In this video I am going to explain to you how a class B power amplifier works, especially the bird class which is the one most used to amplify to amplify alternating current signals.We will first start with a very simple example so that you understand the idea and then already we went to a more complete and more professional circuit so don't miss it from the beginning to the end because learning very interesting things you already know my name is open it chains and this starts right away to start you will remember that the class power amplifier we used alone a transistor for amplifying the semi positive wave and half negative wave is when we introduced a signal alternating transistor had to drive all this it time for it we would just place the transistor at the midpoint and so we add one component continue that I already explained it to you in the previous class and I was able to amplify between a low voltage level that would be this and the highest level s high voltage that would be close to the supply voltage in class b amplifiers the transistor is only going to amplify a half wave, that is to say that we are going to use two transistors, one that is responsible for amplifying the seed, one positive and another that will be responsible to amplify the negative half wave then what we get in this way is an amplifier a much greater amplification a much greater increase because if this transistor amplifies from zero to ub-cc and this amplifier amplifies and this transistor, sorry, amplifies from zero to less ub- cc we get a greater amplitude of voltage between more v cc and less v cc this in theory is worth this in theory so how are you seeing here we need two complementary transistors that is to say one of channel n and another of channel p it is very important that they are complementary then the idea is that during the positive week during the positive wave seed it is clear that this transistor is going to conduct, look at the voltage between ba If emitter, the voltage between base and emitter will be equal to the input attention minus the output voltage, that is, the voltage goes to that emitter, the transistor one, we represent this way by means of an arrow, this voltage will be the potential difference that I have between this point and this point. at this point what I have is a rise and in the output in principle we would have zero because we suppose that these two transistors do not conduct and the output voltage would make then the more emitting voltage would be neither more nor less than the input voltage and therefore So much it is directly polarized, this point is positive like this one and that transistor will conduct the output voltage to whom the output voltage will be equal if we apply Kirch's law of voltages ok we take and leave here I use that more emitter base voltage - I use because I find it in the opposite direction the arrow is equal to zero if we clear from here use s the output voltage would be the input voltage - the base emitter voltage of that transistor ent Once what we see is that this assembly does not amplify the input voltage at all because the output voltage is the input voltage decreased by a small amount, which is the most emitting voltage, however if we are going to have a current if we are going to have a current around here that is going to circulate through the load and that collector current that collector current or that emitter current that circulates through the load resistance is much greater than the intensity that this generator supplies, which is the one that circulates through the base then what we're getting is to amplify the intensity and no stress but this way we are also simplifying it power because we grow if we are amplifying the intensity sure what will happen to that million negative opposite the negative wave seed is arrow this arrow is going to be negative that is to say here we would have less and here we would have more therefore the transistor above does not conduct and the transistor below notice that we have to get on the transistor below it is that this emitter voltage base is positive the positive basis issuer or issuer base or negative emitter gives Ditto notice that now this tension is the tension that I have here to come up less The voltage that I have here, which is pu, then the base emitter voltage rises, as we are assuming that this junction does not conduct, here we would have a voltage of 0, then it would be 0 - and it rises, but as you can see, we are going to write it like this, the base emitter voltage of transistor 2 would be 0 which is the output voltage - the input voltage but be careful the input voltage is negative therefore the base emitter voltage is positive - less is better then this transistor is driving me and the output voltage would be neither more nor less that the input voltage which would be applying to it but notice that now if we write the equation the output voltage the output voltage that to whom it will be the same we write the same as before u So s less less emitter base of transistor 2 and less its v and less in turn is equal to zero if we clear from here in turn the emitter voltage will rise more, but notice that u rises is negative, then what we would have would be this voltage output would be input minus that this small voltage drop that would occur in the base right then issuer as I have already said output signal voltage is input decreased a little bit in a small voltage but clear and current and current through the load as it would have in the sim jong the negative would have this sense the circle would do like this is ok then as you can see we need a symmetric supply a source plus 30 minus 30 for example but we will see later that we can work with an asymmetric source between 0 and 30 Between 0 and 15, the advantage of this assembly in class b is well worth it is that those transistors without input signal do not conduct, that is, no power is consumed , however in class a without input voltage Each transistor was consuming a power is worth and that is why it has a very low performance in this case they do not consume power when we have no input signal but of course there is a problem and that is that for this transistor to conduct the voltage of this generator it has to exceed those 06 volts to conduct the emitter base then notice I have said that the output voltage would be the input voltage this would be the prison of the pagan already but until the input voltage is not greater than 0 6 volts for example it will not drive me To say that here there is going to be a time when that diode will not conduct, okay and then we would have a wave such that it would be animal and less input voltage -0 6 volts approximately the same in the black hemicycle because the same would happen there a period in which the signal notice here is a distortion deformation wave wave but should be continuous so that this deformation is called cross distortion distortion crosses And what obviously has to be eliminated because that penalizes the quality if this is an audio amplifier it penalizes the quality of the audio, it is worth how you eliminate that because the assembly in class b evolved and what we do is that these transistors work instead of cutting Without an input signal what we are going to do is that without an input signal we polarize them at the point that it starts to conduct, that is to say that the voltage goes to semi or initially without input signal it is 06 07 volts put it there so that with a tad input signal the amplifier already conducts ok and then we are in the ave class that differs from the b class simply in that we polarize the transistors so that with a little input signal they start to conduct then how to achieve that there are several methods there are several methods the simplest the simplest would be to place here place a diode here here a diode and place here another day obviously these diodes we have to make them conduct ok we have to make it drive then what we do is swim place it at a resistance there we call r1 we place another resistance here a and r2 and what we get is that at this point at the point or at the point we have zero volts with no input signal the voltage of point a is zero, notice that sometimes this is a bit difficult to achieve and what we do is place an adjustable resistance here to modify this point and place it just right at 0 volts, okay and what will happen to the voltage of this other point what will happen to the output voltage at the point s of shakhtar because it will be at the same tension as the point to how is this possible see if this child is chosen to have an attention equal to the attention base my sun the tension decided to do in conduction it is similar to the base emitter voltage of that transistor that this is sometimes a bit pretentious but it works and well what we are doing is that this voltage this point a and point s will be at the same At the same potential it is valid then look at the tension between a and a and the tension between Jesse that is represented by this arrow here it does not matter if we do the kirch off tension law here the tension between I am going to erase the tension between and that will be this here voltage plus that of the diode plus that of the diode minus the emitter base minus the emitter base of transistor 1 since at the starting point equal to zero if this voltage and this voltage are equal they are subtracting they cancel they cancel out here I have 06 and here I have 06, 07 and 07 are canceled and the voltage v / s will be equal to zero that means that the point to the point is that they are at the same potential, then we are already seeing that the input voltage and the output voltage When the transistor above conducts they are the same with the positive half wave and with the negative seed the same is the same here the same here this voltage on the outside wants it to be we want it to be the same as the base emitter voltage to be worth those diodes fast switching diodes are chosen 1 n 41 48 for example and in this way we get a class a amplifier looks very basic very basic problems of these problems of this the temperature what happens if here the temperature increases if the temperature increases this transistor It is better to drive and if it conducts more the collector current increases and the collector current increases, the temperature increases and we have a problem there, a thermal avalanche effect, what do we do this well because I already explained to you in a previous class when it is polarized go the transistors that were used to put a small resistance here a small resistance we are going to respect here and the same in the one below a small resistance so that if the temperature increases the collector current will increase but there is friend the voltage in this resistance would also increase and as the voltage of the diode is cutting because it is polarized there it is constant because then if this voltage increases this voltage that e misor decreases you simply have to do this mesh if at this time audience it decreases then what we get is a negative feedback effect that increases the temperature increases the current but be careful increases this voltage decreases the base voltage my short and decreases the current of base and therefore the collector current decreases if and in that way we stop that increase in temperature the same way down exactly the same way down well, therefore what we are getting is that if this transistor the base emitter voltage is practically going to have a value without input voltage of those 06 volts that this diode gives it, what we are achieving is eliminating this of this crossover distortion, then what we would have more or less is that the transistor would start to conduct from zero and we would have this signal of input and we would no longer have that crossover distortion, notice that in this setup with symmetrical voltages of plus 30 minus 30 for example more than 24 and less than 24 the tension that you want, obviously the higher the supply voltage because the greater the power that will have on your face because it will circulate more current so we wanted to achieve power without asymmetric ie between 0 and 24 volts is worth the symmetrical supply, what allows me is not to have to place a coupling capacitor here and here a capacitor from that cup because when we work with symmetrical voltages it is not necessary but of course I now put this to mass because we will have here A problem to solve is worth because of course the voltage at this point is not going to be zero, the point voltage is not going to be zero at this point, it would be half of DC or so that to get this transistor to work, imagine that we want 24 volts here we feed with 24 votes then look at the voltage of point s when the input signal is 0 the voltage of point s should be 12 volts half of the power Then when this input voltage increases, what we get is that this transistor starts to conduct more input voltage, more current through the base and more current through the collector , therefore the voltage with the television will decrease until when until the voltage that we have between the emitting collector plus this small voltage is approximately one volt, with which point s would increase to 23 volts 23 volts here plus a vote here make 24 good and what will happen when the input voltage is negative is Negative is the negative cycle that the voltage of the point will decrease below those 12 volts and this transistor will enter the cut- off, however this transistor will begin to conduct and therefore its emitter-collector voltage will decrease, that is to say that here it will begin to decrease 12 11 9 etc. etc. and as a minimum with a minimum here we would have collector inter emitters plus this small voltage one volt, that is, the output voltage is It would oscillate between a minimum value of one volt and a maximum value of 23 volts, that is , we would have a voltage at the output that would oscillate between 23 and a boat, but of course since we only want to have voltage here in the alternating load, let's see what we are going to do. Now what I have is a signal in a supply between 0 and 24 volts, and I said that the voltage in that has to be half, the voltage in that has to be half, because there we would have 12 volts and now I What I do is that the input signal could be all this is worth it will never reach 0 because the transistors I can not bring them to saturation and never the transistor to cut to have that margin notice that the input voltage and the output voltage are practically the same As we said before, this resistance is small, it is a small resistance, and I have a 0 47 or less so now I have a problem and that is that as this voltage is going to be half of the power supply because it is going to be approximately a is ta here I have to place a capacitor I have to place a capacitor okay but not here but there is nothing we do is place a capacitor here and here another capacitor so now if we would place the alternating current generator here, this this generator would be neither more nor less a previous stage a stage in class for example to an amplification class an amplifier in class a that would be the one that will already excite this transistor of this power amplifier then this represents a previous stage you know that in amplification they are done in stages It is worth a pre-amplification and then an amplification and here we would need that to place two capacitors so that to prevent the alternating current to prevent the alternating current from circulating through the generator is valid and if allowing the alternating current to pass and add here and add here then here what I am going to have is a greater tension when this is million is positive and a less tension when it is s emi sling is negative and here the same is true then what will happen that when the half wave is positive here I have a higher voltage and this transistor is not going to conduct me ok because here I have 12 and here I have 12, eat something or 12 12 and 12 this disorder conducts and here yes when it is this week it is positive here I have a higher voltage than here and therefore this transistor conducts I already say the work will conduct when the voltage here is less than the voltage here and the voltage here in this case With this example it is 12 volts well so with this we get but now fix the output voltage has to be the same because what I have in that signal exists and this signal is a direct current signal because only the voltage that takes positive values I have here is a variable direct current ok that we can decompose it into a constant in a constant 12 volt direct current and a variable alternating and a variable alternating then as I just want that the load gets the alternating current, what I have to do is place a capacitor here, it is worth a decoupled capacitor c3 shift and that capacitor that is going to happen with the continuous one blocks it and only lets the alternating through because for the direct component it presents An infinite impedance is as if it were open, so here I do not have DC, however, for alternating current it will present a small impedance if I leave it then this capacitor will be an electrolytic responder because we need maybe 100 micros radios ten micros and where would the positive be where it is that the positive because institute tive would be here ok we would represent this way with this symbol to realize that detail the same as the input capacitors if they are electrolytic it can be positive it would be here ok positive would be there the positive would be positive and this capacitor 6- 3 this capacitor c3 would be permanently charged with what voltage with 12 volts is worth because if the voltage here is this if the capacitor has 12 votes if this we subtract 12 volts what it would be is that this signal drops down those 12 volts and what we would have neither more nor less than this signal would be an alternate signal that is The voltage that we have at the output takes for positive and negative pain that is thanks to the fact that when the top one conducts the capacitor charges and when the bottom one conducts the capacitor discharges and what we get is that it charges and discharges no more no less and so in this simple way you should try the amplifier in class ab but there is still more finally when it is required to feed low resistance loads for example an 8 or mine speaker we need a fairly large current gain and that is why Darlington assembly is used in this case we would be using two arlington assembly transistors, a power transistor for example a 30 55 and a transistor of less power is valid and the same to Here it is interesting and very important that the q1 and q3 and q2 and q4 are complementary, the most equal possible but complementary and obviously we would need four diodes because because now in this assembly talent here for this transistor to conduct conduction The base emitter voltages of each path should conduct the base voltages then from here to here what would have would be the 06 of kilos 06 of means 12 volts is worth that is why we put this diode to compensate this and this diode to compensate here that is, it contributes the voltage that this one needs and this one needs the voltage here is the same for this one and for this one with this that we get because it is a much greater current input from these two transistors a greater resistance we can put larger resistances here because the current we need Because of the bases in the darlington transistors, it is much less worth having that higher gain, larger line, because we need less co Current through the base is ok but darlington assembly is well known but also the assembly to give a complementary touch is very interesting note that now when using darlington transistors here but here the complementary slow assembly is only necessary three fingers of course you will wonder why because very easy because notice that now between here and here we only need to compensate for this emitter base junction and therefore it is the diode that does it if this junction with this diode and these two junctions with these two diodes and in that way we get we get reduce the components and produce the same effect we had amplification but there is another there is another advantage is that it could 54 have to be the same for example a 30 55 12 cn 30 55 1 and 3 complementary complementary we are going to see in a later class a class later a real assembly of an amplifier in class ave with its preamplifier in class and more or less some indications so that you can see how it can be designed r in a very simple way choosing the resistances, the diodes, the transistors to get an amplifier that you can set yourself as a challenge to get worth it in this first class, what I explain to you is how it works and how to take into account these of the amplification in class ab which is the one that is used the most because of the great quality that occurs especially in audio is worth and nothing good the same here the resistors to stop that thermal effect if they always say and recommend that the diodes that the diodes are in the same environment as in the transistors to say in the same area as close as possible so that if the temperature of the fingers increases in the transistors the temperature also increases by two degrees and thus they are compensating because you already know that if the temperature increases the voltage that you wrote down Cathode in a diode decreases and the same happens in an emitter pass because the transistors are still a pn junction between base and emitter so you already know all this you have to Take it into account for the next video that I will teach you how to design a bird class amplifier with components so that you can see how simple it is and nothing up to here all this is everything and nothing to encourage you to subscribe to my channel the sun life those videos that you like them and of course you don't see me at the next premiere which will be shortly and my head is already burning I can't anymore like the song says bye
