 In this lecture, we will talk about power amplifiers. Previously, we have studied about small signal amplifier. So, here small signal means that the output current or voltage swing is small as compared to the maximum collector current. Now, these small signal amplifiers are mostly used in electronic systems for signal processing. So, in most of the electronic systems these small signal amplifiers are used as the starting 2 or 3 stages of amplifiers. Now, if the power delivered to the load is required to be very high, then the amplifier should be selected as power amplifier. So, in general the last 1 or 2 stages of the amplifier are selected as power amplifiers. Here, the one may say that when a amplifier should be considered as power amplifier and when it should be considered as a small signal amplifier. So, when the power is less than 500 milliwatt, then one may say that this can be considered as a small signal amplifier. However, if the power is greater than 500 milliwatt, maybe few watts or few tens of watt or few hundreds of watt, then they can be considered as power amplifiers. So, in this lecture, we will talk about the power amplifiers. We will see the various performance parameters of power amplifiers and we will see how these parameters affect the performance of power amplifier. Then we will see the various classes of power amplifiers which depends upon the output current flow in the amplifier. Then we will see the circuit analysis of power amplifier. We will talk about the DC and AC load line and then we will see the selection of operating point. After that, we will talk about the various classes of power amplifiers in broad way. So, we will talk about class A amplifier, class B amplifier, class AB and class C power amplifier. Now, after discussing these classes of power amplifier, we will take two practical design examples of power amplifier. One will be of 2 watt power and another example will be of 30 watt power amplifier. So, in power amplifier, the power means that how much AC power is delivered to the load. Now, one may ask from where this AC power comes. So, in general, it comes from the supply or from the battery. Now, one may ask that what is the conversion efficiency? So, that means that how effectively one circuit of amplifier converts the DC power into the AC power which is to be delivered to the load. This effectiveness is defined in terms of the conversion efficiency of power amplifier and it is a figure of merit of power amplifier. It is defined as the ratio of the average AC power delivered to the load to the average DC power drawn by the circuit or supplied by the battery. This is one of the consideration of the power amplifier. Next consideration is the distortion in amplifiers. Now, depending upon the output current, the various type of amplifiers provides various types of waveform. So, the distortion is defined as the change in output wave shape from the input wave shape. These distortion could be of various type like amplitude distortion, harmonic distortion and crossover distortion. We will talk about these distortions little later. The next thing is in these power amplifier, the transistor should be designed in such a way that they should be capable of dissipating the high power. So, they dissipate high power. However, in case of a small signal amplifier, the transistors are made of very small size and in the case of power amplifier, the size of the transistor is chosen relatively large. It could be of few centimetres or even more. However, in case of a small signal amplifier, it is of just few millimetre size. Here, in case of power amplifier, the size is made relatively more because they have to dissipate more heat. Now, since the transistor in these power amplifiers dissipate more heat, so, there is a need of some cooling arrangement. So, you might have seen in electronic systems that the fan is associated at the back end of any instrument and this fan provides the cooling to the instrument and that cools the transistor. If the power is not very high, then this cooling is provided through the metal casing of the transistor and it dissipates heat through the metal casing. Now, if the transistor can also be cooled by placing at the back end of the circuitry and in this case, the heat goes through the convection when air flows, then this makes the transistor relatively cool. So, these are the main considerations of power amplifier. In case of power amplifier, the resistance of the collector is chosen relatively low because in these cases, the output current or the voltage swing should be very large and which can happen if the collector resistance of these amplifier is chosen as small. Next, we will talk about the various classes of power amplifiers. Power amplifiers are broadly classified into three categories depending upon for how much portion the output current will flow with respect to the input current or input signal. So, the classes of the amplifier is class A, class B and class C. Now, there are few modifications made for the class B amplifier, then the new type of amplifier is called as the class AB amplifier. This eliminates the distortion that occurs in case of class B amplifier. We will discuss about these features little later. Now, in case of class A amplifier, the output waveform follows the input cycle. So, here output current flows for the whole input cycle. So, the conduction angle in these amplifier is 360 degree. So, here you can see that the output current is the exact replica of the input signal. So, this is class A amplifier. The efficiency of these class A amplifier is relatively low. The next type of amplifier is class B amplifier. In this case, the operating point is chosen at the cutoff. So, the output current flows only for the positive half cycle. So, the output waveform will be between 0 and pi, 2 pi and 3 pi, 4 pi and 5 pi. So, the output waveform will be like this. So, here you can see that the half of the information is lost. One can get the complete waveform if the engineer uses a special type of connection using two transistors that is known as the push-pull connections. Then one can easily get the complete waveform. We will discuss about this configuration little later. Now, one may say why one should choose class B amplifier as it suffers from the distortion. The reason is that in case of class B amplifier, the efficiency of these amplifier is much higher as compared to class A amplifier. Here the efficiency is 78 percent and in case of class A amplifier, it is just 25 percent. So, this is one of the advantage of class B amplifiers. Now, as I mentioned these amplifiers suffers from the distortion. So, another type of amplifier is made by shifting the operating point slightly towards the active region. So, these type of amplifiers are known as the class AB push-pull amplifiers. The conduction angle of these amplifier is greater than 180 degree. In case of class B amplifier, the conduction angle is 180 degree and in case of class AB amplifier, it is little more than 180 degree could be around 200 or 220 degree depending upon the situation. So, in this case, the efficiency is slightly less than the class AB amplifier, but it provides a distortion free output signal. The next type of amplifier is class C amplifier. In these amplifier, output current flows for only very small portion of the input signal. So, you can say if you select somewhere here, then the output current will flow only for this region. So, correspondingly you will get the waveform like this. They provide a very small signal with very high output power. So, these pulses are of very high output power. When these pulses are connected with LC tank circuit, then they provide the carrier wave which is of sinusoidal in nature. So, they are used in generating the sinusoidal carrier wave. Now, if I compare these amplifiers in terms of conduction angle and efficiency. So, the conduction angle is maximum for class A amplifier and it is minimum for class C amplifier. For class A amplifier, it is around 360 degree and for class C, it is of the order of 10 degree or 20 degree depending upon the situation how high pulse or for how much duration you are interested in the pulse that is to be generated by class C amplifier. Now, if I talk about in terms of efficiency, then the efficiency of class A amplifier is minimum that is about 20 percent or 25 percent and it is maximum for class C amplifier. It is around 95 percent. So, they are the most efficient power amplifiers among these class A, B and C amplifiers. Next, we will take an example of class A amplifier. So, here is the circuit of class A amplifier. This is the supply voltage, these are the biasing resistor, this is collective resistor and these are the decoupling capacitors. The input is applied here. So, in this case, these input is coupled to the circuit through this decoupling capacitor and the output resistance is RL and this is connected to the circuit through this decoupling capacitor. We have already seen the DC load line of this circuit. Now, one may say that this circuit looks similar to the small signal amplifier. So, although in appearance they are similar, but the consideration of this transistor is quite different in case of power amplifier because of the dissipation and other considerations as we discussed earlier. Now, we will see that here we are interested in AC power that is delivered to the load. So, we will see the AC load line. Now, one may say that why the AC load line and DC load lines are different. So, if you see here, the impedance seen at this end will be different for the DC circuit and the AC circuit. In case of DC circuit, these capacitors will act like a infinite impedance. So, they will not make connection with the output. So, the impedance seen at this end will be only RC. However, in case of AC analysis, we short circuit all the supply sources and short circuit all the decoupling capacitors. So, if this will be shorted, then the impedance seen at this end will be the parallel combination of RC and RL which is represented by a small RC. So, the impedance seen is different for the AC and DC analysis and that is why the AC load line will be different than the DC load line. Now, just to do the analysis of this circuit, short circuit all the supply sources and short circuit all the decoupling capacitors and then apply the Kirchhoff law in this loop and then superimpose the current achieved in this with the DC current. By simplifying this, you will get this expression IC is equals to minus VCE upon RC plus ICQ plus VCEQ upon RC. This represents the relation between the IC and VCE. It is a straight line. So, one can draw a straight line using this when they want to see the characteristics in IC versus VCE. So, to draw a AC load line, just put VCE equals to 0 to calculate the IC saturation current. So, this will come out to be ICQ plus VCEQ and for the cutoff region put IC equals to 0, then VCE will be VCEQ plus ICQ RC and the line drawn by using these point will be the AC load line. Here the slope of this load line will be minus 1 upon RC which is a parallel combination of RC and RL. Now, the next thing is where should be the operating point? So, in case of class A power amplifiers, the output current and voltage rate should be very large. So, one should choose the operating point in such a way that it should provide the maximum output power without any distortion. So, in general it is suggested that one should choose the operating point in the middle of this AC load line. Why this is chosen? Because if you choose point somewhere here, then there are chances that the positive cycle of this output signal may get clamped. This is known as the saturation clamping and if the point is chosen somewhere here, then there are chances that the negative cycle of the output signal may get clamped. This is known as the cutoff clamping. So, one should choose the operating point at the centre to get the maximum output signal without any distortion. Next, we will be seeing the how much AC power is delivered to the load. So, in order to calculate the power delivered to the load, few assumptions are made that the cutoff point is considered to be here at 0. So, this will corresponds to VCC if you see and this is taken at the saturation. So, this value of the current will be 2 ICQ. Here this represents the maximum voltage and this represents the minimum voltage. So, this will be 0 and maximum voltage over here will be VCC. This represents the minimum current and this represents the maximum current. So, the maximum current will be 2 ICQ and the minimum current will be 0. Now, to calculate the output power delivered to the load, it will be the root mean square value of VCE and ICE. So, it will be the product of RMS value of VCE and IC. Now, this will be the half of peak to peak voltage of VCE and the root mean square value of this. Similar will be in case of current. Now, here we know that VCE peak to peak is equal to VCC. So, just put these values in this expression you will get the VCE RMS value as VCC by 2 root 2 and ICE RMS value as ICQ by 2 root 2 and here 2 and 2 will cancel out. So, it will be ICQ by root 2. So, after simplifying you will get the AC power delivered to the load as VCC into ICQ by 4. Now, how much is the power supplied by the battery in this case? So, in this case the current will be supplied in these 2 branches. So, it is assumed in the analysis that the current drawn by these biasing resistors is negligible as compared to the current drawn by this resistor. So, it has been ignored. So, the supplied power will be VCC into ICQ. Now, to calculate the conversion efficiency we have already studied that the conversion efficiency is the ratio of the output power delivered to the load and the supply power. So, if you put these values then you will get the efficiency of 25 percent. Now, in this case if suppose one want to design a 10 watt power amplifier then one has to supply 40 watt of power through supply. So, in this case the lot of power will be wasted in this circuit. Here half of the power will be wasted in this resistor. This power can be saved if we replace this resistor by a transformer. So, the next type of class A amplifier is transformer coupled class A amplifier. Here the load resistance Rc is replaced by the transformer. So, this is the transformer the output resistance is Rl. Here the number of turns in the primary coil are Np and the number of turns in the secondary coil are Ns. So, the impedance seen at this end at the primary will be represented by this expression Rl dash equals to Np by Ns whole square into Rl. So, in this case the impedance matching can be easily achieved by just changing the turn ratio of the transformer. Now, in this case we know that in case of transformer the resistance is very low or it is 0. So, the DC load line for this configuration will be a vertical line because small resistance or the 0 resistance will corresponds to infinite rho. So, that will make the vertical line. So, in this case this Vceq will be equal to Vcc because there will not be any losses in this transformer. So, if you put the value of Vceq and try to calculate the conversion efficiency then it will come out to be 50 percent. So, by using transformer coupled class A amplifier the efficiency has been improved to 50 percent. So, in case if you want to design a 10 watt of amplifier. So, here you need to supply 20 watt of power in this case. Now, if you see here if the signal is absent. So, in both the amplifier configuration of class A the current will pass through this transistor. So, it will continuously dissipate the heat. So, that is one of the disadvantage of class A configuration that lot of the power is dissipated in the transistor. Now, this configuration is also not desirable because the transformer is relatively bulky and they are frequency sensitive. So, the frequency response is not very good. Now, if the transformer is used for very good quality then it becomes expensive. So, that is why this configuration is not used. The next type of amplifier configuration is class B amplifier. Here the operating point is chosen at the cut off. So, at the cut off the current will be 0. So, in this case the output current will flow only for this half cycle that is positive half cycle. There will not be any conduction in the negative half cycle. So, the output will be similar to the half wave rectifier. Now, if one want to achieve the complete waveform this is also possible using class B power amplifier. This can be made using the special type of arrangement that is known as push-pull connection. In push-pull connection two transistors are used. So, here is the push-pull arrangement of two transistors. The transistors are NPN transistors and these are the common emitter configurations. Here the input is provided by using the input transformer and the output is taken and combined from these transistors by the transformer at the output. So, in this case the input signal is decoupled at the secondary and they are in out of phase. Here you can see. So, they are tapped in such a way that the input to these transistors are out of phase. So, in this case the NPN transistor we know that the emitter base junction for this transistor should be forward biased in order to start the conduction. So, if you see here this cycle this half cycle will provide the forward bias to this transistor. So, output will appear for this much of duration. Since this configuration is common emitter configuration. So, the output will be in opposite phase to that of the input signal. So, that will appear here. Similarly, in this case the output will be appeared for this cycle of input signal. So, it will be in reverse phase for this much of duration. Now using the center tab output transformer these outputs are combined and you will get the output waveform like this. So, this is the push pull amplifier. These are not used conventionally because of the center tab transformer it becomes costly and the size of this amplifier is relatively more. So, the next type of push pull arrangement is using the complementary transistors. Here the transistors are NPN and PNP they are connected back to back here the output is taken at the load and the input is provided here. So, in NPN transistor the emitter base junction will be forward bias if the positive cycle is provided here and in case of PNP transistor if the negative voltage is provided then it will be in the conduction state or in the active region. So, for the positive half cycle the output will appear through this transistor and for the negative half cycle the output will appear through this transistor. Now if a load is connected somewhere here then this output achieved at the load will be a complete waveform. So, this is how the complete waveform is made using the push pull connection of complementary transistors. Now the power delivered to the load in these transistors amplifier will be given by the root mean square value of VCE into IC. Here VCEQ will be half of the VCC because the output current is appeared for only half duration. So, VCEQ will be VCC by 2. So, by putting these values you can calculate the output AC power delivered to the load. So, that will come out to be VCC into ICQ by 4. Now if you want to calculate the power supplied so this will be the product of VCC and the current supplied by the battery. So, this current supplied will be the average value of the current. In this case the output does not appear for the full duration. So, we should take the average over full period. So, the average value will come out to be ICQ by pi. Now for the conversion efficiency it is the ratio of the output power by the supplied power and if you put these values the ratio will come out to be pi by 4. So, the conversion efficiency for these amplifier is 78.5 percent and this is relatively very high as compared to class A amplifier. So, this is one of the advantage of class B amplifiers. Now these amplifiers suffers from the distortion. So, if you see in this push pull arrangement these transistors should be of similar properties. If the properties are not similar then the output signal may contains the high harmonics of the fundamental frequency. So, the output could be containing the fundamental frequency and the second harmonic, third harmonic etcetera. So, if you see here this blue curve is the fundamental frequency and this green one is second harmonic. So, the output waveform will be like this which is not desirable. So, this type of distortion is known as the harmonic distortion. So, the harmonic distortion is given by the value acquired by the harmonics. So, this is given by the root mean square value of the harmonics over the power delivered to the fundamental harmonics. So, here A 1 is the amplitude of the fundamental frequency and A n is the amplitude of the nth harmonic. So, in case of class B amplifier since the output current appeared due to the complementary transistors are in out of phase. So, the even harmonics will be absent in case of class B power amplifiers only odd harmonics will be present. So, this distortion will be relatively very less because the main distortion will be due to this second harmonic. So, this harmonic will be absent in class B amplifier. So, the harmonic distortion will be relatively less in case of class B amplifiers and this will be given by this particular expression. The next type of distortion is the crossover distortion. In the analysis we assumed that these transistors are ideal in nature, but practically they are made of silicon then here the voltage required for the biasing of a meter base junction is 0.7 volt in case of NPN transistor and it should be minus 0.7 volt for PNP transistor. So, there will not be any conduction when the positive half cycle will be there between 0 to 0.7 volt and similarly in this case there will not be any conduction when it will be between minus 0.7 to 0. So, the output will not appear for that duration. So, in this case the output waveform will not be the replica of input or it will not be a complete sinusoidal waveform. So, this type of distortion is known as the crossover distortion. This type of distortion can be reduced by shifting the operating point slightly towards the active region. So, that type of amplifier are known as the class AB amplifiers. So, this is the circuit of resistor biased class AB amplifier here these R2 and R3 are used to provide the biasing to these transistors. So, that they should remain in conduction state for all the time even in case of 0 input signal. Here you can see that the operating point is chosen here just slightly towards the active region this is approximately 5 to 10 percent of the maximum value of the collector current. So, in this case the output will be a complete waveform, but this configuration suffers with the temperature variation because the emitter based junction voltage varies with temperature we know that the base emitter voltage varies by 2.5 millivolt by increasing the temperature by 1 degree centigrade. Now, suppose if the temperature increases by 20 degree then it will corresponds to 50 millivolt variation. Now, the biasing voltage that is required is around 600 millivolt. So, this 50 millivolt variation will be relatively huge. So, there are chances that it may shift the operating point which may not be desirable. Because it will provide the distortion. So, the next type of configuration is the diode biased configuration here these diodes are used to provide the biasing to these NPN and PNP transistors. The properties of these diodes are similar to the emitter based junction of NPN transistor and PNP transistor. So, in these transistors if the temperature changes then similar variations occurs in the emitter based junction of transistors and the similar variation occurs in these diodes. So, they do not suffer with the variation of the operating point. So, they provide the complete waveform without any distortion. So, this class amplifiers gets rid of the distortion. This is one of the biggest advantage of class AB amplifier. So, if one want to use the class B amplifier then it is suggested in general that one should use the class AB amplifier because it provides a distortion free signal and the efficiency of these amplifiers is just slightly less than the class B amplifier. The conduction angle of these amplifier is slightly greater than 180 degree it depends upon the biasing situation. So, it could be of around 200 or 220 degree. Here you can see for this much of duration the output signal will appear in these cases. So, the application of these class AB amplifiers are they can be used in the jammers or signal enhancers and in many other circuits wherever high power is required and efficiency is also one of the considerations. The next type of amplifier is the class C amplifier. In class C amplifiers the output current is present for only very small portion of the input signal. So, here the operating point is chosen away from the cutoff point. So, in this case the bias voltage applied will be negative. So, that is why in the circuit analysis this negative bias voltage is chosen here the output will appear for this much of portion. Now, if you see here here the input signal applied is sinusoidal in nature these are the decoupling capacitors. So, in this case for the negative cycle it will add up to this negative values. So, there will not be any conduction in the transistor and for the positive half cycle in the starting it will try to compensate for this negative bias and for very high value and for the very small portion of the input signal this transistor will go into the active region. So, the collector current will appear. So, here the collector current will be of very less duration with very high power. So, it will be in the form of pulses like this. In standalone configuration these pulses are of not very use. Now, when the tank circuit is connected at the collector end of this particular amplifier then these pulses strikes this tank circuit. Since we know that the time duration for these pulses is very less. So, the frequency will be very high and we know in case of tank circuit the impedance offered by the inductor will be very high for the high frequency range and it will act like a open circuit and in case of capacitor it will provide very low impedance. So, it will act like a short circuit. So, it will start charging. So, the capacitor will charge up to the voltage Vcc or slightly less than Vcc. Then this capacitor will try to discharge by providing current to this inductor and then it will discharge up to the value 0 and then the reverse process will start. So, in this way it will generate the complete sinusoidal waveform. So, this is one of the biggest application of these class C amplifier. So, using the tank circuit one can generate a sinusoidal career wave using the class C amplifiers. The efficiency of these amplifiers is around 95 percent. So, if the supplied power is around 1 kilo watt then the sinusoidal career wave of 900 watt or maybe 950 watt can be achieved using these class C amplifier which is very high value. So, the application of these amplifier is to provide the career signal that is of very high output power. So, they are used in television signals in radar systems to generate the career signals. This is one of the advantage. The limitation of these amplifier is that they can only generate the career signal of sinusoidal in nature. Now, we will take two examples of power amplifier. So, here is the example of 2 watt power amplifier. The number of the IC used in this amplifier is SPB 2026Z. Here the impedance matching is provided. We will not discuss about the impedance matching because it has been already told to you in previous lectures. This is a class AB transistor based power amplifier. The nominal supply voltage for this amplifier is 5 volt and the 1 dB compression point is 33.8 dBm. Here the 1 dB compression point means that for the RF input signal the power gain decreases by 1 dB. So, then RF input corresponding to that variation is known as the 1 dB compression point. So, in this case this is 33.8 dB. The RF input for this amplifier is taken 21.5 dB for measurement. So, here is the plot of this amplifier. The output power is shown here. This amplifier is designed for the GSM 2G band from 1800 to 2000 megahertz or 1.8 gigahertz to 2 gigahertz range. The output power achieved by this amplifier is maximum that is 33.2 dBm at 1.9 gigahertz with the efficiency of 38.1 percent and the current drawn by this amplifier is 1 ampere. The cooling arrangements are made by using these multiple holes and in multiple number of wires. The next type of amplifier is the 30 watt power amplifier. Here you can see that the cooling arrangement are made by placing a copper plate at the back end. So, it will dissipate the heat generated by this transistor of power amplifier. This is also class AB power amplifier. The nominal supply voltage for this amplifier is 28 volt. The 1 dB compression point for this amplifier is 44.8 dBm and the RF input given to this amplifier is 17 dBm and at the output this is connected with the 30 dB attenuator because of the limitation of the instruments that we have for the measurement. So, by considering this the output power achieved by this amplifier is around 44.1 dBm with 2.6 ampere current drawn by the circuit. The efficiency achieved by this amplifier is of the order of around 40 percent. So, these are the examples of the power amplifiers. So, now I would like to conclude in this lecture we talked about the power amplifiers. We talked about the various parameters which takes care of various performance parameters of the power amplifiers. After that we talked about the class A power amplifier. We saw that the efficiency of the class A power amplifier is very less for RC coupled transistor based power amplifier configuration. Then we talked about the transformer coupled class A power amplifiers. The efficiency has been increased to 50 percent, but these class A amplifiers suffers with large heat dissipation when the signal is not present in the amplifier. Then we talked about the class B amplifier. The operating point is chosen at the cutoff frequency. So, there are no dissipation in class B amplifiers in the absence of the signal, but the class B amplifier suffers from the distortion which could be the harmonic distortion or the crossover distortion. Now, these crossover distortions are eliminated by using the different configuration that is known as the class AB amplifier which makes use of the push pull arrangement of two transistors and the biasing is provided with the help of the resistor and the diode. These class AB amplifiers provides slightly less efficiency as compared to class B amplifier. For class B amplifier the efficiency is 78.5 percent. However, for class AB amplifier could be of around between 50 to 70 percent, but they provide the distortion free output signal. Then we talked about the class C power amplifiers. We saw that the output current appears for very less duration of the input signal. So, these amplifier are used to provide the short pulses of very high output power. Now, when these short pulses strikes to the LC tank circuit, then they provides the carrier sinusoidal wave of very high power. These carrier waves are generally used in the radar systems or in radio signals or even in the television signals mostly. So, efficiency of class C amplifier is the highest among class A, B and C amplifiers whatever we discussed in this lecture. So, one should choose the class C operation if one is interested for the maximum efficiency and otherwise if one is interested for the distortion free signal, then that person should go for the class A amplifiers. After that we talked about two practical examples of the power amplifier. We talked about the 2 watt and 30 watt power amplifiers. Then we saw what type of cooling arrangements were made and how they are matched over the frequency range. So, with this I would like to conclude. Thank you very much.