**In 1965, Semiconducting the Fairchild company introduced in the market uA709, the first monolithic op amp widely used. Although it enjoyed a great success, this first generation of op amps had many disadvantages. This fact lead to make an improved op amp, uA741. Because he is very cheap and simple to use, uA741 has been an enormous successful. Other designs of the 741 have appeared from then in the market. For example, Motorola produces the MC1741, Semiconducting National the LM741 and Texas Instruments the SN72741. All these op amps is equivalent to uA741, since they have the same specifications in his leaves of characteristics. In order to simplify the name, most of people it has avoided the area codes and to this op amp of great use it is called 741 simply to him**
## 2. Basic ideas:Op amps are called because we can find circuits mounted with these amplifiers that conduct mathematical operations, like for example sumadores, differentiating, integrating, comparators... Etc. They are elements very used in the analogical electronics, as you can verify in this page, have a pile of applications. The figure shows the typical electronic symbol of an op amp. This concretely it is one fourth part of the LM324 since 4 op amps in a same integrated circuit come. In this example, pins 4 and 11 are of feeding. An op amp can be fed with simple tension or symmetrical tension. The simple tension consists of feeding with two cables, the one positive and the other mass (for example to 12 volts). The symmetrical tension consists of feeding the circuit with three cables, the one positive, another one the one of mass and another one the negative, with the same tension that the negative positive but (for example ±12) The difference between using a type or the other of feeding is in which we want to obtain in the exit: if in the exit we want to obtain positive tensions and negative we will have to use the symmetrical feeding, if single we want to obtain positive tensions we can use simple feeding. Also you will have to consider that neither the entrances nor the exits of the operational one will be able to exceed the limits marked by the feeding, that is to say, if you feed 12 V you do not hope to obtain 15 volts when coming out. Pins 2 and 3 are the entrances, and the 1 is the exit. At the time of analyzing circuits with op amps quédate with this idea because is said well that the current by the entrances investing and noninvesting of the operational one is zero, it is very important. You will observe in the symbol that one of the entrances has the sign + and in the other the sign -. To which it has sign + is called noninvesting entrance to him, and to which it has the sign - entered investing. The operation of the op amp depends on the rest of the circuit. At the moment, the only thing that I can say is that the op amp "reads" the tension in the noninvesting entrance, "reduces" the tension to him of the investing entrance, the result multiplies it by a very great number and that removes it by the exit. Perhaps this does not help much to understand the circuits you with op amps but little by little you will be it understanding. ## 3. Analysis of circuits with negative refeeding:3.1. Introduction One says that a circuit with operational has negative refeeding part of the exit or everything is led back to the investing entrance. When we were with a circuit with negative refeeding we must understand that the operational one, by its form to work, will vary the exit of such form that the tension of the investing entrance will be equal to the one of the noninvesting entrance. This is another important idea that you must consider. The example of simple circuit but that is happened to me with this type of refeeding is the following one: As you see in the figure, the exit tension is introduced directly by the investing entrance, this means that we were before a circuit with negative refeeding, therefore we can say that the tension in the noninvesting entrance is equal to the one of the investing entrance. Therefore we directly found the exit tension, that will be equal to the entrance tension (Vin=Vout). To this circuit tension follower is called to him "" since the exit is identical to the entrance. 3.2. Investing amplifier We are going to be mounting an investing amplifier step by step and you will be it understanding step by step. We left from our op amp: Now we are going to add a R1 resistance to him from entrance + to mass: You must remember that the current that enters by anyone of the two entrances of the operational one is zero, therefore will not circulate current around R1 and the tension in entrance + will be 0 (V=I*R1=0*R1=0). It is just like if we connected entrance + to mass directly, but a resistance is put because the circuit works better can prove not to put it and will see that the circuit also works. Next we put the negative refeeding to him by means of a R2 resistance: We already can say that we are before a circuit with negative refeeding, asi that we can say that the tension in entrance + is equal to the tension of the entrance -, that is to say, 0. But we need to put the entrance of the circuit, the entrance we will put it by means of R3 of the following way: This it is the complete investing amplifier, and everything what we have said until now fulfills, so we happen to analyze it. For it we will lean in the following graph, that shows to all the currents and tensions of the circuit All the circuits with operational are analyzed of very similar form, asi that pays attention. We looked for a mathematical equation that it relates the entrance to us to the exit. First there are the expression of the current of I1 entrance. For it you must consider the tension to which this submissive R3. That will be Vin-0=Vin. The tension in a resistance will always come given according to the direction in which we paint the current, and sera ': the tension of the side of the resistance by where the current except the tension of the side of the resistance by where enters leaves. Therefore according to the equation If we observed the figure and we remembered that by the entrance of the operational one some did not go current we reached the conclusion that I2 = I1, so we will calculate of the same I2 form and we will equal it to I1. According to this we will write Equaling I2 and I1: According to the obtained expression we reached the conclusion that the tension of Vout exit is the one of changed entrance of sign and multiplied by a constant (R2/R3). To this Gain of the circuit is called to him. This circuit has a gain (Av) negative of -(r2/r3) and therefore we can write that: Practical application: So that you see that everything what has said is certain I invite to you to that I mount the following circuit and you verify your same one with a voltmeter that all this is fulfilled actually. The used op amp is the 741: In this circuit the Av = - 56K/27K = -2,07 for the verification you can follow the following points: • To fit Vin by means of the potentiometer to 4 volts • To measure Vout with respect to mass and to verify that Vout = -2,07 * Vin = 8,28V (much comes near, ten in account the tolerances of the resistance) • To measure the tensions in entrances + and - and to verify that volts are 0 • To change Vin and to return to measure Vout, verify that when increasing Vin it arrives a little while in which the exit of the operational one cannot continue lowering, you will have arrived at the "tension of saturation" from the operational one. When diminishing Vin it will end up happening the same with the exit. • Test to clear R1 and directly connects the noninvesting entrance to mass. The circuit continues working correctly, nevertheless the positioning of this R1 resistance is advisable and has to be of value:: 3.3. NonInvesting amplifier: In this type of amplifier, unlike the investor, the entrance I saw will enter directly by the noninvesting entrance of the op amp (entered +): Next we will put the negative refeeding by means of the R1 resistance: In order to finish the circuits we added the R3 resistance of the following form: Now we are going to find the relation enters the exit and the entrance. It remembers once again that the tensions in the noninvesting entrance and the investing entrance are equal and that the current of entrance to the operational one is zero, therefore I1 is equal to I2. So we do not have but that to calculate the two separately and soon to equal them: Tension of R2 = Vi Tension of R1 = Vo-Vi Equaling I1 and I2 Therefore, this circuit has a gain in Av tension = 1 + R1/R2. This means that the exit will be Av times the entrance, without being reversed the signal since Av is positive. Practical application: So that you see that everything what has said is certain I invite to you to that I mount the following circuit and you verify your same one with a voltmeter that all this is fulfilled actually. The used op amp is the 741: In this circuit the Av = 1+ 47K/33K = 2,42 • It fits the tension of entrance to 4 volts by means of the potentiometer of 10K and verifies that the exit is Vo = 4 Av * = 9.7 volts • It measures the tension in the entrances investing and noninvesting and verifies that they are equal. • It varies the potentiometer to your taste and verifies that always one is fulfilled that Vo = Av * Vi • It arrives a little while that Vo cannot raise nor lower but = > saturation tension 3.4. Small amplifier of audio An op amp is not able in case single to give very great currents by the exit, reason why we cannot connect directly a loudspeaker to them and oir music. We have seen until now that the investing and noninvesting amplifiers have a gain in Av. tension This is not the problem, the problem this in the current (Amperes) that are able to give, we needed then to add some device that is able to extend that current. The device able to do this is the transistor, and the form but simple to use it is the following one: In the figure you can distinguish an investing amplifier in which we have made some changes. The used op amp is a LM833, special for audio. You can prove with other operational ones and you will see the difference. The stage of transistors formed by Q1 and Q2 has like only purpose of providing all the current that cannot the operational one. The used transistors are the BC547 and BC557, These are not of great power reason why you will have to use a small loudspeaker. watch where the refeeding has putting, directly in the exit happening over both transistors. This solves some problems of lack of linearity in the stage of the transistors. In addition, if you analyze the circuit just as we did in the section of the investing amplifier you will see that it leaves to you exactly the same: In addition, R2 is a variable resistor, that can do it of 0 to 22K. This does that the gain in tension (Av) of the circuit varies from 0 to -2,2 varying therefore the volume of the loudspeaker, if you want to obtain but volume you can change R2 by the greater one. 3.5. Investing Sumador We can use the op amp to add several signals, with its common mass. An amplifier of this type denominates sumador amplifier. Amplifiers of this type are in any table of mixtures. The basic form of the investing sumador is: If you pay attention a little you will see that it is not but that an investing amplifier with two entrances, and therefore, with two resistance of entrance. In order to facilitate the analysis we will put these two resistance equal (R1). V1 and V2 represent the entrance signals. The circuit is analyzed just as the investing amplifier with the difference that the I3 is the sum of the current months I1 and I2 here: We calculated I1: We calculated I2: Equaling I3 = I1 + I2: But on the other hand we can also calculate I3, like the current that happens through R3 with a tension of 0 - Vout = - Vout: Replacing this value of I3 in the equation before obtained we have: This equation says to us that the exit will be the sum of the two entrances multiplied by a number: Av = -(r3/r1). You can put all the entrances to him that queiras and the exit will be the sum of all the entrances by Av. Practical application: I propose east circuit to you, is a mixer of audio. Fíjate and you will see that this formed by blocks well that already we have studied. The operational employee is the LM833, that is an op amp special double for audio. The two entrances (V1 and V2) happen before being mixed by individual investing amplifiers of variable gain. The gain of V1 will be, according to which we have seen until now, Av1 = -(R3/R1). as R1 is 10K and R3 can vary between 0 and 10K, the gain of V1 will vary between 0 and -1, this means that we will be able to vary the volume of the V1 entrance from 0 to the same level of entrance. And the same it happens with V2. And later the mixer comes. In the circuit that I propose to you has all the equal resistance, so it will have a fixed gain of -1. When coming out you can put the small amplifier of audio which we have seen in the previous section and thus can listen to your mixtures. 4. Comparators 4.1. Introduction Frequently we want to compare a tension with other to see as he is the greater one. In this situation, a comparator can be a perfect solution. This circuit has two terminals of entrance (investing and noninvesting) and an exit terminal. When the tension of the noninvesting entrance (entered +) is greater than the one of the investing entrance (entered -) the comparator produces a tension of exit of high level. When the tension of noninvesting entrance is minor who the one of the investing entrance, the comparator produces a tension of exit of low level. 4.2. Basic circuit The way but simple to construct a comparator consists of connecting an op amp without refeeding resistance. So and as one is in the figure: When we began to study the op amps we said that operational "Lee" the tension in entrance +, reduces the tension to him of the entrance - and the result multiplies it by a very great number to later remove it in form from tension by the exit. In the 741 this number this around the 100000. Sure according to which we finished saying, if Vin is 1 volt the exit would have to be of 100000 volts, this is absurd: A Maxima tension of exit of the operational one of exists which never will happen. To this tension saturation tension is called to him "" (Vsat) This tension of saturation will come determined by the tension from feeding and the type from operational that utilizes. In the case of the 741 fed ±12V the Vsat it is of 10V. Therefore, when Vin is greater than 0 the exit will go off +Vsat, and when he is inferior to 0 the exit goes off - to Vsat: We are comparing Vin with a reference signal that, in this case, is 0. Now we are going to feed the previous circuit with simple tension of 15V and are going to put resistance to be able to vary the reference tension (Vref): Like the current that enters the operational one he is 0 the Vref will come given by the following expression: When the tension of Vin entrance is greater than the one of reference = > Vin - Vref will give a positive number and therefore the exit tension will be at high level (near 12 volts) When the tension of Vin entrance is reference minor who the one = > Vin - Vref will give a negative number and therefore the exit will be at low level (near 0 volts) 4.3. Schmitt Scale If the entrance of a comparator contains noise, the exit can be erratic when Vin is near the reference tension. In order to solve this one resorts to a comparator with Schmitt scale. The scheme is the following one: Fíjate that with this scheme the tension of entrance + is not always the same one, depends on the tension of Vout exit. Therefore not always it exchanges for the same value of Vin For example: if Vout and Vin are zero. You will agree with me in which then the tension of entrance + is zero. If you are increasing Vin, the tension in entrance + will be also increasing, but it will go below Vin When Vin gets to be worth what Vref will not take place the commutation since the tension in entrance + is inferior. You continue increasing Vin until the Vout exit arrives a little while in which the tension of entrance + surpasses to the Vref, then is put at high level. When putting itself the exit to high level suddenly, the tension in the entrance + that was inferior to Vin returns greater than Vin. Now although it lowers slightly to the Vin the comparator will not return to exchange because now the tension in entrance + is greater than Vin. The value of Vin that causes that the exit of the comparator exchanges of 0 at high level (Vsat) comes dice by the expression: And the value of Vin that causes that the exit of the comparator exchanges of the high level (Vsat) at the low level comes dice by this other expression: |