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Power Amplifiers

Text Book:
[1] Jacob Millman, Cristos C. Halkias, “Integrated
Electronics: Analog and Digital Circuits and Systems”,
34th Reprint 2004, Tata-McGraw-Hill. Chapter: 18
[2] Robert Boylestad and Louis Nashelsky, “Electronic
Devices and Circuit theory”, Fifth Edition, Prentice-hall
of India Private Limited 1994. Chapter:16
[3] Albert Paul Malvino, “Electronic Principles”, Tata
McGraw-Hill, 1999.      Chapter: 11
Amplifier:
An amplifier is an electronic device that increases voltage,
  current or power of a signal.
According to the class of operation, the amplifiers can be
  classified as:
• Class A, (ii) Class B, (iii) Class AB, and (iv) Class C.


Class A: A class A
amplifier is one in which
the operating point and the
input signal are such that
the current in the output      The output signal of class
circuit flows at full times.   A amplifiers varies for a
                               full 360o of the cycle.
Class B: A class B amplifier is one in which the operating point is at
an extreme end of its characteristic, so that the quiescent power is
very small. If the signal voltage is sinusoidal, amplification takes
place for only one-half a cycle.
A class B circuit provides an output signal varying over one-half the
input signal cycle, of for 180o of input signal.
Class AB: A class AB amplifier is one operating between the two
extremes defined for class A and class B. Hence the output signal is
zero for part but less than one-half of an input sinusoidal signal.
For class AB operation the output signal swing occurs between 180o
and 360o and is neither class A nor class B operation.
Class C:      A class C amplifier is one in which the
operating point is chosen so that the output current (or
voltage) is zero for more than one-half of the input
sinusoidal signal cycle.
The output of a class C amplifier is biased for operation at
less than 180o of the cycle and will operate only with a
tuned (resonant) circuit which provides a full cycle of
operation for the tuned or resonant frequency.
Series-Fed Class A Amplifier
    The fixed-bias circuit connection shown in Fig. 16.1 can be used
to discuss the main features of a class A amplifier.
    The beta (β) of power transistor is generally less than 100, the
overall amplifier circuit using power transistors being capable of
handle large power or current not providing much voltage gain.
DC Bias
The dc bias set by VCC and RB fixes the dc base-bias current at
                      VCC − VBE
                IB =                  (16.1)
                          RB
with the collector current then being
                         IC = βI B      (16.2)
with the collector-emitter voltage then
                VCE = VCC − I C RC (16.3)
To appreciate the importance of the dc bias the operation of the power
amplifier, consider the collector characteristic shown in Fig. 16.3.
A dc load line is drawn using the values of VCC and RC.
The intersection of the dc bias value of IB with the dc load line then
determines the operating point (Q-point) for the circuit.
The quiescent-point values are those calculated using Eqs. (16.1)
through (16.3).
If the dc bias collector current is set at
one-half the possible signal swing
(between 0 and VCC/RC) the largest
collector current swing will be possible.
Additionally, if the quiescent collector-
emitter voltage is set at one-half the
supply voltage, the largest voltage
swing (between 0 and VCC) will be
possible.
AC Operation
When an input ac signal is applied to the amplifier of Fig. 16.1, the
output will vary from its dc bias operating voltage and current.
A small input signal, as shown in Fig. 16.4(a), will cause the base
current to vary above and below the dc bias point, which will then
cause the collector current (output) to vary from the dc bias point set as
well as the collector-emitter voltage to vary around its dc bias value.

Power Considerations
The input power with no input
signal drawn from the supply is
  Pi (dc) = VCC I CQ   (16.4)
Even with an ac signal applied, the
average current drawn from the
supply remains the same.
So Pi(dc) represent the input power.
Output Power
The output voltage and current varying around the bias point provide
ac power to the load.
The ac power is delivered to the load RC, in the circuit of Fig. 16.1 is
given by

Using rms value :
Po (ac) = VCE (rms) I C (rms)   (16.5a)
            2                             Using peak value :
Po (ac) = I C (rms) RC          (16.5b)
                                                    VCE (p) I C (p)
             2                            Po (ac) =                   (16.5a )
          VCE (rms)                                         2
Po (ac) =                       (16.5c)               2
               RC                                   I C (p) RC
                                          Po (ac) =                    (16.5b)
                                                         2
                                                      2
                                                    VCE (p)
                                          Po (ac) =                    (16.5c)
                                                      2 RC
Using peak - to - peak value :
                                       V (p - p) I C (p - p)
                             Po (ac) = CE                     (16.5a )
                                                   8
                                         2
                                       I C (p - p) RC
                             Po (ac) =                       (16.5b)
                                              8
                                         2
                                       VCE (p - p)
                             Po (ac) =                       (16.5c)
Efficiency                                 8 RC
The efficiency of an amplifier represents the amount of ac power
delivered (transferred) from the dc source.
The efficiency of the amplifier is calculated using
       Po (ac)
 %η =          × 100%     (16.8a )
       Pi (dc)
                                       2
                                     VCE (p - p)
                            %η =                      × 100%      (16.8b)
                                   8RC × VCC I CQ
Maximum Efficiency
For the class A series amplifier the maximum efficiency can be
determined using the maximum voltage and current swings.
For the voltage swing it is   Maximum   VCE (p − p) = VCC
                                                       VCC
                              Maximum   I C ( p − p) =
                                                        RC

Using the maximum voltage swing in Eq. (16.7c) yields

                                                        2
                                                      VCC
                                Maximum Po (ac) =
                                                      8 RC
The maximum power input can be calculated using the dc bias
current set to half the maximum value.
                                                   2
                                                 VCC
Thus                                   V    /2
             Maximum Pi (dc) = VCC CC          =
                                         RC      2 RC
We can then use Eq. (16.8) to calculate the maximum efficiency:
                                               2
                                             VCC 2 RC
                 maximum Po (ac)
 maximum %η =                       × 100% =     ×      × 100%
                 maximum Pi (dc)             8 RC V 2
                                                   CC
            2
           = × 100% = 25%
            8


 The maximum efficiency of a
 class A amplifier is thus seen to
 be 25%.
 A form of class A amplifier
 having maximum efficiency of
 50% uses a transformer to couple
 the output signal to the load as
 shown in Fig. 16.6.
Example 16.1 Calculate the (i) base current, collector current and
collector to emitter voltage for the operating (or Q) point (ii) input
power, output power, and efficiency of the amplifier circuit in the
following figure for an input voltage in a base current of 10 mA peak.
Solution:
(i) the Q-point can be determined by DC
analysis as:

        V − 0.7V 20V − 0.7V
    I = CC      =           =19.3 mA
     B    R        1 KΩ
      Q    B

            I = βI = 25(19.3mA) = 0.48 A
             C    B

        V    = 20V − (0.48A)(20Ω) =10.4 V
         CE
           Q
(ii) When the input ac base current increases from its dc
bias level, the collector current rises by
           I (p) = βI (p) = 25(10mA) = 250 mA peak
            C        B

                 I 2 (p) (250×10− 3)2
         Po(ac) = C Ro =              20 = 0.625 W
                    2         2

            P (dc) =V I    = (20V)(0.48A) = 9.6W
             i       CC C
                         Q

                   Po (ac)         0.625W
              %η =         ×100% =        = 6.5%
                   P (dc)           9.6W
                    i
Class B Amplifier
Class B operation is provided when the dc bias transistor biased just
off, the transistor turning on when the ac signal is applied.
This is essentially no bias and the transistor conducts current for only
one-half of the signal cycle.
To obtain output for the full cycle of signal, it is necessary to use two
transistors and have each conduct on opposite half-cycles, the
combined operation providing a full cycle of output signal.

Since one part of the circuit pushes
the signal high during one-half
cycle and the other part pulls the
signal low during the other half-
cycle, the circuit is referred to as a
push-pull circuit.

     Figure 16.12 shows a diagram for push-pull operation.
The arrangement of a push-pull circuit is shown in Fig. 18-8.
When the signal on transistor Q1 is positive, the signal on Q2 is
negative by an equal amount.
During the first half-cycle of operation, transistor Q1 is driven into
conduction, whereas transistor Q2 is driven off. The current i1 through
the transformer results in the first half-cycle of signal to the load.
During the second half-cycle of operation, transistor Q2 is driven into
conduction, whereas transistor Q1 is driven off. The current i2 through
the transformer results in the second half-cycle of signal to the load.
The overall signal developed across the load then varies over the full
cycle of signal operation.
Consider an input signal (base current) of the form ib1=Ibmcosωt applied
to Q1.
The output current of this transistor is given by Eq. (18.33):
        i = I + B + B cosωt + B cos 2ωt + B cos3ωt + ........... (18.33)
        1 C 0 1                  2           3
The corresponding input signal to Q2 is
                                 i = −i = I cos(ωt + π )
                                  b2 b1 bm
 The output current of this transistor is obtained by replacing ωt by
 ωt+π in the expression for ib1. That is,
                         i = i (ωt + π ) (18.34)
                          2 1
 Hence,
 i = I + B + B cos(ωt + π ) + B cos 2(ωt + π ) + B cos3(ωt + π ) + ...........
  2 C 0 1                       2                 3
     which is
     i = I + B − B cosωt + B cos 2ωt − B cos3ωt + ........... (18.35)
      2 C 0 1               2           3
The total output current is then proportional to the
difference between the collector current in the two
resistors. That is,

    i = k (i − i ) = 2k (B cosωt + B cos3ωt + ...........) (18.36)
           1 2            1         3

This expression shows that a push-pull circuit will balance
out all even harmonics in the output and will leave the
third-harmonic tram is the principle source of distortion.
This conclusion was reached on the assumption that the
two transistors are identical. If their characteristics differ
appreciably, the appearance of even harmonics must be
expected.
The fact that the output current contains no even-
harmonic.
Advantages of Push-Pull System
1) Push-pull circuit give more output per active device: Because no
even harmonics are present in the output of a push-pull amplifier, such
a circuit will give more output per active device for a given amount of
distortion. For the same reason, a push-pull arrangement may be used
to obtain less distortion for a given power output per transistor.
2) Eliminates any tendency toward core saturation and nonlinear
distortion of transformer: The dc components of the collector
current oppose each other magnetically in the transformer core. This
eliminates any tendency toward core saturation and consequent
nonlinear distortion that might arise from the curvature of the
transformer-magnetizing curve.
3) Ripple voltage will not appear in the load: The effects of ripple
voltages that may be contained in the power supply because of
inadequate filtering will be balance out. This cancellation results
because the currents produced by this ripple voltage are in oppose
directions in the transformer winding, and so will not appear in the
load.
Class B amplifier operation
The transistor circuit of Fig. 18-8 operates class B if R2=0 because a
  silicon transistor is essentially at cutoff if the base is shorted to the
  emitter.

Advantages of Class B as compared with class A operation
2. It is possible to obtain greater power output,
3. The efficiency is higher, and
4. There is negligible power loss at no signal.


Disadvantages of Class B as
   compared      with     class  A
   operation
2. The harmonic distortion is
   higher,
3. Self-bias cannot be used, and
4. The supply voltage must have
   good regulation.
The graphical construction from which to determine the output current
and voltage waveshapes for a single transistor operating as a class B
stage is indicated in Fig. 18-9.
Input power: The power supplied to the load by an amplifier is
drawn from the power supply which provides the input or dc power.
The amount of this power can be calculated using
                                           Pi (dc) = VCC I dc (16.17)
In class B operation the current drawn from a single power supply
has the same form of a full-wave rectified signal. So,
                     2                                                   2
        I        =     I ( p ) (16.18)       and Pi (dc) = VCC             I (p) (16.19)
            dc       π                                                   π
Output Power: The output across the load is given by
                                     2
                                   VL (rms)
                         Po (ac) =           (16.20)
                                      RL
                                            2
                                           VL ( p )        2
                                                          VL ( p - p )
                               Po (ac) =              =                  (16.21)
                                            2 RL             8 RL
Efficiency: The efficiency of the class B amplifier can be calculated
using the basic equation
                             Po (ac)
                        %η =         × 100%
                             Pi (dc)

                         2
                        VL ( p )   π
                 %η =                        ×100%
                         2 RL 2VCC I ( p )
                                    V ( p)
                      using, I (p) = L
                                     RL
                      2
                     VL ( p )
                           π   RL
               %η =                  ×100%
                    2 RL 2VCC VL (p)

                        π VL ( p )
                 %η =              ×100% (16.22)
                        4 VCC
Maximum Efficiency: Equation (16.22) shows that the larger the
peak voltage, the higher the circuit efficiency, up to a maximum value
when the , this maximum efficiency then being
                    VCC                                2 VCC
   Maximum, I (p) =                    Maximum, I dc =
                     RL                                π RL
                                      2
                                    2VCC                         2
                                                               VCC
                           2 VCC
Maximum,     Pi (dc) = VCC       =          Maximum, Po (ac) =
                           π RL    πRL                         2 RL


                        2
                      VCC πRL    π
        Maximum, %η =     ×     = ×100% = 78.5%
                      2 RL 2V 2  4
                             CC

   The maximum efficiency of a class B amplifier is thus seen to be
   78.5%.
Crossover distortion
The distortion caused by the nonlinear transistor input characteristic is
indicated in Fig. 18-12. The iB-vB curve for each transistor is drawn,
and the construction used to obtain the output current (assumed
proportional to the base current) is shown.
In the region of small currents (for vB<Vγ) the output is much smaller
than it would be if the response were linear. This effect is called
crossover distortion.

In order to overcome the problem
of crossover distortion, the
transistor must operate in a class
AB mode.
(16.10) Calculate (i) the input power, (ii) the output power, (iii) the
power handled by each transistor, and (iv) the efficiency for an
input of 12 V (rms) of a push-pull Class-B amplifier circuit having
biasing voltage, VCC= 25 V and load resistance, RL= 4 Ω.
Solution: Given, Vi (rms)=12V, VCC= 25 V, load resistance, RL= 4 Ω.
             Vi ( p ) = 2Vi (rms ) = 2 × 12V ≈ 17 V
Since, the operating point is selected in cutoff region, thus
Vi(p)=VL(p)=17V
                              VL ( p ) 17 V
                   I L ( p) =         =     = 4.25A
                               RL       4Ω
                      2           2
                I dc = I L ( p ) = × 4.25A = 2.71A
                      π           π
            Pi (dc) = VCC I dc = 25V × 4.25A = 67.75W
2 ( p)
                      VL         (17 V )2
            Po (ac) =          =          = 36.125W
                       2 RL       2 × 4Ω


Power dissipated by each output transistor is:


            P2Q Pi − Po 67.75 − 36.125
       PQ =    =       =               = 15.8 W
             2     2           2


                  Po        36.125
  Efficiency, %η = × 100% =        × 100% = 53.3%
                  Pi         67.75
Class C Operation
With class B, we need to use a push-pull arrangement. That’s
why almost all class B amplifiers are push-pull amplifiers.
With class C, we need to use a resonant circuit for the load.
This is why almost all class C amplifiers are tuned amplifiers.
With class C operation, the collector current flows for less
than half a cycle.
A parallel resonant circuit can filter the pulses of collector
current and produce a pure sine wave of output voltage.
The main application for class C is with tuned RF amplifiers.
The maximum efficiency of a tuned class C amplifier is 100
percent.
Figure 11-13a shows a tuned RF amplifier.
The ac input voltage drives the base, and an amplified
output voltage appears at the collector.
The amplified and inverted signal is then capacitively
coupled to the load resistance.
Because of the parallel resonant circuit, the output voltage is
maximum at the resonant frequency, given by:



            1
   fr =            (11.14)
          2π LC
On either side of the resonant frequency fr, the voltage gain
drops as shown in Fig. 11-13b.
For this reason, a tuned class C amplifier is always intended
to amplify a narrow band of frequencies.
This makes it ideal for amplifying radio and television
signals because each station or channel is assigned a narrow
band of frequencies on both sides of a center frequency.
END OF
 POWER
AMPLIFIER

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Power amplifiers

  • 1. Power Amplifiers Text Book: [1] Jacob Millman, Cristos C. Halkias, “Integrated Electronics: Analog and Digital Circuits and Systems”, 34th Reprint 2004, Tata-McGraw-Hill. Chapter: 18 [2] Robert Boylestad and Louis Nashelsky, “Electronic Devices and Circuit theory”, Fifth Edition, Prentice-hall of India Private Limited 1994. Chapter:16 [3] Albert Paul Malvino, “Electronic Principles”, Tata McGraw-Hill, 1999. Chapter: 11
  • 2. Amplifier: An amplifier is an electronic device that increases voltage, current or power of a signal. According to the class of operation, the amplifiers can be classified as: • Class A, (ii) Class B, (iii) Class AB, and (iv) Class C. Class A: A class A amplifier is one in which the operating point and the input signal are such that the current in the output The output signal of class circuit flows at full times. A amplifiers varies for a full 360o of the cycle.
  • 3. Class B: A class B amplifier is one in which the operating point is at an extreme end of its characteristic, so that the quiescent power is very small. If the signal voltage is sinusoidal, amplification takes place for only one-half a cycle. A class B circuit provides an output signal varying over one-half the input signal cycle, of for 180o of input signal. Class AB: A class AB amplifier is one operating between the two extremes defined for class A and class B. Hence the output signal is zero for part but less than one-half of an input sinusoidal signal. For class AB operation the output signal swing occurs between 180o and 360o and is neither class A nor class B operation.
  • 4. Class C: A class C amplifier is one in which the operating point is chosen so that the output current (or voltage) is zero for more than one-half of the input sinusoidal signal cycle. The output of a class C amplifier is biased for operation at less than 180o of the cycle and will operate only with a tuned (resonant) circuit which provides a full cycle of operation for the tuned or resonant frequency.
  • 5. Series-Fed Class A Amplifier The fixed-bias circuit connection shown in Fig. 16.1 can be used to discuss the main features of a class A amplifier. The beta (β) of power transistor is generally less than 100, the overall amplifier circuit using power transistors being capable of handle large power or current not providing much voltage gain. DC Bias The dc bias set by VCC and RB fixes the dc base-bias current at VCC − VBE IB = (16.1) RB with the collector current then being IC = βI B (16.2) with the collector-emitter voltage then VCE = VCC − I C RC (16.3)
  • 6. To appreciate the importance of the dc bias the operation of the power amplifier, consider the collector characteristic shown in Fig. 16.3. A dc load line is drawn using the values of VCC and RC. The intersection of the dc bias value of IB with the dc load line then determines the operating point (Q-point) for the circuit. The quiescent-point values are those calculated using Eqs. (16.1) through (16.3). If the dc bias collector current is set at one-half the possible signal swing (between 0 and VCC/RC) the largest collector current swing will be possible. Additionally, if the quiescent collector- emitter voltage is set at one-half the supply voltage, the largest voltage swing (between 0 and VCC) will be possible.
  • 7. AC Operation When an input ac signal is applied to the amplifier of Fig. 16.1, the output will vary from its dc bias operating voltage and current. A small input signal, as shown in Fig. 16.4(a), will cause the base current to vary above and below the dc bias point, which will then cause the collector current (output) to vary from the dc bias point set as well as the collector-emitter voltage to vary around its dc bias value. Power Considerations The input power with no input signal drawn from the supply is Pi (dc) = VCC I CQ (16.4) Even with an ac signal applied, the average current drawn from the supply remains the same. So Pi(dc) represent the input power.
  • 8. Output Power The output voltage and current varying around the bias point provide ac power to the load. The ac power is delivered to the load RC, in the circuit of Fig. 16.1 is given by Using rms value : Po (ac) = VCE (rms) I C (rms) (16.5a) 2 Using peak value : Po (ac) = I C (rms) RC (16.5b) VCE (p) I C (p) 2 Po (ac) = (16.5a ) VCE (rms) 2 Po (ac) = (16.5c) 2 RC I C (p) RC Po (ac) = (16.5b) 2 2 VCE (p) Po (ac) = (16.5c) 2 RC
  • 9. Using peak - to - peak value : V (p - p) I C (p - p) Po (ac) = CE (16.5a ) 8 2 I C (p - p) RC Po (ac) = (16.5b) 8 2 VCE (p - p) Po (ac) = (16.5c) Efficiency 8 RC The efficiency of an amplifier represents the amount of ac power delivered (transferred) from the dc source. The efficiency of the amplifier is calculated using Po (ac) %η = × 100% (16.8a ) Pi (dc) 2 VCE (p - p) %η = × 100% (16.8b) 8RC × VCC I CQ
  • 10. Maximum Efficiency For the class A series amplifier the maximum efficiency can be determined using the maximum voltage and current swings. For the voltage swing it is Maximum VCE (p − p) = VCC VCC Maximum I C ( p − p) = RC Using the maximum voltage swing in Eq. (16.7c) yields 2 VCC Maximum Po (ac) = 8 RC The maximum power input can be calculated using the dc bias current set to half the maximum value. 2 VCC Thus V /2 Maximum Pi (dc) = VCC CC = RC 2 RC
  • 11. We can then use Eq. (16.8) to calculate the maximum efficiency: 2 VCC 2 RC maximum Po (ac) maximum %η = × 100% = × × 100% maximum Pi (dc) 8 RC V 2 CC 2 = × 100% = 25% 8 The maximum efficiency of a class A amplifier is thus seen to be 25%. A form of class A amplifier having maximum efficiency of 50% uses a transformer to couple the output signal to the load as shown in Fig. 16.6.
  • 12. Example 16.1 Calculate the (i) base current, collector current and collector to emitter voltage for the operating (or Q) point (ii) input power, output power, and efficiency of the amplifier circuit in the following figure for an input voltage in a base current of 10 mA peak. Solution: (i) the Q-point can be determined by DC analysis as: V − 0.7V 20V − 0.7V I = CC = =19.3 mA B R 1 KΩ Q B I = βI = 25(19.3mA) = 0.48 A C B V = 20V − (0.48A)(20Ω) =10.4 V CE Q
  • 13. (ii) When the input ac base current increases from its dc bias level, the collector current rises by I (p) = βI (p) = 25(10mA) = 250 mA peak C B I 2 (p) (250×10− 3)2 Po(ac) = C Ro = 20 = 0.625 W 2 2 P (dc) =V I = (20V)(0.48A) = 9.6W i CC C Q Po (ac) 0.625W %η = ×100% = = 6.5% P (dc) 9.6W i
  • 14. Class B Amplifier Class B operation is provided when the dc bias transistor biased just off, the transistor turning on when the ac signal is applied. This is essentially no bias and the transistor conducts current for only one-half of the signal cycle. To obtain output for the full cycle of signal, it is necessary to use two transistors and have each conduct on opposite half-cycles, the combined operation providing a full cycle of output signal. Since one part of the circuit pushes the signal high during one-half cycle and the other part pulls the signal low during the other half- cycle, the circuit is referred to as a push-pull circuit. Figure 16.12 shows a diagram for push-pull operation.
  • 15. The arrangement of a push-pull circuit is shown in Fig. 18-8. When the signal on transistor Q1 is positive, the signal on Q2 is negative by an equal amount. During the first half-cycle of operation, transistor Q1 is driven into conduction, whereas transistor Q2 is driven off. The current i1 through the transformer results in the first half-cycle of signal to the load. During the second half-cycle of operation, transistor Q2 is driven into conduction, whereas transistor Q1 is driven off. The current i2 through the transformer results in the second half-cycle of signal to the load. The overall signal developed across the load then varies over the full cycle of signal operation.
  • 16. Consider an input signal (base current) of the form ib1=Ibmcosωt applied to Q1. The output current of this transistor is given by Eq. (18.33): i = I + B + B cosωt + B cos 2ωt + B cos3ωt + ........... (18.33) 1 C 0 1 2 3 The corresponding input signal to Q2 is i = −i = I cos(ωt + π ) b2 b1 bm The output current of this transistor is obtained by replacing ωt by ωt+π in the expression for ib1. That is, i = i (ωt + π ) (18.34) 2 1 Hence, i = I + B + B cos(ωt + π ) + B cos 2(ωt + π ) + B cos3(ωt + π ) + ........... 2 C 0 1 2 3 which is i = I + B − B cosωt + B cos 2ωt − B cos3ωt + ........... (18.35) 2 C 0 1 2 3
  • 17. The total output current is then proportional to the difference between the collector current in the two resistors. That is, i = k (i − i ) = 2k (B cosωt + B cos3ωt + ...........) (18.36) 1 2 1 3 This expression shows that a push-pull circuit will balance out all even harmonics in the output and will leave the third-harmonic tram is the principle source of distortion. This conclusion was reached on the assumption that the two transistors are identical. If their characteristics differ appreciably, the appearance of even harmonics must be expected. The fact that the output current contains no even- harmonic.
  • 18. Advantages of Push-Pull System 1) Push-pull circuit give more output per active device: Because no even harmonics are present in the output of a push-pull amplifier, such a circuit will give more output per active device for a given amount of distortion. For the same reason, a push-pull arrangement may be used to obtain less distortion for a given power output per transistor. 2) Eliminates any tendency toward core saturation and nonlinear distortion of transformer: The dc components of the collector current oppose each other magnetically in the transformer core. This eliminates any tendency toward core saturation and consequent nonlinear distortion that might arise from the curvature of the transformer-magnetizing curve. 3) Ripple voltage will not appear in the load: The effects of ripple voltages that may be contained in the power supply because of inadequate filtering will be balance out. This cancellation results because the currents produced by this ripple voltage are in oppose directions in the transformer winding, and so will not appear in the load.
  • 19. Class B amplifier operation The transistor circuit of Fig. 18-8 operates class B if R2=0 because a silicon transistor is essentially at cutoff if the base is shorted to the emitter. Advantages of Class B as compared with class A operation 2. It is possible to obtain greater power output, 3. The efficiency is higher, and 4. There is negligible power loss at no signal. Disadvantages of Class B as compared with class A operation 2. The harmonic distortion is higher, 3. Self-bias cannot be used, and 4. The supply voltage must have good regulation.
  • 20. The graphical construction from which to determine the output current and voltage waveshapes for a single transistor operating as a class B stage is indicated in Fig. 18-9.
  • 21. Input power: The power supplied to the load by an amplifier is drawn from the power supply which provides the input or dc power. The amount of this power can be calculated using Pi (dc) = VCC I dc (16.17) In class B operation the current drawn from a single power supply has the same form of a full-wave rectified signal. So, 2 2 I = I ( p ) (16.18) and Pi (dc) = VCC I (p) (16.19) dc π π Output Power: The output across the load is given by 2 VL (rms) Po (ac) = (16.20) RL 2 VL ( p ) 2 VL ( p - p ) Po (ac) = = (16.21) 2 RL 8 RL
  • 22. Efficiency: The efficiency of the class B amplifier can be calculated using the basic equation Po (ac) %η = × 100% Pi (dc) 2 VL ( p ) π %η = ×100% 2 RL 2VCC I ( p ) V ( p) using, I (p) = L RL 2 VL ( p ) π RL %η = ×100% 2 RL 2VCC VL (p) π VL ( p ) %η = ×100% (16.22) 4 VCC
  • 23. Maximum Efficiency: Equation (16.22) shows that the larger the peak voltage, the higher the circuit efficiency, up to a maximum value when the , this maximum efficiency then being VCC 2 VCC Maximum, I (p) = Maximum, I dc = RL π RL 2 2VCC 2 VCC 2 VCC Maximum, Pi (dc) = VCC = Maximum, Po (ac) = π RL πRL 2 RL 2 VCC πRL π Maximum, %η = × = ×100% = 78.5% 2 RL 2V 2 4 CC The maximum efficiency of a class B amplifier is thus seen to be 78.5%.
  • 24. Crossover distortion The distortion caused by the nonlinear transistor input characteristic is indicated in Fig. 18-12. The iB-vB curve for each transistor is drawn, and the construction used to obtain the output current (assumed proportional to the base current) is shown. In the region of small currents (for vB<Vγ) the output is much smaller than it would be if the response were linear. This effect is called crossover distortion. In order to overcome the problem of crossover distortion, the transistor must operate in a class AB mode.
  • 25. (16.10) Calculate (i) the input power, (ii) the output power, (iii) the power handled by each transistor, and (iv) the efficiency for an input of 12 V (rms) of a push-pull Class-B amplifier circuit having biasing voltage, VCC= 25 V and load resistance, RL= 4 Ω. Solution: Given, Vi (rms)=12V, VCC= 25 V, load resistance, RL= 4 Ω. Vi ( p ) = 2Vi (rms ) = 2 × 12V ≈ 17 V Since, the operating point is selected in cutoff region, thus Vi(p)=VL(p)=17V VL ( p ) 17 V I L ( p) = = = 4.25A RL 4Ω 2 2 I dc = I L ( p ) = × 4.25A = 2.71A π π Pi (dc) = VCC I dc = 25V × 4.25A = 67.75W
  • 26. 2 ( p) VL (17 V )2 Po (ac) = = = 36.125W 2 RL 2 × 4Ω Power dissipated by each output transistor is: P2Q Pi − Po 67.75 − 36.125 PQ = = = = 15.8 W 2 2 2 Po 36.125 Efficiency, %η = × 100% = × 100% = 53.3% Pi 67.75
  • 27. Class C Operation With class B, we need to use a push-pull arrangement. That’s why almost all class B amplifiers are push-pull amplifiers. With class C, we need to use a resonant circuit for the load. This is why almost all class C amplifiers are tuned amplifiers. With class C operation, the collector current flows for less than half a cycle. A parallel resonant circuit can filter the pulses of collector current and produce a pure sine wave of output voltage. The main application for class C is with tuned RF amplifiers. The maximum efficiency of a tuned class C amplifier is 100 percent.
  • 28. Figure 11-13a shows a tuned RF amplifier. The ac input voltage drives the base, and an amplified output voltage appears at the collector. The amplified and inverted signal is then capacitively coupled to the load resistance. Because of the parallel resonant circuit, the output voltage is maximum at the resonant frequency, given by: 1 fr = (11.14) 2π LC
  • 29. On either side of the resonant frequency fr, the voltage gain drops as shown in Fig. 11-13b. For this reason, a tuned class C amplifier is always intended to amplify a narrow band of frequencies. This makes it ideal for amplifying radio and television signals because each station or channel is assigned a narrow band of frequencies on both sides of a center frequency.