Class-B Pushpull Amplifier Operation

Class-B Pushpull Amplifier Operation:

Class–B operation is provided when the dc bias leaves the 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 shows a diagram for push-pull operation.An ac input signal is applied to the push-pull circuit, with each half operating on alternate half-cycles, the load then receiving a signal for the full ac cycle.The power transistors used in the push-pull circuit are capable of delivering the desired power to the load, and the class B operation of these transistors provides greater efficiency than was possible using a single transistor in class A operation.

                                                   

                                   

        The circuit of Fig. shown uses a center-tapped input transformer to produce opposite polarity signals to the two transistor inputs and an output transformer to drive the load in a push-pull mode of operation described next.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 the input signal, Q2 conducts whereas Q1 stays off, the current I2 through the transformer resulting in the second half-cycle to the load. The overall signal developed across the load then varies over the full cycle of signal operation.

                                                            

                            

Class B Amplifier Input (DC) Power:

The power supplied to the load by an amplifier is drawn from the power supply (or power supplies; see Fig. 16.13) that provides the input or dc power. The amount of this input power can be calculated using

                                       Pi(dc) = VCC Idc

where Idc is the average or dc current drawn from the power supplies. In class B operation, the current drawn from a single power supply has the form of a full-wave rectified signal, while that drawn from two power supplies has the form of a halfwave rectified signal from each supply. In either case, the value of the average current drawn can be expressed as

                                              Idc = [2/pi]Ip

where Ip is the peak value of the output current waveform

Therefore,

                                      Pi(dc) =  VCC * [2/pi]Ip

Class B Amplifier Output (AC) Power:

The power delivered to the load (usually referred to as a resistance, RL) can be calculated using any one of a number of equations. If one is using an rms meter to measure the voltage across the load, the output power can be calculated as

                                            Po(ac) = (VL(rms))^2/ RL

If one is using an oscilloscope, the peak, or peak-to-peak, output voltage measured

can be used:

                                        Po(ac) =    (VL(p-p))^2/8RL

                                                =    (VL(p))^2/2RL

The larger the rms or peak output voltage, the larger the power delivered to the load.

Class B Amplifier Efficiency:

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

                                 % Efficiency = Po(ac)/Pi(dc)  * 100 %

                                                    = [pi/4]*[VL(p)/VCC]*100 %

Above fig. shows that the larger the peak voltage, the higher the circuit efficiency, up to a maximum value when VL(p)=VCC, this maximum efficiency then being

                                 maximum efficiency =  [pi/4]* 100% = 78.5%

Power Dissipated by Output Transistors:

The power dissipated (as heat) by the output power transistors is the difference between the input power delivered by the supplies and the output power delivered to the load.

                              P2Q = Pi(dc) – Po(ac)

where P2Q is the power dissipated by the two output power transistors. The dissipated power handled by each transistor is then

                               PQ = P2Q/2