Eeg381 electronics iii chapter 2 - feedback amplifiers

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Eeg381 electronics iii chapter 2 - feedback amplifiers

  1. 1. Prof Fawzy IbrahimEEG381 Ch2 Feedback1 of 39EEG381 Electronics IIICHAPTER 2Feedback Amplifiers[[SedraSedra ((Ch.Ch. 88),), BoylestadBoylestad ((Ch.Ch. 1717)])]Prof. Hassan Elghitani Prof. Fawzy IbrahimElectronics and Communication DepartmentMisr International University (MIU)Prof Hassan Elghitani
  2. 2. Prof Fawzy IbrahimEEG381 Ch2 Feedback2 of 39Chapter Contents2.1 Feedback Concepts2.2The Four Basic Feedback Topologies12.2.1 Series - Shunt Feedback or Voltage Amplifiers2.2.2 shunt-series feedback or Current Amplifiers2.2.3 Series-Series Feedback or Transconductance Amplifiers2.2.4 Shunt - Shunt Feedback or Transresistance Amplifiers2.2.5 Summary of Feedback Topologies2.3 Negative Feedback Voltage Amplifiers2.3.1 Gain Calculation2.3.2 Bandwidth Extension2.3.3 Input and output Impedance2.3.4 Noise Reduction2.3.5 Advantages and Disadvantages of negative feedbackProf Hassan Elghitani
  3. 3. FeedbackPositiveRegenerative“Oscillators”NegativeDegenerative“Amplifiers”2.1 Feedback ConceptsProf Fawzy IbrahimEEG381 Ch2 Feedback3 of 39 Prof Hassan Elghitani
  4. 4. Prof Fawzy IbrahimEEG381 Ch2 Feedback4 of 392.1 Feedback Concepts• Most physical systems incorporate some form of feedback. Feedback hasbeen mentioned previously. In particular, feedback was used in op-ampcircuits as described in in chapter 1.• A typical feedback connection is shown in Fig. 2.1. The input signal, Vs, isapplied to a mixer network, where it is combined with a feedback signal, Vf.The sum or difference of these signals, Vi, is then the input voltage to theamplifier. A portion of the amplifier output, Vo, is connected to the feedbacknetwork (), which provides a reduced portion of the output as feedbacksignal to the input mixer network.Prof Hassan ElghitaniFig. 2.1 Block diagram (a) Positive feedback (b) Negative feedback amplifier.(a) (b)
  5. 5. Prof Fawzy IbrahimEEG381 Ch2 Feedback5 of 391.1 Feedback Concepts (Continued)• Depending on the relative polarity of the signal being fed back into a circuit,one may have negative or positive feedback.• Negative feedback results in decreased voltage gain, for which a number ofcircuit features are improved as summarized below.• Positive feedback drives a circuit into oscillation as in various types ofoscillator circuits that will be discussed in The next chapters.• Negative feedback results in reduced overall voltage gain, a number ofimprovements are obtained, among them being:1. Better stabilized voltage gain or desensitize the gain: that is, make thevalue of the gain less sensitive to variations in the value of circuitcomponents, such as might be caused by changes in temperature.2. Improved frequency response or extend the bandwidth of the amplifier.3. Control the input and output impedances or higher input impedance andlower output impedance.4. Reduce the effect of noise: that is, minimize the contribution to the output ofunwanted electric signals generated, either by the circuit componentsthemselves, or by extraneous interference.5. More linear operation or Reduce nonlinear distortion: that is, make theoutput proportional to the input (in other words, make the gain constant,independent of signal level).Prof Hassan Elghitani
  6. 6. 1. Series-Shunt Feedback (Voltage amplifiers)i/p mixing o/p sampling2. Shunt- Series Feedback (Current amplifiers)3. Series-Series Feedback (Transconductance amplifiers)4. Shunt-Shunt Feedback (Transresistance amplifiers)2.2 The Four Basic Feedback TopologiesProf Fawzy IbrahimEEG381 Ch2 Feedback6 of 39 Prof Hassan Elghitani
  7. 7. Prof Fawzy IbrahimEEG381 Ch2 Feedback7 of 392.2 The Four Basic Feedback Topologies• Based on the quantity to be amplified (voltage or current) and on thedesired form of output (voltage or current), amplifiers can be classified intofour categories. These categories Classified as follows:2.2.1 Series - Shunt Feedback or Voltage Amplifiers• Voltage amplifiers are intended to amplify an input voltage signal andprovide an output voltage signal. The voltage amplifier is essentially avoltage-controlled voltage source. A suitable feedback topology is thevoltage-mixing (series connection at the input ) and voltage sampling(parallel or shunt connection at the output) as shown in Fig. 2.2.• Because of the Thevenin representation of the source, the feedback signalVf should be a voltage that can be mixed with the source voltage in series.Prof Hassan ElghitaniFig. 2.2 Series - shunt feedback or voltage amplifier topology.• This topology not only stabilizes thevoltage gain but also results in ahigher input resistance (intuitively, aresult of the series connection at theinput) and a lower output resistance(intuitively, a result of the parallelconnection at the output), which aredesirable properties for a voltageamplifier.
  8. 8. Prof Fawzy IbrahimEEG381 Ch2 Feedback8 of 392.2 The Four Basic Feedback Topologies (Continued)2.2.2 shunt-series feedback or Current Amplifiers• The input signal in a current amplifier is essentially a current, and thus thesignal source is most conveniently represented by its Norton equivalent.• The output quantity of interest is current; hence the feedback networkshould sample the output current.• The feedback signal should be in current form so that it may be mixed inshunt with the source current.• Thus the feedback topology suitable for a current amplifier is the current-mixing current-sampling topology, illustrated in Fig. 2.3.Prof Hassan ElghitaniFig. 2.3 shunt-series feedback or Current amplifier topology.• Because of the parallel (or shunt)connection at the input, and theseries connection at the output, thisfeedback topology is also known asshunt series feedback.• This topology not only stabilizes thecurrent gain but also results in alower input resistance, and a higheroutput resistance, both desirableproperties for a current amplifier.
  9. 9. Prof Fawzy IbrahimEEG381 Ch2 Feedback9 of 392.2 The Four Basic Feedback Topologies (Continued)2.2.3 Series-Series Feedback or Transconductance Amplifiers• In transconductance amplifiers the input signal is a voltage and the outputsignal is a current.• It follows that the appropriate feedback topology is the voltage-mixingcurrent-sampling topology, illustrated in Fig. 2.4.Prof Hassan ElghitaniFig. 2.4 Series-series feedback or transconductance amplifier topology.
  10. 10. Prof Fawzy IbrahimEEG381 Ch2 Feedback10 of 392.2 The Four Basic Feedback Topologies (Continued)2.2.4 Shunt - Shunt Feedback or Transresistance Amplifiers• In Transresistance amplifiers the input signal is current and the outputsignal is voltage.• It follows that the appropriate feedback topology is of the current - mixingvoltage-sampling type, shown in Fig. 2.5. The presence of the parallel (orshunt) connection at both the input and the output makes this feedbacktopology also known as shunt - shunt feedback.Prof Hassan ElghitaniFig. 2.5 Shunt - shunt feedback or Transresistance amplifier topology.
  11. 11. Prof Fawzy IbrahimEEG381 Ch2 Feedback11 of 392.2 The Four Basic Feedback Topologies (Continued)2.2.5 Summary of Feedback Topologies• A summary of the gain, feedback factor, and gain with feedback of Figs. 2.2to 2,5 is provided for reference in Table 2.1.Prof Hassan Elghitani
  12. 12. Prof Fawzy IbrahimEEG381 Ch2 Feedback12 of 392.3 Negative Feedback Voltage Amplifiers2.3.1 Gain Calculation• Fig. 2.6 shows the basic structure or signal-flow diagram of a negativefeedback amplifier. Ao is the Gain of the open-loop amplifier and  is thefeedback factor.• Thus its output voltage Vo is related to the input voltage Vi by:• The output Vo is fed to the load as well as to a feedback network, whichproduces a sample of the output.Prof Hassan ElghitaniFig. 2.6 General structure of negative feedback amplifierioo VAV  (2.1)
  13. 13. Prof Fawzy IbrahimEEG381 Ch2 Feedback13 of 392.3 Negative Feedback Voltage Amplifiers (Continued)2.3.1 Gain Calculation• This sample Vf is related to Vo by the feedback factor  as:• The feedback signal Vf subtracted from the source signal Vs, which is theinput to the complete feedback amplifier, to produce the signal Vi which isthe input to the basic amplifier which is given by:• Here we note that it is this subtraction that makes the feedback negative. Inessence, negative feedback reduces the signal that appears at the input ofthe basic amplifier.• The gain of the feedback amplifier can be obtained by combining Eqs. (2.1)through (2.3) as:• The quantity Af is called the feedback gain or closed loop gain, a name thatfollows from Fig. 2.6, A0 is the open loop gain or gain without feedback,T = A0 is the loop gain, and (1+ A0) is the amount of feedback.Prof Hassan ElghitaniTAAAVVA ooosof11 of VV  (2.2)fsi VVV  (2.3)(2.4)
  14. 14. Prof Fawzy IbrahimEEG381 Ch2 Feedback14 of 392.3 Negative Feedback Voltage Amplifiers (Continued)2.3.1 Gain Calculation• For the feedback to be negative, the loop gain Ao should be positive; thatis, the feedback signal Vf should have the same sign as Vs .• Thus this resulting in a smaller difference signal Vi. Equation (2.4) indicates• that for positive Ao the gain-with-feedback, Af will be smaller than the open-loop gain Ao by the quantity 1+ Ao, which is called the amount of feedback.• In other words, the overall gain will have very little dependence on the gainof the basic amplifier, Ao, a desirable property because the gain Ao isusually a function of many manufacturing and application parameters, someof which might have wide tolerances.• Therefore, the closed-loop gain is almost entirely determined by thefeedback elements.• Equations (2.1) through (2.3) can be combined to obtain the followingexpression for the feedback signal Vf and the output of comparison circuit ormixer, Vf as:Prof Hassan Elghitanisosoosof VTAVAAVVA11ssosoi VTVAVVV1111(2.5)(2.6)
  15. 15. Prof Fawzy IbrahimEEG381 Ch2 Feedback15 of 392.3 Negative Feedback Voltage Amplifiers (Continued)2.3.1 Gain CalculationGain Desensitivity• The gain of the closed-loop amplifier is less sensitive to variation of the gainof the basic amplifier this property can be analytically established as follows:Assume that  is constant. Taking differentials of both sides of Eq. (2.4)results in:Dividing Eq. (2.7) by Eq. (2.4) yields:• which says that the percentage change in Af (due to variations in somecircuit parameter) is smaller than the percentage change in Ao by theamount of feedback.• For this reason the amount of feedback, (1 +  Ao) or (1 + T), is also knownas the Desensitivity factor.Prof Hassan Elghitani(2.7)(2.8)222)1()1()1()()1(TdAAdAAdAAdAAdA oooooooofoooooffAdATAdAAAdA)1(1)1(1
  16. 16. 2.3.1 Gain CalculationExample 2.1:For the Op Amp circuit shown in Fig. 2.7, if the op amp has infinite inputresistance (Ri = ), zero output resistance (Ro = o) and the open-loopvoltage gain Ao = 100 .(a) Find an expression for the feedback factor .(b) Find the ratio R2/R1 to obtain a closed loop voltage gain Af of 10.(c) What is the amount of feedback in decibels?(d) If the source voltage, Vs = 1 V, find the output voltage, Vo, the feedbackvoltage Vf, and the input voltage Vi.(e) If Ao decreases by 20%, what is the corresponding decrease in Af ?Solution(a) The feedback factor,  is given by:(b) From (2.4), closed loop voltage gain Af is:Prof Fawzy IbrahimEEG381 Ch2 Feedback16 of 392.3 Negative Feedback Voltage Amplifiers (Continued)Prof Hassan ElghitaniFig. 2.7 An Op Amp circuit211RRRVVofTAAAVVA ooosof11 
  17. 17. 2.3.1 Gain CalculationExample 2.1 SolutionWhen Af = 10, then(c) The amount of feedback =1 +Ao = 1 + 100 x 9x10-2 = 10 = 20 dB.(d) The output voltage, Vo = Af x Vs = 10 x 1 = 10 V.The feedback voltage Vf =  x Vo = 9x10-2 x 10 = 0.9 V.The input voltage, Vi = Vs - Vf = 1 – 0.9 V = 0.1 V(b) From (2.8), the percentage change in Af is given by:Prof Fawzy IbrahimEEG381 Ch2 Feedback17 of 392.3 Negative Feedback Voltage Amplifiers (Continued)Prof Hassan Elghitani2109100110010  x11.101112211RRRRR%44.280%20)801091(1)1(12 xxAdAAAdAoooff
  18. 18. Prof Fawzy IbrahimEEG381 Ch2 Feedback18 of 392.3 Negative Feedback Voltage Amplifiers (Continued)2.3.2 Bandwidth Extension• Consider an amplifier whose high-frequency response is characterized by asingle pole. Its gain at mid and high frequencies can be expressed as :where Ao denotes the midband gain and b is the upper 3-dBfrequency. Application of negative feedback, with a frequency-independentfactor , around this amplifier results in a closed-loop gain Af(s) given by:• Substituting for A(s) from Eq. (2.9) results, after a little manipulation, in:• Thus the feedback amplifier will have a midband gain of:• and an upper 3-dB frequency bf given by:• It follows that the upper 3-dB frequency is increased by a factor equal to theamount of feedback while midband gain is decreased by the same factor.Cascading amplifier is the solution to increase the voltage gain as in Ch. 1.Prof Hassan Elghitani(2.9)(2.10)bosAsA/1)()(1)()(1)()(sTsAsAsAsAf)1(/1)1/()1(/1)1/()(TsTAAsAAsAbooboof(2.11))1/()1/( TAAAA ooof   (2.12a))1()1( TA bobbf   (2.12b)
  19. 19. 2.3.2 Bandwidth ExtensionExample 2.2:Consider the noninverting op-amp circuit of Fig.2.8. Let the open-loop gain,Ao has a low-frequency value of 104 and a uniform 20 dB/decade rolloff athigh frequencies with a 3-dB frequency, fb of 100 Hz. If R1 = 1 kΩ and R2 =9 kΩ, for the closed-loop amplifier find:(a) The low-frequency gain Af.(b) The upper 3-dB frequency, fbf .Solution(a) From Eq. (2.12a) Af , is given by:(b) From Eq. (2.12a) fbf ) is given by:Prof Fawzy IbrahimEEG381 Ch2 Feedback19 of 392.3 Negative Feedback Voltage Amplifiers (Continued)Prof Hassan ElghitaniFig. 2.8 An Op Amp circuitVVTAAAA ooof /99.91001/10)1/()1/( 4 1.0911211RRRkHzxTfAff bobbf 1.1001001100)1()1(  1001)1()1(  TAFBofamountThe o
  20. 20. Prof Fawzy IbrahimEEG381 Ch2 Feedback20 of 392.3 Negative Feedback Voltage Amplifiers (Continued)2.3.3 Input and output ImpedancesInput Impedance• Consider the series-shunt feedback amplifier connection shown in Fig. 2.9,the input impedance can be determined as follows:• The input impedance with series feedback is increased and has the value ofthe input impedance without feedback multiplied by a factor equal to theamount of feedback (1 +  Ao) = (1 + T).Prof Hassan Elghitani(2.13)iiosiosifsiiiZVAVZVVZVVZVI iioiisiosii ZIAZIVVAVZI  )1()1( TZAZZAZIVZ ioiioiisif  
  21. 21. Prof Fawzy IbrahimEEG381 Ch2 Feedback21 of 392.3 Negative Feedback Voltage Amplifiers (Continued)Prof Hassan ElghitaniFig. 2.9 Voltage amplifier feedback connection model
  22. 22. Prof Fawzy IbrahimEEG381 Ch2 Feedback22 of 392.3 Negative Feedback Voltage Amplifiers (Continued)2.3.3 Input and output ImpedancesOutput Impedance• The series-shunt feedback amplifier connection shown in Fig. 2.9, providessufficient circuit detail to determine the output impedance with feedback.The output impedance is determined by applying a voltage, V, resulting in acurrent, I, with Vs shorted out (Vs = 0). The voltage V is then given by:For Vs = 0so thator• allows solving for the output resistance with feedback:• The output impedance with series feedback is reduced and has the value ofthe output impedance without feedback divided by a factor equal to theamount of feedback (1 +  Ao) or (1 +T).Prof Hassan Elghitani(2.14)ioo VAIZV )1()1( TZAZIVZ ooooffi VV )( VAIZVAIZV ooioo oo IZVAV  )(
  23. 23. 2.3.3 Input and output ImpedancesExample 2.3:Determine the voltage gain, Af, breakdown frequency, fbf, input impedanceZi and output impedance, Zo with feedback for voltage amplifier, shown inFig. 2.9, having Ao = 100, fb = 200 Hz, Ri = 10 kΩ and Ro = 20 kΩ for:(a)  = 0.1 (b)  = 0.5SolutionUsing Eqs. (2.12), (2.13), and (2.14), we obtain(a)(b)Prof Fawzy IbrahimEEG381 Ch2 Feedback23 of 392.3 Negative Feedback Voltage Amplifiers (Continued)Prof Hassan ElghitaniVVAAA oof /09.911/100)1/(   kHzxAobbf 2.211200)1(   kxkAZZ oiif 1101110)1(   kkAZZooof 82.11120)1( 11)1001.01()1(  xAFBofamountThe o51)1005.01()1(  xAFBofamountThe oVVAAA oof /96.151/100)1/(   kHzxAobbf 2.1051200)1(   kxkAZZ oiif 5105110)1(   16.2925120)1(kAZZooof
  24. 24. Prof Fawzy IbrahimEEG381 Ch2 Feedback24 of 392.3 Negative Feedback Voltage Amplifiers2.3.4 Noise Reduction• Negative feedback can be employed to reduce the noise or interference inan amplifier or, more precisely, to increase the ratio of signal to noise.• From Fig. 2.10 (a),the output voltage Vo is related to the input voltage Vi by:where D is the distortion generated by the amplifier.• With the negative feedback amplifier of Fig. 2.10 (b), when Vs =0, to studythe effect of noise, the output voltage Vo is related to the input voltage Vi by:• Feedback distortion < open loop distortionProf Hassan ElghitaniFig. 2.10 Voltage amplifier noise model: (a) Open loop; (b) With feedbackDVAV ioo (2.15)(a) (b))1()1( TDADdDdADVAdooioThe distortion with feedback is reduced andhas the value of distortion without feedback,D, divided by a factor equal to the amount offeedback (1 +  Ao) or (1 + T).
  25. 25. Prof Fawzy IbrahimEEG381 Ch2 Feedback25 of 392.3 Negative Feedback Voltage Amplifiers (Continued)1. Negative feedback results in reduced overall voltage gain, the solution is todo cascading of amplifiers.Eq. (2.4):2. Better stabilized voltage gain or desensitize the gain: that is, make thevalue of the gain less sensitive to variations in the value of circuitcomponents..Eq. (2.8):3. Improved frequency response or extend the bandwidth of the amplifier.Eq. (2.12):3. Higher input impedance and Lower output impedance.Eq. (2.13): Eq. (2.14):4. Reduce the effect of noise: that is, minimize the contribution to the output ofunwanted electric signals generated, either by the circuit componentsthemselves, or by extraneous interference.Eq. (2.15):Prof Hassan Elghitani2.3.5 Advantages and Disadvantages of negative feedbackTAAAVVA ooosof11 oooooffAdATAdAAAdA)1(1)1(1)1( obbf A )1()1( TZAZZ ioiif  )1()1( TZAZZ oooof)1()1( TDADdo 
  26. 26. • Fig. 2.11 illustrates two Common Emitter (CE) connected in cascade.• The input and output of the overall amplifier are ac coupled through capacitorsC1 and C5.• Bypass capacitors C2 and C4 are used to obtain maximum voltage gain from thetwo inverting amplifiers.• Interstage coupling capacitor C3 transfers the ac signals between the amplifiersbut provides isolation at dc. Thus, the individual Q-points of the transistors arenot affected by connecting the stages together.• Fig. 2.12 gives the dc equivalent circuit for the amplifiers in which the capacitorshave been removed.• The amplifier is characterized to determine its voltage gain, Av = Ao, inputresistance Ri and output resistance Ro by using the small signal models of thetransistors Q1 and Q2 by applying exactly the same procedure as discussed inElectronics II Course.Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps26 of 392.4 Feedback in Multistage BJT Amplifiers2.4.1 Two Stage ac-Coupled Amplifier StructuresProf Hassan Elghitani
  27. 27. Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps27 of 392.4 Feedback in Multistage BJT Amplifiers (Continued)Fig. 2.11 Two stage cascaded common-emitter amplifiers.2.4.1 Two Stage ac-Coupled Amplifier StructuresProf Hassan Elghitani
  28. 28. Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps28 of 392.4 Feedback in Multistage BJT Amplifiers (Continued)Fig. 2.12 dc equivalent circuit for the two stage cascaded common-emitter amplifiers:(a) dc circuit; (b) Thevenin equivalent for each section.2.4.1 Two Stage ac-Coupled Amplifier Structures(a) (b)Prof Hassan Elghitani
  29. 29. Fig. 2.13 shows the small signal model for the amplifier circuits. We note thefollowing:1. The input resistance of the amplifier, Ri is equal to the input resistance of the firststage.2. The load resistance of the first stage (RL1) is equal to the input resistance of thesecond stage, Rin2.3. The output resistance of the amplifier, Rout is equal to the output resistance of thesecond stage.4. The output voltage of the first stage (vo1) is the input to the second stage.5. The total voltage gain is given by:Therefore, the overall voltage gain is equal to the product of the gains ofindividual single transistor amplifier stages.Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps29 of 392.4 Feedback in Multistage BJT Amplifiers (Continued)2.4.2 Analysis of two stage (CE) Amplifier2111vvoososovo xAAvvxvvvvAA (2.16)(2.18)11 // rRR Bi 22 // oCo rRR  (2.17)Prof Hassan Elghitani
  30. 30. 2.4.1 Two Stage ac-Coupled Amplifier StructuresProf Fawzy IbrahimEEG284 Ch.3 Multistage Amps30 of 392.4 Feedback in Multistage BJT Amplifiers (Continued)Fig. 2.13 Small signal equivalent circuit for two Stage ac-Coupled Amplifiers211 // RRRB TCmVIg 11 111mBTgIVr CAoIVr 11 Where:432 // RRRB TCmVIg 22 222mBTgIVr CAoIVr 22 and11 // rRR Bi 22 // oCo rRR Prof Hassan Elghitani
  31. 31. • Fig. 2.14 illustrates two Common Emitter (CE) connected in cascade. Theresistors Rx and Ry represent the feedback loop that take part of the outputvoltage, Vo and feed it back to the input.• The input voltage vi is given by:• The feedback voltage vf is related to the output voltage vo by:• In order to have a negative feedback, vf must be phase with vs and vo, thereforetwo stages or any other even numbed stages are used in cascade since eachstage cause 180o shift. If it is a single stage of CE amplifier (or odd number ofstages), it would be positive feedback.• Since  is defined, we use Equs. (2.4) and (2.13) to calculate the feedback gain,Af, input resistance Ri and the output resistance Ri as already explained in thischapter.Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps31 of 392.4 Feedback in Multistage BJT Amplifiers (Continued)2.4.3 Analysis of two stage (CE) Amplifier with negative Feedbackfsebbei vvvvvv  111yxxofoyxxofRRRvvvRRRvv (2.19)(2.20)Prof Hassan Elghitani
  32. 32. Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps32 of 392.4 Feedback in Multistage BJT Amplifiers (Continued)Fig. 2.14 Two stage cascaded common-emitter amplifiers with negative feedback.2.4.3 Analysis of two stage (CE) Amplifier with negative FeedbackProf Hassan Elghitani
  33. 33. Example 2.4For the two stage identical common-emitter amplifiers shown in Fig. 2.14, letVcc = 9V, R1 = R3 = 27 kΩ, R2 = R4 = 15 kΩ, RE1 = RE2 =1.2 kΩ, Rc1 = Rc2 = 2.2kΩ. The transistors have  = 100 and VA = 100 V. If the amplifier operatesbetween a source for which Rs = 10 kΩ and a load of RL = 2 kΩ, For theamplifier without feedback (neglect Rx and Ry) do:(a) Perform the dc analysis to calculate the currents IE1, IB1, IC1, IE2, IB2 and IC2.(b) Determine the transistors small signal model parameters r1, gm1, ro1, r1, gm2and ro2.(c) Replace the transistors with their models and find the overall values of:(i) The input resistance Ri . (ii) the output resistance Ro .(iii) voltage gain Ao = AV = v0/vs (in ratio and dB).(d) If Rx = 1 kΩ and Ry = 9 kΩ, calculate the feedback factor , amplifier gain Af,input resistance Rif and output resistance Rof.Solution:(a) For dc analysis C1 through C5 are open circuit or of infinite impedance, so theresulting circuit is shown in Fig. 2.15 (a). When the voltage divider is replacedwith its Thevenin equivalent the equivalent circuit is shown in Fig. 2.15(b).Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps33 of 392.4 Feedback in Multistage BJT Amplifiers (Continued)2.4.3 Analysis of two stage (CE) Amplifier with negative FeedbackProf Hassan Elghitani
  34. 34. Example 2.4 Solution:Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps34 of 392.4 Feedback in Multistage BJT Amplifiers (Continued)(a) (b)Fig. 2.15 Two Stage Amplifier: (a) dc equivalent circuit ; (b) Thevenin equivalent circuits.2.4.3 Analysis of two stage (CE) Amplifier with negative FeedbackProf Hassan Elghitani
  35. 35. Since the amplifiers are typical, VBB and RB are given by:Since the amplifiers are typical, the currents IE, IB and IC are given by:Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps35 of 392.4 Feedback in Multistage BJT Amplifiers (Continued)VxRRRVVV CCBBBB 21.3152715921221   mARRVVIIBBEBEBBEE 94.110164.92.17.021.3)1(21  kxRRRRRRRR BB 64.915271527//21212121AIII EBB 1910194.1121 mAxIII ECC 92.194.199.021  99.010110012.4.3 Analysis of two stage (CE) Amplifier with negative FeedbackExample 2.4 Solution:Prof Hassan Elghitani
  36. 36. (b) The transistors small signal model parameters r, gm and ro. are given by:The small signal models for both transistors are shown in Fig. 2.15 (c).(c) From in Fig. 2.15 (c), we can determine the following:(i) The input resistance Ri = RB1 // r 1 = 9.64 // 1.3 = 1.15 kΩ.(ii) the output resistance Ro. = Rout1. = RC2 // r 02 = 2.21 // 39.1 = 2.11 kΩ.The input resistance of the second stage or the load of the first stage is:Rinput2 = RL1 = RB2 // r 2 = 9.64 // 1.3 = 1.15 kΩ.Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps36 of 392.4 Feedback in Multistage BJT Amplifiers (Continued)VmAxVIggTCmm /8.76102592.13121   kxgIVrrmBT3.1108.7610031121 kxIVrrCAoo 1.521092.110031212.4.3 Analysis of two stage (CE) Amplifier with negative FeedbackExample 2.4 Solution:Prof Hassan Elghitani
  37. 37. Example 2.4Solution:Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps37 of 392.4 Feedback in Multistage BJT Amplifiers (Continued)Fig. 2.15 (c) The small signal models for Two stage ac-coupled amplifierVmAgg mm /8.7621  krr 3.121  krr oo 1.52212.4.3 Analysis of two stage (CE) Amplifier with negative FeedbackProf Hassan Elghitani
  38. 38. (c) From in Fig. 2.15 (c), we can determine the following:(iii) The overall voltage gain using Eq. (3.1) is as follows:ThendBVVxvvRRgxRRRvvxvvvvAininLoutminsigininosiginsigov41.15/896.5)15.1//11.2(8.7615.11015.1)//( 11111Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps38 of 392.4 Feedback in Multistage BJT Amplifiers (Continued)2111vvoosigosigov xAAvvxvvvvA dBVVvvRRgvvAooLoutmoov 94.37/86.78)2//11.2(8.76)//(1112 dBVVxxAAA vvv 35.53/88.46486.78895.521 2.4.3 Analysis of two stage (CE) Amplifier with negative FeedbackExample 2.4 Solution:Prof Hassan Elghitani
  39. 39. (d) From Equ. (2.20) the feedback factor  is given by:From Equ. (2.4), the amplifier feedback gain Af, is given by:From Equ. (2.13) The amplifier feedback input resistance Rif, is given by:From Equ. (2.8) The amplifier feedback output resistance Rof is given by:Prof Fawzy IbrahimEEG284 Ch.3 Multistage Amps39 of 392.4 Feedback in Multistage BJT Amplifiers (Continued)2.4.3 Analysis of two stage (CE) Amplifier with negative FeedbackExample 2.4 Solution:1.0911kkkRRRyxx79.9488.4788.46411TAAAA ooof kTRARR ioiif 6.54)488.47(15.1)1()1(  4.44488.4711.2)1()1( TRARR oooof488.47)88.4641.01()1()1(  xTAFBofamountThe oProf Hassan Elghitani

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