Push Pull Class B Power Amplifier

Push Pull Class B Power Amplifier

• The push pull circuit requires two transformers, one as input transformer called driver transformer and the other to connect the load called output transformer.

• Both the transformers are centre tapped transformers. The push pull class B amplifier circuit is shown in the Fig. 8.17.1.

 

• In the circuit, both Qx and Q2 transistors are of n-p-n type.

• The circuit can use both Q1 and Q2 of p-n-p type. In such a case, the only change is that the supply voltage must be - VCC, the basic circuit remains the same.

• The input signal is applied to the primary of the driver transformer.

• The centre tap on the secondary of the driver transformer is grounded.

• The centre tap on the primary of the output transformer is connected to the supply voltage + V

• With respect to the centre tap, for a positive half cycle of input signal, the point A shown on the secondary of the driver transformer will be positive. While the point B will be negative.

• Thus the voltages in the two halves of the secondary of the driver transformer will be equal but with opposite polarity.

• Hence the input signals applied to the base of the transistors Q₁ and Q₂ will be 180° out of phase.

Operation of the circuit :

• For positive half cycle of the input, A is positive and B is negative.

• Due to this, Q1 gets driven into active region while Q2 is cut-off.

• Thus i_b1 flows while i _b2 = 0, as shown in the Fig. 8.17.2.

 • Hence icl flows through upper part of primary of the output transformer while i_c2 =0.

• Thus positive half cycle is produced across the load.

• For negative half cycle of the input, B is positive and A is negative.

•  Due to this, Q2 gets driven into active region while Q1 is cut-off.

• Thus ib2 flows while iM = 0 as shown in the Fig. 8.17.3.

• Hence i_c2 flows through lower part of primary of the output transformer while ic1 = 0.

• Thus negative half cycle is produced across the load.

• The full cycle is thus obtained across the load for full input cycle.

• The waveforms of the input current, base currents, collector currents and the load current are shown in the Fig. 8.17.4.

Important Concept

For the output transformer, the number of the turns of each half of the primary is N₁ while the number of the turns on the secondary is N₂. Hence the total number of primary turns is 2N₁. So turns ratio of the output transformer is specified as 2N₂ : N₂.

 

1. D.C. Operation

• The d.c. biasing point i.e. Q point is adjusted on the X-axis such that V_CEQ = V_CC and I_CEQ is zero.

• Hence the co-ordinates of the Q point are (VCC, 0). There is no d.c. base bias voltage.

 

2. D.C. Power Input

• Each transistor output is in the form of half rectified waveform.

• Hence if Im is the peak value of the output current of each transistor, the d.c. or average value is 1m due to half rectified waveform.

• The two currents, drawn by the two transistors, form the d.c. supply are in the same direction. Hence the total d.c. or average current drawn from the supply is the algebraic sum of the individual average current drawn by each transistor.

3. AC Operation 

• When a.c. signal is applied to the driver transformer, for positive half cycle of input Q1 conducts while for negative half cycle of input Q2 conducts.

• When Qi conducts, lower half of the primary of the output transformer does not carry any current. Hence only N1 number of turns carry the current.

• While when Q2 conducts, upper half of the primary does not carry any current. Hence again only N number of turns carry the current. Hence the reflected load on the primary can be written as,

• Though the turns ratio is 2N₁ : N₂, while operation the ratio becomes n=N₂ / N₁ which decides the reflected load.

• The slope of the a.c. load line is -1/RL while the d.c. load line is the vertical line passing through the operating point Q on the X-axis. The load lines are shown in th Fig. 8.17.5.

• The slope of the a.c. load line (magnitude of slope) can be represented interms of V_m and I_m as,

where I_m = Peak value of the collector current

 

4. A.C. Power Output

• As Im and Vm are the peak values of the output current and the output voltage respectively, then

V_rms = V_m / √2 and I_rms = I_m / √2

• Hence the a.c. power output is expressed as,

• While using peak values it can be expressed as,

 

5. Efficiency

• The efficiency of the class B amplifier can be calculated using the basic equation.

 

6. Maximum Efficiency

• From the equation (8.17.7), it is clear that as the peak value of the collector voltage Vm increases, the efficiency increases.

• The maximum value of Vm possible is equal to VCC as shown in the Fig. 8.17.6.

Important Concept

Thus the maximum possible theoretical efficiency in case of push pull class B amplifier is 78.5 % which is much higher than the transformer coupled class A amplifier.

For practical circuits it is upto 65 to 70 %.

Important Concept

Practically the collector-emitter voltage of transistor is neglected as small. But if V_CE(min) is given then maximum collector voltage V_m reduces by V_CE min and becomes V_m = V_CC - V_{CE (min)} under maximum efficiency condition.

 

7. Advantages

1. The efficiency is much higher than the class A operation.

2. When there is no input signal, the power dissipation is zero.

3. The even harmonics get cancelled. This reduces the harmonic distortion.

4. As the d.c. current components flow in opposite direction through the primary winding, there is no possibility of d.c. saturation of the core.

5. Ripples present in supply voltage also get eliminated.

6. Due to the transformer, impedance matching is possible.

 

8. Disadvantages

1. Two centre tap transformers are necessary.

2. The transformers, make the circuit bulky and hence costlier.

3. Frequency response is poor.

Review Questions

1. Draw the circuit diagram of a class B push-pull power amplifier and discuss its operation, merits and demerits.

2. Find out the expressions for the following parameters for the class B push-pull amplifier :

i) d.c. power input ii) a.c. power output iii) efficiency iv) maximum efficiency v) power dissipation.

3. Show that the maximum conversion efficiency of the ideal class B push-pull circuit is 78.5 %.

4. Explain with circuit diagram class B power amplifier and derive for its efficiency.

AU : Dec.-15, Marks 8