kω) (.4 kω) ma 38. DQ 6 ma DSS 44

48 4.3 kω 39. B ( 8 ) kω+ 4.3 kω E ( 9.7 ) E 4.6 ma.8 kω 40. E Z R E BE kω 3.67 ma 4. CC 0 C 4.67 ma sat RC.4 kω From characteristics B max 3 μa i BE B 5.67 μa R 80 kω B 5.67 μa 3 μa, well saturated o 0 (0. ma)(.4 kω) C sat 8 ma 5 RC 5 R C 0.65 kω 8 ma C 8 ma sat B max 80 μa β 00 Use. (80 μa) 96 μa R B kω 96 μa Standard values: R B 43 kω R C 0.6 kω 45

49 43. (a) From Fig. 3.3c: C ma: t f 38 ns, t r 48 ns, t d 0 ns, t s 0 ns t on t r + t d 48 ns + 0 ns 68 ns t off t s + t f 0 ns + 38 ns 48 ns (b) C 0 ma: t f ns, t r 5 ns, t d ns, t s 0 ns t on t r + t d 5 ns + ns 37 ns t off t s + t f 0 ns + ns 3 ns The turn-on time has dropped dramatically 68 ns:37 ns 4.54: while the turn-off time is only slightly smaller 48 ns:3 ns.: 44. (a) Open-circuit in the base circuit Bad connection of emitter terminal Damaged transistor (b) Shorted base-emitter junction Open at collector terminal (c) Open-circuit in base circuit Open transistor 45. (a) The base voltage of 9.4 reveals that the 8 kω resistor is not making contact with the base terminal of the transistor. f operating properly: B 8 k Ω(6 ) 8 kω+ 9 kω.64 vs. 9.4 As an emitter feedback bias circuit: CC BE B R + ( β + ) RE 9 k Ω+ (00 + ). kω 7. μa B CC B (R ) 6 (7. μa)(9 kω)

50 (b) Since E > B the transistor should be off 8 k Ω(6 ) With B 0 μa, B 8 kω + 9 kω.64 Assume base circuit open The 4 at the emitter is the voltage that would exist if the transistor were shorted collector to emitter.. k Ω(6 ) E. kω+ 3.6 kω (a) R B, B, C, C (b) β, C (c) Unchanged, C sat not a function of β (d) CC, B, C (e) β, C,, R C R E, CE ETh BE ETh BE 47. (a) B RTh + ( β + ) RE RTh + βre ETh BE ETh BE C β B β R R Th + β RE Th + RE β RTh As β, β, C, C CC and C R C R C (b) R open, B, C CE CC C (R C + R E ) and CE (c) CC, B, E, E, C (d) B 0 μa, C CEO and C (R C + R E ) negligible with CE CC 0 (e) Base-emitter junction short B but transistor action lost and C 0 ma with CE CC (a) R B open, B 0 μa, C CEO 0 ma and C CC 8 (b) β, C,,, CE R C R E (c) R C, B, C, E (d) Drop to a relatively low voltage 0.06 (e) Open in the base circuit 47

51 49. B CC R B BE μa 50 kω 50 kω C β B (00)(.6 μa).6 ma C CC + C R C + (.6 ma)(3.3 kω) 4.69 CE C βr E 0R (0)(0.75 kω) 0(6 kω) 65 kω 60 kω (checks) Use approximate approach: 6 k Ω ( ) B 6 k Ω + 8 kω 3.59 E B C E E /R E.89/0.75 kω 3.85 ma 3.85 ma B C 7.5 μa β 0 C CC + C R C + (3.85 ma)(. kω) E R E BE ma 3.3 kω 3.3 kω C CC + C R C + (. ma)(3.9 kω) (a) S( CO ) β + 9 (b) S( BE ) C β kω S R B.93 ma (c) S(β) β A (d) Δ C S( CO )Δ CO + S( BE )Δ BE + S(β)Δβ (9)(0 μa 0. μa) + ( S)( ) + ( A)(.5 90) (9)(9.8 μa) + ( S)(0. ) + ( A)(.5) A A A A.66 ma 48

52 53. For the emitter-bias: ( + RB / RE) ( + 50 k Ω/.5 k Ω) (a) S( CO ) (β + ) (00 + ) ( β + ) + RB / RE (00 + ) + 50 k Ω/.5 kω 78. (b) S( BE ) R B β 00 + ( β + ) R 50 k Ω+ (00 + ).5 kω E S C ( + R / ).9 ma( + 340) B RE (c) S(β) β ( + β + R / R ) 00( ) B A E (d) Δ C S( CO )Δ CO + S( BE )Δ BE + S(β)Δβ (78.)(9.8 μa) + ( S)( 0. ) + ( A)(5) ma ma ma.33 ma 54. (a) R Th 6 kω 9. kω 7.94 kω + RTh / RE ( k Ω/ 0.68 k Ω) S( CO ) (β + ) (80 + ) ( β + ) + RTh / RE (80+ ) k Ω/0.68 kω (8)( +.68) β 80 (b) S( BE ) RTh + ( β + ) RE 7.94 k Ω+ (8)(0.68 k Ω) kω kω S C ( + R / ).7 ma( k / 0.68 k ) Th RE Ω Ω (c) S(β) β ( + β + R / R ) 80( k Ω/ 0.68 k Ω) Th.7 ma(.68) 80(.68) E A (d) Δ C S( CO )Δ CO + S( BE ) Δ BE + S(β)Δβ (.08)(0 μa 0. μa) + ( S)( ) + ( A)(00 80) (.08)(9.8 μa) + ( S)( 0. ) + ( A)(0) A A A A 0.4 ma 49

53 55. For collector-feedback bias: ( + RB / RC) ( k Ω/3.9 k Ω) (a) S( CO ) (β + ) ( ) ( β + ) + RB / RC ( ) k Ω/3.9 kω (97.3) ( ) (b) S( BE ) R B β ( β + ) R 560 k Ω+ ( )3.9 kω C S C ( R ).56 ma(560 k 3.9 k ) B + RC Ω+ Ω (c) S(β) β ( R + R ( β + )) 96.3(560 kω+ 3.9 k Ω ( )) B C A (d) Δ C S( CO )Δ CO + S( BE ) Δ BE + S(β)Δβ (83.69)(9.8 μa) + ( S)( 0. ) + ( A)(49.) A A A A.087 ma 56. Type S( CO ) S( BE ) S(β) Collector feedback S A Emitter-bias S A oltage-divider S A Fixed-bias S A S( CO ): Considerably less for the voltage-divider configuration compared to the other three. S( BE ): The voltage-divider configuration is more sensitive than the other three (which have similar levels of sensitivity). S(β): The voltage-divider configuration is the least sensitive with the fixed-bias configuration very sensitive. n general, the voltage-divider configuration is the least sensitive with the fixed-bias the most sensitive. 57. (a) Fixed-bias: S( CO ) 9, Δ C 0.89 ma S( BE ) S, Δ C ma S(β) A, Δ C ma (b) oltage-divider bias: S( CO ).08, Δ C ma S( BE ) S, Δ C ma S(β) A, Δ C ma 50

54 (c) For the fixed-bias configuration there is a strong sensitivity to changes in CO and β and less to changes in BE. For the voltage-divider configuration the opposite occurs with a high sensitivity to changes in BE and less to changes in CO and β. n total the voltage-divider configuration is considerably more stable than the fixed-bias configuration. 5

55 Chapter 5. (a) f the dc power supply is set to zero volts, the amplification will be zero.. (b) Too low a dc level will result in a clipped output waveform. (c) P o R (5 ma). kω 55 mw P i CC (8 )(3.8 ma) 68.4 mw Po (ac) 55 mw η % P (dc) 68.4 mw i 3. x C 5.9 Ω π fc π( khz)(0μf) f 00 khz: x C 0.59 Ω Yes, better at 00 khz 4. i 0 m 5. (a) Z i i 0.5 ma 0 Ω (r e ) (b) o c R L α c R L (0.98)(0.5 ma)(. kω) o (c) A v i 58.8 (d) Z o Ω m o (e) A i i α e e α 0.98 (f) b e c 0.5 ma 0.49 ma 0 μa 5

56 6. (a) r e i i 48 m 5 Ω 3. ma (b) Z i r e 5 Ω (c) C α e (0.99)(3. ma) 3.68 ma (d) o C R L (3.68 ma)(. kω) 6.97 (e) A v o m 45. i (f) b ( α) e ( 0.99) e (0.0)(3. ma) 3 μa 7. (a) r e (b) b 6 m 6 m 3 Ω E (dc) ma Z i βr e (80)(3 Ω).04 kω C α β e e e β β β + β β + ma μa o L (c) A i i b ro( β b) L ro + RL ro β b ro + RL ro A i β r + R (d) A v b o L 40 kω (80) 40 kω+. kω RL ro. kω 40 kω r 3 Ω.65 kω 3 Ω 89.6 e 53

57 8. (a) Z i βr e (40)r e 00 r e Ω i 30 m (b) b Z 5 μa. kω i (c) c β b (40)(5 μa) 3.5 ma (d) L A i r o c (50 k Ω)(3.5 ma) r + R 50 kω+.7 kω o L i L 3.3 ma 5 μa ma o AR i L (e) A v Z i i (.7 k Ω) (3.84). k Ω CC BE (a) r e : B 5.36 μa RB 0 kω E (β + ) B (60 + )(5.36 μa) 3.3 ma 6 m 6 m r e 8.3 Ω 3.3 ma E Z i R B βr e 0 kω (60)(8.3 Ω) 0 kω Ω Ω r o 0R C Z o R C. kω (b) A v RC. kω r 8.3 Ω e (c) Z i Ω (the same) Z o r o R C 0 kω. kω.98 kω (d) A v RC ro.98 kω r 8.3 Ω 38.7 e A i A v Z i /R C ( 38.7)( Ω)/. kω

58 RC 0. A v r e r r e. (a) B e RC 4.7 kω 3.5 Ω A ( 00) v 6 m 6 m 6 m E.06 ma E r e 3.5 Ω.06 ma B E.5 μa β + 9 CC BE B CC B R B + BE RB (.5 μa)( MΩ) CC R B BE μa 390 kω E (β + ) B (0)(3.85 μa).4 ma 6 m 6 m r e 0.79 Ω E.4 ma C β B (00)(3.85 μa).38 ma (b) Z i R B βr e 390 kω (00)(0.79 Ω) 390 kω.08 kω.08 kω r o 0R C Z o R C 4.3 kω (c) A v (d) A v RC 4.3 kω r 0.79 Ω e RC ro (4.3 k Ω) (30 k Ω) 3.76 kω r 0.79 Ω 0.79 Ω e. (a) Test βr E 0R? (00)(. kω) 0(4.7 kω) 0 kω > 47 kω (satisfied) Use approximate approach: R 4.7 k (6 ) B CC Ω R + R 39 kω+ 4.7 kω.7 E B BE E.0 E R ma. kω r e E 6 m 6 m Ω ma E 55

59 (b) Z i R R β r e 4.7 kω 39 kω (00)(30.56 Ω).768 kω r o 0R C Z o R C 3.9 kω (c) A v RC 3.9 kω r Ω 7.6 e (d) r o 5 kω (b) Z i (unchanged).768 kω Z o R C r o 3.9 kω 5 kω 3.37 kω (c) A v ( RC ro) (3.9 k Ω) (5 k Ω) 3.37 kω r Ω Ω e 0.8 (vs. 7.6)? 3. βr E 0R (00)( kω) 0(5.6 kω) 00 kω > 56 kω (checks!) & r o 0R C Use approximate approach: RC RC 3.3 kω A v re 0.65 Ω r A 60 e v 6 m 6 m 6 m r e E.6 ma E re 0.65 Ω E E R E E R E (.6 ma)( kω).6 E B BE + E kω B CC 5.6 kω+ 8 kω kω CC (.96 )(87.6 kω) CC Test βr E 0R? (80)(. kω) 0(56 kω) 396 kω < 560 kω (not satisfied) Use exact analysis: (a) R Th 56 kω 0 kω kω 56 k Ω(0 ) E Th 0 kω + 56 kω ETh BE B R + ( β + ) R k Ω+ (8)(. k Ω) Th E 56

60 7.58 μa E (β + ) B (8)(7.58 μa).37 ma 6 m 6 m r e 8.95 Ω.37 ma E (b) E E R E (.37 ma)(. kω) 3.0 B E + BE C CC C R C 0 β B R C 0 (80)(7.58 μa)(6.8 kω) 0.7 (c) Z i R R βr e 56 kω 0 kω (80)(8.95 kω) kω 3.4 kω 3.7 kω RC ro r o < 0R C A v re (6.8 k Ω) (50 k Ω) 8.95 Ω CC BE (a) B RB + ( β + ) RE 390 k Ω+ (4)(. k Ω) μa 559. kω E (β + ) B (40 + )(34.5 μa) ma 6 m 6 m r e 5.34 Ω ma E (b) Z b βr e + (β + )R E (40)(5.34 kω) + (40 + )(. kω) Ω kω kω Z i R B Z b 390 kω kω 8.37 kω Z o R C. kω (c) A v β R Z C b (40)(. k Ω) kω.8 (d) Z b βr e + ( β + ) + RC / ro + ( R + R )/ r C E o R (4) +. k Ω/ 0 kω Ω + (3.4 k Ω)/ 0 kω. kω E 57

61 747.6 Ω kω kω Z i R B Z b 390 kω kω kω Z o R C. kω (any level of r o ) β R C r e RC + + Z o b ro ro A v R i C + r o (40)(. k Ω) 5.34 Ω. kω kω 0 k 0 k Ω Ω. kω + 0 kω Even though the condition r o 0R C is not met it is sufficiently close to permit the use of the approximate approach. β RC β RC RC A v 0 Zb β RE RE RC 8. kω R E 0.8 kω m 6 m E 6.84 ma 3.8 Ω r e E E R E (6.84 ma)(0.8 kω) 5.6 B E + BE ma B E μa ( β + ) R B CC B and R B 4.09 kω μa B 7. (a) dc analysis the same r e 5.34 Ω (as in #5) B (b) Z i R B Z b R B βr e 390 kω (40)(5.34 Ω) Ω vs kω in #5 Z o R C. kω (as in #5) (c) A v RC. kω r 5.34 Ω e 4.99 vs.8 in #5 (d) Z i Ω vs kω for #5 Z o R C r o. kω 0 kω.98 kω vs.. kω in #5 58

62 8. (a) B RC ro.98 kω A v vs..8 in #5 re 5.34 Ω Significant difference in the results for A v. CC BE R + ( β + ) R B E k Ω+ (8)(. kω k Ω) kω μa E (β + ) B (8)(45.78 μa) 3.7 ma 6 m 6 m r e 7 Ω 3.7 ma E (b) r o < 0(R C + R E ) ( β + ) + RC / ro Z b βr e + + ( R + R )/ r C E o R E (80)(7 Ω) + (8) k Ω/ 40 kω k Ω/ 40 kω. kω 560 Ω kω (note that (β + ) 8 R C /r o 0.4) 560 Ω + [8.4 /.7]. kω 560 Ω kω kω Z i R B Z b 330 kω kω 66.8 kω β R C r e RC + + Zb ro ro A v RC + ro (80)(5.6 k Ω) 7 Ω 5.6 kω kω 40 kω 40 kω k Ω/40 kω (5.35) (a) B CC BE R + ( β + ) R 70 k Ω+ ()(.7 k Ω) kω B E 59

63 6.86 μa E (β + ) B (0 + )(6.86 μa).98 ma 6 m 6 m r e 8.7 Ω E.98 ma βr e (0)(8.7 Ω) 959. Ω (b) Z b βr e + (β + )R E 959. Ω + ()(.7 kω) kω Z i R B Z b 70 kω kω 4.5 kω Z o R E r e.7 kω 8.7 Ω 8.69 Ω (c) A v RE.7 kω R + r.7 kω Ω E e 0. (a) B CE BE R + ( β + ) R 390 k Ω+ ()5.6 kω B E E (β + ) B ()(6.84 μa) 0.88 ma 6 m 6 m r e 3.4 Ω E 0.88 ma r o < 0R E : Z b βr e + ( β + ) RE + RE / ro ()(5.6 k Ω) (0)(3.4 Ω) k Ω/40 kω 3.77 kω kω kω Z i R B Z b 390 kω kω 36. kω Z o R E r e 5.6 kω 3.4 Ω 3. Ω 6.84 μa (b) A v ( β + ) R / E Z + R / r E o ()(5.6 k Ω) / kω k Ω/ 40 kω (c) A v i b o A v i (0.994)( m) m 60

64 . (a)? Test βr E 0R (00)( kω) 0(8. kω) 400 kω 8 kω (checks)! Use approximate approach: 8. k Ω(0 ) B 8. kω+ 56 kω.5545 E B BE E.855 E R 0.97 ma E kω 0.97 ma B E 4.6 μa ( β + ) (00 + ) C β B (00)(4.6 μa) 0.9 ma (b) r e 6 m 6 m 8.05 Ω 0.97 ma E (c) Z b βr e + (β + )R E (00)(8.05 Ω) + (00 + ) kω 5.6 kω + 40 kω kω Z i 56 kω 8. kω kω 7.5 kω kω 7.03 kω Z o R E r e kω 8.05 Ω 7.66 Ω (d) A v. (a) E r e RE kω R + r kω Ω E EE E e R BE ma 6.8 kω 6 m 6 m Ω ma E (b) Z i R E r e 6.8 kω Ω 33. Ω Z o R C 4.7 kω α RC (0.998)(4.7 k Ω) (c) A v re Ω α β β

65 EE BE E. ma RE 3.9 kω 3.9 kω 6 m 6 m r e 3.58 Ω E. ma RC (0.9868)(3.9 k Ω) A v α 63. r 3.58 Ω e CC BE (a) B RF + β RC 0 kω+ 0(3.9 k Ω) 6.4 μa E (β + ) B (0 + )(6.4 μa).987 ma r e 6 m 6 m 3.08 Ω.987 ma E (b) Z i βr e R A F v Need A v! RC 3.9 kω A v re 3.08 Ω 98 Z i (0)(3.08 Ω) 0 k Ω kω 738 Ω Ω Z o R C R F 3.9 kω 0 kω 3.83 kω (c) From above, A v 98 RC 5. A v r e 60 R C 60(r e ) 60(0 Ω).6 kω A i β RF R + β R F C RF R + 00(.6 k Ω) F 9R F R C 00R F R F 3800 R C 3800(.6 k Ω) kω CC BE B R + β R F B (R F + βr C ) CC BE C 6

66 and CC BE + B (R F + βr C ) 6 m 6 m with E.6 ma r e 0 Ω.6 ma B E.94 μa β CC BE + B (R F + βr C ) (.94 μa)(33.59 kω + (00)(.6 kω)) (a) A v : i b βr e + (β + ) b R E o + C β b but i + b and i b Substituting, o + ( i b ) β b and o (β + ) b i Assuming (β + ) b i o (β + ) b and o o R C (β + ) b R C o ( β + ) brc Therefore, βr + ( β + ) R β R b C β r b e+ β R b E o RC RC and A v r + R R (b) i β b (r e + R E ) For r e R E i β b R E i b e b E i e E E Now i + b R i o F + b 63

67 Since o i o i + R o or b i + RF and i β b R E o i βr E i + β R E RF but o A v i β AR v i E and i βr E i + R Avβ REi or i R F F b F βr E i Avβ R E i [ β RE ] i RF i β RE β RERF so Z i Avβ RE R + β ( A ) R R i Z i with Z i Z i Z o : Set i 0 i F v E i F x y where x βr E and y R F / A v x y ( β RE )( RF / Av ) x + y β R + R / A β RERF β R A + R E v F E F v i b βr e + (β + ) b R E i β b (r e + R E ) 0 since β, r e + R E 0 b 0 and β b 0 64