 In today's presentation, we are going to learn about Class A-B power amplifiers. At the end of this session, students can analyze the operation of Class A-B power amplifiers. These are the contents of today's presentation. Let us get acquainted with Class A-B power amplifier. In Class A-B power amplifier, each transistor in the amplifier circuit is biased in such a way that the output collector current flows or varies for more than 180 degree but less than 360 degree of input AC signal cycle. Since DC operating point or QC point is located on load line just above the cut-off region so that output current for each transistor flows for more than half cycle but less than full cycle of input AC signal. Hence, the conduction angle for each transistor is greater than 180 degree and less than 360 degrees of input AC signal cycle. To get the output for full cycle of input signal for both positive and negative half cycle and increase the efficiency of power amplifier, Class A-B power amplifier is connected in push-pull arrangement. Since each transistor in the amplifier conducts for just more than half cycle but less than full cycle of input AC signal so power dissipation of each transistor is much less. Therefore, efficiency of Class A-B power amplifier is much higher. This figure shows the circuit diagram for Class A-B power amplifier. Here, Class B power amplifier is modified to Class A-B power amplifier. The crossover distortion produced in the Class B power amplifier is eliminated by slightly forward biasing base symmetry junctions of both transistor Q1 and transistor Q2. Using the resistors R1 and R2 by the amount equal to VB that is cut in voltage of transistor Q1 and Q2. Therefore, as soon as input AC signal is applied the output collector current immediately starts flowing. The input AC signal is applied to driver transformer T1. Resistors R1 and R2 are used to set a DC operating point or Q-send point for transistor Q1 and Q2 much closer to cut-off region that is Class B operation. Resistor RE is used to stabilize the DC operating point of transistor Q1 and Q2 and reduce distortion in the output signal. The input AC signal is coupled to the input of Class A-B power amplifier using driver transformer T1 whose second is center tapped. So, driver transformer T1 produces two voltage signals of equal magnitude and opposite polarity to drive transistor Q1 and Q2. During positive half cycle of input AC signal, transistor Q1 conducts and during negative half cycle of input AC signal transistor Q2 conducts. When Q1 conducts the current flows through upper half part of primary of transformer T2 when Q2 conducts the current flows through lower half part of primary winding of transformer T2. Due to transformer action, voltage and currents are provided on the secondary of transformer T2 and thus AC power is delivered to load registrarial. Now this figure shows a DC and AC condition for Class A-B power amplifier. The DC operating point for each transistor used in Class A-B power amplifier is selected much closer to a cut-off region. So, each transistor conducts for more than half cycle and less than full cycle of input AC signal. The collector current IC and collector to emitter voltage varies in response to variation of input base voltage and base current. And this figure shows a load line for Class A-B power amplifier. The variation of output current and output voltage in response to variation of input base voltage and base current when input AC signal is present. This figure shows the circuit diagram for Class A-B complementary symmetry amplifier. This consists of output complementary symmetry amplifier stage and a common emitter driver stage to stabilize the operation of amplifier and reduce the distortion in the output signal. Instead of driver transformer, a common emitter transistor amplifier as a driver stage is used at the input side of Class A-B power amplifier. It is designed using the resistor R1, R2, R3, R4 and R5 and transistor Q1 with emitter current bias drives the transistor Q2 and Q3. Q2 and Q3 are of opposite type NPN and PNP. The resistor R4 connected to base of transistor Q2 and Q3 provides required biasing for transistor Q2 and Q3. The input driver stage common emitter transistor amplifier Q1 decides major part of voltage gain of overall complementary symmetry amplifier. The resistor R4 which provides required biasing for transistor Q2 and Q3 acts as a short circuit when input AC signal is present. Due to capacitor C2, since capacitor C2 is connected in parallel to resistor R4 in presence of AC signal, C2 provides very negligible reactance so it behaves like a short circuit so therefore R4 acts as a short circuit. The resistor R6 and R7 are used to limit the amount of current flowing through transistor Q2 and Q3 whenever the output voltage is developed whenever output is developed across RL as well as the common point of resistor R6 and R7 which are connected to transistor Q2 and Q3 emitter is biased at half of DC power supply voltage VCC. So that the output applied to capacitor C4 can swing by equal amount in both positive and negative direction in presence of input AC signal. During conduction of Q2, capacitor C4 charges through VCC and RL and supplies a load current so during positive half cycle of input AC signal the current is drawn from DC power supply and provided to load resistor RL. When Q3 is conducting, load current flows through Q3 and RL and this time the current is not drawn from DC power supply VCC. Instead of that capacitor C4 which is acting as energy storage device supplies load current so C4 discharges through RL and Q3 so C4 provides current to the load during negative half cycle of input AC signal. So the maximum VCA for output transistor is supply voltage VCC. The maximum current drawn from DC power supply for each transistor is peak load current plus the DC collector current at a Q point. The transistor current IM is equal to peak load current plus ICQ. ICQ is a collector current at DC operating point or Q point or Q point so that is equal to IP plus 0.1 IP. Generally the DC collector current at Q point is 10% of peak load current so maximum current drawn towards each transistor is equal to 1.1 IP. The current drawn from DC power supply plus VCC has half your rectified form. Since the current flows towards load from DC power supply only during positive half cycle of input AC signal. In negative half cycle of input AC signal the load current is supplied from capacitor C4. Therefore the current drawn from DC power supply has a half your rectified form. So the average current from DC power supply is equal to IM by pi. So that is equal to 1.1 IP divided by pi. Now student can pause video here and think over this question. How crossover distortion reduces in class AB power amplifier? In class AB power amplifier each transistor is slightly power biased by amount equal to AB that is cut in voltage and its DC operating point is set just above the cutoff region. So as soon as input AC signal is applied the transistor conducts and output collector current immediately starts flowing. So in this way the crossover distortion is eliminated in class AB power amplifier. These are the references. Thank you.