General Amplifiers. Analog Electronics Circuits Nagamani A N. Lecturer, PESIT, Bangalore 85. Cascade connection - FET & BJT

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1 Analg lectrnics Circuits Nagamani A N Lecturer, PST, Bangalre 85 mail nagamani@pes.edu General Amplifiers Cascade cnnectin - FT & BJT Numerical Cascde cnnectin arlingtn cnnectin Packaged arlingtn cnnectin c bias f arlingtn cnnectin AC equivalent ac utput impedance f arlingtn cnnectin AC vltage gain Feedback cncept Feedback cnnectin type Practical feedback circuits Practical feedback circuits Numerical

2 Cascade cnnectin FT Cascade cnnectin is a series cnnectin wit te utput f ne stage ten applied as input t te secnd stage. Cascade cnnectin prvides a multiplicatin f te gain f eac stage fr a larger verall gain. Gain f verall cascade amplifier is te prduct f stage gains A 1 and A 2 A v = A v1 A 2 = (-g m 1 ) (-g m 2 ) Te input impedance f te cascade amplifier is tat f stage 1, Z i = G1 Output impedance is tat f stage 2, Z 0 = 2 Te main functin f cascading te stages is te larger verall gain acieved.

3 Numerical Calculate dc bias, vltage gain, input impedance, utput impedance,als calculate te lad vltage if a 10K Ω lad is cnnected acrss te utput ata fr numerical C1=C2=C3=0.05uF G1 = G2 =3.3 MΩ S1 = S2 =680 Ω 1 = 2 =2.4 KΩ SS =10mA; P =-4 fr bt stages Slutin Step 1: frm te dc bias details we can find ut GSQ = -1.9, Q =2.8mA

4 Step 2: bt transistrs ave g m =2 SS / p =2(10mA)/4 =5mS At dc bias pint, g m =g m (1- GS / P ) gm =5m(1-(-1.9)/(-4) = 2.6mS Step 3: te vltage gain f eac stage A 1 =A 2 =-g m =-2.6m x 2.4K = -6.2 Step 4: Overall gain f cascaded stage is A v =A v1 A v2 =-6.2 x -6.2 = 38.4 (utput is in pase wit input) Step 5: utput vltage is =A v i =384 m Cascade amplifier input impedance is Z i = G = 3.3 MΩ Output impedance (wit rd=very ig) Z = = 2.4 KΩ Lad vltage if lad resistance is 10 KΩ L = [ L /( L +Z )] =[10K/(10K+2.4)] 384m=310m

5 Cascade amplifier BJT C cupled cascade amplifier is taken ere fr example Advantage f cascading is increase in te verall vltage gain. c bias is btained by prcedure fllwed fr single stage amplifier. Gain f eac stage: A = -( C װ L )/r e Amplifier input impedance is tat f stage 1: Zi= 1 װ 2 װ βr e Output impedance is tat f stage 2 : Z=c װ r

6 Numerical Calculate vltage gain, utput impedance, input impedance fr cascaded BJT amplifier f fig abve. Calculate utput vltage resulting if 10K ms lad is cnnected t lad. Given, 1=15KΩ; 2=4.7KΩ;c=2.2KΩ;=1KΩ C1=C2=C3=10uF β=200 fr bt transistrs nput vltage vi= 25u Slutin: c analysis yields

7 B =4.7; =4.0; C =11; =4.0mA At bias pint, re= T / =26m/4.0m=6.5 Ω ltage gain f stage 1 is ten, A1= -{C װ (1 װ 2 װ βre)}/re = /6.5= A2= -c/re = -2.2K/6.5 = Overall gain f A =A 1 A 2 = x =34,624 Output vltage is : =A i=34624 x 25u =0.866 Amplifier input impedance is Zi= 1 װ 2 װ βre =4.7K װ 15 K װ 200x6.5 =953.6 ms. L= {L/Z+L} ={10K/2.2K+10K}0.866 = 0.71

8 Cascde cnnectin A cascde cnnectin as ne transistr n tp in series wit anter. Figure belw sws C stage feeding a CB stage. Tis arrangement is designed t prvide a ig input impedance wit lw vltage gain t ensure tat te input Miller capacitance is at a minimum wit te CB stage prviding gd ig frequency peratin. Cascade cnnectin cnfiguratin fig:1

9 Cascade cnnectin cnfiguratin fig:2 Numerical Calculate te vltage gain fr te cascde amplifier f fig abve.. Slutin: c analysis: B1=4.9 ; B2=10.8; C1=c2=3.8mA ynamic resistance f eac transistr is ten re=26/3.8=6.8 ms ltage gain f stage 1 is

10 Av1= -c/re= -re/re = -1 ltage gain f stage 2 is Av2=c/re =1.8K/6.8 = 265 esulting in an verall cascde amplifier gain f Av=Av1 x Av2 =-1 x 265 =-265 C stage wit a gain f -1 prvides te iger input impedance f C stage. Wit gain f -1, miller capacitance is kept very small. A large gain is ten prvided by te CB stage, resulting in large verall gain f arlingtn cnnectin Ppular cnnectin perates as super beta transistr is arlingtn cnnectin. Main feature f te arlingtn cnnectin is tat te cmpsite acts as a single unit wit a current gains f individual transistrs. arlingtn cnnectin prvides a current gain f β= β1+ β2 f β1= β2= β ten β= β 2 Tis cnfiguratin prvides a transistr aving a very large current gain, typically a few tusands.

11 Packaged arlingtn transistr Specificatin infrmatin f 2N999 arlingtn transistr package C bias f arlingtn circuits B B B B B B cc B ltages 1) (

12 Numerical Calculate dc bias vltages and currents fr te arlingtn cnnectin. Given B=3.3MΩ;=390 Ω;βd=8000;CC=18;B=1.6 ltages B B C B ( 18 B cc 1) M 8000(390) 20.48m(390) 8 B B B uA 8000(2.56u) 20.48mA

13 AC equivalent circuit quivalent mdel

14 nput impedance Te ac base current trug ri is b=i-/ri Since =( b +β b ) Substituting b in expressin, b ri=i-=i-b(1+ β ) slving fr i, i=b[ri+(1+ β ) ]=b(ri+ β ) Ac input impedance lking int te transistr base is ten i/b= ri+ β Zi=B װ (ri+ β ) ac utput impedance f arlingtn cnnectin Tis can be determined fr ac circuit swn in fig belw

15 Output impedance Te utput impedance can be determined by applying a vltage and measuring te current wit s setting t zer. Slutin fr yields.. Z = װ r i װ r i /β i i i i B i r r r r r 1 1 i i r r Z / 1/ 1/ 1

16 ac vltage gain Gain expressin On simplificatin Numerical Fr te arlingtn pair, given =390 ms and β=8000. Calculate gain if ri=5kω b b b b b bri i ) ( ) ( ) ( 1 ) ( ) ( ) ( ) ( ri i Av r r i i i b i ] 8000 [ x K x Av

17 Feedback cncepts epending n te relative plarity f fed back signal in t te circuit, tere are tw types f feedback > Negative feedback > Psitive feedback Negative feedback results in educed gain Psitive feedback are used in scillatrs. Feedback amplifier Negative feedback circuits educes te gain ncreases input impedance Better stabilized frequency respnse Lwer utput impedance educed nise

18 Mre linear peratin Feedback cnnectin types ltage series feedback ltage sunt feedback Current series feedback Current sunt feedback Here vltage refers t small part f vltage as input t te feedback netwrk Current refers t tapping sme part f utput current trug feedback netwrk. Series refers t cnnecting feedback signal in series wit te input signal vltage. Sunt refers t cnnecting feedback signal in sunt wit te input signal vltage Series feedback cnnectins increases te input resistance Sunt feedback cnnectins decreases te input resistance.

19 ltage series feedback Af=/s ltage sunt feedback Af=/s

20 Current series feedback Af=/s Current sunt feedback Af=/s

21 Gain wit feedback Gain witut feedback is A Feedback factr β Gain wit feedback is (1+A β) Parameter ltage series ltage sunt Current series Current sunt Gain wit feedback A /i /i /i /i Feedback factr β f/v f/ f/ f/ Gain wit feedback Af /s /s /s /s ltage series feedback Wit zer feedback ten f=0 te vltage gain f amplifier stage is A=/s=/i f feedback f f is cnnected ten, i=s-f =Ai=A(s-f)=As-Af=A(s-A(β) Ten, (1+ βa)=as Overall gain wit feedback is Af=/i=A/(1+A β) Tis sws tat gain f feedback as reduced by factr (1+A β)

22 ltage sunt feedback Af=/s=A i / (i+f)=ai/(i+ βai) Af=A/(1+ βa) nput impedance wit FB ef t fig(1) i=i/zi=(s-f) / Zi = (s- β) / Zi i Zi= s- βai s=i Zi+ β A i = i Zi+ β A i Zi Zif = s/i=zi+(βa)zi=zi(1+ βa) mprved circuit features f feedback eductin in frequency distrtin Wen Aβ» 1, ten Af=A/(1+A β) 1/ β Here feedback is cmpletely resistive and tus frequency distrtin arising because f varying gain wit frequency is cnsiderably reduced. Bandwidt variatin Wen Aβ» 1, ten Af=A/(1+A β) 1/ β Terefre, ere we can see tat, practical circuits, pen lp gain drps at ig frequencies. Terefre Aβ n lnger» 1, ence Af=1/ β N lnger lds gd. Here reductin in gain as prvided imprvement in te Bandwidt. Prduct f gain and Bandwidt remains same it s a tradeff between gain and BW Gain stability fr Aβ»1,

23 Tis sws tat magnitude f relative cange in daf/a is reduced by te factr Aβ cmpared t tat witut feedback da/a Numerical f a amplifier wit gain f and feedback f β=-0.1 as a gain cange f 20% due t temperature, calculate te cange in gain f te feedback amplifier. Slutin: Af 1 A daf da 1 A da A daf Af 1 A % da A 1 20% 0.1( 1000) Practical feedback circuits ltage series feedback

24 Here part f utput vltage () is btained using a feedback netwrk f resistrs 1 and 2. Te feedback vltage f is cnnected in series wit te surce signal s. teir difference being te input signal i. Gain witut feedback A=/i=-g m L Were L =parallel cmbinatin f,,( ) Te feedback netwrk prvides a feedback factr r β= f / = - 2 / Using values f A and β in abve equatin, Af is Af A 1 A 1 if.. A 1, ten Af gml /( L 1 2 ) g m Numerical: Calculate te gain witut and wit feedback fr te FT amplifier swn in fig. circuit values are given t be 1 =80KΩ, 2 =20KΩ, =10KΩ and g m =4000uS Slutin : L=5K Ω A=-20 β=-0.2 and Af=-4

25 Series feedback cnnectin Here gain f p-amp is reduced by factr β=2/1+2 Numerical f pen lp gain f p-amp is 100,000 and feedback resistrs are 1=1.8K Ω and 2=200 Ω ten calculate te gain wit feedback. Slutin β=0.1 Af= Here Aβ>>1, Af=1/ β=1/0.1=10

26 mitter fllwer circuit Te utput vltage is als te feedback vltage in series wit te input vltage. Operatin f te circuit witut te feedback f=0 ten, A i f 1 fe b s fe ( s / s ie ) fe s Te peratin wit feedback ten prvides tat, Af s A fe / ie 1 A 1 (1)( fe / ie)

27 ie fe fe fe 1, Af 1 Current series feedback Feedback tecnique is t sample te utput current () and return a prprtinal vltage in series wit te input. t stabilizes te amplifier gain, te current series feedback cnnectin increases te input resistance. n tis circuit, emitter f tis stage as an un bypassed emitter, it effectively as current-series feedback. Te current trug results in feedback vltage tat ppses te surce signal applied s tat te utput vltage is reduced. T remve te current-series feedback, te emitter resistr must be eiter remved r bypassed by a capacitr (as is dne in mst f te amplifiers)

28 Te fig belw sws te equivalent circuit fr current series feedback Gain, input and utput impedance fr tis cnditin is, Numerical Calculate te vltage gain f te circuit.. fe ie C fe C f c s s c s vf ie fe c f ie fe ie i if ie fe ie fe s f A A A feedback wit A Z Z A Z Z A A A ;.. 1 ) (1 1 ) (1 ) ( 1 / 1

29 Wit B=470Ω,C=2.2KΩ,=510 Ω, fe=120,ie=900ω. Slutin: fe 120 A i ie f Te factr (1+Aβ) is ten, 1+(-0.085)(-510) =44.35 Te gain wit feedback is Af=/vi=A/(1+A β) =-.085/44.35 = -1.92x10e-3 ltage gain wit feedback is Avf=/s=AfC=(-1.92x10e-3)(2.2x10e3)=4.2 Witut feedback (=0) te vltage gain is

30 Av=-C/re=-2.2x10e3/7.5= ltage sunt feedback Cnstant gain p-amp circuit prvides vltage sunt feedback. ef t fig belw. Te input impedance f a ideal p-amp is taken t be infinite. Hence i=0,vi=0 and vltage gain is infinity. e., A=/i=infinity And β=f/= -1/ Tis is transfer resistance gain. ltage sunt negative feedback amplifier 1.Cnstant gain circuit 2.quivalent circuit ltage gain wit feedback, Avf s s 1 0 ( ) 1 1 1

31 ltage sunt feedback using FT quivalent circuit Wit n feedback A=/i=-g m S Te feedback factr is β= f / = -1/ F

32 Wit feedback, gain f te circuit is, Numerical Calculate vltage gain wit and witut feedback fr te circuit f FT f/b. Wit te values, gm=5ms, =5.1KΩ, s=1kω, F=20KΩ Slutin : Use abve frmulae Av=-gm=-25.5 Feedback gain Avf=-11.2 S m F F m S m F F m S S m F F S m s s s vf S m F S m s f g g g g g g A is witfeedback gain vltage g g A A A ) ( 1,.. ) )( 1/ ( 1 1

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