EC205 ELECTRONIC CIRCUITS POWER AMPLIFIERS -PART 2 QUICK REVISION

In the last lecture, we discussed about the different types of power amplifiers. Today we are going to discuss about the transformer coupled power amplifiers. The limitations of previous circuits are eliminated. There are two types of class A power amplifiers. One is series fed class A power amplifier. This is not in the syllabus. This is a series fed class A power amplifier. It provides a fixed bias to work the circuit as a class A power amplifier. This is not in the syllabus. There are some limitations of this series fed class A power amplifier. Low conversion efficiency. The flexibility in obtaining impedance matching condition is limited because of resistive load. The DC power dissipated in the load is wasted. The conversion efficiency is low. We are using resistive load for impedance matching. The DC power dissipated in the load is wasted. We can use transformer coupled power amplifiers to overcome these limitations. What is the difference between this and the previous series fed power amplifiers? In the previous series fed power amplifiers, we used a resistor. After replacing the resistor, we are using a transformer. We are using a transformer as a load. After replacing the load, we are using a transformer. We are using a transformer as a load. The name of the circuit diagram is single ended. Let us see the load line characteristics. We are using a transformer. The impedance of this transformer is low. current will be infinity when the impedance is minimum, so we can take 0 impedance. DC load line is a straight line between Vcc and Ic. Voltage is Vcc, but current is infinity. That is why we have drawn DC load line as a vertical line. Voltage is Vcc. To find the Q point, we have to draw AC load line. DC load line and AC load line meet at a point. That is known as Q point. AC load line is 2Vcc and 2Vcc by RL dash. This point is the quiescent point. Voltage is Vcc. RL dash divided by RL is V1 by I1 divided by V2 by I2 or V1 by V2 into I2 by I1. RL dash is equal to N1 by N2 the whole square into RL. We are deriving all the terms in this equation. Vce is equal to Vcc. Ic is equal to 2Vcc by RL dash. Icq is equal to Vcc by RL dash. Vce is equal to Vce peak to peak into Ic peak to peak divided by 8Vcc into Icq. Ic peak to peak is equal to 2Vcc by RL dash. 8Vcc into Icq is equal to Vcc by RL dash. Ic peak to peak is equal to Vcc by RL dash. Since RL is zero, PL depends on Pout AC. Pout AC is equal to IRMS square into RL dash. IRMS is equal to Vcc by root 2 RL dash the whole square into RL dash. Vcc square divided by 2 RL dash. Collector dissipation is the dissipation in the circuit. Vcc into Icq is equal to Vcc into Icq. Vcc into Icq is equal to Vcc by RL dash. Icq is equal to Vcc by RL dash into PL. PL is equal to Vcc square divided by 2 RL dash. Figure of Merit. Figure of Merit can be studied without seeing the equation. Substitute these values. Vce peak to peak, Ic peak to peak and Vcc into Icq. Since RL is zero, PL is equal to Vcc by RL dash. Figure of Merit. Main Limitations. Total harmonic distortion is high mainly due to second harmonic. To reduce the distortion, we may have to design an amplifier with a load. In that case, we have to sacrifice AC power output. To reduce the distortion, we may have to design an amplifier with a load. In that case, we have to sacrifice AC power output. The DC current flowing through the primary causes core saturation and magnetic distortion. These are the three limitations of transformer coupled class A power amplifier. The next configuration to overcome these limitations is the push-pull amplifier. This is the diagram of class A push-pull amplifier. During positive half cycle, Q1 conducts and current Ic1 flows in this direction. During negative half cycle, Q2 conducts and current Ic2 flows in this direction. The resistance reflected on the primary side of the transistors is N1 by N2 divided by N2 the whole square into R. This is N1 by N2 the whole square into RL by 4. we have already studied the effect of distortion we have studied distortion in general we have taken a linear case and made it non-linear and we have found the effect of distortion you can understand this by looking at the previous lecture the first session of power amplifier we can write IC1 as IC1 is equal to B0 plus B1 cos omega t plus B2 cos 2 omega t plus etc this is the fundamental component this is the harmonic component the collector current here IC1 is equal to B0 plus B1 cos omega t plus B2 cos 2 omega t plus etc but the collector current IC2 the collector current in IC2 comes with a 180 degree phase shift look here IC1 this direction IC2 is a collector current with a 180 degree phase shift then look here then this equation IC2 B0 plus B1 cos omega t plus 180 plus B2 into cos 2 of omega t plus 180 plus etc cos omega t plus 180 is equal to minus cos omega t when we substitute this IC2 is equal to B0 plus B0 minus B1 cos omega t plus B2 cos 2 omega t minus etc then look here the total current here what is the difference? total current is the difference between this current and this current that is I current I is proportional to IC1 minus IC2 or I is equal to a constant into IC1 minus IC2 when we subtract this we get B0 when we subtract this we get B1 cos omega t minus minus B1 cos omega t becomes 2 B1 cos omega t when we subtract this we get B1 cos omega t minus minus B1 cos omega t when we multiply by K 2K B1 cos omega t plus these two terms are cancelled these two terms B2 cos 2 omega t and B2 cos 2 omega t are cancelled then 2K B3 cos 3 omega t we get an equation like that what we understood in this is this push pull circuit eliminates all even harmonics this push pull circuit eliminates all even harmonics B3 is the main component of harmonic distortion what is the main component here? B3 is the main component of harmonic distortion the important advantage of this push pull circuit is even harmonics are absent in the output since even harmonics are absent the circuit gives more output power per transistor for a given amount of distortion DC components for the collector current of two transistors oppose each other magnetically in core of the output transformer this eliminates the tendency to saturate and consequent non-linear distortion the effect of ripple voltage contained in the power supply caused by inadequate filtering are balanced out because the current produced by ripple voltages are in opposite direction in the transformer winding mainly to eliminate harmonic distortion what did we use? we used this push pull configuration this is the two configurations of class A power amplifier which are transformer coupled class A and class A push pull amplifier now let's see class B power amplifier in class B power amplifier this is a push pull configuration because two transistors are used this is a push pull configuration transistors are biased at cut-off region here transistors are biased at cut-off region in class A power amplifier voltage divider bias is used which region are transistors biased? in active region in class B power amplifier transistors are biased at cut-off region what is special about class A power amplifier? its conduction angle is 180 degree to ensure 360 degree operation we used this push pull configuration in class B power amplifier transistors are biased at cut-off region in cut-off region only one half cycle conducts to ensure 360 degree operation we used this push pull configuration now let's see the load line how to draw this line? what is AC load line? VCC by RL dash is the line between VCC and AC DC load line is always a vertical line from VCC because we applied transformer here so RL is zero because RL is zero current is infinity so we are drawing a vertical line parallel to IC DC load line and AC load line meet at VCC from here we can write what we need VCE is VCC IC is VCC by RL dash VCE is equal to VCC IC is equal to VCC by RL dash VCE is VCC so what is VCE peak to peak? VCE is 2VCC IC peak to peak is 2VCC by RL dash now let's see another thing in this case we learned the diagram of class B positive half cycle and negative half cycle everything is in the same direction it is like a full wave rectifier output positive half cycle and negative half cycle will conduct but here it is like a full wave we get output waveform like a full wave so we have to pay attention here full wave rectifier operation is performed here ICQ is 2 IM by Pi because in case of full wave DC current value is 2 IM by Pi we substituted this value for 2 IM IM is VCC by RL dash IM is IC peak value that is VCC by RL dash instead of IM we got 2 VCC by Pi RL dash if you learn to write this much you can do the rest by yourself VCE is equal to VCC VCE peak to peak is equal to 2VCC IC peak to peak is equal to 2VCC by RL dash ICQ is equal to 2VCC by Pi RL dash if you write this much you can do the general equation eta is equal to VCE peak to peak into IC peak to peak divided by 8VCC into ICQ if you substitute these values you will get efficiency conversion efficiency is equal to 78.5 similarly for load power equation for load power is ICQ square RL plus Pout AC in this case RL is the transformer output is coming in load RL is 0 so we don't need this unit we only need Pout AC Pout AC is IRM square into RL dash if we substitute this we get equation VCC square by 2 RL dash collector power dissipation collector power dissipation is PD is equal to total input power minus PL total input power is VCC into ICQ minus load power if you substitute this you will get the value there is nothing special we know all these terms we just need to substitute this in general equation if we want to find power dissipation in one transistor we need to divide this by 2 because there are two transistors here power dissipation in the transistor is we get power dissipation in two transistors if it is in one transistor then half of it will be in one transistor so when we find power dissipation here to find power dissipation in transistor we need to divide by 2 write the equation of figure of merit in the equation of figure of merit substitute the value we know the value of everything substitute the value we get figure of merit as 0.39 for high fidelity applications the transformers are used at the input and output amplifiers and are not suitable because there is frequency limitation moreover the transformer increases the weight and cost of the amplifier the main disadvantage of this is we are using transformer here because it is a transformer the main problem is frequency limitation moreover the weight and cost of the amplifier increases next is complementary class symmetry class B push-pull amplifier what is complementary symmetry? here we have class B amplifier this transistor and this transistor both are NMOS transistors not NMOS, BJN here we have NPN transistors Q1 and Q2 are NPN transistors at the same time in complementary symmetry class B push-pull amplifier the transistors are one is PNP and the other is NPN here we don't need two transformers here we are using a transformer that is not required here we are giving only one input here is positive and here is negative in positive half cycle Q1 is ON when it is NPN in positive half cycle Q1 is OFF when it is PNP in positive half cycle Q1 is ON when it is NPN in negative half cycle Q2 is ON when it is NPN Q1 is ON when it is PNP Q2 is ON when it is NPN the advantage of this is what we can avoid? we can avoid the transformer at the input side when we use the transformer at the input side we can reduce the weight of the transformer so that is the advantage of this type complementary symmetry class B power amplifier keep this in mind next is crossover distortion this is the main problem I see crossover distortion this is seen in all class B power amplifiers it is important there is no question of crossover distortion what is crossover distortion? the signal is cut off at the zero crossing point this is the diagram this is class B power amplifier Q1 and Q2 are biased the Q point is fixed at the cut off region we give an input here positive half cycle comes when the positive half cycle comes this transistor does not conduct Q1 does not conduct when the negative half cycle comes Q2 does not conduct to conduct this to turn on a transistor it needs a certain voltage 0.7 voltage this transistor does not work at that time at that point when it comes to the output at that point distortion occurs that's all the transistor is biased at the cut off region because the transistor is biased at the cut off region until a certain cut in voltage reaches that transistor does not conduct when it comes to that point at that point the signal is not amplified properly that is what we call crossover distortion it is very simple to draw at the zero crossing point you can draw it distorted how to overcome crossover distortion? to overcome crossover distortion the circuit we use is class AB push pull amplifier we can draw the same circuit we drew here same circuit we can draw the same circuit there here this circuit is drawn here R1 R2 R1 R2 is drawn but this transistor both are operated at the cut off region at the same time when it comes to class AB push pull this R1 R2 this voltage divider is biased Q1 and Q2 are at little above the cut off region understood? Q1 and Q2 are at little above the cut off region that is, cut off region and active region are at intermediate bias When you bias like that, the transistors will start to conduct when the input is coming. What can we avoid by doing this? We can avoid crossover distortion. Just remember one thing. We are biasing this transistor. That is, we are biasing the cut-off above. When that happens, we can reduce the situation where there is no signal at the zero-crossing point. That is class A B power amplifier. Next is class C power amplifier. Class C power amplifier operates beyond the cut-off region. The cut-off region and the tail are what makes it work. Beyond the cut-off region, there is a circuit diagram. This is a B-I capacitor. This is a resistor. This is an emitter load. This is a transistor's emitter load. This is a tank circuit. This is what class C power amplifier does. This is the input signal. The transistor itself is biased beyond the cut-off region. Since it is beyond the cut-off region, a negative bias has been provided to the transistor. See, a negative bias has been provided to the transistor through these six ways. That is, even if the input increases, the transistor will turn on only when the input reaches a certain value. That is, only in this particular region, the current flows through the transistor. There is nothing in this region. Very little. What is special about this is that the efficiency is high. Distortion is the highest. The higher the value, the more distorted the signal becomes. This is class C power amplifier. In the case of class C power amplifier, the condition angle will be less than 180 degrees. The efficiency will be high. The efficiency will be around 85%. This is to ensure that the transistor is operating beyond the cut-off region. A resistor is connected to the ground. The base emitter voltage will be a negative voltage. That's all about power amplifiers. The most important topic in power amplifiers is the derivation of class B. Class B derivation, crossover distortion, push-pull configuration. How to eliminate harmonic distortion in push-pull configuration? It is important to know how to eliminate harmonic distortion in push-pull configuration. The derivation of class B is also important. Thank you.