Many analog circuits use operational amplifiers.

What is an operational amplifier and how does it work?

In this article on the site, the site will discuss exactly this. When studying digital electronics, we are accustomed to perceiving logical elements as some kind of “black boxes”, “waders” from which circuits are made up. We know their properties, but do not think about their inner content.

Indeed, it has long been accepted that a digital circuit consists of logic elements, and not of transistors and diodes. This attitude has also formed towards operational amplifiers, such elements of analog technology.

When designing a circuit on an op-amp or assembling a finished design, we perceive an operational amplifier (op-amp) as a “box” with known properties, and rarely think about its contents. That is why, on diagrams with an op-amp, for a long time no one draws the diagrams of the op-amps themselves, but only their graphic designation (Fig. 1).

If the operational amplifier is described in detail by clicking on the theory, then you will get good material, if not for a dissertation, then for thesis University (however, as in the case of a simple transistor switch). We are pursuing another task, to understand how it works and what we can get from it. If, however, someone does not have enough theory, then you can turn to university textbooks.

And so, the operational amplifier

This is an element of analog electronics, so we will study it. And as a "guinea pig" we will take the most common "model" today - KRN0UD608 (Fig. 1). The KR140UD608 case looks like a 16-pin case of some digital microcircuit broken in half (Figure 1).

It is just two times shorter than, for example, K561IE10. There are four pins on each side. The key (point, recess, groove) is located at the end of the first pin. Every op amp has two inputs and one output. The inputs are bipolar, our "rabbit" has a DIRECT input on pin 3, and an INVERSE input on pin 2 (output - pin 6).

The operational amplifier amplifies the signal applied to one of its inputs relative to the other, or rather, it turns out that the input signal is the potential difference between its inputs (or the current between its inputs). To understand how this looks in practice, you can assemble the circuit shown in Figure 2.

We use two "flat" batteries of 4.5 V each as batteries; turning them on in series, and making a conclusion from the middle (bipolar power supply). We will control the voltage with the same multimeter, and even better with two multimeters (such as M-838 or others).

And so, in the circuit in Figure 2, the inverse input (pin 2) is connected to a common wire (with the middle pin of the power source), and we apply voltage from the variable resistor R1 to the direct input (pin 3). By rotating R1 and measuring the voltage at the output of the op-amp and on the R1 engine, you can understand that setting R1 in such a position that the output is 0V is very difficult (almost impossible).

If the voltage on the R1 engine is slightly more than 0V (slightly more than the voltage at the inverted input), then the output will be about +4V, and if this voltage is slightly less than 0V (negative relative to the common point of the power source), then the output will be (-4V). Now let's change the point of connection of the negative wire of the multimeter (Fig. 3).

Now it turns out that if the voltage on the resistor R1 engine is slightly more than 4.5 V, then the output will be approximately 8.5V, and if the voltage on the R1 engine is slightly less than 4.5V, then the output will be approximately 0.5V. Thus, we returned back to digital technology, if the voltage at the direct input is greater than at the inverse, then the output is a logical unit, and if the voltage at the direct input is lower than at the inverse, then the output is a logical zero (like this from analog to digital one step).

Now, for the sake of the purity of the experiment, you can swap the connections of the op-amp inputs, and check everything again. The dependence stated above will be confirmed. Thus, if the voltage at the direct input is greater, then it is also greater at the output, and if the voltage at the inverse input is greater, then it is less at the output. This is the difference between direct and inverse inputs.

This is how an analog comparator works, it serves to compare different voltages applied to its inputs. In such an inclusion (Fig. 2, Fig. 3), the gain of the op amp tends to infinity (about 30,000). But to work in analog circuits, it is usually not a comparator that is required, but an amplifier, and it is necessary that the gain of this amplifier can be set “to taste”.

In order for the operational amplifier to cease to be a comparator, it is necessary to introduce negative feedback between its output and the inverse input. So let's do it, disconnect the inverted input from the common wire of the power source and connect it to the output (Figure 4).

Now there is no trace of the huge gain. The gain in the circuit in fig. 4 is equal to 1. That is, the output voltage changes in the same way as the voltage at the direct input. The op-amp only repeats the input signal and does not amplify it by voltage. The thing is that OOS is one hundred percent.

In order to be able to set any desired gain, you need to turn on the op amp according to the circuit shown in Figure 5 (or Figure 6). And the gain will be determined by the ratio of the left and right (according to the diagram) parts of the variable resistor R3 (Fig. 5, Fig. 6) relative to the location of its engine. That is, the gain of the op amp will be equal, for Figure 5:

Ku \u003d 1 * (R3np / R3n)
for figure 6:
Ku \u003d - (R3np / R3n)
Where R3np is the resistance of the right side of R3, and R3n is the resistance of the left side of R3.

The input impedance of the amplifier according to fig. 6 will be equal to R3n.
The input impedance of the amplifier according to fig. 5 is determined mainly by the input impedance of the direct input of the op-amp. And so, two typical circuits for switching on any operational amplifier, Figure 7.

These circuits are designed to operate with a DC input voltage applied to ground, although of course they will work with an AC input voltage as long as it does not have a DC component.

If the alternating input voltage has a constant component (for example, it is taken from the collector of the transistor of the preliminary amplifying stage), it must be removed by turning on the isolation capacitor at the input (Fig. 8).

A significant disadvantage of the circuits shown in figures 7 and 8 is the need for a bipolar power supply. To power the op-amp from a unipolar source, you need to “deceive” it a little, make such a circuit in which there will be some constant voltage equal to half the supply voltage, and connect its inputs to this voltage, as if to a common power wire. If you need to amplify only the alternating voltage, then such a "deception" completely passes.

Figure 9 shows a diagram of an inverting amplifier based on an op amp operating with a single supply. Resistors R3 and R4 have the same resistance, and the voltage at their connection point will be equal to half the supply voltage. We connect this point to the direct input of the op-amp, and the capacitor C2 suppresses various interferences that may occur in this circuit.

If we need a non-inverting amplifier, the circuit will be as in Figure 10. In this case, the input resistance will be almost equal to the resistance of each of the resistors R3 and R4.
Capacitor C2 acts as a separator. It passes alternating current, and the OOS depends on the resistances R1 and R2. AC current by setting the desired AC gain.

For direct current, R1 seems to be absent, and the inverted input is connected to the output through R2, therefore, the depth of DC feedback is almost equal to 100%, and, therefore, the DC gain of such a circuit is equal to 1. It should be noted that in the circuits in Fig. . 9 and 10, the gain depends not only on the ratio of R1 and R2, but also on the capacitance of the isolation capacitor (C1 for Fig. 9, C2 for Fig. 10), since C2 (Fig. 10)

has a reactance that adds up to the resistance R1, so that the gain will depend on the frequency of the input signal, increase with its increase and decrease with its decrease. The comparator can also be unipolar (Fig. 11).

For experiments, in addition to the operational amplifier KR140UD608, you can use other op-amps, Figure 12 shows the pinouts of other popular op-amps. As a power source, you can use two "batteries" of 4.5 V each, for example, 312S. 3R12. Variable resistors can be from 100 kilo ohms to 1 megohm.

The journey of ten thousand miles begins with the first step.
(Chinese proverb)

It was in the evening, there was nothing to do ... And so suddenly I wanted to solder something. Sort of ... Electronic! .. Solder - so solder. The computer is available, the Internet is connected. We choose a scheme. And suddenly it turns out that the schemes for the conceived subject are a wagon and a small cart. And everyone is different. No experience, little knowledge. Which one to choose? Some of them contain some kind of rectangles, triangles. Amplifiers, and even operational ones ... How they work is not clear. Stra-a-ashno! .. What if it burns down? We choose what is simpler, on familiar transistors! Chose, soldered, turned on ... HELP !!! Does not work!!! Why?

Yes, because "Simplicity is worse than theft"! It's like a computer: the fastest and most sophisticated - gaming! And for office work, the simplest is enough. It's the same with transistors. Soldering a circuit on them is not enough. You still need to know how to set it up. Too many "pitfalls" and "rake". And this often requires experience that is by no means an entry level. So what, quit an exciting activity? By no means! Just do not be afraid of these "triangles-rectangles". It turns out that in many cases it is much easier to work with them than with individual transistors. IF YOU KNOW - HOW!

This is it: we will now deal with understanding how an operational amplifier (op-amp, or OpAmp in English) works. At the same time, we will consider his work literally “on the fingers”, practically without using any formulas, except perhaps, except for Ohm’s grandfather’s law: “Current through a circuit section ( I) is directly proportional to the voltage across it ( U) and inversely proportional to its resistance ( R)»:
I=U/R. (1)

To begin with, in principle, it is not so important how exactly the op-amp is arranged inside. Let's just take as an assumption that it is a "black box" with some stuffing there. At this stage, we will not consider such parameters of the op-amp as “bias voltage”, “shift voltage”, “temperature drift”, “noise characteristics”, “common-mode suppression coefficient”, “supply voltage ripple suppression coefficient”, “bandwidth " etc. All these parameters will be important at the next stage of its study, when the basic principles of its work “settle down” in the head, because “it was smooth on paper, but forgot about the ravines” ...

For now, let's just assume that the parameters of the op-amp are close to ideal and consider only what signal will be at its output if some signals are applied to its inputs.

So, the operational amplifier (op-amp) is a DC differential amplifier with two inputs (inverting and non-inverting) and one output. In addition to them, the op-amp has power leads: positive and negative. These five conclusions are found in almost any OS and are fundamentally necessary for its operation.

The op-amp has a huge gain, at least 50,000 ... 100,000, but in reality - much more. Therefore, as a first approximation, we can even assume that it is equal to infinity.

The term "differential" ("different" is translated from English as "difference", "difference", "difference") means that the output potential of the op-amp is affected solely by the potential difference between its inputs, regardless from them absolute meaning and polarity.

The term "DC" means that the op-amp amplifies the input signals starting from 0 Hz. The upper frequency range (frequency range) of the signals amplified by the op-amp depends on many factors, such as the frequency characteristics of the transistors of which it consists, the gain of the circuit built using the op-amp, etc. But this issue is already beyond the scope of the initial acquaintance with his work and will not be considered here.

Op-amp inputs have a very high input impedance equal to tens/hundreds of MegaOhm, or even GigaOhm (and only in the memorable K140UD1, and even in K140UD5 it was only 30...50 kOhm). Such a high impedance of the inputs means that they have almost no effect on the input signal.

Therefore, with a high degree of approximation to the theoretical ideal, we can assume that current does not flow into the inputs of the op-amp . This is - first an important rule that is applied in the analysis of the operation of the OS. Please remember well what it concerns only the OU itself, but not schemes with its use!

What do the terms "inverting" and "non-inverting" mean? In relation to what is the inversion determined and, in general, what kind of “animal” is this - signal inversion?

Translated from Latin, one of the meanings of the word "inversio" is "wrapping", "coup". In other words, inversion is a mirror image ( mirroring) signal relative to the horizontal axis X(time axis). On Fig. 1 shows a few of the many options signal inversion, where the direct (input) signal is marked in red and the inverted (output) signal is in blue.

Rice. 1 Concept of signal inversion

It should be especially noted that to the zero line (as in Fig. 1, A, B) the signal inversion not tied! Signals can be inverse and asymmetrical. For example, both are only in the region of positive values ​​(Fig. 1, B), which is typical for digital signals or with unipolar power supply (this will be discussed later), or both are partially in the positive and partially in the negative regions (Fig. 1, B, D). Other options are also possible. The main condition is their mutual specularity relative to some arbitrarily chosen level (for example, an artificial midpoint, which will also be discussed later). In other words, polarity signal is also not a determining factor.

Depict OU on circuit diagrams in different ways. Abroad, OS were previously depicted, and even now they are very often depicted in the form of an isosceles triangle (Fig. 2, A). The inverting input is marked with a minus symbol, and the non-inverting input is marked with a plus symbol inside a triangle. These symbols do not mean at all that the potential at the respective inputs must be more positive or more negative than at the other. They simply indicate how the output potential reacts to the potentials applied to the inputs. As a result, they are easy to confuse with power leads, which can be an unexpected "rake", especially for beginners.

Rice. 2 Variants of conditional graphic images (UGO)
operational amplifiers

In the system of domestic conditional graphic images (UGO) before the entry into force of GOST 2.759-82 (ST SEV 3336-81), OUs were also depicted as a triangle, only the inverting input - with an inversion symbol - a circle at the intersection of the output with a triangle (Fig. 2, B), and now - in the form of a rectangle (Fig. 2, C).

When designating the op-amp on the diagrams, the inverting and non-inverting inputs can be interchanged if it is more convenient, however, traditionally, the inverting input is shown at the top, and the non-inverting input is at the bottom. Power pins, as a rule, always have one way (positive at the top, negative at the bottom).

Op-amps are almost always used in negative feedback (NFB) circuits.

Feedback is the effect of applying a portion of the output voltage of an amplifier to its input, where it is algebraically (subject to sign) added to the input voltage. The principle of signal summation will be discussed below. Depending on which input of the op-amp, inverting or non-inverting, the OS is fed, there is a negative feedback (NFB), when part of the output signal is applied to the inverting input (Fig. 3, A) or positive feedback (PIC), when a part the output signal is fed, respectively, to the non-inverting input (Fig. 3, B).

Rice. 3 The principle of feedback formation (OS)

In the first case, since the output is the inverse of the input, it is subtracted from the input. As a result, the overall gain of the stage is reduced. In the second case, it is added to the input, the overall gain of the cascade is increased.

At first glance, it may seem that POS has a positive effect, and OOS is a completely useless undertaking: why reduce the gain? This is exactly what U.S. patent examiners thought when, in 1928, Harold S. Black tried patent the OS. However, sacrificing amplification, we significantly improve other important parameters of the circuit, such as its linearity, frequency range, etc. The deeper the OOS, the less the characteristics of the entire circuit depend on the characteristics of the op-amp.

But the PIC (given its own huge gain of the op-amp) has the opposite effect on the characteristics of the circuit and the most unpleasant thing is that it causes its self-excitation. It, of course, is also used consciously, for example, in generators, comparators with hysteresis (more on this later), etc., but in general view its influence on the operation of amplifier circuits with op amps is rather negative and requires a very thorough and reasonable analysis of its application.

Since the OS has two inputs, the following main types of its inclusion using the OS are possible (Fig. 4):

Rice. 4 Basic schemes for switching on the OS

a) inverting (Fig. 4, A) - the signal is applied to the inverting input, and the non-inverting one is connected directly to the reference potential (not used);

b) non-inverting (Fig. 4, B) - the signal is applied to the non-inverting input, and the inverting one is connected directly to the reference potential (not used);

in) differential (Fig. 4, B) - signals are fed to both inputs, inverting and non-inverting.

To analyze the operation of these schemes, one should take into account second the most important rule, to which the operation of the OS is subject: The output of an op-amp tends to have zero voltage difference between its inputs..

However, any wording must be necessary and sufficient to limit the entire subset of cases that obey it. The above formulation, for all its “classicism”, does not give any information about which of the inputs the output “seeks to influence”. Based on it, it turns out that the op-amp seems to equalize the voltages at its inputs, applying voltage to them from somewhere “from the inside”.

Looking closely at the diagrams in Fig. 4, you can see that the OOC (through Rooc) in all cases is started from the exit only to the inverting input, which gives us reason to reformulate this rule as follows: Voltage on the output of the op-amp, covered by the OOS, tends to ensure that the potential at the inverting input is equal to the potential at the non-inverting input.

Based on this definition, the “leading” at any inclusion of the OA with OOS is the non-inverting input, and the “slave” is the inverting one.

When describing the operation of an op amp, the potential at its inverting input is often referred to as "virtual zero" or "virtual midpoint". The translation of the Latin word "virtus" means "imaginary", "imaginary". A virtual object behaves close to the behavior of similar objects of material reality, i.e., for input signals (due to the action of the FOS), the inverting input can be considered connected directly to the same potential as the non-inverting input. However, "virtual zero" is just a special case that takes place only with bipolar power supply of the op-amp. When using a unipolar power supply (which will be discussed below), and in many other switching circuits, there will be no zero on either the non-inverting or inverting inputs. Therefore, let's agree that we will not use this term, since it interferes with the initial understanding of the principles of operation of the OS.

From this point of view, we will analyze the schemes shown in Fig. 4. At the same time, to simplify the analysis, we will assume that the supply voltages are still bipolar, equal to each other in value (say, ± 15 V), with a midpoint (common bus or “ground”), relative to which we will count the input and output voltages. In addition, the analysis will be carried out in direct current, because. a changing alternating signal at each moment of time can also be represented as a sample of direct current values. In all cases, feedback through Rooc is connected from the output of the op-amp to its inverting input. The difference is only in which of the inputs the input voltage is applied.

BUT) inverting switching on (Fig. 5).

Rice. 5 The principle of operation of the op-amp in an inverting connection

The potential at the non-inverting input is zero, because it is connected to the midpoint ("ground"). An input signal equal to +1 V relative to the midpoint (from GB) is applied to the left terminal of the input resistor Rin. Let us assume that the resistances Rooc and Rin are equal to each other and amount to 1 kOhm (their total resistance is 2 kOhm).

According to Rule 2, the inverting input must have the same potential as the non-inverting input, i.e., 0 V. Therefore, a voltage of +1 V is applied to Rin. According to Ohm's law, a current will flow through it Iinput= 1 V / 1000 ohms = 0.001 A (1 mA). The direction of flow of this current is shown by an arrow.

Since Rooc and Rin are connected by a divider, and according to Rule 1, the op-amp inputs do not consume current, in order for the voltage to be 0 V at the midpoint of this divider, a voltage must be applied to the right output of Rooc minus 1 V, and the current flowing through it Ioos should also be equal to 1 mA. In other words, a voltage of 2 V is applied between the left terminal Rin and the right terminal Rooc, and the current flowing through this divider is 1 mA (2 V / (1 kΩ + 1 kΩ) = 1 mA), i.e. I input = I oos .

If a negative polarity voltage is applied to the input, the output of the op-amp will be a positive polarity voltage. Everything is the same, only the arrows showing the flow of current through Rooc and Rin will be directed in the opposite direction.

Thus, if the values ​​​​of Rooc and Rin are equal, the voltage at the output of the op-amp will be equal to the voltage at its input in magnitude, but inverse in polarity. And we got inverting repeater . This scheme is often used if you need to invert the signal received using circuits that are fundamentally inverters. For example, logarithmic amplifiers.

Now let's keep Rin equal to 1 kOhm and increase the Rooc resistance to 2 kOhm with the same input signal +1 V. The total divider resistance Rooc+Rin has increased to 3 kOhm. In order for a potential of 0 V (equal to the potential of the non-inverting input) to remain at its midpoint, the same current (1 mA) must flow through Rooc as through Rin. Therefore, the voltage drop across Rooc (voltage at the output of the op-amp) should already be 2 V. At the output of the op-amp, the voltage is minus 2 V.

Let's increase the value of Rooc to 10 kOhm. Now the voltage at the output of the op-amp under the same other conditions will already be 10 V. Wow! Finally we got inverting amplifier ! Its output voltage is greater than the input voltage (in other words, the gain Ku) as many times as the resistance Rooc is greater than the resistance Rin. No matter how I swore not to use formulas, let's still display this as an equation:
Ku \u003d - Uout / Uin \u003d - Rooc / Rin. (2)

The minus sign in front of the fraction on the right side of the equation only means that the output signal is inverse with respect to the input. And nothing more!

And now let's increase the resistance Rooc to 20 kOhm and analyze what happens. According to formula (2), with Ku \u003d 20 and an input signal of 1 V, the output should have been a voltage of 20 V. But it wasn’t there! We previously assumed that the supply voltage of our op-amp is only ± 15 V. But even 15 V cannot be obtained (why so - a little lower). "You can't jump above your head (supply voltage)"! As a result of such abuse of the ratings of the circuit, the output voltage of the op-amp “rests” on the supply voltage (the output of the op-amp enters saturation). The balance of current equality through the divider RoocRin ( Iinput = Ioos) is violated, a potential appears at the inverting input, which is different from the potential at the non-inverting input. Rule 2 no longer applies.

Input resistance inverting amplifier is equal to the resistance Rin, since all the current from the input signal source (GB) flows through it.

Now let's replace the constant Rooc with a variable, with a nominal value of, say, 10 kOhm (Fig. 6).

Rice. 6 Variable gain inverting amplifier circuit

With the right (according to the circuit) position of its slider, the gain will be Rooc / Rin \u003d 10 kOhm / 1 kOhm = 10. By moving the Rooc slider to the left (decreasing its resistance), the gain of the circuit will decrease and, finally, at its extreme left position it will become equal to zero, since the numerator in the above formula will become zero at any the value of the denominator. The output will also be zero for any value and polarity of the input signal. Such a scheme is often used in audio signal amplification circuits, for example, in mixers, where you have to adjust the gain from zero.

B) non-inverting switching on (Fig. 7).

Rice. 7 The principle of operation of the op-amp in a non-inverting inclusion

The left pin of Rin is connected to the midpoint ("ground"), and the input signal equal to +1 V is applied directly to the non-inverting input. Since the nuances of the analysis are “chewed” above, here we will pay attention only to significant differences.

At the first stage of the analysis, we also take the resistances Rooc and Rin equal to each other and equal to 1 kOhm. Because at the non-inverting input, the potential is +1 V, then according to Rule 2, the same potential (+1 V) must be at the inverting input (shown in the figure). To do this, there must be a voltage of +2 V on the right terminal of the Rooc resistor (output of the op-amp). Currents Iinput and Ioos, equal to 1 mA, now flow through the resistors Rooc and Rin in the opposite direction (shown by arrows). We got it non-inverting amplifier with a gain of 2, since an input of +1V produces an output of +2V.

Strange, isn't it? The ratings are the same as in the inverting connection (the only difference is that the signal is applied to another input), and the gain is obvious. We'll look into this a little later.

Now we increase the value of Rooc to 2 kOhm. To keep the balance of currents Iinput = Ioos and the potential of the inverting input is +1 V, the output of the op-amp should already be +3 V. Ku \u003d 3 V / 1 V \u003d 3!

If we compare the values ​​of Ku with a non-inverting connection with an inverting one, with the same ratings Rooc and Rin, it turns out that the gain in all cases is greater by one. We derive the formula:
Ku \u003d Uout / Uin + 1 \u003d (Rooc / Rin) + 1 (3)

Why is this happening? Yes, very easy! The NFB works exactly the same as in an inverting connection, but according to Rule 2, the potential of the non-inverting input is always added to the potential of the inverting input in a non-inverting connection.

So, with a non-inverting inclusion, it is impossible to obtain a gain equal to 1? Why not, why not. Let's reduce the value of Rooc, similar to how we analyzed Fig. 6. When its value is zero - by short-circuiting the output with the inverting input (Fig. 8, A), according to Rule 2, the output will have such a voltage that the potential of the inverting input is equal to the potential of the non-inverting input, i.e., +1 V. We get: Ku \u003d 1 V / 1 V \u003d 1 (!) Well, since the inverting input does not consume current and there is no potential difference between it and the output, then no current flows in this circuit.

Rice. 8 Scheme of switching on the op-amp as a voltage follower

Rin becomes generally superfluous, because it is connected in parallel with the load on which the output of the op-amp should work, and its output current will flow through it in vain. And what happens if you leave Rooc, but remove Rin (Fig. 8, B)? Then in the gain formula Ku = Roos / Rin + 1, the resistance Rin theoretically becomes close to infinity (in reality, of course, not, because there are leaks on the board, and the input current of the op-amp, although negligible, is still zero is still not equal), and the ratio Rooc / Rin is equated to zero. Only one remains in the formula: Ku \u003d + 1. Can the gain be less than one for this circuit? No, less will not work under any circumstances. You can’t go around the “extra” unit in the gain formula on a crooked goat ...

After we removed all the "extra" resistors, we get a circuit non-inverting repeater shown in Fig. 8, V.

At first glance, such a scheme does not make practical sense: why do we need a single, and even non-inverse "amplification" - what, you can’t just send a signal further ??? However, such schemes are used quite often and here's why. According to Rule 1, current does not flow into the inputs of the op-amp, i.e., input impedance the non-inverting follower is very large - the same tens, hundreds and even thousands of MΩ (the same applies to the circuit according to Fig. 7)! But the output resistance is very small (fractions of Ohm!). The output of the op-amp “chuffs with all its might”, trying, according to Rule 2, to maintain the same potential at the inverting input as at the non-inverting one. The only limitation is the permissible output current of the op-amp.

But from this place we will wag a little to the side and consider the issue of the output currents of the op-amp in a little more detail.

For most general purpose op amps, the technical specifications state that the resistance of the load connected to their output should not be smaller 2 kOhm More - as much as you want. For a much smaller number, it is 1 kOhm (K140UD ...). This means that under the worst-case conditions: the maximum supply voltage (e.g. ±16 V or a total of 32 V), a load connected between the output and one of the power rails, and the maximum output voltage of opposite polarity, a voltage of about 30 V will be applied to the load. In this case, the current through it will be: 30 V / 2000 Ohm = 0.015 A (15 mA). Not so little, but not too much either. Fortunately, most general purpose op amps have built-in overcurrent protection - typical maximum output current is 25 mA. Protection prevents overheating and failure of the op-amp.

If the supply voltages are not the maximum allowable, then the minimum load resistance can be proportionally reduced. Say, with a power supply of 7.5 ... 8 V (total 15 ... 16 V), it can be 1 kOhm.

AT) differential switching on (Fig. 9).

Rice. 9 The principle of operation of the op-amp in a differential connection

So, let's assume that with the same ratings Rin and Rooc equal to 1 kOhm, the same voltages equal to +1 V are applied to both inputs of the circuit (Fig. 9, A). Since the potentials on both sides of the resistor Rin are equal to each other (the voltage across the resistor is 0), no current flows through it. This means that the current through the resistor Rooc is also zero. That is, these two resistors do not perform any function. In fact, we actually got a non-inverting follower (compare with Fig. 8). Accordingly, we will get the same voltage at the output as at the non-inverting input, i.e., +1 V. Let's change the polarity of the input signal at the inverting input of the circuit (turn GB1 over) and apply minus 1 V (Fig. 9, B). Now a voltage of 2 V is applied between the terminals Rin and a current flows through it Iin\u003d 2 mA (I hope that it is no longer necessary to describe in detail why this is so?). In order to compensate for this current, a current of 2 mA must also flow through Rooc. And for this, the output of the op-amp must have a voltage of +3 V.

That's where the malicious "grin" of an additional one appeared in the formula for the gain of a non-inverting amplifier. It turns out that with such simplified In differential switching, the difference in gain constantly shifts the output signal by the potential at the non-inverting input. A problem with! However, "Even if you were eaten, you still have at least two exits." This means that we somehow need to equalize the gains of the inverting and non-inverting inclusions in order to “neutralize” this extra one.

To do this, let's apply the input signal to the non-inverting input not directly, but through the divider Rin2, R1 (Fig. 9, B). Let's take their denominations also for 1 kOhm. Now, at the non-inverting (and therefore also at the inverting) input of the op-amp, there will be a potential of +0.5 V, a current will flow through it (and Rooc) Iin = Ioos\u003d 0.5 mA, to ensure which the output of the op-amp must have a voltage equal to 0 V. Phew! We got what we wanted! With equal magnitude and polarity signals at both inputs of the circuit (in this case +1 V, but the same will be true for minus 1 V and for any other digital values), zero voltage will be maintained at the output of the op-amp, equal to the difference in input signals .

Let's check this reasoning by applying a signal of negative polarity minus 1 V to the inverting input (Fig. 9, D). Wherein Iin = Ioos= 2 mA, for which the output should be +2 V. Everything was confirmed! The output level corresponds to the difference between the inputs.

Of course, if Rin1 and Rooc are equal (respectively, Rin2 and R1), we will not get amplification. To do this, you need to increase the values ​​​​of Rooc and R1, as was done when analyzing previous inclusions of the op-amp (I will not repeat it), and it should strictly respect the ratio:

Rooc / Rin1 = R1 / Rin2. (4)

What useful do we get from such an inclusion in practice? And we get a remarkable property: the output voltage does not depend on the absolute values ​​of the input signals, if they are equal to each other in magnitude and polarity. Only the difference (differential) signal is output. This makes it possible to amplify very small signals against the background of noise acting equally on both inputs. For example, a signal from a dynamic microphone against the background of a 50 Hz industrial frequency mains pickup.

However, in this barrel of honey, unfortunately, there is a fly in the ointment. First, equality (4) must be observed very strictly (up to tenths and sometimes hundredths of a percent!). Otherwise, there will be an unbalance of the currents acting in the circuit, and therefore, in addition to the difference ("anti-phase") signals, the combined ("common-mode") signals will also be amplified.

Let's understand the essence of these terms (Fig. 10).

Rice. 10 Signal phase shift

The phase of the signal is a value that characterizes the offset of the origin of the signal period relative to the origin of time. Since both the origin of time and the origin of the period are chosen arbitrarily, the phase of one periodical signal has no physical meaning. However, the phase difference between the two periodical signals is a quantity that has a physical meaning, it reflects the delay of one of the signals relative to the other. What is considered the beginning of the period does not matter. For the point of the beginning of the period, you can take a zero value with a positive slope. It is possible - maximum. Everything is in our power.

On Fig. 9, red indicates the original signal, green - shifted by ¼ period relative to the original, and blue - by ½ period. If we compare the red and blue curves with the curves in Fig. 2, B, it can be seen that they are mutually inverse. Thus, “in-phase signals” are signals that coincide with each other at each of their points, and “anti-phase signals” are inverse relative to each other.

At the same time, the concept inversions broader than the concept phases, because the latter applies only to regularly repeated, periodic signals. And the concept inversions applicable to any signals, including non-periodic ones, such as an audio signal, a digital sequence, or a constant voltage. To phase is a consistent value, the signal must be periodic at least over a certain interval. Otherwise, both phase and period turn into mathematical abstractions.

Secondly, the inverting and non-inverting inputs in the differential connection, with equal ratings Rooc = R1 and Rin1 = Rin2, will have different input resistances. If the input resistance of the inverting input is determined only by the value Rin1, then the non-inverting input is determined by the values successively included Rin2 and R1 (have not forgotten that the op-amp inputs do not consume current?). In the example above, they will be 1 and 2 kΩ, respectively. And if we increase Rooc and R1 to obtain a full-fledged amplifying stage, then the difference will increase even more significantly: with Ku \u003d 10 - up to, respectively, the same 1 kOhm and as much as 11 kOhm!

Unfortunately, in practice, the ratings Rin1 = Rin2 and Rooc = R1 are usually set. However, this is only acceptable if the signal sources for both inputs are of very low output impedance. Otherwise, it forms a divider with the input impedance of this amplifying stage, and since the division factor of such “dividers” will be different, the result is obvious: a differential amplifier with such resistor values ​​will not perform its function of suppressing common-mode (combined) signals, or perform this function poorly .

One of the ways to solve this problem can be the inequality of the values ​​of the resistors connected to the inverting and non-inverting inputs of the op-amp. Namely, so that Rin2 + R1 = Rin1. Another important point is to achieve exact observance of equality (4). As a rule, this is achieved by splitting R1 into two resistors - a constant, usually 90% of the desired value, and a variable (R2), whose resistance is 20% of the required value (Fig. 11, A).

Rice. 11 Differential amplifier balancing options

The path is generally accepted, but again, with this method of balancing, albeit slightly, the input impedance of the non-inverting input changes. A much more stable option with the inclusion of a tuning resistor (R5) in series with Rooc (Fig. 11, B), since Rooc does not participate in the formation of the input resistance of the inverting input. The main thing is to keep the ratio of their denominations, similar to option "A" (Rooc / Rin1 = R1 / Rin2).

Since we talked about differential switching and mentioned repeaters, I would like to describe one interesting circuit (Fig. 12).

Rice. 12 Switched inverting/non-inverting follower circuit

The input signal is applied simultaneously to both inputs of the circuit (inverting and non-inverting). The ratings of all resistors (Rin1, Rin2 and Rooc) are equal to each other (in this case, let's take their real values: 10 ... 100 kOhm). The non-inverting input of the op-amp with the SA key can be closed to a common bus.

In the closed position of the key (Fig. 12, A), the resistor Rin2 does not participate in the operation of the circuit (only current “uselessly” flows through it Ivx2 from the signal source to the common bus). We get inverting follower with a gain equal to minus 1 (see Fig. 6). But with the SA key in the open position (Fig. 12, B), we get non-inverting follower with gain equal to +1.

The principle of operation of this scheme can be expressed in a slightly different way. When the SA key is closed, it works as an inverting amplifier with a gain equal to minus 1, and when it is open - simultaneously(!) And as an inverting amplifier with a gain, minus 1, and as a non-inverting amplifier with a gain of +2, from where: Ku = +2 + (–1) = +1.

In this form, this circuit can be used if, for example, the polarity of the input signal is unknown at the design stage (say, from a sensor that is not accessible until the device is set up). If, however, a transistor (for example, a field-effect transistor) is used as a key, controlled from the input signal using comparator(which will be discussed below), we get synchronous detector(synchronous rectifier). The specific implementation of such a scheme, of course, goes beyond the initial acquaintance with the operation of the OS, and we will not consider it in detail here again.

And now let's consider the principle of summing the input signals (Fig. 13, A), and at the same time we will figure out what values ​​​​of the resistors Rin and Rooc should be in reality.

Rice. 13 The principle of operation of the inverting adder

We take as a basis the inverting amplifier already discussed above (Fig. 5), only we connect not one, but two input resistors Rin1 and Rin2 to the input of the op-amp. So far, for "educational" purposes, we accept the resistance of all resistors, including Rooc, equal to 1 kOhm. We supply input signals equal to +1 V to the left terminals Rin1 and Rin2. Currents equal to 1 mA flow through these resistors (shown by arrows pointing from left to right). To maintain the same potential at the inverting input as at the non-inverting one (0 V), a current equal to the sum of the input currents (1 mA + 1 mA = 2 mA) must flow through the Rooc resistor, shown by an arrow pointing in the opposite direction (from right to left ), for which the output of the op-amp must have a voltage of minus 2 V.

The same result (output voltage minus 2 V) can be obtained if +2 V is applied to the input of the inverting amplifier (Fig. 5), or the value of Rin is halved, i.e. up to 500 Ohm. Let's increase the voltage applied to the resistor Rin2 up to +2 V (Fig. 13, B). At the output we get a voltage of minus 3 V, which is equal to the sum of the input voltages.

There can be not two inputs, but as many as you like. The principle of operation of this circuit will not change from this: the output voltage in any case will be directly proportional to the algebraic sum (taking into account the sign!) of the currents passing through the resistors connected to the inverting input of the op-amp (inversely proportional to their ratings), regardless of their number.

If, however, signals equal to +1 V and minus 1 V are applied to the inputs of the inverting adder (Fig. 13, B), then the currents flowing through them will be in different directions, they will cancel each other out and the output will be 0 V. Through the resistor Rooc in this case no current will flow. In other words, the current flowing through Rooc is algebraically summed with input currents.

An important point also follows from this: while we were operating with small input voltages (1 ... 3 V), the output of a widely used op-amp could well provide such a current (1 ... 3 mA) for Rooc and something else remained for the load connected to the output of the op-amp. But if the voltages of the input signals are increased to the maximum allowable (close to the supply voltages), then it turns out that the entire output current will go to Rooc. Nothing left to load. And who needs an amplifying stage that works "for itself"? In addition, input resistor values ​​of only 1 kΩ (respectively, determining the input resistance of the inverting amplifier stage) require excessively high currents to flow through them, heavily loading the signal source. Therefore, in real circuits, the resistance Rin is chosen not less than 10 kOhm, but it is also desirable not more than 100 kOhm, so that at a given gain, Rooc is not set too high. Although these values ​​\u200b\u200bare not absolute, but only estimates, as they say, "in the first approximation" - it all depends on the specific circuit. In any case, it is undesirable that a current flowing through Rooc exceeds 5 ... 10% of the maximum output current of this particular op-amp.

The summed signals can also be applied to the non-inverting input. It turns out non-inverting adder. In principle, such a circuit will work in exactly the same way as an inverting adder, the output of which will be a signal that is directly proportional to the input voltages and inversely proportional to the values ​​of the input resistors. However, in practice it is used much less frequently, because. contains a "rake" that should be taken into account.

Since Rule 2 is valid only for the inverting input, which has a “virtual zero potential”, then the non-inverting input will have a potential equal to the algebraic sum of the input voltages. Therefore, the input voltage available at one of the inputs will affect the voltage supplied to the other inputs. There is no “virtual potential” at the non-inverting input! As a result, additional circuitry tricks have to be applied.

So far, we have considered circuits based on OS with OOS. What happens if feedback is removed altogether? In this case, we get comparator(Fig. 14), i.e., a device that compares the absolute value of two potentials at its inputs (from English word compare- compare). At its output, there will be a voltage approaching one of the supply voltages, depending on which of the signals is greater than the other. Typically, the input signal is applied to one of the inputs, and to the other - a constant voltage with which it is compared (the so-called "reference voltage"). It can be anything, including zero potential (Fig. 14, B).

Rice. 14 Scheme of switching on the op-amp as a comparator

However, not everything is so good "in the kingdom of Denmark" ... And what happens if the voltage between the inputs is zero? In theory, the output should also be zero, but in reality - never. If the potential at one of the inputs even slightly outweighs the potential of the other, then this will already be enough for chaotic voltage surges to occur at the output due to random disturbances induced at the inputs of the comparator.

In reality, any signal is "noisy", because ideal cannot be by definition. And in the area close to the point of equality of the potentials of the inputs, a burst of output signals will appear at the output of the comparator instead of one clear switching. To combat this phenomenon, the comparator circuit is often introduced hysteresis by creating a weak positive PIC from the output to the non-inverting input (Figure 15).

Rice. 15 The principle of operation of the hysteresis in the comparator due to the PIC

Let's analyze the operation of this scheme. Its supply voltage is ± 10 V (for an even account). The resistance Rin is 1 kOhm, and Rpos is 10 kOhm. The midpoint potential is chosen as the reference voltage applied to the inverting input. The red curve shows the input signal coming to the left pin Rin (input scheme comparator), blue - the potential at the non-inverting input of the op-amp and green - the output signal.

While the input signal has a negative polarity, the output is a negative voltage, which, through Rpos, is added to the input voltage in inverse proportion to the values ​​of the corresponding resistors. As a result, the potential of the non-inverting input over the entire range negative values 1 V (absolute value) higher than the input signal level. As soon as the potential of the non-inverting input is equal to the potential of the inverting one (for the input signal, this will be + 1 V), the voltage at the output of the op-amp will begin to switch from negative to positive polarity. The total potential at the non-inverting input will start like an avalanche become even more positive, supporting the process of such a switch. As a result, the comparator simply “will not notice” insignificant noise fluctuations of the input and reference signals, since they will be many orders of magnitude smaller in amplitude than the described “step” of the potential at the non-inverting input when switching.

When the input signal decreases, the reverse switching of the comparator output signal will occur at an input voltage of minus 1 V. This difference between the input signal levels leading to the switching of the comparator output, which in our case is equal to a total of 2 V, is called hysteresis. The greater the resistance Rpos with respect to Rin (the smaller the POS depth), the smaller the switching hysteresis. So, with Rpos \u003d 100 kOhm, it will be only 0.2 V, and with Rpos \u003d 1 MΩ, it will be 0.02 V (20 mV). The hysteresis (PIC depth) is selected based on the actual operating conditions of the comparator in a particular circuit. In which 10 mV will be a lot, and in which - and 2 V will be small.

Unfortunately, not every op amp and not in all cases can be used as a comparator. Specialized comparator microcircuits are produced for matching between analog and digital signals. Some of them are specialized for connecting to digital TTL microcircuits (597CA2), some - to digital ESL microcircuits (597CA1), but most are so-called. "comparators for general use" (LM393/LM339/K554CA3/K597CA3). Their main difference from the op amps lies in the special device of the output stage, which is made on an open collector transistor (Fig. 16).

Rice. 16 Comparator output stage for general applications
and its connection to the load resistor

This requires the mandatory use of an external load resistor(R1), without which the output signal is simply not physically capable of generating a high (positive) output level. The voltage +U2 to which the load resistor is connected may be different from the supply voltage +U1 of the comparator chip itself. This allows simple means provide the output signal of the desired level - be it TTL or CMOS.

Note In most comparators, an example of which can be dual LM393 (LM193 / LM293) or exactly the same in circuitry, but quad LM339 (LM139 / LM239), the emitter of the output stage transistor is connected to the negative power terminal, which somewhat limits their scope. In this regard, I would like to draw attention to the comparator LM31 (LM111 / LM211), the analogue of which is the domestic 521 / 554CA3, in which both the collector and the emitter of the output transistor are separately output, which can be connected to other voltages than the supply voltage of the comparator itself. Its only and relative disadvantage is that it is only one in an 8-pin (sometimes 14-pin) package.

So far, we have considered circuits in which the input signal was fed to the input(s) through Rin, i.e. they were all converters input voltage in day off voltage same. In this case, the input current flowed through Rin. What happens if its resistance is taken equal to zero? The circuit will work in exactly the same way as the inverting amplifier discussed above, only the output impedance of the signal source (Rout) will serve as Rin, and we get converter input current in day off voltage(Fig. 17).

Rice. 17 Scheme of the current-to-voltage converter at the op-amp

Since the potential at the inverting input is the same as at the non-inverting one (in this case it is “virtual zero”), the entire input current ( Iin) will flow through Rooc between the output of the signal source (G) and the output of the op-amp. The input resistance of such a circuit is close to zero, which makes it possible to build micro/milliammeters on its basis, which practically do not affect the current flowing through the measured circuit. Perhaps the only limitation is the permissible input voltage range of the op-amp, which should not be exceeded. With its help, you can also build, for example, a linear photodiode current-to-voltage converter and many other circuits.

We have considered the basic principles of operation of the OS in various schemes for its inclusion. One important question remains: nutrition.

As mentioned above, an op amp typically has only 5 pins: two inputs, an output, and two power pins, positive and negative. In general, bipolar power is used, that is, the power supply has three outputs with potentials: + U; 0; -U.

Once again, carefully consider all the above figures and see that a separate output of the midpoint in the op-amp NO ! It is simply not needed for their internal circuitry to work. In some circuits, a non-inverting input was connected to the midpoint, however, this is not the rule.

Hence, overwhelming majority modern op amps are designed to power UNIPOLAR tension! A logical question arises: “Why then do we need bipolar power,” if we depicted it so stubbornly and with enviable constancy in the drawings?

It turns out it's just very comfortably for practical purposes for the following reasons:

A) To ensure sufficient current and output voltage swing through the load (Fig. 18).

Rice. 18 The flow of output current through the load with various options for supplying the op-amp

For now, we will not consider the input (and OOS) circuits of the circuits shown in the figure (“black box”). Let's take it for granted that some input sinusoidal signal is applied to the input (black sinusoid on the graphs) and the output is the same sinusoidal signal, amplified with respect to the input colored sinusoid on the graphs).

When connecting the load Rload. between the output of the op-amp and the midpoint of the connection of the power supplies (GB1 and GB2) - Fig. 18, A, the current flows through the load symmetrically with respect to the midpoint (respectively, red and blue half-waves), and its amplitude is maximum and the voltage amplitude at Rload. also the maximum possible - it can reach almost supply voltages. The current from the power source of the corresponding polarity is closed through the OS, Rload. and a power source (red and blue lines showing current flow in the respective direction).

Since the internal resistance of the op-amp power supplies is very low, the current through the load is only limited by its resistance and the op-amp's maximum output current, which is typically 25 mA.

When the op-amp is powered by a unipolar voltage as common bus the negative (negative) pole of the power source is usually selected, to which the second output of the load is connected (Fig. 18, B). Now the current through the load can only flow in one direction (shown by the red line), the second direction simply has nowhere to come from. In other words, the current through the load becomes asymmetrical (pulsating).

It is impossible to say unequivocally that this option is bad. If the load is, say, a dynamic head, then for it it is bad unambiguously. However, there are many applications where connecting a load between the output of the op-amp and one of the power rails (usually negative polarity) is not only acceptable, but also the only possible one.

If, nevertheless, it is necessary to ensure the symmetry of the current flow through the load with a unipolar supply, then it is necessary to galvanically decouple it from the output of the op-amp with a galvanic capacitor C1 (Fig. 18, B).

B) To ensure the required current of the inverting input, as well as bindings input signals to some arbitrarily selected level accepted for the reference (zero) - setting the operation mode of the OS for direct current (Fig. 19).

Rice. 19 Connecting the input signal source with various options for supplying the op-amp

Now consider the options for connecting input signal sources, excluding from consideration the connection of the load.

Connecting the inverting and non-inverting inputs to the midpoint of the power supply connection (Fig. 19, A) was considered when analyzing the previously given diagrams. If the non-inverting input does not consume current and simply accepts the midpoint potential, then through the signal source (G) and Rin, connected in series, the current flows, closing through the corresponding power source! And since their internal resistances are negligible compared to the input current (many orders of magnitude less than Rin), it practically does not affect the supply voltage.

Thus, with a unipolar supply of the op-amp, you can quite easily form the potential supplied to its non-inverting input using the R1R2 divider (Fig. 19, B, C). Typical resistor values ​​of this divider are 10 ... 100 kOhm, and it is highly desirable to shunt the lower one (connected to a common negative bus) with a capacitor by 10 ... 22 microfarads in order to significantly reduce the effect of supply voltage ripples on the potential of such artificial middle point.

But it is extremely undesirable to connect the signal source (G) to this artificial midpoint because of the same input current. Let's guess. Even with the ratings of the divider R1R2 = 10 kOhm and Rin = 10…100 kOhm, the input current Iin will be at best 1/10, and at worst - up to 100% of the current passing through the divider. Consequently, the potential at the non-inverting input will “float” by the same amount in combination (in phase) with the input signal.

To eliminate the mutual influence of the inputs on each other when amplifying DC signals with such a connection, for the signal source it is necessary to organize a separate potential of the artificial midpoint, formed by resistors R3R4 (Fig. 19, B), or, if the AC signal is amplified, galvanically isolate the signal source from the inverting input by capacitor C2 (Fig. 19, B).

It should be noted that in the diagrams above (Fig. 18, 19) we assumed by default that the output signal should be symmetrical about either the midpoint of the power supplies or the artificial midpoint. In reality, this is not always necessary. Quite often, you want the output signal to have predominantly either positive or negative polarity. Therefore, it is not necessary that the positive and negative polarities of the power supply be equal in absolute value. One of them can be much smaller in absolute value than the other - only in such a way as to ensure the normal functioning of the OS.

A logical question arises: “Which one exactly?” To answer it, let's briefly consider the allowable voltage ranges of the input and output signals of the op-amp.

For any op amp, the output potential cannot be higher than the potential of the positive power rail and lower than the potential of the negative power rail. In other words, the output voltage cannot go beyond the limits of the supply voltages. For example, for an OPA277 op amp, the output voltage at a load resistance of 10 kΩ is 2 V less than the positive power rail and 0.5 V less than the negative power rail. The width of these "dead zones" of the output voltage, which the op amp output cannot reach, depends on the series factors such as output stage circuitry, load resistance, etc.). There are op amps that have minimal dead zones, for example, 50 mV to the supply rail voltage at a load of 10 kΩ (for OPA340), this feature of the op amp is called "rail-to-rail" (R2R).

On the other hand, for general-purpose op-amps, the input signals should also not exceed the supply voltage, and for some, be less than 1.5 ... 2 V. However, there are op-amps with specific input stage circuitry (for example, the same LM358 / LM324) , which can work not only from the negative power level, but even “negative” by 0.3 V, which greatly facilitates their use with unipolar power supply with a common negative bus.

Let's finally look at and feel these "spider bugs". You can even sniff and lick. I allow. Consider their most common options available to novice radio amateurs. Especially if you have to solder the op amp from the old equipment.

For op-amps of old designs, without fail requiring external circuits for frequency correction, in order to prevent self-excitation, it was typical to have additional conclusions. Because of this, some op-amps did not even “fit” into an 8-pin package (Fig. 20, A) and were made in 12-pin round metal-glass, for example, K140UD1, K140UD2, K140UD5 (Fig. 20, B) or in 14-pin DIP packages, for example, K140UD20, K157UD2 (Fig. 20, B). The abbreviation DIP is an abbreviation of the English expression "Dual In line Package" and translates as "double-sided package".

The round metal-glass case (Fig. 20, A, B) was used as the main one for imported op-amps until about the mid-70s, and for domestic op-amps - until the mid-80s and is now used for the so-called. "military" applications ("5th acceptance").

Sometimes domestic op-amps were placed in currently rather "exotic" cases: a 15-pin rectangular metal-glass for the hybrid K284UD1 (Fig. 20, D), in which the key is an additional 15th pin from the case, and others. True, I personally have not met planar 14-pin packages (Fig. 20, E) for placing an op-amp in them. They were used for digital circuits.

Rice. 20 Cases of domestic operational amplifiers

Modern op-amps, for the most part, contain corrective circuits right on the chip, which made it possible to get by with a minimum number of pins (as an example, a 5-pin SOT23-5 for a single op-amp - Fig. 23). This made it possible to place two to four completely independent (except for common power outputs) op-amps made on a single chip in one case.

Rice. 21 Two-row plastic cases of modern op amps for output mounting (DIP)

Sometimes you can find op-amps placed in single-row 8-pin (Fig. 22) or 9-pin packages (SIP) - K1005UD1. The abbreviation SIP is an abbreviation of the English expression "Single In line Package" and translates as "housing with one-way pinout."

Rice. 22 Single-row plastic case of double op-amps for through-hole mounting (SIP-8)

They were designed to minimize the space occupied on the board, but, unfortunately, they were "late": by this time, surface mount packages (SMD - Surface Mounting Device) by soldering directly to the board tracks (Fig. 23) had become widespread. However, for beginners, their use presents significant difficulties.

Rice. 23 Cases of modern imported op amps for surface mounting (SMD)

Very often, the same microcircuit can be "packed" by the manufacturer in different packages (Fig. 24).

Rice. 24 Placement options for the same chip in different packages

The conclusions of all microcircuits have a sequential numbering, counted from the so-called. "key", indicating the location of the output at number 1. (Fig. 25). AT any if you arrange the case with leads Push, their numbering goes in ascending order against clockwise!

Rice. 25 Operational amplifier pin assignment
in various cases (pinout), top view;
numbering direction shown by arrows

In round metal-glass cases, the key has the form of a side protrusion (Fig. 25, A, B). Here, from the location of this key, huge “rakes” are possible! In domestic 8-pin cases (302.8), the key is located opposite the first pin (Fig. 25, A), and in imported TO-5, opposite the eighth pin (Fig. 25, B). In 12-pin cases, both domestic (302.12) and imported, the key is located between the first and 12th conclusions.

Typically, the inverting input, both in round glass-metal and DIP packages, is connected to the 2nd pin, the non-inverting input to the 3rd pin, the output to the 6th pin, the power minus to the 4th pin, and the power plus to the pin 4. 7th. However, there are exceptions (another possible "rake"!) In the pinout of the OU K140UD8, K574UD1. In them, the numbering of the conclusions is shifted by one counterclockwise compared to the generally accepted for most other types, i.e. they are connected to the terminals, as in imported cases (Fig. 25, B), and the numbering corresponds to domestic ones (Fig. 25, A).

AT last years most of the OS "domestic purposes" began to be placed in plastic cases (Fig. 21, 25, C-D). In these cases, the key is either a recess (dot) opposite the first pin, or a cutout in the end of the case between the first and 8th (DIP-8) or 14th (DIP-14) pins, or a chamfer along the first half of the pins (Fig. 21, middle). The pin numbering in these cases also goes against clockwise when viewed from above (with conclusions away from you).

As mentioned above, internally corrected op amps have only five leads, of which only three (two inputs and an output) belong to each individual op amp. This made it possible to place two completely independent (with the exception of plus and minus power, which require two more pins) op amps on one chip in one 8-pin package (Fig. 25, D), and even four in a 14-pin package (Fig. 25, D). As a result, at present, most op amps are produced at least dual, for example, TL062, TL072, TL082, cheap and simple LM358, etc. Exactly the same in internal structure, but quad - respectively, TL064, TL074, TL084 and LM324.

With regard to the domestic analogue of the LM324 (K1401UD2), there is another “rake”: if in the LM324 the plus of the power supply is connected to the 4th output, and the minus to the 11th, then in K1401UD2 it is the other way around: the plus of the power is connected to the 11th output, and minus - on the 4th. However, this difference does not cause any difficulties with wiring. Since the pinout of the op-amp pins is completely symmetrical (Fig. 25, E), you just need to turn the case 180 degrees so that the 1st pin takes the place of the 8th. Yes, that's all.

A few words about the labeling of imported OUs (and not only OUs). For a number of developments of the first 300 digital designations, it was customary to designate the quality group with the first digit of the digital code. For example, LM158/LM258/LM358 op amps, LM193/LM293/LM393 comparators, TL117/TL217/TL317 adjustable three-pin stabilizers, etc. are completely identical in internal structure, but differ in temperature operating range. For LM158 (TL117) the operating temperature range is from minus 55 to +125 ... 150 degrees Celsius (the so-called "combat" or military range), for LM258 (TL217) - from minus 40 to +85 degrees ("industrial" range) and for LM358 (TL317) - from 0 to +70 degrees ("household" range). At the same time, the price for them may be completely inappropriate for such a gradation, or differ very slightly ( inscrutable ways of pricing!). So you can buy them with any marking available “for the pocket” of a beginner, without particularly chasing the first “troika”.

After the first three hundred digital markings were exhausted, reliability groups began to be marked with letters, the meaning of which is deciphered in datasheets (Datasheet literally translates as “data table”) for these components.

Conclusion

So we studied the "alphabet" of the operation of the op-amp, capturing a little and comparators. Then you need to learn how to put words, sentences and whole meaningful “compositions” (workable schemes) out of these “letters”.

Unfortunately, "It is impossible to grasp the immensity." If the material presented in this article helped to understand how these "black boxes" work, then further deepening into the analysis of their "stuffing", the influence of input, output and transient characteristics, is the task of a more advanced study. Information about this is described in detail and thoroughly in a variety of existing literature. As grandfather William of Ockham used to say: "Entities should not be multiplied beyond what is necessary." There is no need to repeat what has already been well described. All you need to do is not be lazy and read it.

11. http://www.texnic.ru/tools/lekcii/electronika/l6/lek_6.html

Therefore, let me take my leave, with respect, etc., the author Alexey Sokolyuk ()

An inverting amplifier is one of the simplest and most commonly used analog circuits. With just two resistors, we can set the gain we need. Nothing prevents us from making the coefficient less than 1, thereby weakening the input signal.

Often, another R3 is added to the circuit, the resistance of which is equal to the sum of R1 and R2.

To understand how an inverting amplifier works, let's simulate a simple circuit. We have a voltage of 4V at the input, the resistance of the resistors is R1 \u003d 1k and R2 \u003d 2k. One could, of course, substitute all this into the formula and immediately calculate the result, but let's see how exactly this scheme works.

Let's start with a reminder of the basic principles of operation of an operational amplifier:

Rule No. 1 - the operational amplifier affects its output through the NOS (negative feedback), as a result of which the voltages at both inputs, both inverting (-) and non-inverting (+), are equalized.

Please note that the non-inverting input (+) is connected to ground, that is, the voltage on it is 0V. In accordance with rule #1, the inverting input (-) should also be 0V.

So, we know the voltage at the terminals of the resistor R1 and its resistance 1k. Thus, with the help we can perform the calculation, and calculate how much current flows through the resistor R1:

IR1 \u003d UR1 / R1 \u003d (4V-0V) / 1k \u003d 4mA.

Rule #2 - Amplifier Inputs Don't Draw Current

Thus, the current flowing through R1 flows further through R2!

Again, we use Ohm's law and calculate what voltage drop occurs across the resistor R2. We know its resistance and we know what current through it, therefore:

UR2 = IR2R2 = 4mA *2k = 8V.

It turns out that we have 8V at the output? Not certainly in that way. Let me remind you that this is an inverting amplifier, that is, if we apply a positive voltage to the input, and remove the negative voltage at the output. How does it happen?

This is due to the fact that the feedback is set at the inverting input (-), and to equalize the voltages at the input, the amplifier reduces the potential at the output. Connections of resistors can be considered as simple, therefore, in order for the potential at the point of their connection to be equal to zero, the output must be minus 8 volts: Uout. = -(R2/R1)*Uin.

There is another catch associated with the 3rd rule:

Rule number 3 - the voltages at the inputs and outputs must be in the range between the positive and negative supply voltage of the op-amp.

That is, you need to check that the voltages calculated by us can actually be obtained through the amplifier. Often beginners think that the amplifier works as a source of free energy and generates voltage from nothing. But we must remember that the amplifier also needs power to work.
Classic amplifiers operate on voltages of -15V and +15V. In such a situation, our -8V, which we calculated, is the real voltage, as it is in this range.

However, modern amplifiers often operate at 5V and below. In such a situation, there is no chance that the amplifier will give us minus 8V at the output. Therefore, when designing circuits, always remember that theoretical calculations must always be supported by reality and physical capabilities.

It should be noted that the inverting amplifier has one drawback. We already know what does not load the signal source, because the inputs of the amplifier have a very high resistance, and draw so little current that in most cases it can be ignored (rule # 2).

The inverting amplifier has an input impedance equal to the resistance of the resistor R1, in practice it ranges from 1k ... 1M. For comparison, an amplifier with field-effect transistor inputs has a resistance of the order of hundreds of megaohms and even gigaohms! Therefore, it may sometimes be advisable to install a voltage follower in front of the amplifier.

In this article we will talk about the operational amplifier. An example of work and use.

Operational amplifier- an electronic circuit of an amplifier based on semiconductors, in an integrated design, having two balanced inputs - direct and inverse, with a high gain. Integrated design implies a complete design of the amplifier, placed in a single integrated circuit (IC) package. The use of operational amplifiers (op-amps) is the most diverse - in amplifiers of various signals, in signal generators, in frequency filters in the audio range, in control circuits physical quantities(temperature, light, humidity, wind), etc.

The direct input of the op amp is marked with a "+" sign, and the inverted input is marked with a "-". You should be aware that in various literature there is another designation: the inverse input is indicated by a circle. This is a typical designation of the inversion sign, which is also found in digital electronics - logical elements. Direct entry does not have a circle in the designation.

Pharmaceutical scales are not able to show how much the weight of the weight of one bowl differs from the weight of the weight of the other bowl. For an approximate observation of the difference in weights, sometimes special plumb lines combined with an arrow are used in technochemical scales, which at the same time reduce the "sensitivity" of the scales to small loads. In the same way, a negative feedback is introduced into the op-amp, which reduces its sensitivity to the input signal - a feedback resistor connecting the output to the inverse input of the op-amp, as shown in the figure above.

An example of the use and operation of an operational amplifier

Consider the operation of an operational amplifier using the example of a circuit that controls the temperature of the air, or some other object on which a thermistor is fixed - a temperature-sensitive radio element that decreases its resistance as the temperature rises. An operational amplifier circuit that measures temperature and signals when a predetermined temperature threshold is exceeded is shown in the figure.

The inputs of the operational amplifier are connected to two resistive supply voltage dividers, only one of them is made on linear elements - resistors, and the second one contains a non-linear element that changes its resistance depending on temperature. What is a voltage divider, you can find out in the article Voltage divider. By design, these four resistors perform the function of a measuring bridge.

When the temperature is “normal”, at the midpoint “A” of the divider R1 and R2 (op-amp inverted input), the voltage is greater than at the midpoint “B” of the divider R3 and R4 (direct input of the op-amp), therefore, at the output of the operational amplifier, the signal low level- the voltage is minimal, the transistor is closed, and the VL1 light is off.

As the temperature rises, the resistance of the resistor R2 decreases, so the voltage at the midpoint "A" of the divider R1 and R2 also decreases. When, with increasing temperature, the resistance of the thermistor drops to such a value at which the voltage at the midpoint "A" of the divider R1 and R2 (op-amp inverted input) becomes lower than at the midpoint "B" of the divider R3 and R4 (direct input of the op-amp), on A high level signal will appear at the output of the operational amplifier - the voltage will become maximum, the transistor will open and the light will turn on.

The temperature control circuit shown in the figure is a real-life circuit, and properly assembled, it works right away. The response temperature threshold is set using resistor R4. It can be powered both from batteries and from power supply rectifiers. The supply voltage range can be from 6 to 30 volts.

If the thermistor R2 is fixed on any surface, for example, a cooling radiator of a powerful transistor, instead of a light bulb, an ordinary computer fan (cooler) for a voltage of 12 volts is used, then the circuit can be used as an automatic cooling device for something, for example, a powerful transistor. The fan will start when a certain temperature is reached, and stop after the “control object” cools down.

To reduce the sensitivity of the operational amplifier, like special plumb lines in pharmaceutical scales, negative feedback (NFB) is used, which is performed on a resistor (in the diagram this is R5). The resistor connects the output of the amplifier to the inverted input. As the output voltage of the amplifier increases, the output voltage is passed through the resistor to the negative input of the amplifier, causing it to lower the output voltage. The lower the resistance of the negative feedback resistor, the higher the feedback, which means the worse the gain of the operational amplifier. The value of the feedback resistor R5 for the type of microcircuit proposed in the diagram can range from 10 kilo ohms to 1.5 mega ohms. Negative feedback makes the graph of output voltage versus input voltage flatter. This relationship is shown in the left graph.

If an operational amplifier is used to control a relay of an automation system, or other equipment that “does not tolerate” frequent voltage drops, then to eliminate frequent switching, or “bounce” of contacts, not negative, but positive feedback (POS) can be used. In this case, the feedback resistor connects the output of the amplifier not to the inverted input, but to the direct one. Then, when the voltage at the output of the amplifier increases, the output voltage is transferred through the resistor to the positive input of the amplifier, causing it to increase the output voltage even faster. With this connection, the operation, both to “turn on” and to “turn off” the operational amplifier, occurs with a greater voltage difference at the input voltage dividers - unbalancing of the measuring bridge than with negative feedback. The switching behavior of the amplifier becomes more "sharp" - it has a steeper edge when "turned on" and a steep decline when "turned off". The smaller the resistance of the positive feedback resistor, the higher the feedback, and hence the greater the gain of the operational amplifier. But be aware that excessive positive feedback causes distortion in the output signal and self-excitation of the operational amplifier.

With positive feedback (POS) appears by-effect- "hysteresis loop", in which the amplifier is turned on at a larger difference in input voltages, and turned off - at a much smaller one, compared to an amplifier with negative feedback. The stronger the POS, the more “rectangular” the hysteresis loop (right graph in the figure). The presence of strong positive feedback turns the circuit into a Schmitt trigger. Therefore, this type of feedback allows a significant temperature spread in the automatic temperature control system and is not suitable, for example, for an incubator, in which a large temperature spread is not acceptable.

Op-amps can be powered by a single supply, as shown earlier, but in general they are designed for dual supply. Bipolar supply is mandatory in those circuits in which the operational amplifier measures both positive and negative voltages, or the measured voltages are comparable to "zero", for example, in harmonic signal amplifier circuits. In the case of a bipolar supply, the output voltage of the operational amplifier, depending on the input signal, can vary from “-” supply to “+” supply.

In some types of operational amplifiers with a bipolar supply, it is possible to adjust the "zero balance" - a state when, in the absence of an input signal at both inputs, its output is neither positive nor negative voltage, but equal to zero. For this, there are special conclusions of the op-amp microcircuits, where a tuning resistor is connected to regulate the zero balance.

To all operational amplifiers operating in the mode of amplifying harmonic signals to eliminate non-linear distortions, additional elements can be connected - filters, usually consisting of capacitors and resistors. For each type of operational amplifier, the filter circuit is different. As a rule, it is given in reference books.

Especially for you now we are developing operational amplifier workshop so that everyone can practice working with this useful view microchips.

Something often began to ask me questions on analog electronics. Did the session of the students take the balls? ;) Okay, it's time to move a little educational program. In particular, on the operation of operational amplifiers. What is it, what is it eaten with and how to calculate it.

What is it
An op-amp is a two-input amp, nevie... uhm... big signal gain and one output. Those. we have U out \u003d K * U in, and K is ideally equal to infinity. In practice, of course, there are more modest numbers. Let's say 1000000. But even such numbers explode the brain when trying to apply them directly. Therefore, as 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 one. The account there goes to hundreds of Megaohms, and even to gigaohms. Those. it measures the voltage at the input, but it is minimally affected. And we can assume that the current in the op-amp does not flow.

The output voltage in this case is calculated as:

U out \u003d (U 2 -U 1) * K

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

Of course, in a real circuit, there will be no plus and minus infinity, and they will be replaced by the highest and lowest power 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 of the signals is greater. Already interesting. You can think of 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 AVR as an example in articles about creating . Also, the comparator is wonderfully used to create .

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 the signal from the output and send it directly to the input, then feedback will occur.

positive feedback
Let's take and drive the signal directly from the output into the direct input.

• Voltage U1 Above zero- at the output -15 volts
• The voltage U1 is less than zero - at the output +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 by one electron, then this is already enough to roll the potential to the output at an infinite amplification. And at the output, a shaped hell will begin - signal jumps here and there at the speed of random disturbances induced at the inputs of the comparator.

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

We consider that at the inverse input at this moment +10 volts. At the output from the op-amp, 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 inverted input drops below -1.4 volts, the output of the op-amp will not change its voltage. And as soon as the voltage drops below -1.4, then the output of the op-amp will jump sharply to +15 and there will already be a +1.4 volt bias at the direct input.

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

There is a kind of gap where there is no sensitivity, between 1.4 and -1.4 volts. The gap width is controlled by the ratios of the 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, the op-amp is more often used in the 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 everything 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 \u003d U 1.

The question is what for we are such happiness? It was possible to throw the wire directly and no op-amp would be needed!

It is possible, but not always. Imagine such a situation, there is a sensor made in the form of a resistive divider:

The lower resistance changes its value, the layout of the output voltage 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, if we do not want a voltmeter, but want a light bulb to change brightness? There is no way to connect a light bulb here! Therefore, the output is buffered by an operational amplifier. Its input resistance is huge and it will have a minimal effect, and the output can provide a quite tangible current (tens of milliamps, or even hundreds), which is quite enough for the light bulb to work.
In general, applications for the repeater can be found. Especially in precision analog circuits. Or where the circuitry of one stage can affect the operation of another, to separate them.

Amplifier
And now let's do a feint with our ears - let's take our feedback and put it on the ground through a voltage divider:

Now half the output voltage is applied to the inverted input. And 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 to compensate for the divider that has arisen.

Now there will be U 1 on the straight line. On the inverse U out /2 \u003d U 1 or U out \u003d 2 * U 1.

Let's put a divisor with a different ratio - the situation will change in the same way. In order not to turn the voltage divider formula in your mind, I will immediately give it:

U out \u003d U 1 * (1 + R 1 / R 2)

It is mnemonically remembered what is divided into what is very simple:

It turns out that the input signal goes through the circuit of resistors R 2 , R 1 in U out . In this case, the direct input of the amplifier is set to zero. We recall the habits of the op-amp - it will try by hook or by crook to make sure that a voltage equal to the direct input is formed at its inverse input. Those. zero. The only way to do this is to lower the output voltage below zero so that zero occurs at point 1.

So. Imagine that U out =0. While equal to zero. And the input voltage, for example, is 10 volts relative to U out. The divisor of R 1 and R 2 will divide it in half. Thus, at point 1 there are five volts.

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

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

U out \u003d - U in * R 1 / R 2

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

Let's say U 2 and U 1 will be 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 it also becomes 5 volts. That is, zero. So it turns out that 10 volts minus 10 volts equals zero. Everything is right :)

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

Thus, our op-amp solved the arithmetic problem from 10 subtracted 20, getting -10 volts.

Moreover, in the problem there are coefficients determined by resistors. It's just that, for simplicity, the resistors are of the same value, and therefore all the coefficients are equal to one. But in fact, if we take arbitrary resistors, then the dependence of the output on the input will be as follows:

U out \u003d U 2 * K 2 - U 1 * K 1

K 2 \u003d ((R 3 + R 4) * R 6) / (R 6 + R 5) * R 4
K 1 \u003d R 3 / R 4

The mnemonics for memorizing the coefficient calculation formula is as follows:
Straight to the diagram. The numerator of the fraction is at the top, so we add the upper resistors in the current flow circuit and multiply by the lower one. The denominator is at the bottom, so add 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 drew only two.

U out \u003d -1 (R 3 * U 1 / R 1 + R 3 * U 2 / R 2)

The input resistors (R 1 , R 2 ) determine the amount of current, and hence 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 \u003d -1 (U 1 + U 2)

Adder non-inverting
Everything is a bit more complicated, but it seems.

Uout \u003d U 1 * K 1 + U 2 * K 2

K 1 \u003d R 5 / R 1
K 2 \u003d R 5 / R 2

Moreover, the feedback resistors must be such that the equation R 3 / R 4 \u003d K 1 + K 2 is observed

In general, on operational amplifiers, you can create any math, add, multiply, divide, count derivatives and integrals. And almost instantly. At the OU, analog computers are made. I even saw one of these on the fifth floor of SUSU — a fool the size of a room floor. Several metal cabinets. The program is typed by connecting different blocks with wires :)