 In today's session, we are going to learn about Class B power amplifiers. Learning outcome of today's session, at the end of the session, student can analyze Class B power amplifier. These are the contents of today's presentation. Let us get introduction about Class B power amplifier. In Class B power amplifier, the output collector current flows or varies only for either 180 degrees of input AC voltage signal waveform. Since a DC operating point or quiescent point of amplifier using amplifying device transistor is located on a load line in the cutoff region. Output current flows only either for positive half cycle or negative half cycle of input AC voltage signal. Therefore, for triester, the conduction angle will be 180 degrees, so as to get the output for full cycle of input voltage signal and to increase efficiency of power amplifier. Power amplifier is connected in push-pull arrangement using two triesters. Each triester will conduct for each half cycle of input AC voltage signal waveform. In power amplifier, a biasing circuit is not connected. Since triesters are biased in the cutoff region, so in Class B power amplifier, a distortion is produced called as a cross-over distortion. It produces a cross-over distortion, so it is due to a clipping of input voltage signal during the switching action of triesters between on and off. Since triesters are off at zero input signal condition, some part of input AC voltage signal is consumed in turning on the triesters. So output voltage, that is collector voltage and collector current of a transistor does not follow the input base voltage and base current. So distortion is caused in Class B power amplifier. Since only one transistor is conducting in each half cycle and power dissipation is very less, so power wastage in power amplifier is very less. So efficiency for this power amplifier will be higher since the conduction angle for each transistor is only 180 degrees. This figure shows the circuit diagram for basic Class B power amplifier. To get output voltage and current, that is output power to the load for full cycle of input AC voltage signal, the circuit is designed and connected in push-full arrangement using two triesters, transistor Q1 and Q2. At the input side of amplifier, transformer T1 is a driver transformer. It is used to couple input AC voltage signal to the base of the two triesters Q1 and Q2. As well as at the output side, transformer coupling is used. So transformer T2, that is a driver transformer, it is used to connect output load RL. And input driver transformer T1 is used to produce two voltage signals of equal magnitude and opposite polarity and it is provided to transistor Q1 and Q2. If as Q1 and Q2 are not biased at 0, the base diameter of cut-in voltage, if it is neglected, it is very small. It provides a push-full operation, Class B push-full operation. During positive half cycle of input AC voltage signal, the secondary of transformer T1 turns on transistor Q1, so transistor Q1 conducts at the same time transistor Q2 in cut-off state. So current flows through upper half part of primary of transformer T2 and voltage is produced across upper half part of transformer T2 primary. In the negative half cycle of input AC voltage signal, the transistor Q2 conducts at the same time transistor Q1 will be in the cut-off. So current flows through lower half part of primary of transformer T2 and voltage is produced across lower half part of transformer T2. So due to transformer action, the output voltage and current is provided to the load RL is connected to secondary of transformer T2. Now this figure shows the AC load line for Class B power amplifier. So output current and output voltage varies in only half cycle of input AC voltage signal for each transistor. So output collector current varies from a zero value to its peak value in positive direction as well as in collector to emitter voltage is also varies from its cosine point value that is to maximum positive or negative voltage. DC and AC conditions for Class B power amplifier when input signal is zero, both transistors remains off. So collector current IC equal to zero and collector to emitter voltage is equal to a biasing voltage VCC. AC load resistance for each transistor RL AC is designated as RP is nothing but the resistance of each half of primary of transformer T2. When collector to emitter voltage is minimum, so collector current will be maximum. It is given by VCC upon RL AC that is the resistance of each half of primary winding of transformer T2. This figure shows a load line for Class B power amplifier. It shows a variation of output collector current and collector to emitter voltage in response to a variation in small base voltage and base current. So output collector current varies or swings both positively as well as negatively so that collector voltage is also varies positively and negatively and due to transformer reaction collector voltage and collector current are provided to output load resistance RL which is connected to secondary of transformer T2. Let us find out the efficiency of this Class B power amplifier. So convergent efficiency of Class B power amplifier is defined as it is ratio of AC power delivered to load upon DC power provided in the Class B power amplifier in the collector circuit. So DC input power PDC is given by DC voltage VCC into the DC current flowing in the collector circuit of a transistor IDC. So IDC is the average current drawn by each half of primary of transformer T2. So IDC is equal to 2IP upon pi where IP stands for peak value of DC current collector current output current that is equal to 2IM by pi. So PDC is equal to 2VCC into IP divided by pi so that is equal to 2VPIP divided by pi. Since VP equal to VCC maximum peak voltage is equal to DC voltage VCC. So similarly AC output power delivered to load is given by VRMS into IRMS. So PAC load delivered to load is equal to VP upon root 2 multiplied by IP upon root 2. So AC power delivered to load is equal to VP into IP divided by 2. So PAC is equal to VCC into IDC so divided by 2. Percentage efficiency of class B power amplifier is equal to VP IP upon 2 divided by 2VP IP divided by pi so multiplied by 100. So this is equal to pi upon 4 multiplied by 100 so 78.5%. So thus 78.5% of DC power converted into AC power and remaining power is dissipated across transistor and transformer primary resistors. So this figure shows a basic circuit diagram for complementary symmetry amplifier. The input driver transformer in the basic class B amplifier is used to produce two signal voltages of equal magnitude and opposite polarity have a poor frequency response. So the frequency response of class B amplifier can be improved by eliminating input driver transformer by using two transistors of two opposite transistors Q1 and Q2 that is N-pane transistor and PNP transistors. The driver transformer has a poor frequency response at lower and higher side of audio frequency range. So the audio driver transformer can be eliminated using two opposite types of transistors Q1 and Q2 connected in push-pull arrangement. So in positive half cycle of input AC voltage signal, transistor Q1 conducts and load resistance RL will get current and voltage. At the same time transistor Q2 will remain in cutoff state. In negative half cycle of input signal, Q1 is in cutoff state and Q2 conducts and current flows through collector emitter circuit of load resistance RL. So load resistance RL is directly connected in collector emitter circuit of power transistor. So we can eliminate the use of input driver transformer and output coupling transformer using complementary symmetry push-pull amplifier. A student can pause video here and think over this question. How cross-volt distribution can be reduced in class B amplifier? In class B amplifier, since transistors are not biased, they are biased at zero voltage. Based on emitter cut-in voltage, which is very small, if it is not overcome, transistor will remain off. So some part of input AC voltage signal is used to turn on transistors, to overcome barrier potential of base emittering snob to transistors. So at that time, the transistor remains off. So output voltage and current will be zero. So this produces a cross-volt distortion. This cross-volt distortion can be reduced by biasing the two transistors using voltage divided biasing circuit to provide a turn-on voltage between base emitter junction of the transistors, so that output will follow the input. So in this way, a cross-volt distortion can be minimized in class B amplifier by biasing the two transistors slightly above the cut-off region using voltage divided biasing circuit. These are the references. Thank you.