Well, in this video I am going to explain a super interesting amplifier, it is an integrated audio amplifier based on this IC8002, it is a very small integrated that is used in low voltage audio applications also in both laptop and desktop computer systems. It is interesting that I am going to explain to you step by step how an internal operation is like so that you can make the most of the possibilities, do not miss it, you already know my name is Aurelio Cadenas and without further ado, this starts first, as you can see, this integrated circuit has eight terminals internally this equivalent circuit based on two operational amplifiers has a resistive voltage divider and another divider here with 40K, how are we going to analyze this this circuit? Very simple, look what I notice is that around here we feed with a voltage between 0 and 5 volts which is the most common but it also admits smaller voltages later I will tell you more details about this interesting integrated circuit it is a very good class AB amplifier of very good quality on the internet you have demonstrations of how it sounds let's see then look around here we feed. We are going to assume with 5 volts and we place a capacitor to stabilize that supply, to prevent voltage peaks that may come from that source from affecting the operation of the integrated or also the operation of the integrated itself make the output of the source oscillate and generate variations of high frequency voltage, then we place a capacitor there to stabilize (to dampen those voltage variations) and we are also going to place another capacitor here. note that here internally there is a voltage divider that is to say that between here we have 5 volts because in this terminal in terminal 2 we would have 2.5 volts that would be in permanent regime but at the initial moment of connecting the power supply this capacitor is discharged and therefore the voltage at this point in the 2 in the positive input of the two are zero volts, so why do we place this capacitor? To avoid that click produced when connecting or disconnecting the power supply and that can appear in the speaker or in the earphone that we then place that one microfarad capacitor (more than enough) is to avoid that the voltage here appears suddenly then when having this capacitor what we do is retard and make the voltage at this input slowly rise to 2.5 volts not suddenly, well, so for our analysis this capacitor is already charged and we forget about it, okay? What this other capacitor forms is a high pass filter, that is to say that here we would connect our audio signal and here what we are doing is letting the high frequencies pass, the low frequencies are going to dampen them, it is going to eliminate them. This operational amplifier, note that it is an inverting amplifier, we are introducing the signal through the negative input and we have negative feedback, therefore assuming that this capacitor, sorry with this amplifier is not going to be saturated because the voltage here and here will be practically the same, then the voltage at point 4 is going to be 2.5 volts due to this voltage divider. OK, so what we have is that the tension in this resistor is the tension that I'm going to have at point 5, okay? What tension can point 5 have? that of this resistance plus the voltage that I have here with respect to ground, which are 2.5 volts. Then notice that what we are doing is mounting an audio signal on top of a 2.5 volt direct component in order to amplify an alternating audio signal with its positive and negative half cycle with a direct current supply. For a simple analysis we are going to apply the superposition theorem, okay? That is to say, this amplified one, I am going to consider it excited with two signals, one alternating and one continuous, then from the point of view of direct current that I have here, look here I have 2.5 volts this capacitor from the point of view of direct current, how does it behave ? as an open circuit because its impedance is 1 divided by 2 by PI by the frequency, the frequency is zero, then it presents an infinite impedance therefore it is as if it were open, that is to say that the current will not circulate here, that is, if here I have 2.5 volts here I also have 2.5 volts. Does current flow through this resistance? not because the current that circulates through here is very small (an operational amplifier for an input remembers that it is very small, negligible) then here I do not have voltage either, that is to say that the direct current voltage of 2.5 V I also have it here that is that from the point of view of direct current the voltage at point 5 we are going to call it this way with this line is 2.5 volts we are going to put it like this Vcc divided by 2. and from the point of view of alternating current we are going to see from the point From an alternating current point of view, now this point of this mass is valid because it eliminates the component, then he applied his proposition annulled the with direct current that is to say that here we would not have voltage here we would not have voltage this capacitor to connect alternating is like a conductor and what I have What I have is that the voltage at the output, what I have is an inverting amplifier, it is worth less this resistance / than this other one and x look but less well then its pr is nothing opposition now what we do is add that is to say that the voltage of point 5 would be the sum of this plus this rf king's match on the ground well then we already know that in signal we have there what signal we have there what we have is a signal what it has a continuous component of ufc media etc media and to that is added the input signal multiplied by that quotient between this value and this value and with a minus sign that is, and here what we would have would be I'm going to paint it blue we would have this signal like this And that is what we have right away if we go now with the one below, we now go with the amplifier, notice that it is an amplifier, also look at the two resistors, the input voltage here, here, of these two resistors, the cards with a direct current voltage. What I'm going to do is also apply superposition, okay, so what I'm going to do now is see what voltage I have an output due to this input and see what voltage I have quality of life to this other input Ok then when I am with this entry this entry I mass it to 0 the cloud and when I am with this one what I do is set it to better since I am not now in incorrect direct current t no what is analysts of this circuit is the first one took an input and no longer the other one and then took they are no longer right then canceling this input what I have here is an inverting amplifier this would be the input is worth this would be the gain which is 140 article 40 therefore in the output 2 already taking the input the input and person considering the inverting input what I have is that the attention in point 8 I am going to call it that way, it would be less the attention that was here that is the attention of point 5 5.5 if not less than It would be this because if not it would be less it raises etc means + rf the king's party for its sake and now I am going to consider the positive input the positive input the positive input and what I do is cancel this input there is here is 0 then what is it what you I engo what I have is a non-inverse amplifier I am putting an ub-cc voltage here means this I have to steel mass then the gain in an amplifier the verse reminds attention renewal would be gain the tension behind the pensions etc etc divided by 2 x The win and the profit is one more, this is divided by this 40 houses, 40 houses, then nothing, notice this is one plus one, two games, one ub-cc ub-cc well, how am I applying your proposition then I will say that the tension in 8 who is it? the sum of these two the sum of this - v cc means to do this plus this well look at ub-cc minus ub-cc means what I have is ub-cc means plus rf king's match for an envelope and this is the voltage that I have at point 8 that is at point 8 as we did here what I have note that it is the same the same as vdc media v from media a continuing education and the same there is but be careful that here it is not displaced 180 degrees like here this but less is that here is the signal like this, then this is the signal that I have at 8 and this is in turn events and now that the speaker is reaching me, the speaker is reaching me here, what is reaching the speaker is the potential difference, the voltage between 8 and 5 that is, as this is the positive terminal of the speaker, since the voltage that I have between 8 and 5, the difference would be this minus this and look at vfc minus, in turn, there would be less means, sc health rises to zero and this minus minus this is two Sometimes this, so what I have here is twice the king's match, in turn, that is the gain I have from this amplifier integrated with those two with those external components and so simple of brno transition from betis if you have compression problems it I invite you to the one to start is the electronics course because surely there are gaps and then all this of the signs and equations is basic and you have to start from scratch from the beginning taking into account all this tension and from zero, well then, notice that we do not need that capacitor that we sometimes place in the outputs of a speaker to eliminate that continuous component, here we do not have it here, we are eliminated here, I do not know if it is not eliminated because here and here we have the The same component of knowledge is to make the difference, it is canceled, and the signal here looks at them here as they are the same but they are displaced to close the subtraction that the two times that it gives twice well a detail that I forgot is that this integrated audio amplifier class ab has a psuv down input if we have it set at a high level because it is ub-cc this does not work that is, it does not work it is in a minimum consumption saving mode so for it to work you have to set the input to more is To say that what we would do would be to put this input through either a signal that comes to me from a micro that presents a signal of a small voltage of zero votes or of 02 in a low signal then it could start to work and now It could work and however if we put this signal on high because it would go into saving mode in standby mode and they would not amplify it would not work it is an interesting input also this can this amplifier you can try it we can simulate it I will do it next but On the internet on the internet they test it with the output of a cell phone from a mobile phone and mobile phones, the audio output is an amplified signal, and then some dispense with this resistance, they connect it directly here and what we get is that the gain is practically only two because so that would be 222 but if we connect here a microphone that gives a signal of a thousand volts of 50,000 volts for example we can have a gain of 100 the gain said well a gain of 50 of what you need is worth and if here we put an adjustable resistance we can regulate the gain well let's move on to the simulation well then there you have the simulation let's do r a pause in the simulator to take more precise values and as you can see we have a low frequency generator that is providing an amplitude of 30 millivolts of the peak value and a frequency of one third we look here at what the signal would be in red color and we can Make a more approximate measurement of 29.4 more or less the maximum value and here we would have the signal in green color that we would be measuring at the output of this operational amplifier, which would be terminal 5 of the integrated amplifier, and as you can see, it has a continuous component which is in 2.5 volts since we have the scale the value of each height of each division of a volt 1 2 2 and a half is worth and here the same in the signal in blue color would be the output of this operational amplifier while the output in color Red is the difference between this output and this output. Well, let's measure the amplitude of this output, which would be around 140 and 57, that is, the gain 50 games out of 20 then Like 5 and by 2 a gain of 5 is to say that if we multiply 29 with 4 by 5 approximately, we would get one hundred almost 146 millivolts well more things this capacitor is going to see what cutoff frequency it is going to provide us for this we have here the diagram of bode and as you can see we have a gain that would be around 14 decibels that with 3 decibels below about 11 over there this would be the cut-off point that would be at 21 hertz more or less and the upper cut-off point the point higher cut would be out there out there more or less more or less over 16 kilos is worth that would be the bandwidth of this amplifier with this capacitor is worth well finally as the most remarkable characteristics of this integrated circuit is that with power supply 5 volts we achieve a distension of less than 10 percent and a power of 3 watts on a load on a speaker of 3 or mine or in case of using a speaker of 8 or mine we would reach up to 1.8 watts os and a total harmonic distortion of 10 percent the maximum power ranges that would be up to 6 volts the input voltage we could feed from minus 0.3 volts to 03 volts above the power supply well with respect to consumption as you see there If we do not have input voltage and there is no load, we have a maximum consumption of 10,000 amps in case of having the 8 or mine load connected even though we do not have input voltage, it keeps that consumption at 10,000 amps, the typical thing is that they are 6 and medium and 7000 amps nothing a very small consumption that better the power that I had already mentioned and the total harmonic distortion plus the noise that practically nothing has 0 2% is a very very small distortion I already tell you that it is an amplifier with enough quality this would be the audio pins of the integrated circuit op capsule as we can see the manufacturer also provides the gain in closed block twice resistance resale tion divided by the input resistance, the maximum dissipation power is what you see here in this formula provided by the manufacturer and we must always try in the case of needing to place a heat sink to avoid reaching 150 degrees centigrade of course It would be necessary to limit the temperature much earlier so as not to shorten the life of the integrated, well, nothing up to here is all I hope you liked that you have enjoyed that you have learned something else this can be done with discrete components with amplifiers and check it is ok and if not use This integrated or super interesting high quality that you will see that it is also a price has a very low price that can be obtained in the Chinese market and well we enjoy an application that at a given moment we need to amplify we see ourselves better said you see me in the next class bye
