Tuned amplifire

Introduction
Band Pass Amplifiers
Series & Parallel Resonant Circuits & their Bandwidth
Analysis of Single Tuned Amplifiers
Analysis of Double Tuned Amplifiers
Primary & Secondary Tuned Amplifiers with BJT & FET
Merits and de-merits of Tuned Amplifiers

1. TUNED AMPLIFIER Rakesh Mandiya https://in.linkedin.com/in/rakeshmandiya [email protected]

2. TUNED AMPLIFIER Introduction Band Pass Amplifiers Series & Parallel Resonant Circuits & their Bandwidth Analysis of Single Tuned Amplifiers Analysis of Double Tuned Amplifiers Primary & Secondary Tuned Amplifiers with BJT & FET Merits and de-merits of Tuned A mplifiers

3. DEFINITION:- An amplifier circuit in which the load circuit is a tank circuit such that it can be tuned to pass or amplify selection of a desired frequency or a narrow band of frequencies, is known as Tuned Circuit Amplifier.

4. Characteristics of Tuned Amplifier Tuned amplifier selects and amplifies a single frequency from a mixture of frequencies in any frequency range. A Tuned amplifier employs a tuned circuit. It uses the phenomena of resonance, the tank circuit which is capable of selecting a particular or relative narrow band of frequencies. The centre of this frequency band is the resonant frequency of the tuned circuit . Both types consist of an inductance L and capacitance C with two element connected in series and parallel.

5. RESONANCE CIRCUITS: When at particular frequency the inductive reactance became equal to capacitive reactance and the circuit then behaves as purely resistive circuit. This phenomenon is called the resonance and the corresponding frequency is called the resonant frequency.

6. Classification of Tuned Amplifier

7. CLASSIFICATION OF TUNED AMPLIFIERS Small Signal Tuned Amplifiers :- They are used to amplify the RF signals of small magnitude. They are further classified as: (a) Single Tuned Amplifiers :- In this we use one parallel tuned circuit in each stage. (b) Double Tuned Amplifiers :- In this we use two mutually coupled tuned circuits for every stage both of tuned circuits are tuned at same freq. (c) Stagger Tuned Amplifiers :- It is a multistage amplifier which has one parallel tuned circuit for every stage but tuned frequency for all stages are slightly different from each other.

8. (2) Large signal tuned amplifiers :- They are meant for amplifying large signals in which large RF power is involved & distortion level is also higher. But tuned circuit itself eliminates most of the harmonic distortion.

9. BAND PASS AMPLIFIER: An amplifier designated to pass a definite band of frequencies with uniform response. The new band pass amplifier perform both function of low noise amplifier (LNA) & band pass filter is proposed for application of 900Mhz RF Front – end in wireless receivers .

10. BAND PASS AMPLIFIER: It is having two differential stage comprising two transistor.

11. Main function of band pass filter to remove the band noise ,which also contributes to the rejection of image signals. Finally a band pass amplifier amplifies only a band of frequency which lie in bandwidth of amplifier & thus named as band pass amplifier . BAND PASS AMPLIFIER

12. BAND PASS AMPLIFIER

13. BAND PASS FILTER

14. SERIES RESONANT CIRCUIT It is the circuit in which all the resistive and reactive components are in series.

15. SERIES RESONANT LC

16. SERIES RESONANT CIRCUIT Impedance Of The Circuit: - Z = { R 2 + (X L – X c ) 2 } 1/2 Z = { R 2 + ( ωL – 1/ ωC ) 2 } 1/2 For resonant frequency:- (X L = X C ) XL = ωL = 2 π f r L X C = 1/ ωC = 1 / 2 π f r C

17. SERIES RESONANT CIRCUIT Since at resonance, X L = X c 2 π f r L = 1 / 2Пf r C f r = 1 / 2 π √LC ω r = 1 / √LC

18. RESONANCE CURVE OF SERIES RESONANT CIRCUIT :

19. QUALITY FACTOR It is voltage magnification that circuit produces at resonance is called the Q factor. Voltage Magnification = I max X L / I min R = X L / R At Resonance X L /R = X C /R ω r L / R = 1 / ω r RC

20. Thus Q r = ω r L / R = 1/ ω r C R = 2 π f r L / R = (2 π L / R) * (1 / 2 π √LC ) = √(L/C) / R = tanФ { tan Ф = power factor of coil }

21. IMPORTANT POINTS (1) Net reactance , X = 0. (2) Impedance Z = R . (3) Power factor is unity. (4) Power expended = 6 watt. Current is so large & will produce large voltage across inductance & capacitance will be equal in magnitude but opposite in phase. Series resonance is called an acceptor circuit because such a circuit accepts current at one particular frequency but rejects current at other frequencies these circuit are used in Radio – receivers .

22. REACTANCE CURVE SERIES RESONANT CIRCUIT X L = 2 Πf L X = X L - X C X C = 1 2 Πf C R f r current

23. PARALLEL OR CURRENT RESONANCE

24. PARALLEL OR CURRENT RESONANCE When an inductive reactance and a capacitance are connected in parallel as shown in figure condition may reach under which current resonance (also known as parallel or anti- resonance ) will take palace. In practice, some resistance R is always present with the inductor. Such circuit is said to be in electrical resonance when reactive(watt less) components of line current becomes zero. The frequency at which this happened is known as resonant frequency. Current will be in resonance if reactive component of R-L branch I R-L sinФ R-L = Reactive component of capacitive branch, neglecting leakage reactance of capacitor C

25. FREQUENCY V/S IMPEDANCE CURVE FOR LCR CIRCUIT

26. CURRENT AT RESONANCE Capacitive Current I c = 2 π f r CV Coil Current I R-L = V/Z = V / √ R 2 + ( ω r L ) 2 Ф R-L = Cos -1 (R/Z) Ф R-L = Sin -1 (2Пf r L/Z)

27. ( V / √ R 2 + ( ω r L ) 2 ) * ( ω r L / √ R 2 + ( ω r L ) 2 ) = ω r CV C = L / (R 2 + ( ω r L ) 2 ω r = (1 / √LC) * (( 1- CR 2 /L) ) 1/2 ω r = √(1/LC – R 2 /L 2 ) f r = ω r /2 π = (1/2 π ) * ((1/LC)- (R 2 / L 2 ))

28. RESONANCE CURVE OF PARALLEL RESONANT CIRCUIT : With low resistance With high resistance current Resonant frequency f R

29. Active component of coil I A = I R- L cosФ R -L = (V/Z) * (R/Z) = VR/Z 2 Reactive component of coil I R = I R-L sinФ = (V/Z ) * (2 π f r L /Z )

30. Since at resonance Reactive component of coil current = Capacitive current (V / Z ) * (2 π f r L / Z) = 2 π f r CV Z = √(L/C) ………..(1) Line current I L = Active component of coil current = I A = I R-L cosФ R -L = VR/Z 2 [using (1)] = VR(C/L) I L = [ V / (L/RC) ] (L/RC) = Effective or equivalent dynamic impedance of parallel circuit at resonance.

31. IMPORTANT POINTS FOR CURRENT OR PARALLEL RESONANCE: (1) Net susceptance is zero (1 / X C ) = ( X L / Z 2 ) (2) Admittance = Conductance (3) Power factor is unity as reactive ( wattles) components of the current is zero (4) Impedance is purely resistive Z Max = (L / CR) (5) I Line(Min) = V / ( L/CR ) ( in phase with applied voltage)

32. (6) f = (1/2 П) * ( √(1/LC) – (R 2 / L 2 )) Hz The frequency at which the net susceptance curve crosses the frequency axis is called the resonant frequency . At this point impedance is maximum or admittance is minimum & is equal to G , consequently (I) Line is minimum .

33. Band with of parallel resonant circuit B.W. = (f 2 – f 1 ) Quality Factor Q = X L / R = 2Пf r L / R Quality factor determines sharpness of resonance curve and selectivity of circuit. Higher the value of quality factor more selective the tuned circuit is.

34. CHARACTERSTICS OF PARALLEL OR CURRENT RESONANCE Admittance is equal to conductance. Reactive or watt less component of line current is zero hence circuit power factor is unity. Impedance is purely resistive , maximum in magnitude and is equal to L/CR. Line current is minimum and is equal to V / (L/CR) in magnitude and is in phase with the applied voltage.

35. REACTANCE CURVE PARALLEL RESONANT CIRCUIT X L = 2 Πf L X = X L - X C current 1 X C = 2 Πf C R

36. (1) SINGLE TUNED AMPLIFIER

37. (1) SINGLE TUNED AMPLIFIER O/P of this amplifier may be taken either with the help of Capacitive. A parallel tuned circuits is connected in the collector circuit. Tuned voltage amplifier are usually employed in RF stage of wireless communication , where such circuits are assigned the work of selecting the desired carrier frequency and of amplifying the permitted pass-band around the selected carrier frequency.

38. SINGLE TUNED AMPLIFIER Tuned amplifier are required to be R 1 , R 2 , & R e = For biasing & stabilization circuit. C e = By pass capacitor L-C = Tuned circuit connected in collector, the impedance of which depend upon frequency, act as a collector load. If i /p signal has same frequency as resonant frequency of L-C circuit . Large amplification will be obtain because of high impedance of L-C ckt .

39. SINGLE TUNED AMPLIFIER USING FET

40. SINGLE TUNED AMPLIFIER USING FET In the shown figure the single tuned amplifier is depicted using a field effect transistor. The value of L and C is selected as per the desired frequency level. One of the components either L or C is variable so as to adjust the variable frequency.

41. The high frequency signal to be applied between base & emitter. The resonant frequency of circuit is made equal to frequency of i/p signal by varying L or C . Now tuned ckt will offer very high impedance to the signal freq. & thus large o/p appear across it. A v = ( β R ac )/ r in { R ac = Tuned circuit impedance} = β (L/CR)/ r in A V = β L / ( CRr in ) Bandwidth = (f 2 - f 1 ) The amplifier will amplify any freq. well within this range. CIRCUIT OPERATION

42. LIMITATION This tuned amplifier are required to be highly selective. But high selectivity required a tuned circuit with a high Q- factor . A high Q- factor circuit will give a high A v but at the same time , it will give much reduced band with because bandwidth is inversely proportional to the Q- factor . It means that tuned amplifier with reduce bandwidth may not be able to amplify equally the complete band of signals & result is poor reproduction . This is called potential instability in tuned amplifier.

43. DOUBLE TUNED CIRCUIT :

44. DOUBLE TUNED CIRCUIT The problem of potential instability with a single tuned amplifiers overcome in a double tuned amplifier which consists of independently coupled two tuned circuit : (1) L 1 C 1 in collector circuit (2) L 2 C 2 in output circuit A change in the coupling of two tuned circuit results in change in the shape of frequency response . By proper adjustment of coupling between two coils of two tuned circuits, the required results are : High selectivity High voltage gain Required bandwidth

45. CIRCUIT OPERATION The resonant freq. of tuned circuit connected in collector circuit is made equal to signal freq. by varying the value of C 1 . Tuned circuit (L 1 C 1 ) Offer very high impedance to signal frequency & this large o/p is developed across it. The o/p of (L 1 C 1 ) is transferred to (L 2 C 2 ) through mutual inductance. Thus the freq. response of double tuned circuit depends upon magnetic coupling of L 1 & L 2 . Most suitable curve is when optimum coefficient of coupling exists between two tuned circuit .The circuit is then highly selective & also provides sufficient amount of gain for a particular band of frequency.

46. SHUNT PEAKED CIRCUITS FOR INCREASED BANDWIDTH For expanding bandwidth we use various combinations of BJT & FET(MOS) in series or shunt so that we can use Stagger tuned amplifiers. Shunt Peaking If a coil is placed in parallel (shunt) with the output signal path, the technique is called SHUNT PEAKING. R1 is the input-signal-developing resistor. R2 is used for bias and temperature stability. C1 is the bypass capacitor for R2. R3 is the load resistor for Q1 and develops the output signal. C2 is the coupling capacitor which couples the output signal to the next stage.

47. SHUNT PEAKED CIRCUITS FOR INCREASED BANDWIDTH :

48. STAGGER TUNED AMPLIFIERS It is a multistage amplifier which has one parallel resonant circuit for every stage, while resonant frequency of every stage is slightly different from previous stages. From circuit diagram it is clear that first stage of this amplifiers has a resonant circuit formed by L 1 & C 1 that f 1 = 1 / (2 Π √L 1 C 1 ) The o/p of stage is applied to second stage which is tuned to slightly higher frequency. f 2 = 1 / (2 Π √L 2 C 2 )

49. Second stage amplifiers the signals of frequency f 2 by maximum amplitude while other frequency signal are amplified by less quantity . Thus frequency response Curve of second stage has a peak of f 2 which is slightly higher than f 1 .

50. STAGGER TUNED AMPLIFIERS :

51. Over all response Freq. response of first stage Freq. response of second stage f 1 f f 2 Voltage Frequency STAGGER TUNED AMPLIFIER

52. STAGGER TUNED AMPLIFIERS Over all response of these two stage is obtained by combining individual response & it exhibits a maximum flatness around the center frequency f . Thus overall bandwidth is better than individual stage. Since two stages are in parallel (shunt) & overall bandwidth is increased thus, it behaves like shunt circuits for the increased bandwidth.

53. LARGE SIGNAL (NARROW BAND AMPLIFIER) TUNED AMPLIFIER Single & double stage amplifier are not suitable for applications involving larger RF power , because of lower of efficiency of class A operation (single double) such as for excitation of transmitting antenna. For such application larger signal tuned amplifier are employed because they are operation in class C operation that has high efficiency & capable of delivering more power in comparison to that of class A operation .

54. CIRCUIT DIAGRAM OF LARGE SIGNAL (NARROW BAND AMPLIFIER) TUNED AMPLIFIER :

55. The resonant tuned circuit is tuned to freq. of i /p signal . When circuit has a high Q- factor , parallel resonance occur approximate freq. : f = 1 / (2 π √ LC) At resonant freq. the impedance of parallel circuit is very large & purely resistive. LARGE SIGNAL (NARROW BAND AMPLIFIER) TUNED AMPLIFIER

56. LARGE SIGNAL (NARROW BAND AMPLIFIER) TUNED AMPLIFIER Higher the Q of circuit faster gain drops on either side of resonance freq. A large Q leads to small bandwidth equal top sharp tuning this amplifier has Q>> 10,This means Bandwidth is less than 10% of f r & for this reason , it is called as narrow band amplifier.

57. COMPARISON BETWEEN TUNED AND AF AMPLIFIER It has to amplify narrow band of frequencies defined by the tuned load at the collector They are bulky and costlier Used in radio transmitters and receivers, and television receivers circuits . Works with a complete audio frequency range More compact Amplifies sound signals and act as drive for loud speakers Tuned Amplifier AF Amplifier

58. APPLICATIONS OF TUNED AMPLIFIER Tuned amplifiers serve the best for two purposes: a)Selection of desired frequency. b)Amplifying the signal to a desired level. USED IN : Communication transmitters and receivers. In filter design :--Band Pass, low pass, High pass and band reject filter design.

59. ADVANTAGES It provides high selectivity. It has small collector voltage. Power loss is also less. Signal to noise ratio of O/P is good. They are well suited for radio transmitters and receivers .

60. DISADVANTAGES They are not suitable to amplify audio frequencies. If the band of frequency is increase then design becomes complex. Since they use inductors and capacitors as tuning elements, the circuit is bulky and costly.

61. Numerical

62. Q-1) A single tuned amplifier consist of tuned circuits having R=5ohm,L=10mh,c=0.1mf. Determine a)resonant frequency. b)quality factor of tank circuit c)band width of amplifier Ans - Given data-: R=50ohm; L=10mh; C = 0.1mf

63. We know: Resonant frequency = 1/ 2 π √ [ (1/LC) – (R 2 / L 2 ) ] = 1 / 2 π √ [(1/10*10 -3 - 25/100*10 -6 ) ] = 5.034 KHz Q = 2 π * 5.034 * 10 -3 * 10 * 10 -3 / 5 = 63.227 BW = F R /Q = 5.034 / 63.227 = 79.62KHz RESULT:- F r = 5.034 KHz Q = 63.227 BW = 79.62KHz

64. Q2.In a class c amplifier ckt C=300pf,L=50mH,R=40ohm, R L =4M ohm. Determine:- a)Resonant frequency b)D.C load c)A.C load d)Quality factor Ans :- Given: C=300pf L=50mH R=40ohm R L =4M ohm

65. F r = 1/2 π √ LC = 1/2 π √ (50 * 10 -6 * 300 * 10 -12 ) = 1.3 MHz R dc = 40 ohm X L = 2 π f r L = 2* 3.14 * 1.3 * 10 6 * 50 * 10 -6 = 408.2 Q dc = 408.2/ 40 = 10.205

66. R ac = R p ll R L R p = Q dc * X L = 10.205 * 408.2 = 4165.681 R L = 4* 10 6 R ac = 4161.34 Q ac = [ R ac / X L ] = 4161.34/408.2 = 10.194 Result: - F r = 1.3 MHz Q dc = 10.205 Q ac = 10.194

67. Q. A circuit is resonant resonant at 455 khz and has a 10khz bandwidth. The inductive reactance is 1255ohm. What is the parallel impedance of the circuit at resonance? Solution : Given that: f r =455 khz Frequency BW=10khz and X L Let z p be the value of impedance at resonance We know that the value of bandwidth (BW)= f r /Q So, 10*10 3 =455*10 3 /Q Q=45.5 Q=X L /R X L =1255 So , R=1255/45.5=27.6 ohm

68. 1255=2∏f r L=2859*10 3 L L=1255/(2859*10 3 ) L=.439*10 3 H Value of capacitance reactance at resonance: X C =X L =1255 Ω 1/2∏f r = 1255 Therefore, C=278.7*10 -12 F And value of circuit impedance at resonance z p =L/CR =0.439*10 3 / (278.7*10 -12 )*27.6 =57*10 3 =57 k Ω RESULT: the parallel impedance at resonance is 57k Ω .

69. Q: A FET has gm=6 mA /v, has a tuned anode load consisting of a 400 microH inductance of 5 ohm in parallal with a capacitor of 2500pf. Find:- The resonant frequency Tuned circuit dynamic resistance Gain at resonance The signal bandwidth Solution:- 1. F r => resonant frequency = 1 / (2 π√ (LC)) = 0.159/ √ (400*2500) = 1.59*10 5 Hz = 0.159 MHz 2. R d => tuned circuit dynamic resistance = L/CR = 400/(2500*5)

70. = 10 6 * 80/2500 = 0.032 * 10 6 = 32 k ohm 3. A v = -g m r d = 6*32 = -192 4. BW= f r / Q Q = W r L/ R = (2 π * 0.159 * 10 -6 * 400 * 10 6) / 5 = 79.92 BW = 0.159/ 79.92 = 1.98 KHz RESULT: Resonant frequency=0.159MHz Dynamic Resistance=32k Ω Resonance gain=-192 Bandwidth=1.98khz

71. Q: A tuned voltage amplifier, using FET with r d = 100 k ohm and g m = 500 micro s has tuned circuit, consisting of L= 2.5 mH , C = 200 pF , as its load. At its resonant frequency , the circuit offers an equivalent shunt resistance of 100 kohm . For the amplifier determine:- Resonant gain The effective Q The Bandwidth Solution: - Given that: g m= 500 Shunt resistance=100 kohm Resonant gain:- A v = -g m (r d ll R d )/(1 + jf / f r ) A v = 500(100 ll 100)/(1 + j1) A v = 17.68

72. 2. Effective Q: - Q eff = L/CR R = 100 ll 100 Q = 2.5* 10 3 /(200*10 -12 * 50 * 10 3 ) = 2.5 * 10 2 3. Bandwidth: - BW = f r / Q eff f r = 1/2 π √ LC = 225 kHz BW = 225/ 2.5* 10 2 = 900 Hz RESULT: Resonant gain=17.68 Q eff =2.5*10 2 BW = 900 Hz

73. Submitted By: Rakesh mandiya