Module 3 Unit 3 Small Signal BJT Amplifiers

Transcription

1 Modul 3 Unit 3 Small Signal BJT mplifirs viw Qustions: 1. Dfin h paramtrs for a transistor. Why ar ths calld hybrid paramtrs? What ar thir units? 2. Out of four h paramtrs, two ar most important. Which ar ths? nd why th othr two hav lss significanc? 3. h i and Z i both rprsnt input impdanc in h - and Z systms of paramtrs but thy ar most qual. Why? 4. What ar r paramtrs and how ar thy suprior to h paramtrs? 5. Common bas (CB) amplifir has limitd applications. Why? 6. mong BJT amplifirs, common-mittr amplifir is most favourd. Giv rasons. 7. What is an mittr followr? Discuss its main applications. 8. What ar, in gnral, prformanc paramtrs of an amplifir?

2 Problms: 3.1 Calculat dynamic mittr rsistanc r (also calld ac mittr rsistanc) of th transistor in th circuit of fig.( B = 0.7) +5(+ CC ) 12k k B I 2k I C + C - 4.3k -5(- )

3 Solution:- Th dynamic mittr rsistanc, ' r, is xprssd as, r' 25mv I Whr I is dc mittr currnt. Th dc mittr currnt in th circuit (i., voltag drop across rsistor, dividd by th rsistanc) is I 0.7 B Thrfor, 5 m r' 25m 1m 25 That is, ' r = 25 Ω

4 3.2 How much is th voltag gain of th amplifir (fig) if th dynamic mittr rsistanc, r, is 25 Ω. Th currnt gain of th transistor is 80. How much is input impdanc of th amplifir. Th coupling and bypass capacitors may b assumd of ngligibl impdanc at th signal frquncy. Tak B = k C 5k C 2 v 0 v i C 1 B 6k k 2.3k C 3 L 5k

5 Solution:- Th voltag gain, for a common mittr amplifir with rsistor by- passd is rc r ' Whr r c is ffctiv (ac) impdanc sn by th collctor and it is r c = c L = 5k 5k = 2.5 kω nd ' r = 25Ω (givn) Thrfor, First w find out impdanc btwn bas (point in Fig 3.2) and ground, Z i(bas) and w know, it is givn by Z i(bas) = β ' r = 80 x 25 or Z i(bas) = 2 kω nd, input impdanc of th amplifir Z i(amp) is, Z i(amp) = B Z i(bas) Whr B is ffctiv biasing rsistanc, and B = 1 2 = 12k 6k = 4 kω Thrfor, Z i(amp) = 4k 2k or Z i(amp) = 1.33 kω

6 3.3 Calculat th valu of rsistor C so that th voltag gain of th amplifir in fig is 100. Capacitors C 1, C 2, and C 3 may b assumd short at signal frquncy. +12 (+ CC ) C C 2 v o v i C 1 10k k 5k C 3 L 4k -12 (- )

7 Solution:- In cas, is by-passd, th voltag gain v of th amplifir is, rc r ' Whr r c is th ffctiv (ac) rsistanc sn by th collctor of th transistor, and th dynamic rsistanc of th mittr. W know, ' r is r' 25mv I Now, B I 2. 26m Thrfor, r' nd, 25m 2.26m k c = C L 4 C C 4 Whr rsistancs hav bn takn in kω Thrfor,

8 rc r' 4 c c s v = 100 (givn), c (in kω) is, 100 or c ( c 1.51k 4c 4)(0.011)

9 3.4 Find out th smallst valu of load L in th amplifir circuit shown in fig so that th voltag gain is at last 40. Th dynamic mittr rsistanc of th transistor is 25 Ω. Th coupling and by-pass capacitors may b assumd short at signal frquncy. +12k 8k C 3k v o v i 4k k 4k L

10 Solution:- Th voltag gain of th common mittr amplifir with rsistor by-passd (s fig.) is xprssd as rc r' 40 ( rquird gain) Whr r c is ffctiv ac impdanc sn by th collctor, Which is r c = C L C C L L Taking rsistancs in kω r c 3L 3 L nd, or L 3L 3 L 1.5k

11 3.5 Th silicon transistor in th common- bas amplifir has th currnt gain α of Find th input impdanc and voltag gain of th amplifir in fig. ( B = 0.7) -9 (- ) +9 (+ cc ) 6k C 4k ~ v i B 4k L

12 Solution:- Th dc voltag sourcs hav to b groundd for ac analysis of th amplifir. Thn th input impdanc of th CB amplifir (in fig.) is Z i(amp) = r r Whr r is dynamic mittr rsiatanc and r << lso, r' 25m I Whr I is dc mittr currnt in th circuit I B or, I m Thrfor, r' 25m 1.38m 18 Thus, th input impdanc Z i(amp) is, Z i(amp) r = 18Ω Th low valu of input impdanc is th main rason for limitd applications of CB amplifir. Th voltag gain of CB amplifir is xprssd as, r r' c Sinc, r c = C L = 4k 4k = 2kΩ = 2000 Ω Thn,

13 3.6 Th transistor in th amplifir circuit shown in fig, has h paramtrs, h i = 2kΩ and h fl = 80. Th valus of h o and h r ar ngligibl. Calculat th voltag gain and input impdanc Z i(amp) of th amplifir. Capacitors C 1, C 2, and C 3 may b assumd short at signal frquncy du to small impdancs k C 6k C 2 v o v i C k k 2k C 3 L 6k

14 Solution:- Th magnitud of voltag gain with h-paramtrs h o and h r droppd, and mittr rsistanc by-passd by capacitor C 3 is h f h. Z i i 80 3k 2k 120 Bcaus th ac load at collctor, Z l, is Z l = C L = 6k 6k = 3kΩ Furthr, th impdanc btwn bas and ground, Z i(bas) is, Z i(bas) = h i = 2kΩ nd, input impdanc of amplifir, Z i(amp) is, Z i(amp) = 1 2 Z i(bas) = 20k 20k Z i(bas) = 10k Z i(bas) = 10k 2k = 1.6kΩ or Z i(amp) = 1.6kΩ

15 3.7 For th amplifir circuit shown in fig. calculat th voltag gain and input impdanc of th amplifir whn by-pass capacitor C 3 is rmovd from th circuit k C 6k C 2 v o v i C k k 2k C 3 L 6k

16 Solution:- With by-pass capacitor C 3 (in fig.) rmovd, th gain of a amplifir falls and input impdanc of th amplifir incrass. In cas is not by-passd, th magnitud of voltag gain is Z l 3k 2k 1.5 nd, Z i(bas) = h i + (1 + h f ) = 2k + (1 + 80) X 2k Or, Z i(bas) = 164 kω nd, input impdanc of amplifir, Z i(amp) is Z i(amp) = 1 2 Z i(bas) = 20k 20k 164 k 10k 164k = 9.4 kω or Z i(amp) = 9.4 kω

17 3.8 Th mittr followr (common collctor amplifir) shown in fig. uss a transistor with h-paramtrs h i = 4.5 kω, h f = 120. Othr paramtrs h o and h r hav ngligibl ffct on amplifir prformanc. Calculat voltag gain and input impdanc of th amplifir. Th coupling and by-pass capacitors may b assumd short at signal frquncy k ~ 30k 3k L 6k Z i (amp) Z i (bas)

18 Solution:- Nglcting th ffct of h o and h r on amplifir prformanc, th voltag gain of mittr followr may b xprssd as, h ic h fc h Z fc. Z Whr Z is th ffctiv load sn by th mittr, and it is Z = L = 3k 6k = 2 kω nd using h fc h f and h ic = h i, W hav, h i 4k h 0.98 f h Z f Z 120 2k (120 2k ) Th input impdanc as sn at th bas w.r.t ground is, Z i(bas) = h fc. Z = h f. Z = 120 X 2 kω = 240 kω Th input impdanc of th amplifir (that is, aftr taking th ffct of biasing rsistors), Z i(amp) = B Z i(bas) nd th ffctiv bas rsistanc B is, B = 1 2 = 60 k 30k = 20 kω Thrfor, Z i(amp) = 20k 240k = kω or Z i(amp) = kω

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