Probe for checking the suitability of operational amplifiers. Operational Amplifier Test Methods

Greetings, dear friends! Finally I got to my computer, made myself some tea and cookies and off I went...

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And today we will talk about such an electronic device as an operational amplifier. These amplifiers are used everywhere, wherever it is necessary to amplify the signal in terms of power, there is work for the op-amp.

The use of operational amplifiers is especially common in audio engineering. Every audiophile strives to enhance the sound of his music speakers and therefore tries to make the amplifier more powerful. This is where we come across operational amplifiers, because many audio systems are simply loaded with them. Thanks to the ability of the operational amplifier to amplify the signal by power, we feel more powerful pressure on our eardrums when we listen to songs on our audio speakers. This is how in everyday life we evaluate the quality of an operational amplifier by ear.

In e In this article, we will not evaluate anything by ear, but we will try to consider everything in detail and put everything on the shelves so that even the most samovar teapot can understand it.

What is an operational amplifier?

Operational amplifiers are microcircuits that can look different.

For example, this picture shows two Russian-made operational amplifiers. On the left is the K544UD2AR operational amplifier in a plastic DIP case, and on the right is an op-amp in a metal case.

At first, before I got acquainted with op-amps, I constantly confused microcircuits in such metal cases with transistors. I thought that these were such clever multi-emitter transistors :)

Conventional graphic designation (UGO)

The symbol for an operational amplifier is as follows.

So an operational amplifier (op-amp) has two inputs and one output. There are also pins for connecting power, but they are usually not indicated on the graphical symbols.

For such an amplifier there are two rules that will help you understand the operating principle:

The output of the op-amp tends to ensure that the voltage difference at its inputs is equal to zero
The operational amplifier inputs do not consume current.

Input 1 is designated as “+” and is called non-inverting and input 2 is designated as “-” and is inverting.

The op-amp inputs have a high input impedance, or otherwise called high impedance.

This means that the inputs of the operational amplifier consume almost no current (literally some nanoamps). The amplifier simply evaluates the voltage at the inputs and, depending on this, produces a signal at the output, amplifying it.

The gain of the operational amplifier is simply enormous, it can reach a million, and this is a very large value! This means that if we apply a small voltage to the input, at least 1 mV, then at the output we will immediately get a maximum, a voltage almost equal to the voltage of the op-amp’s power supply. Because of this property, opamps are almost never used without feedback (OS). Really, what’s the point of an input signal if we always get the maximum voltage at the output, but we’ll talk about that a little later.

The inputs of the op-amp work in such a way that if the value at the non-inverting input is greater than at the inverting input, then the output will have a maximum positive value of +15V. If the voltage at the inverting input turns out to be more positive, then at the output we will observe a maximum negative value, somewhere around -15V.

Indeed, an operational amplifier can produce voltage values of both positive and negative polarity. A beginner may wonder how this is possible? But this is really possible and this is due to the use of a power source with a split voltage, the so-called bipolar power supply. Let's look at the power supply of the op-amp in a little more detail.

Proper OU power supply

It will probably not be a secret that in order for the op-amp to work, it needs to be powered, i.e. connect it to a power source. But there is an interesting point, as we saw a little earlier, an operational amplifier can output voltages of both positive and negative polarity. How can this be?

But this can happen! This is due to the use of a bipolar power supply; of course, it is also possible to use a unipolar source, but in this case the capabilities of the operational amplifier will be limited.

In general, when working with power supplies, a lot depends on what we took as the starting point, i.e. for 0 (zero). Let's figure this out.

Battery-powered example

Usually it’s easiest to give examples on fingers, but in electronics I think AA batteries are also suitable :)

Let's say we have a regular AA battery. It has two poles, positive and negative. When we take the negative pole as zero, we consider it a zero reference point, then accordingly the positive pole of the battery will show + 5V (the value with a plus).

We can see this with the help of a multimeter (by the way, it will help), just connect the negative black probe to the minus of the battery and the red probe to the plus and voila. Everything here is simple and logical.

Now let’s complicate the task a little and take exactly the same second battery. Let's connect the batteries in series and consider how the readings of measuring instruments (multimeters or voltmeters) change depending on different points of application of the probes.

If we take the negative pole of the extreme battery as zero and connect the measuring probe to the positive of the battery, then the multimeter will show us a value of +10 V.

If the positive pole of the battery is taken as the reference point and the measuring probe was connected to the negative, then any voltmeter will show us -10 V.

But if the point between two batteries is taken as the starting point, then as a result we can obtain a simple bipolar power source. And you can verify this; a multimeter will confirm to us that this is so. We will have both positive polarity voltage +5V and negative polarity voltage -5V available.

Bipolar power supply circuits

I gave examples on batteries as an example to make it more clear. Now let's look at a few examples of simple split-power supply circuits that you can use in your amateur radio designs.

Circuit with a transformer, with a tap from the “middle” point

And the first circuit of the power supply for the op-amp is in front of you. It is quite simple, but I will explain a little how it works.

The circuit is powered from our usual home network, so it is not surprising that an alternating current of 220V is supplied to the primary winding of the transformer. Then the transformer converts 220V alternating current into the same alternating current but 30V. This is the kind of transformation we wanted to make.

Yes, there will be an alternating voltage of 30V on the secondary winding, but pay attention to the tap from the middle point of the secondary winding. A branch is made on the secondary winding, and the number of turns before this branch is equal to the number of turns after the branch.

Thanks to this branch, we can obtain an alternating voltage of both 30 V and 15 V at the output of the secondary winding. We take this knowledge into account.

Next, we need to straighten the variable and turn it into a constant, therefore. The diode bridge coped with this task and at the output we got a not very stable constant of 30V. This voltage will be shown to us by the multimeter if we connect the leads to the output of the diode bridge, but we need to remember about the branch on the secondary winding.

We have achieved a zero reference point between the poles of potentials of positive and negative polarity. As a result, at the output we have a fairly stable voltage of both +15V and -15V. This circuit, of course, can be further improved by adding zener diodes or integrated stabilizers, but nevertheless, the above circuit can already cope with the task of powering operational amplifiers.

This circuit, in my opinion, is simpler, simpler in the sense that there is no need to look for a transformer with a tap from the middle or to form a secondary winding yourself. But here you will have to fork out for a second diode bridge.

The diode bridges are connected so that the positive potential is formed from the cathodes of the diodes of the first bridge, and the negative potential comes from the anodes of the diodes of the second bridge. Here the zero reference point is drawn between the two bridges. I will also mention that isolation capacitors are used here; they protect one diode bridge from influences from the second.

This circuit is also easily subject to various improvements, but most importantly, it solves the main problem - it can be used to power an operational amplifier.

Op amp feedback

As I already mentioned, operational amplifiers are almost always used with feedback (OS). But what is feedback and what is it for? Let's try to figure this out.

We encounter feedback all the time: when we want to pour tea into a mug or even go to the toilet for a small need :) When a person drives a car or a bicycle, feedback also works here. Indeed, in order to drive easily and naturally, we are forced to constantly monitor the controls depending on various factors: the situation on the road, the technical condition of the vehicle, and so on.

What if the road became slippery? Yeah, we reacted, made a correction and are moving forward more carefully.

In an operational amplifier, everything happens in a similar way.

Without feedback, when a certain signal is applied to the input, we will always get the same voltage value at the output. It will be close to the supply voltage (since the gain is very large). We do not control the output signal. But if we send part of the signal from the output back to the input, what will this give?

We will be able to control the output voltage. This control will be so effective that you can simply forget about the gain; the opamp will become obedient and predictable because its behavior will depend only on feedback. Next I will tell you how to effectively control the output signal and how to control it, but for this we need to know some details.

Positive feedback, negative feedback

Yes, feedback is used in operational amplifiers and very widely. But feedback can be both positive and negative. We need to figure out what the point is.

Positive Feedback this is when part of the output signal is fed back to the input and it (part of the output) is summed with the input.

Positive feedback in opamps is not used as widely as negative feedback. Moreover, positive feedback is often an undesirable side effect of some schemes, and people try to avoid positive feedback. It is undesirable because this coupling can increase distortion in the circuit and ultimately lead to instability.

On the other hand, positive feedback does not reduce the gain of the op-amp, which can be useful. And instability also finds its application in comparators, which are used in ADCs (Analog-to-Digital Converters).

Negative Feedback this is a connection where part of the output signal is fed back to the input, but at the same time it is subtracted from the input

But negative feedback is simply created for operational amplifiers. Although it does contribute some gain attenuation, it brings stability and controllability to the circuit. As a result, the circuit becomes gain independent, its properties being completely controlled by negative feedback.

By using negative feedback, an op-amp gains one very useful property. The opamp monitors the states of its inputs and strives to ensure that the potentials at its inputs are equal. The op-amp adjusts its output voltage so that the resulting input potential (the difference between In.1 and In.2) is zero.

The overwhelming majority of op-amp circuits are built using negative feedback! So in order to understand how negative coupling works, we need to look at the op-amp switching circuits.

Operational amplifier circuits

The circuits for connecting operational amplifiers can be very diverse, so it’s unlikely that I will be able to talk about each one, but I will try to consider the main ones.

Op-amp comparator

The formulas for the comparator circuit will be as follows:

Those. the result will be a voltage corresponding to a logical unit.

Those. the result will be a voltage corresponding to logical zero.

The comparator circuit has a high input resistance (impedance) and a low output.

Let's first consider this circuit for switching on the op-amp in comparator mode. This switching circuit is devoid of feedback. Such circuits are used in digital circuitry when it is necessary to evaluate input signals, find out which is larger and produce the result in digital form. As a result, the output will be logical 1 or logical zero (for example, 5V is 1 and 0V is zero).

Let's say the stabilization voltage of the zener diode is 5V, we applied 3V to input one and 1V to input 2. Next, the following happens in the comparator: the voltage at direct input 1 is used as is (simply because it is a non-inverting input) and the voltage at inverse input 2 is inverted. As a result, where there was 3B remains 3B and where there was 1B there will be -1B.

As a result, 3V-1V = 2V, but thanks to the gain of the op-amp, the output will receive a voltage equal to the voltage of the power source, i.e. about 15V. But the zener diode will work and 5V will go to the output, which corresponds to a logical one.

Now imagine that we threw 3V at input 2 and applied 1V at input 1. The opamp will chew through all this, leave the direct input unchanged, and change the inverse (inverting) input to the opposite of 3V and make -3V.

As a result, 1V-3V = -2V, but according to the operating logic, the minus of the power supply will go to the output, i.e. -15V. But we have a zener diode and it will not miss this and at the output we will have a value close to zero. This will be logical zero for the digital circuit.

Schmitt trigger on op-amp

A little earlier we considered such a circuit for connecting an op-amp as a comparator. A comparator compares two input voltages and produces a result at the output. But in order to compare the input voltage with zero, you need to use the circuit presented just above.

Here the signal is fed to the inverting input and the direct input is grounded to zero.

If we have a voltage greater than zero at the input, then we will have -15V at the output. If the voltage is less than zero, then the output will be +15V.

But what happens if we want to apply a voltage equal to zero? It will never be possible to create such a voltage, because there is no ideal zero and the input signal, even by fractions of a microvolt, will definitely change in one direction or another. As a result, the output will be complete chaos, the output voltage will jump from maximum to minimum many times, which in practice is completely inconvenient.

To get rid of such chaos, a hysteresist is introduced - this is a certain gap within which the output signal will not change.

This gap allows this circuit to be implemented through positive feedback.

Let's imagine that we applied 5V to the input, and at the first moment the output will produce a signal with a voltage of -15V. Then the positive feedback begins to work out. The feedback is formed by a voltage divider, as a result of which a voltage of -1.36V appears at the direct input of the op-amp.

At the inverse input, the signal is more positive, so the operational amplifier will work as follows. Inside it, the 5V signal is inverted and becomes -5V, then the two signals are added and a negative value is obtained. A negative value due to the gain will become -15V. The output signal will not change until the input signal drops below -1.36V.

Let the input signal change and become -2V. Inside, this -2V is inverted and becomes +2V, and -1.36V will remain as it was. Then all this is added up and a positive value is obtained, which at the output will turn into +15V. At the direct input, the value -1.36V will turn into +1.36V due to feedback. Now, to change the output value to the opposite one, you need to apply a signal of more than 1.36V.

Thus, we have a zone with zero sensitivity with a range from -1.36V to +1.36V. This dead zone is called hysteresis.

Repeater

The simplest owner of negative feedback is a repeater.

The repeater outputs the voltage that was applied to its input. It would seem why this is needed because nothing changes from this. But this makes sense, because let us remember the property of the op-amp: it has a high input impedance and a low output impedance. In circuits, repeaters act as a buffer that protects weak outputs from overload.

To understand how it works, rewind a little back to where we discussed negative feedback. There I mentioned that in the case of negative feedback, the opamp strives in every possible way to achieve equal potential across its inputs. To do this, it adjusts the voltage at its output so that the potential difference at its inputs is zero.

So let’s say we have 1B at the input. In order for potentials at the inputs to be present, the inverting input must also be 1V. That's why he's a repeater.

The non-inverting amplifier circuit is very similar to the repeater circuit, only here the feedback is represented by a voltage divider and connected to ground.

Let's see how it all works. Let's say 5V is applied to the input, resistor R1 = 10 Ohm, resistor R2 = 10 Ohm. In order for the voltages at the inputs to be equal, the opamp is forced to raise the voltage at the output so that the potential at the inverse input is equal to the direct one. In this case, the voltage divider divides in half, it turns out that the output voltage should be twice the input voltage.

In general, to apply this switching scheme you don’t even need to rack anything in your head, just use the formula where it’s enough to find out the coefficient K.

And now we will look at the operation of such a switching circuit as an inverting amplifier. For an inverting amplifier there are the following formulas:

An inverting amplifier allows you to amplify a signal while simultaneously inverting (changing its sign). Moreover, we can set any gain. We generate this gain through negative feedback, which is a voltage divider.

Now let's try it in action, let's say we have a 1V signal at the input, resistor R2 = 100 Ohm, resistor R1 = 10 Ohm. The signal from the input goes through R1, then R2 and to the output. Let's say the output signal incredibly becomes 0V. Let's calculate the voltage divider.

1V/110=X/100, hence X = 0.91V

It turns out that at point A the potential is 0.91V, but this contradicts the operational amplifier rule. After all, the opamp strives to equalize the potentials at its inputs. Therefore, the potential at point A will be zero and equal to the potential at point B.

How to make it so that there is 1V at the input and 0V at point A?

To do this, you need to reduce the output voltage. And as a result we get

Unfortunately, the inverting amplifier has one obvious drawback - low input resistance, which is equal to resistor R1.

And this connection circuit allows you to add multiple input voltages. Moreover, voltages can be both positive and negative. In truth, analog computers can be built using op-amps. So let's figure it out.

The basis of the adder is the same inverting amplifier with only one difference: instead of one input, it can have as many of these inputs as you like. Let's remember the formula for the inverting amplifier. The potential of point X will be equal to zero, so the sum of the currents entering from each input will look like this: If our goal is pure addition of input voltages, then all resistors in this circuit are chosen to be of the same value. This also leads to the fact that the gain for each input will be equal to 1. Then the formula for the inverting amplifier takes the form:

Well, I think it’s not difficult to understand the operation of the adder and other switching circuits on op-amps. It is enough to practice a little and try to assemble these circuits and see what happens with the input and output signals.

And I’ll probably stop here because when working with operational amplifiers, a lot of different switching circuits are used, these are various current-voltage converters, adders, integrators and logarithmic amplifiers, and all of them can be considered for a very long time.

If you are interested in other switching schemes and want to understand them, I advise you to look through them, everything will definitely fall into place.

And with this I will conclude, especially since the article turned out to be quite voluminous and after writing it needs to be slightly polished and put into place.

Friends, do not forget to subscribe to blog updates, because the more readers subscribe to updates, the more I understand that I am doing something important and useful and this damn motivates me for new articles and materials.

By the way, friends, I came up with a cool idea and it’s very important for me to hear your opinion. I'm thinking of releasing training material on operational amplifiers, this material will be in the form of a regular pdf book or video course, I haven't decided yet. It seems to me that despite the large abundance of information on the Internet and in the literature, there is still a lack of visual, practical information, something that everyone can understand.

So, please write in the comments what information you would like to see in this training material so that I can provide not just useful information, but information that is actually in demand.

And that’s all for me, so I wish you good luck, success and a great mood, even despite the fact that it’s winter outside!

With n/a Vladimir Vasiliev.

P.S. Friends, be sure to subscribe to updates! By subscribing, you will receive new materials directly to your email! And by the way, everyone who signs up will receive a useful gift!

They often started asking me questions about analog electronics. Did the session take the students for granted? ;) Okay, it’s high time for a little educational activity. In particular, on the operation of operational amplifiers. What is it, what is it eaten with and how to calculate it.

What is this
An operational amplifier is an amplifier with two inputs, neve... hmm... high signal gain and one output. Those. we have U out = K*U in and K ideally equals infinity. In practice, of course, the numbers are more modest. Let's say 1,000,000. But even such numbers blow your mind when you try to apply them directly. Therefore, like in kindergarten, one Christmas tree, two, three, many Christmas trees - we have a lot of reinforcement here;) And that’s it.

And there are two entrances. And one of them is direct, and the other is inverse.

Moreover, the inputs are high-impedance. Those. their input impedance is infinity in the ideal case and VERY high in the real case. The count there goes into hundreds of MegaOhms, or even gigaohms. Those. it measures the voltage at the input, but has minimal effect on it. And we can assume that no current flows in the op-amp.

The output voltage in this case is calculated as:

U out =(U 2 -U 1)*K

Obviously, if the voltage at the direct input is greater than at the inverse input, then the output is plus infinity. Otherwise it will be minus infinity.

Of course, in a real circuit there will not be infinity plus and minus, and they will be replaced by the highest and lowest possible supply voltage of the amplifier. And we will get:

Comparator
A device that allows you to compare two analog signals and make a verdict - which signal is larger. Already interesting. You can come up with a lot of applications for it. By the way, the same comparator is built into most microcontrollers, and I showed how to use it using the example of AVR in articles about creation. The comparator is also great for creating .

But the matter is not limited to one comparator, because if you introduce feedback, then a lot can be done from the op-amp.

Feedback
If we take a signal from the output and send it straight to the input, then feedback will arise.

Positive Feedback
Let’s take and drive the signal directly from the output into the direct input.

The voltage U1 is greater than zero - the output is -15 volts
The voltage U1 is less than zero - the output is +15 volts

What happens if the voltage is zero? In theory, the output should be zero. But in reality, the voltage will NEVER be zero. After all, even if the charge of the right one outweighs the charge of the left one by one electron, then this is already enough to drive the potential to the output at an infinite gain. And at the output all hell will begin - the signal jumps here and there at the speed of random disturbances induced at the inputs of the comparator.

To solve this problem, hysteresis is introduced. Those. a kind of gap between switching from one state to another. To do this, positive feedback is introduced, like this:

We assume that at this moment there is +10 volts at the inverse input. The output from the op-amp is minus 15 volts. At the direct input it is no longer zero, but a small part of the output voltage from the divider. Approximately -1.4 volts Now, until the voltage at the inverse input drops below -1.4 volts, the op-amp output will not change its voltage. And as soon as the voltage drops below -1.4, the output of the op-amp will sharply jump to +15 and there will already be a bias of +1.4 volts at the direct input.

And in order to change the voltage at the output of the comparator, the U1 signal will need to increase by as much as 2.8 volts to reach the upper level of +1.4.

A kind of gap appears where there is no sensitivity, between 1.4 and -1.4 volts. The width of the gap is controlled by the ratios of resistors in R1 and R2. The threshold voltage is calculated as Uout/(R1+R2) * R1 Let's say 1 to 100 will give +/-0.14 volts.

But still, op-amps are more often used in negative feedback mode.

Negative Feedback
Okay, let's put it another way:

In the case of negative feedback, the op-amp has an interesting property. It will always try to adjust its output voltage so that the voltages at the inputs are equal, resulting in a zero difference.
Until I read this in the great book from comrades Horowitz and Hill, I could not get into the work of the OU. But it turned out to be simple.

Repeater
And we got a repeater. Those. at the input U 1, at the inverse input U out = U 1. Well, it turns out that U out = U 1.

The question is, why do we need such happiness? It was possible to directly connect the wire and no op-amp would be needed!

It is possible, but not always. Let's imagine this situation: there is a sensor made in the form of a resistive divider:

The lower resistance changes its value, the distribution of output voltages from the divider changes. And we need to take readings from it with a voltmeter. But the voltmeter has its own internal resistance, albeit large, but it will change the readings from the sensor. Moreover, what if we don’t want a voltmeter, but want the light bulb to change brightness? There’s no way to connect a light bulb here anymore! Therefore, we buffer the output with an operational amplifier. Its input resistance is huge and its influence will be minimal, and the output can provide quite a noticeable current (tens of milliamps, or even hundreds), which is quite enough to operate the light bulb.
In general, you can find applications for a repeater. Especially in precision analog circuits. Or where the circuitry of one stage can affect the operation of another, in order to separate them.

Amplifier
Now let’s do a feint with our ears - take our feedback and connect it to the ground through a voltage divider:

Now half of the output voltage is supplied to the inverse input. But the amplifier still needs to equalize the voltages at its inputs. What will he have to do? That's right - raise the voltage at your output twice as high as before in order to compensate for the resulting divider.

Now there will be U 1 on the straight line. On inverse U out /2 = U 1 or U out = 2*U 1.

If we put a divisor with a different ratio, the situation will change in the same way. So that you don’t have to turn the voltage divider formula in your mind, I’ll give it right away:

U out = U 1 *(1+R 1 /R 2)

It is mnemonic to remember what is divided into what is very simple:

It turns out that the input signal goes through a chain of resistors R 2, R 1 in U out. In this case, the direct input of the amplifier is set to zero. Let us remember the habits of the op-amp - it will try, by hook or by crook, to ensure that a voltage equal to the direct input is generated at its inverse input. Those. zero. The only way to do this is to lower the output voltage below zero so that a zero appears at point 1.

So. Let's imagine that U out =0. It's still zero. And the input voltage, for example, is 10 volts relative to U out. A divisor of R 1 and R 2 will divide it in half. Thus, at point 1 there are five volts.

Five volts is not zero and the op amp lowers its output until point 1 is zero. To do this, the output should become (-10) volts. In this case, relative to the input, the difference will be 20 volts, and the divider will provide us with exactly 0 at point 1. We have an inverter.

But we can also choose other resistors so that our divider produces different coefficients!
In general, the gain formula for such an amplifier will be as follows:

U out = - U in * R 1 / R 2

Well, a mnemonic picture for quickly memorizing xy from xy.

Let's say U 2 and U 1 are 10 volts each. Then at the 2nd point there will be 5 volts. And the output will have to become such that at the 1st point there is also 5 volts. That is, zero. So it turns out that 10 volts minus 10 volts equals zero. That's right :)

If U 1 becomes 20 volts, then the output will have to drop to -10 volts.
Do the math yourself - the difference between U 1 and U out will be 30 volts. The current through resistor R4 will be (U 1 -U out)/(R 3 +R 4) = 30/20000 = 0.0015A, and the voltage drop across resistor R 4 will be R 4 *I 4 = 10000 * 0.0015 = 15 volts . Subtract the 15 volt drop from the 20 input drop and get 5 volts.

Thus, our op-amp solved an arithmetic problem from 10 subtracted 20, resulting in -10 volts.

Moreover, the problem contains coefficients determined by resistors. It’s just that, for simplicity, I have chosen resistors of the same value and therefore all coefficients are equal to one. But in fact, if we take arbitrary resistors, then the dependence of the output on the input will be like this:

U out = U 2 *K 2 - U 1 *K 1

K 2 = ((R 3 +R 4) * R 6) / (R 6 +R 5)*R 4
K 1 = R 3 / R 4

The mnemonic technique for remembering the formula for calculating coefficients is as follows:
Right according to the scheme. The numerator of the fraction is at the top, so we add up the upper resistors in the current flow circuit and multiply by the lower one. The denominator is at the bottom, so we add up the lower resistors and multiply by the upper one.

Everything is simple here. Because point 1 is constantly reduced to 0, then we can assume that the currents flowing into it are always equal to U/R, and the currents entering node number 1 are summed up. The ratio of the input resistor to the feedback resistor determines the weight of the incoming current.

There can be as many branches as you like, but I only drew two.

U out = -1(R 3 *U 1 /R 1 + R 3 *U 2 /R 2)

Resistors at the input (R 1, R 2) determine the amount of current, and therefore the total weight of the incoming signal. If you make all the resistors equal, like mine, then the weight will be the same, and the multiplication factor of each term will be equal to 1. And U out = -1(U 1 +U 2)

Non-inverting adder
Everything is a little more complicated here, but it’s similar.

Uout = U 1 *K 1 + U 2 *K 2

K 1 = R 5 / R 1
K 2 = R 5 / R 2

Moreover, the resistors in the feedback must be such that the equation R 3 / R 4 = K 1 + K 2 is observed

In general, you can do any math using operational amplifiers, add, multiply, divide, calculate derivatives and integrals. And almost instantly. Analog computers are made using op-amps. I even saw one of these on the fifth floor of SUSU - a fool the size of half a room. Several metal cabinets. The program is typed by connecting different blocks with wires :)

There is a wide variety of these microcircuits, and they are incompatible with each other in terms of pin locations. These microcircuits can be checked by setting the operating mode, which can be done on a stand specially assembled for a particular case, where the microcircuit is connected using a universal contact socket, or the test can be carried out already as part of a circuit assembled on them. The second is more convenient, as it requires less time.

Now about the verification itself. First of all, you need to measure the levels of supply voltages, voltages at the inputs of the microcircuit, as well as at the output (with a digital voltmeter). Usually, if the values of the negative feedback resistors are known, then by calculating the gain, you can draw conclusions about what should be at the output and with what sign, of course, if it is a linear amplifier.

Doubts may arise when checking more complex circuits (integrators, autogenerators, etc.). In this case, you can use another method. As you know, any operational amplifier can easily be made to work in comparator mode. To do this, we can temporarily apply a small voltage alternately to the direct and inverse inputs of the microcircuit from an external source through a current-limiting resistor (Fig. 6.17). The voltage at the output of the “op-amp” must be monitored with a digital voltmeter or oscilloscope (during normal operation we will see the output switching).

Rice. 6.17. Principle of testing operational amplifiers

An oscilloscope is more convenient for carrying out such measurements, since it makes it possible to detect not only changes in output levels, but also the presence of unintended self-excitation of cascades (self-generation).

Source: Radio amateurs: useful diagrams. Book 6. - M / SOLON-Press, 2005. 240 p.

In table 1.4 and 1.5 present the electrical characteristics of solar modules and batteries. Table 1.4. Electrical characteristics of domestically produced solar modules FSM-50 50 21 2.95 10720 FSM-55 55 21 3.15 1028x450x28…….

In a working element, during continuity between the power circuits there should be an infinitely large resistance, and between the control electrode and one of the terminals (the cathode of the thyristor) there should be a small resistance (from 30 to .......

To quickly tighten small screws in hard-to-reach places (and not lose them when unscrewing), you will need a screwdriver with a magnet. It is easy to make such a screwdriver from an ordinary one. It is enough to wrap 100-200...... around the screwdriver shaft.

Be careful to protect yourself from electric shock or disconnection of connectors and parts of the washing machine when troubleshooting. Observe electrical safety measures! First of all, check the connections of each connector. If you are replacing PWB…….

remote control (RC) 90% of remote controls are occupied by two types of defects: 1) some buttons do not work (usually those that are frequently pressed). In this case, you need to cut out a piece of foil and…….

The other day I bought an operational amplifier (op-amp) in a store for $1.5, came home, soldered it, silence. There was no doubt that the operability of the op-amp circuit was to blame, so I unsoldered the purchased op-amp and decided to check it. I connected the inverting input to the output, applied power and voltage to the direct input (1V), a working op-amp at the output should have given out what it was fed to the input, in fact this is what checking the op-amp is all about, but my output is zero.

Interesting, I thought then, either I overheated it when soldering, which is unlikely, or I bought a faulty one. I went to the store again, bought another one, but decided to check it before sealing it and lo and behold, this one is also faulty, but now at least you can return it to the seller, apparently, he has a whole lot of them...

But I didn’t have time to figure it out, I went to another store and bought the same op-amp, but for $4. When I bought it, we agreed that if it didn’t work, I’d bring it back. I came home, checked - it works, soldered it - it works. The following conclusion can be drawn from this: after purchasing a part, before soldering it, it is advisable to check it, and the seller, most likely, ordered a batch of these op-amps from China and when he received it, did not check it, this is understandable when you have a whole store with radio components to check everything you'll get tired.

Why did I write all this, after that I looked for these op-amps on Ali and when I found them I was pleasantly surprised, with the money that I spent in my city to buy a working op-amp ($4), I could have bought 5 pieces in China, but they were in the soic8 building, and having the negative experience described above, of course, I wanted to check them when they arrived. This issue could be solved in several ways, by etching a breadboard into which the op-amp could be soldered each time, on the other hand, in order not to solder it, you could simply press the op-amp to the board with a clothespin, this is already better, but there is an even more interesting option, since you often have to have deal with soic8, I decided to look ZIF adapter soic8 – dip8, then it will be possible to assemble the circuit using breadboard, which will significantly speed up the process.

In general, I found such an adapter on Ali for $1.7 and this includes delivery. When the op-amps arrived, the adapter was already in hand, and since I have a signal generator in my arsenal, I checked them according to the diagram from the datasheet.

Operational amplifier (op-amp) English. Operational Amplifier (OpAmp), popularly known as an operational amplifier, is a direct current amplifier (DCA) with a very high gain. The phrase "DC amplifier" does not mean that the op amp can only amplify DC current. This means, starting from a frequency of zero Hertz, and this is direct current.

The term “operational” has been strengthened for a long time, since the first samples of op-amps were used for various mathematical operations such as integration, differentiation, summation, etc. The gain factor of an op-amp depends on its type, purpose, structure and can exceed 1 million!

Operational amplifier circuit

In the diagrams, the operational amplifier is designated like this:

or so

Most often, op-amps on diagrams are indicated without power pins

An input with a plus sign is called non-inverting, and an input with a minus sign is called inverting. Do not confuse these two signs with power polarity! They do NOT say that it is necessary to supply a signal with negative polarity to the inverting input, but to a non-inverting signal with positive polarity, and then you will understand why.

Op amp power supply

If the power pins are not indicated, then it is assumed that the op-amp is supplied with bipolar power +E and -E Volts. It is also labeled +U and -U, V CC and V EE, Vc and V E. Most often it is +15 and -15 Volts. Bipolar nutrition is also called bipolar nutrition. How do you understand this - bipolar nutrition?

Let's imagine a battery

I think you all know that a battery has a “plus” and a “minus”. In this case, the “minus” of the battery is taken as zero, and the batteries are counted relative to zero. In our case, the battery voltage is 1.5 Volts.

Let's take another such battery and connect them in series:

So, our total voltage will be 3 Volts, if we take the minus of the first battery as zero.

But what if you take the minus of the second battery to zero and measure all the voltages relative to it?

This is where we got bipolar power supply.

Ideal and real operational amplifier model

In order to understand the essence of the operation of the op-amp, consider it perfect And real models.

1) the ideal op-amp is infinitely large.

In real op-amps, the value of the input resistance depends on the purpose of the op-amp (universal, video, precision, etc.), the type of transistors used and the circuit design of the input stage and can range from hundreds of ohms to tens of megohms. A typical value for general purpose op amps is a few megohms.

2) The second rule follows from the first rule. Since the input impedance of an ideal op-amp is infinitely large, the input impedance will be zero.

In fact, this assumption is quite valid for op-amps with input currents, whose input currents can be less than picoamps. But there is also an op-amp with an input. Here the input current can already be tens of microamps.

3) The output impedance of an ideal op-amp is zero.

This means that the voltage at the output of the op-amp will not change when the load current changes. In real general use op amps it is tens of ohms (usually 50 ohms).
In addition, the output impedance depends on the signal frequency.

4) The gain in an ideal op-amp is infinitely large. In reality, it is limited by the internal circuitry of the op-amp, and the output voltage is limited by the supply voltage.

5) Since the gain is infinitely large, therefore, the voltage difference between the inputs of an ideal op-amp is zero. Otherwise, even if the potential of one input is greater or less than at least the charge of one electron, then the output will have an infinitely large potential.

6) The gain in an ideal op-amp does not depend on the signal frequency and is constant at all frequencies. In real op-amps, this condition is met only for low frequencies up to a certain cutoff frequency, which is individual for each op-amp. Typically, the cutoff frequency is taken to be a gain drop of 3 dB or up to 0.7 of the gain at zero frequency (DC).

The circuit of the simplest op-amp using transistors looks something like this:

Operating principle of an operational amplifier

Let's look at how an op-amp works

The operating principle of the op-amp is very simple. It compares the two voltages and produces a negative or positive supply potential at the output. It all depends on which input has the greatest potential. If the potential at the NON-inverting input U1 is greater than at the inverting U2, then the output will be +Upit, but if the potential at the inverting input U2 is greater than at the NON-inverting U1, then the output will be -Upit. That's the whole principle ;-).

Let's look at this principle in the Proteus simulator. To do this, we will select the simplest and most common operational amplifier LM358 (analogs 1040UD1, 1053UD2, 1401UD5) and assemble a primitive circuit showing the principle of operation

Let's apply 2 Volts to the non-inverting input, and 1 Volt to the inverting input. Since the potential is greater at the non-inverting input, therefore, at the output we should get +Upit. We got 13.5 Volts, which is close to this value

But why not 15 Volts? The internal circuitry of the op-amp itself is to blame for everything. The maximum value of the op-amp may not always be equal to the positive or negative supply voltage. It can deviate from 0.5 to 1.5 Volts depending on the type of op-amp.

But, as they say, every family has its blacks, and therefore op-amps have long appeared on the market that can produce an acceptable supply voltage at the output, that is, in our case, these are values close to +15 and -15 Volts. This feature is called Rail-to-Rail, which is literally translated from English. “from rail to rail”, and in the language of electronics “from one power bus to the other”.

Let's now apply a potential greater to the inverting input than to the non-inverting input. We apply 2 Volts to the inverting one, and 1 Volt to the non-inverting one:

As you can see, at the moment the output is “low” at -Upit, since the potential at the inverting input was greater than at the non-inverting input.

In order not to download the Proteus software package once again, you can simulate the operation of an ideal op-amp online using the Falstad program. To do this, select the Circuits—Op-Amps—>OpAmp tab. As a result, the following diagram will appear on your screen:

On the right control panel you will see sliders for adding voltage to the inputs of the op-amp and you can already visually see what happens at the output of the op-amp when the voltage at the inputs changes.

So, we have considered the case when the voltage at the inputs may differ. But what happens if they are equal? What will Proteus show us in this case? Hmm, showed +Upit.

What will Falstad show? Zero Volt.

Who to believe? No one! In real life, it is impossible to do this in order to drive absolutely equal voltages to the two inputs. Therefore, such a state of the op-amp will be unstable and the output values can take values of either -E Volt or +E Volt.

Let's apply a sinusoidal signal with an amplitude of 1 Volt and a frequency of 1 kilohertz to the non-inverting input, and put the inverting one to ground, that is, to zero.

Let's see what we have on the virtual oscilloscope:

What can be said in this case? When the sinusoidal signal is in the negative region, we have -Upit at the output of the op-amp, and when the sinusoidal signal is in the positive region, then we have +Upit at the output. Also note that the voltage at the output of the op-amp cannot change its value suddenly. Therefore, the op-amp has such a parameter as the rate of rise of the output voltage V Uout .

This parameter shows how quickly the op-amp's output voltage can change when operating in pulsed circuits. Measured in Volts/sec. Well, as you understand, the higher the value of this parameter, the better the op-amp behaves in pulsed circuits. For LM358 this parameter is 0.6 V/µs.

With input from Jeer