CHAPTER 13. Exercises. E13.1 The emitter current is given by the Shockley equation:

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1 HPT 3 xercses 3. The emtter current s gen by the Shockley equaton: S exp VT For operaton wth, we hae exp >> S >>, and we can wrte VT S exp VT Solng for, we hae ln 78.4 mv V VT ln 4 S β 50 α β + 5 α m. β 96 µ α β α α β m α / β / The base current s gen by quaton 3.8: 6 ( α ) S exp.96 0 exp VT 0.06 whch can be plotted to obtan the nput characterstc shown n Fgure 3.6a. For the output characterstc, we hae β proded that 43

2 approxmately 0. V. For 0. V, falls rapdly to zero at 0. The output characterstcs are shown n Fgure 3.6b. 3.5 The load lnes for n 0.8 V and V are shown: s shown on the output load lne, we fnd V V, V 5 V, andv.0 V. max 9 Q mn 43

3 3.6 The load lnes for the new alues are shown: s shown on the output load lne, we hae V.8 V, V 7 V, andv 3.0 V. max 9 Q mn 433

4 3.7 efer to the characterstcs shown n Fgure 3.7 n the book. Select a pont n the acte regon of the output characterstcs. For example, we could choose the pont defned by 6 V and.5 m at whch we fnd 50 µ. Then we hae β / 50. (For many transstors the alue found for β depends slghtly on the pont selected.) 3.8 (a) Wrtng a KV equaton around the nput loop we hae the equaton for the nput load lnes: 0.8 n ( t ) The load lnes are shown: Then we wrte a K equaton for the output crcut: The resultng load lne s: From these load lnes we fnd 434

5 max 48 µ, Q 4 µ, mn 5 µ, V.8 V, V 5.3 V, V 8.3 V max Q mn (b) nspectng the load lnes, we see that the maxmum of n corresponds to mn whch n turn corresponds to V mn. ecause the maxmum of n corresponds to mnmum V, the amplfer s nertng. Ths may be a lttle confusng because V takes on negate alues, so the mnmum alue has the largest magntude. 3.9 (a) utoff because we hae V 0.5 V andv V V 4.5 V whch s less than 0.5 V. (b) Saturaton because we hae < β. < (c) cte because we hae 0 andv > 0. V. > 3.0 (a) n ths case ( β 50) the JT operates n the acte regon. Thus the equalent crcut s shown n Fgure 3.8d. We hae V µ β m V V.43 V ecause we hae V > 0., we are justfed n assumng that the transstor operates n the acte regon. (b) n ths case ( β 50),the JT operates n the saturaton regon. Thus the equalent crcut s shown n Fgure 3.8c. We hae V 0.7 V 0. V 0. V 7.5 µ 4.8 m ecause we hae β >, transstor operates n the acte regon. we are justfed n assumng that the 3. For the operatng pont to be n the mddle of the load lne, we want V V V V / 0 V and m. Then we hae V 0.7 (a) / β 0 µ 965 kω (b) V 0.7 / β µ.985 MΩ 435

6 3. Notce that a pnp JT appears n ths crcut. (a) For β 50, t turns out that the JT operates n the acte regon µ β m V (b) For β 50, V t turns out that the JT operates n the saturaton regon V 0. V 9.3 µ.98 m ecause we hae β >, n the acte regon. 3.3 V V 5 V + β we are assured that the transstor operates V V + ( β + ) β V V + ) (µ) ( V (m) (V) For the larger alues of and used n ths xercse, the rato of the collector currents for the two alues of β s.05, whereas for the smaller alues of and used n xample 3.7, the rato of the collector currents for the two alues of β s.03. n general n the four-resstor bas network smaller alues for and lead to more nearly constant collector currents wth changes n β Ω + k V V 5 V + V V Q 4.3 µ Q β Q 4.39 m + β + βvt r Q ( ) 300( 6 mv) 4.38 m 840 Ω 436

7 β Ω r β oc 63.0 Z n r + + r Z n G Ω o Z kω o n s Z n sn( ωt ) Z + n s 3.5 Frst, we determne the bas pont: kω V V 0 V + + V V Q 4.6 µ Q β Q 4.79 m + β + ( ) Now we can compute r and the ac performance. r ( ) βv mv 4.79 m 83 Ω T Ω Q + r + ( β + ) ( β + ) ( β + ) ( β + ) oc r Zn Zn 40.0 kω [ + ( + ) r β ] G s + s kω Zo 33.8 Ω ( β + ) + + r s Problems P3. 437

8 P3. The sketch should resemble Fgure 3.a n the book. n normal operaton, current flows nto the base, nto the collector, and out of the emtter. P3.3 To forward bas a pn juncton, the p-sde of the juncton should be connected to the poste oltage. n normal operaton of a JT, the emtter-base juncton s forward based and the collector-base juncton s reerse based. P3.4 The emtter current s gen by: S exp VT P3.5 For a JT, the parameters are defned as: α β α α t s assumed that the base-emtter juncton s forward based and that the collector-base juncton s reerse based. P3.6* + α β m P3.7* The emtter current s gen by the Shockley equaton: S exp VT For operaton wth, we hae exp >> S >>, and we can wrte: VT S exp VT Solng for, we hae 438

9 V 0 6ln mv V T ln 3 S β 00 α β + 0 α 9.90 m.0 β 99 µ P3.8 From quaton 3.9, we hae α β α Solng for α, we hae β 00 α β + 0 P ( 0 µ ) m β +.0 m P m α β P3. Solng the Shockley equaton, we hae S exp VT lso, we hae V T kt / q Substtutng alues for 300 K, we obtan V mv T S t 30 K, we hae V 6.74 mv T S

10 For the 0 K ncrease n temperature, we fnd that S has ncreased by a factor of P3. From the crcut, we can wrte: m.03 µ 0 kω 680 kω Then, we hae β P3.3 We hae S exp VT Solng for the saturaton current: S 0.7 exp exp V 0.06 T For m, we hae ln VT S For µ, we hae V ln ln T ln 4 S V V P3.4 Wrtng a current equaton at the collector of Q, we hae: + m + Howeer, and β, so we hae β + m µ µ + β m m s n the soluton to Problem P3.7, we hae: ln 0 3 ln VT 4 S V 440

11 P3.5 The Shockley equaton states: S exp VT Thus, s proportonal to, and we can wrte: 0 S ecause α and β are the same for both transstors, we also hae: 0 and 0 Wrtng a current equaton at the collector of Q, we hae + m + β µ µ β β + m m m m m s n the soluton to Problem 3.3, we hae ln 0 3 ln VT 4 S P3.6* We hae: S exp VT S exp VT ddng the respecte sdes of these equatons, we hae: ( ) eq + S + S exp VT Thus, we hae: Seq S S lso, we hae, and we can wrte: eq β eq + + eq V 44

12 P3.7 We hae eq + β + β β + β( β + ). Then we can wrte eq + ( ) β + β β + β β + β( β + eq ) eq P3.8* t 80 o and.7 0. m, the base-to-emtter oltage s approxmately: ( 30) 0.4 V P3.9* We select a pont on the output characterstcs n the acte regon and compute β /. For example on the cure for 0 µ, we hae 8 m n the acte regon. Thus, β (8 m)/(0 µ ) 400. Then, we hae α β /( β + ) P3.0 n the acte regon (whch s for > 0. V ), we hae β. n saturaton, the current falls to zero. The sketches are: 44

13 P3. P3. Dstorton occurs n JT amplfers because the nput characterstc of the JT s cured (rather than straght) and because the output characterstc cures are not unformly spaced. P3.3 The slope of the load lne does not change as V changes. 443

14 P3.4* Followng the approach of xample 3., we construct the load lnes shown. We estmate that V.4 V, V 5.6 V, and V V. 8 max Q mn Thus, the oltage gan magntude s ( 8.4 ) P3.5 Followng the approach of xample 3., we construct the load lnes shown. From the nput load lne, we hae 0. Thus, we mn Q max hae V V 0. Thus, the oltage gan magntude s: max mn 444

15 ( 0 0) P3.6 The nput load lne s the same as n Problem P3.4. Thus, we agan hae µ, 5.5 µ and 0. The output load lne s mn Q max µ shown below. ecause V, we know that the output waeform s Q V mn seerely dstorted. t s not approprate to compute oltage gan for such a seerely dstorted output waeform. 445

16 P3.7 P3.8* 446

17 P3.9* (a) and (b) The characterstcs and the load lne are: (c) We are gen ( t ) ( t ) sn(000t ). From whch we s mn 5 Q µ max determne that µ, 0, and 5 µ. Then at the ntersectons of the load lne wth the respecte characterstcs, we determne that and.5 mn 0.5 m, Q.0 m, max m and that V 5 V, V V, 5 V. 0 V mn Q max (d) The sketches of (t) are: (e) We are gen ( t ) ( t ) sn(000t ). From whch we s mn 5 µ, Q 0 µ, and max 5 µ determne that. Then at the ntersectons of the load lne wth the respecte characterstcs, we determne that.5 m,.0 m, and max.0 m and that mn Q V 5 V, V V, V 0. V. sketch of (t) s shown mn Q aboe. 0. max 447

18 P3.30 α / β / /( ) 99. o P3.3 The magntude of s decreased by about mv for each ncrease n temperature. Thus at 80 o, we hae: ( 30) 0.4 V P3.3 We can wrte eq ( β + ) ( β + ) β βeq β( β eq + ) P3.33 See Fgure 3.6 n the text. P3.34 See Fgure 3.6 n the text. P3.35 P3.36* n the acte regon, the base-collector juncton s reerse based and the base-emtter juncton s forward based. n the saturaton regon, both junctons are forward based. n the cutoff regon, both junctons are reerse based. (ctually, cutoff apples for slght forward bas of the base-emtter juncton as well, proded that the base current s neglgble.) 448

19 P3.37 (a) cte regon. (b) utoff regon. (c) utoff regon (because 0 forv 0.4 V at room temperature). (d) Saturaton regon (because s less than β ). P3.38 (a) utoff regon (because 0 forv 0.3 V at room temperature). (b) Saturaton regon (because s less than β ). (c) cte regon. P3.39 (a) 0.5 m The transstor s n saturaton because we hae Thus, we hae m,.5 m, andv 0. V. < β (b) The transstor s n the acte regon because we hae V > 0. and > 0. Thus, we hae β m, 0. m, andv 0 5 V. (c) Ths pnp transstor s n the acte regon because we hae V < 0. and > 0. Thus, we hae β m,.0 m, andv 4.6 V. (d) Ths pnp transstor s n cutoff because both junctons are reerse based. Thus, we hae 0, andv 3V. P3.40 For the Darlngton par, we hae V V + V. V. eq For the Szkla par, we hae V V 0.6 V. eq P3.4*. ssume operaton n saturaton, cutoff, or acte regon.. Use the correspondng equalent crcut to sole for currents and oltages. 3. heck to see f the results are consstent wth the assumpton made n step. f so, the crcut s soled. f not, repeat wth a dfferent assumpton. 449

20 P3.4 The fxed base bas crcut s: t s unsutable for mass producton because the alue of ares by typcally a factor of 3: between JTs of the same type. onsequently, the bas pont ares too much from one crcut to another. P3.43 See Fgure 3.a n the text. β P3.44* The results are gen n the table: egon of rcut β operaton (m) V (olts) (a) 00 acte (a) 300 saturaton (b) 00 acte (b) 300 saturaton.8 0. (c) 00 cutoff 0 5 (c) 300 cutoff 0 5 (d) 00 acte (d) 300 saturaton P3.45 The results are gen n the table: egon of operaton (m) V (olts) rcut β (a) 00 acte (a) 300 saturaton (b) 00 cutoff 0 0 (b) 300 cutoff 0 0 (c) 00 acte (c) 300 acte (d) 00 Q acte Q acte (d) 300 Q acte Q saturated 450

21 P3.46 V V 5 V + / + / V V.07 µ + ( β + ) kω β 0.44 m V V ( + ) V P3.47* The JT operates n the acte regon. We can wrte the oltage equaton: Howeer, we can substtute usng the relatons: β + and β β Thus, we hae: 4.3 β β + + β For β 00, we requre 4 m. Furthermore, for β 300, we requre 5 m. Thus, we hae two equatons: ( ) + ( ) 3 3 ( ) + ( ) Solng, we fnd: 3.5 kω and 753 Ω P3.48 t turns out that for the alues gen the JT operates n the saturaton regon and the equalent crcut s: 45

22 nclosng the transstor n a closed surface (supernode) and wrtng a K equaton, we hae V V V V Substtutng the resstance alues and solng we obtan V V. Then we hae 5 ( V + 0.) m m. s a check we note that we hae β > as requred for operaton n the saturaton regon. P3.49* From quatons 3.0 through 3.3, we hae: V V β β + + V V + ( β + ) Maxmum occurs for β βmax 00, kω mn 4465 Ω, mn kω 95 kω, and max kω kω. Wth these alues, we calculate: 3.48 kω V 5.8 V 0.95 m and max Mnmum occurs for β β 50, kω max kω, kω 05 kω, and max mn mn kω kω. Wth these alues, we calculate: 3.33 kω V V m mn 5 V P µ 6 β m ( + ) + V (0 + ) (0 ) β m β + ) 3.93 m V V µ V 4.63 V V ( 45

23 P3.5 The crcut s: P3.5 We hae: µ 00 kω β 0 µ + 77 µ V V kω m 5 V 4.59 kω m 50 kω Substtutng β and solng for, we obtan: ( ) ( ) 7 453

24 ( 5700) ( β + ) µ β.79 m V ( ) 6.06 V 5 P3.53 See Fgure 3.6 n the text. P3.54 r βvt Q P3.55 r βvt Q r Q µ.6 MΩ 0. m 6 kω m.6 kω P3.56* We use the same approach as n Secton 3.7. We can wrte 5 ( t ) 0 ( t ) 5 Q + b t ) 0 [ V + Q Q ( ( t )] be ( t ) 0 V + 0 V ( t ) + 0 ( t ) () b Q 5 Howeer for the Q-pont, we hae Q V. Therefore, the frst Q be 0 Q term on each sde of quaton () can be canceled. Furthermore, the last term on the rght hand sde of quaton () s neglgble for small sgnals [.e., be (t ) much less than V Q at all tmes]. Thus, quaton () becomes 5 ( t ) 0 V ( t ) b We hae defned be ( t ) r ( t ) b and we hae r 0 Q be 5 0 For Q m, we obtan r V 5 5 Q 0 Q 0 Q /00 50 kω. be 58 Q 454

25 P3.57 The equalent crcut s: be eq r b + r ( b + βb ) r eq r + r ( + beq b β ) P3.58 The equalent crcut s: r eq beeq beq r b b r P3.59 ouplng capactors are used to prode a path for ac sgnals whle mantanng an open crcut for dc. n connectng the sgnal source or load, couplng capactors preent the source or load from affectng the bas pont of the amplfer. When t s desred to amplfy dc sgnals, couplng capactors should not be used. P3.60 The common-emtter amplfer s nertng. oth the oltage gan and current gan magntudes are potentally greater than unty. 455

26 P3.6 See Fgure 3.7a n the text. P3.6 The dc crcut s: The bas pont calculatons are: 3.97 kω V V V + + V V Q 39.3 µ Q β Q 3.93 m + ( β + ) r βvt Q ( ) 00 6 mv 3.93 m 66 Ω Then, we compute the amplfer performance: β 500 Ω r 5 β oc 5.0 Z n 548 Ω r + + r Z n 4.4 G 34 Z kω o P3.63* The soluton s smlar to that of Problem P3.6. The results are: 456

27 Hgh mpedance amplfer (Problem 3.63) ow mpedance amplfer (Problem 3.6) Q m 3.93 m r 66. kω 66 Ω oc -5-5 Z n 54.8 kω 548 Ω G Z o 00 kω kω P3.64 (a) The small-sgnal equalent crcut s: n whch we hae defned / + / (b) From the equalent crcut we can wrte: n r b + ( β + ) b () β o Then we hae o n b β b β r + ( β + ) r + ( β + ) b b (c) From the equalent crcut we can wrte: n n + b Then solng quaton () for b and substtutng, we hae n n n + r + ( β + ) n n n / + / [ r + ( β + ) ] 457

28 (d) ecause the couplng capactors are open crcuts for dc, we can gnore the sgnal source and the load n the dc analyss. To attan Q m, we must hae Q ( m)/β 50 µ. Wrtng a oltage equaton we hae V + V + ( β + ) Q Q Substtutng alues and solng, we obtan kω. Q (e) Frst, we hae r βv t / ( ) / Ω. Then Q usng the equatons from parts (a) and (b) we determne that and n 0.33 kω. P3.65 See Fgure 3.30a n the text. P3.66 The oltage gan of an emtter follower s poste and less than unty n magntude. The current gan and power gans are potentally greater than unty. P3.67* The dc crcut s: The bas pont calculatons are: 5 Ω + k V V 7.5 V + V V Q 64. µ + β + β Q Q ( ) 6.4 m Now we can compute r and the ac performance. βvt r Q ( ) 00 6 mv 6.4 m 405 Ω 458

29 + r + ( β + ) ( β + ) ( β + ) ( β + ) 333 Ω 0.98 oc r + Z n [ r + ( β + ) ] 4.36 kω Zn 8.6 G 8. 5 Zo ( β + ) + + r Z o s. Ω n whch s + s 833 Ω P3.68 The soluton s smlar to that of Problem P3.67. The results are: Hgh mpedance amplfer (Problem 3.68) ow mpedance amplfer (Problem 3.67) Q 64. µ 6.4 m r 40.5 kω 405 Ω oc Z n 436 kω 4.36 kω G Z o 0 Ω. Ω 459

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