Non-inverting operational amplifier circuit. Inverting amplifier on the op amp. Principle of operation

A non-inverting amplifier is perhaps one of the three most elementary circuits of analog electronics, along with circuits of an inverting amplifier and a voltage follower. It is even simpler than an inverting amplifier, since bipolar power is not needed for the circuit to work.

Pay attention to the unit contained in the formula. This tells us that a non-inverting amplifier always has a gain greater than 1, which means that with this circuit you cannot attenuate the signal.

To better understand how a non-inverting amplifier works, let's look at the circuit on and think about what the output voltage will be.

First of all, we should think about what voltages are present at both inputs of our operational amplifier. Recall the first of the rules that describes the operation of an operational amplifier:

Rule number 1 - the operational amplifier exerts its output to the input through the OOS (negative feedback), as a result of which the voltages at both inputs, both inverting (-) and non-inverting (+) are equalized.

That is, the voltage at the inverting input is 3V. In the next step, let's look at a 10k resistance. We know what voltage is on it and its resistance, which means we can calculate from what current flows through it:

I \u003d U / R \u003d 3V / 10k \u003d 300μA.

This current, according to rule 2, cannot be taken from the inverting input (-), so it comes from the output of the amplifier.

Rule number 2 - the inputs of the amplifier do not consume current

A current of 300 µA also flows through a 20k resistor. We can easily calculate the voltage on it using Ohm's law:

U \u003d IR \u003d 300μA * 20k \u003d 6V

It turns out that this voltage is the output voltage of the amplifier? No, it is not. Recall that a 20k resistor at one of its terminals has a voltage of 3V. Pay attention to the direction of voltage across both resistors.

Current flows in the opposite direction to the arrow, symbolizing a point with a higher voltage. Therefore, to the calculated 6V, you need to add another 3V at the input. In this case, the final result will be 9V.

It is worth noting that the resistors R1 and R2 form a simple one. Remember that the sum of the voltages on the individual resistors of the divider must be equal to the voltage supplied to the divider - the voltage cannot disappear without a trace and arise out of nowhere.

In conclusion, we must verify the result with the last rule:

Rule number 3 - the voltage at the inputs and output must be in the range between the positive and negative supply voltage of the OS.

That is, it is necessary to verify that the voltage calculated by us can be obtained realistically. Beginners often think that the amplifier works like a Perpetuum Mobile, and generates voltage from nothing. But we must remember that for the amplifier to work, you also need power.

Classic amplifiers operate on voltage -15V and + 15V. In such a situation, the 9V rated by us is the real voltage, since 9V is in the range of the supply voltage. However, modern amplifiers often operate with voltages of 5V or lower. In this situation, there is no chance for the amplifier to output 9V.

Therefore, when developing circuits, it must always be remembered that theoretical calculations should always be checked against the reality and physical capabilities of the components.

A ten thousand li road begins with the first step.
  (Chinese proverb)

It was evening, there was nothing to do ... And so suddenly I wanted to solder something. Sort of ... Electronic! .. Solder - so solder. There is a computer, the Internet is connected. We select the scheme. And suddenly it turns out that the schemes for the planned subject are a wagon and a small cart. And all are different. No experience, not enough knowledge. Which one to choose? Some of them contain some rectangles, triangles. Amplifiers, and even operating ones ... How they work is not clear. Stra-a-ashno! .. What if it burns out? We choose that simpler, on familiar transistors! Chose, solder, turn 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. So it is with transistors. It’s not enough to solder a circuit on them. We must also be able to configure it. Too many pitfalls and rakes. And this often requires experience by no means at the initial level. So, quit an exciting activity? Not at all! 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 TO KNOW - HOW!

Here is this: understanding now how the operational amplifier (op-amp, or in English OpAmp) works, we are now going to do it. 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: “The current through the circuit I) is directly proportional to the voltage on it ( U) and is inversely proportional to its resistance ( R)»:
I \u003d U / R. (1)

For starters, in principle, it is not so important how exactly the op-amp is arranged inside. Just take it as an assumption that it is a “black box” with some stuffing there. At this stage, we will not consider such op-amp parameters as “bias voltage”, “shear stress”, “temperature drift”, “noise characteristics”, “common-mode component rejection coefficient”, “power voltage ripple suppression coefficient”, “passband " etc. All these parameters will be important at the next stage of his study, when the basic principles of his work “settle down” in his head for “it was smooth on paper, but forgot about the ravines” ...

For now, we simply assume that the op-amp parameters are close to ideal and consider only what signal will be at its output if some signals are fed to its inputs.

So, the operational amplifier (OA) is a differential DC amplifier with two inputs (inverting and non-inverting) and one output. In addition to them, the op-amp has power outputs: positive and negative. These five findings are available in nearly   any op amp and fundamentally necessary for its work.

Shelter has a huge gain of at least 50,000 ... 100,000, but in reality - much more. Therefore, in 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, whatever   from their absolutevalues \u200b\u200band polarity.

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

The inputs of the op-amp have a very large input resistance 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 large input resistance means that they practically do not affect the input signal.

Therefore, with a large degree of approximation to the theoretical ideal, we can assume that current the op amp does not flow into the inputs . It - the first   an important rule that is applied when analyzing the operation of an op amp. I ask you to remember well that it concerns only the opamp itself, but not schemes   with its application!

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). In Fig. Figure 1 shows several of the many possible options for signal inversion, where red indicates a direct (input) signal and blue indicates an inverted (output) signal.

Fig. 1 The concept of signal inversion

It should be especially noted that to the zero line (as in Fig. 1, A, B) the signal inversion not attached! Signals can be inverse and asymmetric. For example, both are only in the region of positive values \u200b\u200b(Fig. 1, B), which is typical for digital signals or with unipolar power (we will discuss this later), or both partially in the positive and partially in the negative regions (Fig. 1, B, D). Other options are 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 opamps on concepts in different ways. Abroad, OSs were previously depicted, and even now they are often depicted as an isosceles triangle (Fig. 2, A). The inverting input is indicated by the minus symbol, while the non-inverting input is indicated by the plus symbol inside the triangle. These symbols do not mean at all that the potential must be more positive or more negative on the corresponding inputs than on the other. They simply indicate how the output potential reacts to the potentials supplied to the inputs. As a result, they are easily confused with the power leads, which may turn out to be unexpected “rakes”, especially for beginners.

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

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

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

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

Feedback is the effect of supplying a part of the output voltage of an amplifier to its input, where it is algebraically (taking into account the sign) summed with the input voltage. The principle of summing signals will be discussed below. Depending on which input of the op-amp, inverting or non-inverting, the OS is supplied, negative feedback (OOS) is distinguished when part of the output signal is fed to the inverting input (Fig. 3, A) or positive feedback (PIC) when the output signal is supplied, respectively, to a non-inverting input (Fig. 3, B).

Fig. 3 The principle of forming feedback (OS)

In the first case, since the output signal is inverse with respect to the input, it is subtracted from the input. As a result, the overall cascade gain is reduced. In the second case, it is summed with the input, the overall gain of the cascade increases.

At first glance, it may seem that PIC has a positive effect, and OOS is a completely useless venture: why reduce the gain? That is exactly what the U.S. patent examiners thought when Harold S. Black tried to   patent OOS. However, sacrificing gain, 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 opamp.

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

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

   Fig. 4 Basic schemes of inclusion of OS

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

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

To analyze the operation of these schemes, one should take into account second   the most important the rule, which obeys the work of the OS: The output of the operational amplifier tends to ensure that the voltage difference between its inputs is zero.

However, any wording should be necessary and sufficientto limit the entire subset of cases subordinate to it. The above wording, with all its “classicality”, does not provide any information about which of the inputs the output “seeks to influence”. Proceeding from it, it turns out that the op-amp seems to equalize the voltage at its inputs, supplying voltage to them from somewhere "inside".

If you carefully consider the diagrams in Fig. 4, it can be noted that the OOS (through Roos) in all cases is started from the exit only   to an inverting input, which gives us reason to reformulate this rule as follows: Voltage on the output of the OS, covered by the OOS, seeks 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 turn-on of the op-amp with the OOS is the non-inverting input, and the “slave” is the inverting input.

When describing the operation of an op-amp, the potential at its inverting input is often called a “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 OOS), the inverting input can be considered connected directly to the same potential as the non-inverting input is connected to. However, the “virtual zero” is just a special case, which takes place only with bipolar supply of the op-amp. When using unipolar power supply (which will be discussed below), and in many other switching circuits, there will be no zero either on 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 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 assume that the supply voltages are still bipolar, equal in magnitude (say ± 15 V), with a midpoint (common bus or ground), against which we will count the input and output voltages. In addition, the analysis will be carried out by direct current, as a varying 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 established 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.

A) Inverting   inclusion (Fig. 5).

   Fig. 5 The principle of operation of the op-amp in inverting inclusion

The potential at the non-inverting input is zero, because it is connected to a midpoint (“earth”). An input signal equal to +1 V relative to the midpoint (from GB) is applied to the left terminal of the input resistor Rin. Suppose that the resistance Rooos and Rin are equal to each other and amount to 1 kOhm (in total, their resistance is 2 kOhm).

According to Rule 2, the inverting input must have the same potential as the zeroed non-inverting one, 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 Iin.   \u003d 1 V / 1000 Ohms \u003d 0.001 A (1 mA). The direction of flow of this current is shown by an arrow.

Since Rooc and Rin are turned on by the divider, and according to Rule 1, the op amp inputs do not consume current, so that the voltage at the midpoint of this divider is 0 V, the voltage must be applied to the right terminal 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 of Rin and the right terminal of Rooc, and the current flowing through this divider is 1 mA (2 V / (1 kΩ + 1 kΩ) \u003d 1 mA), i.e. I in. = I oos .

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

Thus, with the equality of the values \u200b\u200bof Rooos and Rin, 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 circuit is often used if you need to invert the signal obtained using circuits that are fundamentally inverters. For example, logarithmic amplifiers.

Now, keeping the nominal value of Rin equal to 1 kOhm, let's increase the resistance Roohs to 2 kOhm with the same input signal +1 V. The total resistance of the divider Roohs + \u200b\u200bRinuh increased to 3 kOhm. In order for the potential 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. Consequently, the voltage drop at Roos (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.

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

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

Now let's increase the resistance of Rooos 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 there it was! We previously made the assumption 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 (voltage supply)!” As a result of such abuse over the circuit ratings, the op-amp output voltage “rests” on the supply voltage (the op-amp output enters saturation). The balance of the equality of currents through the divider RoocRvh ( Iin. = Ioos) is violated, a potential appears at the inverting input, different from the potential at the non-inverting input. Rule 2 ceases to apply.

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

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

   Fig. 6 Variable gain inverting amplifier circuit

With the right (according to the scheme) position of its engine, the gain will be Rоос / Rin \u003d 10 kОм / 1 кОм \u003d 10. Moving the Роос engine to the left (reducing its resistance), the gain of the circuit will decrease and, finally, at its leftmost position it will become zero, since the numerator in the above formula becomes zero at any   denominator value. The output will be zero also for any value and polarity of the input signal. Such a scheme is often used in amplification schemes of audio signals, for example, in mixers, where it is necessary to adjust the gain from zero.

B) Non inverting   inclusion (Fig. 7).

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

The left terminal Rin is connected to the midpoint (ground), and an input signal of +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 assume that the resistance Rooos and Rin equal to each other and amount to 1 kOhm. Because at a non-inverting input, the potential is +1 V, then according to Rule 2, the same potential (+1 V) should be at the inverting input (shown in the figure). To do this, the right terminal of the Rooc resistor (OU output) must have a voltage of +2 V. Currents Iin.and Ioosequal to 1 mA, now flow through the resistors Rooc and Rin in the opposite direction (shown by arrows). We have succeeded non inverting amplifier with a gain of 2, since an input signal of +1 V generates an output signal of +2 V.

Strange, isn't it? The ratings are the same as in the inverting inclusion (the only difference is that the signal is applied to another input), and the gain is obvious. We'll figure it out a bit later.

Now we increase the nominal Rooc to 2 kOhm. To maintain current balance Iin. = 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 \u200b\u200bof Ku at a non-inverting turn-on with an inverting one, at the same values \u200b\u200bof Rooc and Rin, then it turns out that the gain in all cases is greater by one. We derive the formula:
   Ku \u003d Uout / Uin + 1 \u003d (Rooos / Rin) + 1 (3)

Why is this happening? Yes, very easy! OOS acts in exactly the same way as with an inverting switch, but according to Rule 2, the potential of a non-inverting input is always added to the potential of an inverting input in a non-inverting switch.

So, with a non-inverting turn on, you can not get a gain equal to 1? Why not - you can. Let's reduce the nominal value of Rooc, in the same way as we analyzed Fig. 6. At its zero value - shorting the output with the inverting input shortly (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, that is, +1 V. We get: Ku \u003d 1 V / 1 V \u003d 1   (!) Well, since the inverting current input does not consume and there is no potential difference between it and the output, then no current flows in this circuit.

   Fig. 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 op-amp output must work and through it its output current will flow in vain. And what will happen if we leave Rooc but remove Rin (Fig. 8, B)? Then, in the amplification formula Ku \u003d Rooos / Rin + 1, the resistance Rin theoretically becomes close to infinity (in reality, of course, no, because there are leaks on the board, and even the input op-amp current is negligible, but everything is zero it’s not equal), wherein the ratio of Rooos / Rin equals zero. Only one unit remains in the formula: Ku \u003d + 1. And can the gain be less than unity for this circuit? No, less will not work under any circumstances. The "extra" unit in the gain formula on the curve of a goat can not go around ...

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

At first glance, such a scheme does not have practical meaning: why do we need a single and even non-inverse “gain” - what, can’t you just give a signal further ??? However, such schemes are used quite often and that is why. According to Rule 1, the current does not flow into the opamp inputs, i.e., input impedance   the non-inverting repeater is very large - the very tens, hundreds and even thousands of megohms (the same applies to the circuit in Fig. 7)! But the output impedance is very small (Ohm fraction!). The output of the op-amp is "puffing with all its might", trying, according to Rule 2, to maintain the same potential on the inverting input as on the non-inverting one. The limitation is only the permissible output current of the op-amp.

And here from this place we are a little vil to the side and we will consider the issue of op-amp output currents in a little more detail.

For most opamps of widespread use, the technical parameters indicate that the load resistance connected to their output should not be smaller   2 kOhm. More - as much as you like. For a much smaller number, it is 1 kOhm (K140UD ...). This means that under the worst conditions: the maximum supply voltage (for example, ± 16 V or a total of 32 V), the load connected between the output and one of the supply buses and the maximum output voltage of the 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 \u003d 0.015 A (15 mA). Not so much, but not too much. Fortunately, most general-purpose opamps have built-in protection against overcurrent - the typical maximum output current is 25 mA. Protection prevents overheating and failure of the op-amp.

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

IN) Differential   inclusion (Fig. 9).

   Fig. 9 The principle of op-amp operation in differential switching

So, suppose that with the same values \u200b\u200bof Rin and Rooos equal to 1 kOhm, the same voltage equal to +1 V is 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 (voltage across the resistor is 0), no current flows through it. So, it is equal to zero and the current through the resistor Rooc. That is, these two resistors do not perform any function. In fact, we actually got a non-inverting repeater (compare with Fig. 8). Accordingly, at the output we get the same voltage as at the non-inverting input, that is, +1 V. Change the polarity of the input signal at the inverting input of the circuit (turn GB1 upside down) and apply minus 1 V (Fig. 9, B). Now a voltage of 2 V is applied between the terminals of Rin and a current flows through it Iin   \u003d 2 mA (I hope that it is no longer necessary to paint in detail why this is so?). In order to compensate for this current, a current of 2 mA must also flow through Rooos. And for this, the output of the op-amp should be a voltage of +3 V.

This is where the malicious “grin” of an additional unity appeared in the formula of the gain of a non-inverting amplifier. It turns out that with this simplified   differential inclusion, the difference in the gain constantly shifts the output signal by the value of the potential at the non-inverting input. A problem with! However, "Even if you were eaten, you still have at least two exits." So, somehow we need to equalize the gain of the inverting and non-inverting inclusions in order to “neutralize” this extra unit.

To do this, we apply the input signal to the non-inverting input not directly, but through the divider Rin2, R1 (Fig. 9, B). We also accept their nominal values \u200b\u200bof 1 kOhm. Now at the non-inverting (and therefore also inverting) input of the op-amp there will be a potential of +0.5 V, current will flow through it (and Rooc) Iin = Ioos   \u003d 0.5 mA, to ensure that the output of the op-amp must have a voltage equal to 0 V. Fu-uh! We achieved what we wanted! If the signals at both inputs of the circuit are equal in magnitude and polarity (in this case +1 V, but the same will be true for minus 1 V and for any other digital values), a zero voltage equal to the difference in input signals will be stored at the output of the op-amp .

Let us verify this reasoning by applying a negative polarity signal minus 1 V to the inverting input (Fig. 9, D). Wherein Iin = Ioos   \u003d 2 mA, for which the output should be +2 V. Everything was confirmed! The output level corresponds to the difference between the input.

Of course, with the equality of Rin1 and Rooc (respectively, Rin2 and R1) we will not receive amplification. To do this, it is necessary to increase the values \u200b\u200bof Rooos and R1, as was done in the analysis of previous inclusions of the OS (I will not repeat), and strictly   the ratio is observed:

Roox / Rin1 \u003d R1 / Rin2. (4)

What usefulness do we get from this inclusion in practice? And we get a wonderful property: the output voltage does not depend on the absolute values \u200b\u200bof the input signals, if they are equal to each other in magnitude and polarity. Only the difference (differential) signal is output. This allows amplification of very small signals against a background of interference acting equally on both inputs. For example, a signal from a dynamic microphone against a background of a 50 Hz power frequency network.

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 imbalance of currents acting in the circuit, and therefore, in addition to difference ("out-of-phase") signals, combined ("in-phase") signals will also be amplified.

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

   Fig. 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 the time. Since both the time reference and the period reference are selected arbitrarily, the phase of one periodic   the signal does not have a physical meaning. However, the phase difference of the two periodic   signals is a quantity that has physical meaning; it reflects the delay of one of the signals relative to the other. What is considered the beginning of a period does not matter. For the beginning point of the period, you can take a zero value with a positive slope. You can - maximum. Everything is in our power.

In Fig. 9, the source signal is red, green is shifted by ¼ of the period relative to the source, and blue is ½ of the period. If we compare the red and blue curves with the curves in Fig. 2B, it can be seen that they are mutually inverse. Thus, "in-phase signals" are signals that coincide with each other at each of its points, and "antiphase signals" - inverse   relative to each other.

At the same time, the concept inversions   broader than concept phasebecause the latter applies only to regularly repeating, periodic signals. A concept inversions applicable to any signals, including non-periodic ones, such as an audio signal, a digital sequence, or a constant voltage. To phase   was a consistent quantity, the signal should be periodic at least on some interval. Otherwise, both phase and period turn into mathematical abstractions.

Secondly, the inverting and non-inverting inputs in the differential inclusion with the equality of the values \u200b\u200bof Roo \u003d R1 and Rin1 \u003d Rin2 will have different input resistances. If the input resistance of the inverting input is determined only by the value of Rin1, then non-inverting - by the values consistently   included Rin2 and R1 (have not forgotten that the op amp inputs do not consume current?). In the above example, they will be, respectively, 1 and 2 kOhm. And if we increase Rooc and R1 to get a full-fledged amplification stage, then the difference will increase even more significantly: at Ku \u003d 10 - up to, respectively, the same 1 kOhm and as much as 11 kOhm!

Unfortunately, in practice, the values \u200b\u200bRin1 \u003d Rin2 and Roox \u003d R1 are usually set. However, this is acceptable only if the signal sources for both inputs are very low. output impedance. Otherwise, it forms a divider with the input resistance of this amplifier stage, and since the division coefficient of such “dividers” will be different, the result is obvious: a differential amplifier with such resistor ratings will not perform its function of suppressing common-mode (combined) signals, or it will not perform this function well .

One of the ways to solve this problem may be the inequality of the resistors connected to the inverting and non-inverting inputs of the op-amp. Namely, that Rin2 + R1 \u003d Rin1. Another important point is the achievement of strict observance of equality (4). As a rule, this is achieved by dividing R1 into two resistors - a constant, usually 90% of the desired value and a variable (R2), whose resistance is 20% of the desired value (Fig. 11, A).

   Fig. 11 Differential Amplifier Balancing Options

The path is generally accepted, but again, with this method of balancing, even if the input impedance of a non-inverting input changes slightly, it changes. A much more stable option is 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 maintain the ratio of their denominations, similarly to option “A” (Roox / Rin1 \u003d R1 / Rin2).

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

   Fig. 12 Switchable inverting / non-inverting repeater circuit

The input signal is applied simultaneously to both inputs of the circuit (inverting and non-inverting). The values \u200b\u200bof all resistors (Rin1, Rin2 and Rooc) are equal to each other (in this case, we take their real values: 10 ... 100 kOhm). The non-inverting input of the op-amp with the SA key can be shorted 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 (through it only current is “useless” Ivkh2   from the signal source to the common bus). We get inverting repeater   with a gain equal to minus 1 (see. Fig. 6). But when the key SA is open (Fig. 12, B) we get non-inverting repeater   with a gain of +1.

The principle of operation of this circuit can be expressed in a slightly different way. When the SA key is closed, it works as an inverting amplifier with a gain of minus 1, and when open, at the same time   (!) And as an inverting amplifier with gain, minus 1, and as a non-inverting amplifier with gain +2, whence: Ku \u003d +2 + (–1) \u003d +1.

In this form, this circuit can be used if, for example, at the design stage, the polarity of the input signal is unknown (say, from a sensor that cannot be accessed before starting the device setup). If, however, you use a transistor (for example, a field-effect) controlled by the input signal using comparator   (we will talk about it below), we get synchronous detector   (synchronous rectifier). The concrete implementation of such a scheme, of course, goes beyond the initial familiarization with the work of the OS and we again will not consider it in detail here.

And now let's look at the principle of summing the input signals (Fig. 13, A), and at the same time we will figure out what values \u200b\u200bof resistors Rin and Rooc should be in reality.

   Fig. 13 The principle of operation of the inverting adder

We take as a basis the inverting amplifier already considered above (Fig. 5), we connect only one, but two input resistors Rin1 and Rin2 to the input of the op-amp. So far, for "educational" purposes, we take the resistance of all resistors, including Rooc, equal to 1 kOhm. To the left pins Rin1 and Rin2 we give input signals equal to +1 V. Currents equal to 1 mA flow through these resistors (shown by arrows pointing from left to right). In order to maintain the same potential at the inverting input as at the non-inverting (0 V), a current equal to the sum of the input currents (1 mA + 1 mA \u003d 2 mA) must flow through the Rooc resistor, indicated by an arrow pointing in the opposite direction (from right to left) ), for which there should be a voltage of minus 2 V.

The same result (output voltage minus 2 V) can be obtained if +2 V voltage is applied to the input of the inverting amplifier (Fig. 5), or the Rin value is halved, i.e. up to 500 ohms. Increase the voltage applied to the resistor Rin2 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: in any case, the output voltage 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 oppositely directed, they will cancel each other and the output will be 0 V. In this case, through the Rooc resistor no current will flow. In other words, the current flowing along Rooc is algebraically summed with input   currents.

An important point also comes from here: while we were operating with small input voltages (1 ... 3 V), the output of an op-amp of wide application could well provide such a current (1 ... 3 mA) for Roos and something else remained for the load connected to the output of the op-amp. But if the voltage of the input signals is increased to the maximum allowable (close to the supply voltage), it turns out that the entire output current will go to Rooc. There is nothing left for the load. And who needs an amplification cascade that works "on its own"? In addition, the values \u200b\u200bof the input resistors, which are equal to only 1 kOhm (respectively, determining the input resistance of the inverting amplifier stage), require excessively large currents flowing through them, strongly loading the signal source. Therefore, in real circuits, the resistance Rin is selected not less than 10 kOhm, but preferably not more than 100 kOhm, so that at a given gain not to put Rooo too high a nominal value. Although these values \u200b\u200bare not absolute, but only approximate, as they say, "in the first approximation" - it all depends on the particular scheme. In any case, it is undesirable for a current exceeding 5 ... 10% of the maximum output current of this particular op-amp to flow through Rooos.

Summed signals can also be applied to a non-inverting input. It turns out non-inverting adder. Fundamentally, 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 nominal values \u200b\u200bof the input resistors. However, in practice it is used much less often, because contains a "rake" to be considered.

Since Rule 2 is valid only for an inverting input, on which the “virtual potential of zero” acts, there will be a potential on the non-inverting 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, we have to apply additional circuitry tricks.

So far, we have considered schemes for OS with environmental protection. And what will happen if feedback is removed altogether? In this case, we get comparator   (Fig. 14), i.e., a device comparing the two potentials at their inputs by absolute value (from the 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 larger 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 any, including equal to zero potential (Fig. 14, B).

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

However, not everything is so good “in the Danish kingdom” ... 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 outweighs at least a little the potential of the other, then this will be enough to cause chaotic voltage surges due to random disturbances pointing at the inputs of the comparator.

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

   Fig. 15 Principle of hysteresis in the comparator due to PIC

Let us analyze the operation of this circuit. The voltage of its power is ± 10 V (for an even account). Resistance Rin is 1 kOhm, and Rpos is 10 kOhm. The potential of the midpoint is selected as the reference voltage supplied to the inverting input. The red curve shows the input signal coming to the left pin Rin (input schemes   comparator), blue - the potential at the non-inverting input of the op-amp, and green - the output signal.

As long as the input signal has a negative polarity, the output is a negative voltage, which, through Rpos, is summed with the input voltage in inverse proportion to the values \u200b\u200bof the corresponding resistors. As a result, the potential of a non-inverting input in the entire range of negative values \u200b\u200bby 1 V (in absolute value) exceeds the level of the input signal. As soon as the potential of the non-inverting input equals 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 polarity to positive. The total potential at the non-inverting input will start avalanche become even more positive, supporting the process of such a switch. As a result, the comparator simply “does 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 potential at the non-inverting input during switching.

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

Unfortunately, not every op amp, and not in all cases, can be used as a comparator. Specialized comparator circuits are available for matching between analog and digital signals. Some of them are specialized for connecting to digital TTL-microcircuits (597CA2), part - to digital ESL-microcircuits (597CA1), however the majority is the so-called "Comparators of wide application" (LM393 / LM339 / K554CA3 / K597CA3). Their main difference from the op-amp is a special device of the output stage, which is made on an open collector transistor (Fig. 16).

   Fig. 16 Output stage for wide range comparators
   and its connection to a load resistor

This requires the use of an external load resistor   (R1), without which the output signal is simply physically unable to form a high (positive) output level. The voltage + U2 to which the load resistor is connected may be different than the supply voltage + U1 of the comparator chip itself. This allows simple means to provide the output signal of the desired level - whether it is TTL or CMOS.

Note In most comparators, for example, dual LM393 (LM193 / LM293) or exactly the same in circuitry but quadruple LM339 (LM139 / LM239), the emitter of the output stage transistor is connected to the negative power output, 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 emitter of the output transistor are separately output, which can be connected to voltages other than the supply voltage of the comparator itself. Its only and relative drawback is that in the 8-pin (sometimes in 14-pin) package it is only one.

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

   Fig. 17 Scheme of the Converter of current to voltage on the OS

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 impedance of such a circuit is close to zero, which allows one to build micro / milliammeters based on it, practically not affecting the current flowing along the measured circuit. Perhaps the only limitation is the permissible input voltage range of the op-amp, which should not be exceeded. Using it, you can also build, for example, a linear converter of the photodiode current into voltage and many other circuits.

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

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

Once again, we carefully consider all the above figures and see that a separate output of the midpoint in the OS NOT ! For the operation of their internal circuit, it is simply not needed. On some circuits, a non-inverting input was connected to the midpoint, however, this is not the rule.

Consequently, overwhelming most   modern op amps are designed to power SINGLE POLAR energized! A logical question arises: “Why then do we need bipolar nutrition” if we so persistently and with enviable constancy depicted it in the drawings?

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

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

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

For now, we will not consider the input (and OOS) circuits of the circuits shown in the figure (“black box”). Let us take it for granted that some input sinusoidal signal (black sinusoid on the graphs) is fed to the input and the same sinusoidal signal is amplified relative to the input color sinusoid on the graphs).

When connecting the load R between the output of the op-amp and the midpoint of the connection of power supplies (GB1 and GB2) - Fig. 18A, the current flows through the load symmetrically with respect to the midpoint (respectively, the red and blue half-waves), and its amplitude is maximum and the voltage amplitude is 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 op-amp, Rnag. and a power source (red and blue lines showing the flow of current in the corresponding direction).

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

When supplying an op-amp with unipolar voltage as common bus   usually the negative (minus) pole of the power source is selected, to which the second load terminal is connected (Fig. 18, B). Now the current through the load can flow in only 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 asymmetric (pulsating).

It is impossible to assert unequivocally that such an option is bad. If the load is, say, a dynamic head, then for her this is poorly unambiguous. However, there are many applications where connecting the load between the output of the op-amp and one of the supply buses (usually of negative polarity) is not only permissible, but also the only possible.

If, nevertheless, it is necessary to ensure the symmetry of the current flowing through the load with unipolar power supply, then you have to galvanically decouple it from the output of the op-amp galvanically by 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 op-amp operating mode for direct current (Fig. 19).

   Fig. 19 Connecting an input source with various op-amp power options

Now we consider the connection options for input sources, excluding the load connection from consideration.

The connection of the inverting and non-inverting inputs to the midpoint of the connection of the power sources (Fig. 19, A) was considered in the analysis of the above schemes. If the non-inverting current input does not consume and simply accepts the potential of the midpoint, 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 the unipolar supply of the op-amp, it is possible to perfectly form the potential supplied to its non-inverting input using the R1R2 divider (Fig. 19, B, C). Typical resistor ratings of this divider are 10 ... 100 kOhm, and the lower (connected to the common negative bus) it is highly desirable to shunt the capacitor by 10 ... 22 microfarads in order to significantly reduce the effect of power supply ripples on the potential artificial   midpoint.

But the signal source (G) is extremely undesirable to connect to this artificial midpoint due to the same input current. Let's estimate. Even with the divider ratings R1R2 \u003d 10 kOhm and Rin \u003d 10 ... 100 kOhm, the input current Iin   at best it will be 1/10, and at worst up to 100% of the current passing through the divider. Therefore, the potential at the non-inverting input in combination (in phase) with the input signal will “float” as much.

To eliminate the influence of inputs on each other during amplification of DC signals during this switching on, a separate artificial midpoint potential formed by R3R4 resistors should be arranged for the signal source (Fig. 19, B), or if the AC signal is amplified, galvanically decouple the signal source from the inverting input by capacitor C2 (Fig. 19, B).

It should be noted that in the above schemes (Fig. 18, 19), we assumed by default that the output signal must be symmetrical with respect to either the midpoint of the power sources or the artificial midpoint. In reality, this is not always necessary. Quite often, it is necessary that the output signal has predominantly either positive or negative polarity. Therefore, it is absolutely not necessary that the positive and negative polarities of the power source are equal in absolute value. One of them can be much smaller in absolute value than the other - only so as to ensure the normal functioning of the OS.

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

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

On the other hand, for op amps of wide application, 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 level of negative power, but even “minus” it by 0.3 V, which greatly facilitates their use with unipolar power with a common negative bus.

Finally, let's take a look and feel for these “spider bugs”. You can even sniff and lick. I allow it. Consider their most common options available to beginner hams. Especially if you have to solder the op-amp from old equipment.

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

The round metal-glass case (Fig. 20, A, B) was used as the main one for imported op amps until around 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 housed in currently quite “exotic” cases: a 15-pin rectangular metal-glass for the hybrid K284UD1 (Fig. 20, D), in which the key is the additional 15th output from the case, and others. True, planar 14-pin cases (Fig. 20, D) for placing the OS in them I have not personally met. They were used for digital circuits.

   Fig. 20 Housing domestic operational amplifiers

Modern opamps for the most part contain corrective circuits directly on the chip, which made it possible to dispense with a minimum number of conclusions (as an example, the 5-pin SOT23-5 for a single opamp - Fig. 23). This made it possible to place two or four completely independent (except for common power leads) op amps made on a single chip in one housing.

   Fig. 21 Double-row plastic enclosures of modern op amp for output mounting (DIP)

Sometimes you can find op-amps located 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 “single-sided pin package”.

   Fig. 22 Single-row plastic housing of dual op-amps for output mounting (SIP-8)

They were designed to minimize the space occupied on the board, but, unfortunately, they were “late”: by this time, the SMD (Surface Mounting Device) was widely used by soldering directly to the board tracks (Fig. 23). However, for beginners, their use presents significant difficulties.

   Fig. 23 Shells of modern imported surface mount op amps (SMD)

Very often, the same microcircuit can be “packaged” by the manufacturer in different cases (Fig. 24).

   Fig. 24 Options for placing the same microchip in different cases

The findings of all microcircuits have sequential numbering, counted from the so-called "Key" indicating the location of the output at number 1. (Fig. 25). IN any   if you place the case conclusions push, their numbering is increasing against clockwise!

   Fig. 25 Pinout of operational amplifiers
   in various cases (pinout), top view;
   numbering direction is indicated by arrows

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

Typically, an inverting input in both round metal-glass and DIP cases is connected to the 2nd output, non-inverting input to the 3rd, output to the 6th, minus supply to the 4th and plus power to the 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 in comparison with the generally accepted for most other types, i.e. they are connected to the conclusions, as in import buildings (Fig. 25, B), and the numbering corresponds to domestic ones (Fig. 25, A).

In recent years, the majority of public facilities for public use have been placed in plastic cases (Fig. 21, 25, V-D). In these cases, the key is either a recess (point) opposite the first terminal, or a cutout in the end of the case between the first and 8th (DIP-8) or 14th (DIP-14) terminals, or a chamfer along the first half of the terminals (Fig. 21, in the middle). The numbering of conclusions in these cases also goes against clockwise   when viewed from above (conclusions from yourself).

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

Regarding the domestic analogue of LM324 (K1401UD2), there is another “rake”: if in LM324 plus power is output to the 4th output, and minus to the 11th, then in K1401UD2 it is the other way round: plus power is output 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 terminals 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. And that’s it.

A few words regarding the marking of imported op-amps (and not only op amps). For a number of developments of the first 300 digital designations, it was customary to designate a quality group as the first digit of a digital code. For example, the LM158 / LM258 / LM358 op-amp, 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 inconsistent with such a gradation, or differ very slightly ( inscrutable pricing paths!). So you can buy them with any marking available "for the pocket" of a beginner, not particularly chasing the first "three".

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

Conclusion

So we studied the "alphabet" of the op-amp operation, capturing the comparators a bit. Next, you need to learn to add words, sentences and whole meaningful “compositions” (workable schemes) from these “letters”.

Unfortunately, "It is impossible to embrace the immense." If the material presented in this article helped to understand how these "black boxes" work, then further deepening into the analysis of their "filling", the influence of input, output and transitional characteristics, is the task of a more advanced study. Information about this is detailed and thoroughly set forth in a variety of existing literature. As grandfather William Ockham used to say: "You should not multiply entities beyond what is necessary." There is no need to repeat the already well-described. You just need not to be lazy and read it.

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

Then let me take my leave, with respect, etc., by Alexey Sokolyuk ()