The ideal op amp is a non-inverting amplifier. Operational amplifier

An operational amplifier (op-amp) is usually called an integrated DC amplifier with a differential input and a push-pull output, designed to work with feedback circuits. The name of the amplifier is due to its original area of ​​application - performing various operations on analog signals (addition, subtraction, integration, etc.). Currently, op-amps serve as multifunctional units in the implementation of a variety of electronic devices for various purposes. They are used for amplification, limiting, multiplication, frequency filtering, generation, stabilization, etc. signals in continuous and pulsed devices.

It should be noted that modern monolithic op-amps differ slightly in size and price from individual discrete elements, for example, transistors. Therefore, the implementation of various devices on an op-amp is often much simpler than on discrete elements or on amplification ICs.

An ideal op-amp has an infinitely large voltage gain ( K and op-amp=∞), infinitely large input impedance, infinitely small output impedance, infinitely large CMRR and infinitely wide operating frequency band. Naturally, in practice, none of these properties can be fully realized, but they can be approached to a degree sufficient for many areas.

Figure 6.1 shows two versions of op-amp symbols - simplified (a) and with additional terminals for connecting power circuits and frequency correction circuits (b).

Figure 6.1. OS symbols

Based on the requirements for the characteristics of an ideal op-amp, it is possible to synthesize its internal structure, presented in Figure 6.2.

Figure 6.2. Block diagram of the op-amp

A simplified electrical circuit of a simple op-amp, implementing the block diagram of Figure 6.2, is shown in Figure 6.3.

Figure 6.3. Simple op-amp circuit

This circuit contains an input remote control (VT 1 and VT 2) with a current mirror (VT 3 and VT 4), intermediate stages with OK (VT 5) and with OE (VT 6), and an output current booster on transistors VT 7 and VT 8 . The op-amp may contain frequency correction circuits (Ccor), power supply and thermal stabilization circuits (VD 1, VD 2, etc.), IST, etc. Bipolar power supply allows for galvanic communication between the stages of the op-amp and zero potentials at its inputs and output in the absence of a signal. In order to obtain a high input impedance, the input remote control can be performed on a DC. It should be noted that there is a wide variety of op-amp circuit solutions, but the basic principles of their construction are quite fully illustrated in Figure 6.3.

6.2. Main parameters and characteristics of the op-amp

The main parameter of the op-amp is the voltage gain without feedback K u op-amp, also called total voltage gain. In the bass and midrange regions it is sometimes designated K u Op-amp 0 and can reach several tens and hundreds of thousands.

Important parameters of the op-amp are its accuracy parameters, determined by the input differential stage. Since the accuracy parameters of the remote control were considered in subsection 5.5, here we limit ourselves to listing them:

◆ zero offset voltage U cm;

◆ temperature sensitivity of zero offset voltage dU cm/dT;

◆ bias current Δ I input;

◆ average input current I input wed.

The input and output circuits of the op-amp are represented by the input R input and weekends R out of the op amp resistances given for op-amps without OOS circuits. For the output circuit, parameters such as maximum output current are also given I output OU and minimum load resistance R n min and sometimes the maximum load capacity. The input circuit of the op amp may include capacitance between the inputs and the common bus. Simplified equivalent circuits of the input and output circuits of the op-amp are presented in Figure 6.4.

Figure 6.4. A simple linear macromodel of an op-amp

Among the parameters of the op-amp, it is worth noting the CMRR and the coefficient of attenuation of the influence of instability of the power source KOVNP=20lg·(Δ E/Δ U in). Both of these parameters in modern op-amps have their values ​​within (60...120) dB.

The energy parameters of the op-amp include the voltage of the power supplies ±E, the current consumption (quiescent) I P and power consumption. Usually, I P amounts to tenths - tens of milliamps, and the power consumption is uniquely determined I P, units - tens of milliwatts.

The maximum permissible parameters of the op-amp include:

◆ maximum possible (undistorted) output signal voltage U out max (usually slightly less than E);

◆ maximum permissible power dissipation;

◆ operating temperature range;

◆ maximum supply voltage;

◆ maximum input differential voltage, etc.

Frequency parameters include absolute cut-off frequency or unity gain frequency f T (F 1), i.e. frequency at which K u op-amp=1. Sometimes the concept of slew rate and settling time of the output voltage is used, determined by the response of the op-amp to the impact of a voltage surge at its input. For some op amps, additional parameters are also provided that reflect their specific area of ​​application.

The amplitude (transfer) characteristics of the op-amp are presented in Figure 6.5 in the form of two dependencies U out=f(U in) for inverting and non-inverting inputs.

When at both inputs of the op-amp U in=0, then an error voltage will be present at the output U osh, determined by the precision parameters of the op-amp (in Figure 6.5 U osh not shown due to its small size).

Figure 6.5. AH OU

The frequency properties of an op-amp are represented by its frequency response, performed on a logarithmic scale, K u op-amp=φ(log f). This frequency response is called logarithmic (LAFC), its typical form is shown in Figure 6.6 (for the K140UD10 op amp).

Figure 6.6. LFC and LFCH OU K140UD10

Frequency dependence K u op-amp can be represented as:

Here τ V time constant of the op-amp, which at M in=3 dB determines the coupling (cutoff) frequency of the op-amp (see Figure 6.6);

ω V= 1/τ V= 2π f in.

Replacing in the expression for K u op-amp τ V by 1/ω V, we get the entry LACHH:

On bass and midrange K u Op-amp=20lg K u Op-amp 0, i.e. The LFC is a straight line parallel to the frequency axis. With some approximation, we can assume that in the HF region the decrease K u Op-amp occurs at a rate of 20 dB per decade (6 dB per octave). Then for ω>>ω V you can simplify the expression for LAC:

K u op-amp= 20lg K u Op-amp 0 – 20log(ω/ω V).

Thus, the LFC in the HF region is represented by a straight line with a slope to the frequency axis of 20 dB/dec. The intersection point of the considered straight lines representing the LFC corresponds to the conjugation frequency ω V (f in). The difference between the real and ideal LFC at frequency f in is about 3 dB (see Figure 6.6), however, for the convenience of analysis this is tolerated, and such graphs are usually called Bode diagrams .

It should be noted that the LFC decay rate of 20 dB/dec is typical for corrected op-amps with external or internal correction, the basic principles of which will be discussed below.

Figure 6.6 also shows the logarithmic phase response (LPFC), which is the dependence of the phase shift j of the output signal relative to the input signal on frequency. The real LFFC differs from the presented one by no more than 6°. Note that for a real op-amp j=45° at frequency f in, and at frequency f T- 90°. Thus, the intrinsic phase shift of the working signal in the corrected op-amp in the HF region can reach 90°.

The parameters and characteristics of the op-amp discussed above describe it in the absence of OOS circuits. However, as noted, op-amps are almost always used with OOS circuits, which significantly affect all of its indicators.

6.3. Inverting amplifier

Op-amps are most often used in inverting and non-inverting amplifiers. A simplified circuit diagram of an op-amp inverting amplifier is shown in Figure 6.7.

Figure 6.7. Op amp inverting amplifier

Resistor R 1 represents the internal resistance of the signal source E g, by means of R os the OU is covered by ∥OOSN.

With an ideal op-amp, the voltage difference at the input terminals tends to zero, and since the non-inverting input is connected to the common bus through resistor R2, the potential at the point a must also be null (“virtual zero”, “apparent ground”). As a result, we can write: I g=I os, i.e. E g/R 1 =–U out/R os. From here we get:

K U inv = U out/E g = –R os/R 1 ,

those. with ideal op amp K U inv is determined by the ratio of the values ​​of external resistors and does not depend on the op-amp itself.

For a real op-amp, it is necessary to take into account its input current I input, i.e. I g=I os+I input or ( E g–U in)/R 1 =(U in–U out)/R os+U in/U input, Where U in- signal voltage at the inverting input of the op-amp, i.e. at the point a. Then for a real op-amp we get:

It is easy to show that when the OOS depth is more than 10, i.e. K u op-amp/K U inv=F>10, calculation error K U inv for the case of an ideal op-amp, it does not exceed 10%, which is quite sufficient for most practical cases.

Resistor values ​​in op-amp devices should not exceed several megohms, otherwise unstable operation of the amplifier may occur due to leakage currents, op-amp input currents, etc. If, as a result of the calculation, the value R os exceeds the maximum recommended value, then it is advisable to use a T-shaped OOS chain, which, with moderate resistor values, allows it to perform the function of an equivalent high-resistance R os(Figure 6.7b) . In this case, you can write:

In practice it is often believed that R OS 1 =R OS 2 >>R OS 3 and the value R 1 is usually given, so R OS 3 is determined quite simply.

Op-amp inverting amplifier input impedance R input inv has a relatively small value determined by parallel OOS:

R input inv = R 1 +(R os/K u op-amp + 1)∥R input≈ R 1 ,

those. at large K u op-amp input resistance is determined by the value R 1 .

Inverting amplifier output impedance R out inv in a real op-amp it is different from zero and is defined as R out op amp, and the depth of environmental protection F. For F>10, we can write:

R out inv = R out op amp/F = R out op amp/K U inv/K u op-amp.

Using the LFC of the op-amp, you can represent the frequency range of the inverting amplifier (see Figure 6.6), and

f OC = f T/K U inv.

In the limit you can get K U inv=1, i.e. get an inverting follower. In this case, we obtain the minimum output impedance of the op-amp amplifier:

R out = R out op amp/K u op-amp.

In an amplifier using a real op-amp at the output of the amplifier at U in=0 error voltage will always be present U osh, generated U cm and Δ I input. In order to reduce U osh strive to equalize the equivalent resistors connected to the inputs of the op-amp, i.e. take R 2 =R 1 ∥R os(See Figure 6.7a). If this condition is met for K U inv>10 can be written:

U osh ≈ U cm K U inv + Δ I in R os.

Decrease U osh possible by applying additional bias to the non-inverting input (using an additional divider) and reducing the values ​​of the resistors used.

Based on the considered inverting UPT, it is possible to create an AC amplifier by connecting separating capacitors to the input and output, the ratings of which are determined based on a given frequency distortion factor M n(see subsection 2.5).

6.4. Non-inverting amplifier

A simplified circuit diagram of a non-inverting op-amp amplifier is shown in Figure 6.8.

Figure 6.8. Non-inverting op-amp amplifier

It is easy to show that in a non-inverting amplifier the op-amp is covered by the POSN. Because the U in And U os are supplied to different inputs, then for an ideal op-amp we can write:

U in = U out R 1 /(R 1 + R os),

whence the voltage gain of the non-inverting amplifier:

K U noninv = 1 + R os/R 1 ,

K U noninv = 1 + |K U inv|.

For a non-inverting amplifier based on a real op-amp, the obtained expressions are valid at a feedback depth of F>10.

Input impedance of a non-inverting amplifier R input noninv is large and is determined by deep consistent OOS and high value R input:

R input noninv = R input· F = R input· K U OU/K U noninv.

The output impedance of a non-inverting op-amp amplifier is determined as for an inverting amplifier, because in both cases, the voltage protection system applies:

R out non-inv = R out of the op amp/F = R out of the op amp/K U noninv/K U OU.

The expansion of the operating frequency band in a non-inverting amplifier is achieved in the same way as in an inverting amplifier, i.e.

f OC = f T/K U noninv.

To reduce the current error in a non-inverting amplifier, similar to an inverting amplifier, the following condition must be met:

R g = R 1 ∥R os.

A non-inverting amplifier is often used for large R g(which is possible due to the large R input noninv), therefore, fulfilling this condition is not always possible due to restrictions on the value of resistor values.

The presence of a common-mode signal at the inverting input (transmitted through the circuit: non-inverting op-amp input ⇒ op-amp output ⇒ R os⇒ inverting input of the op-amp) leads to an increase U osh, which is a disadvantage of the amplifier in question.

By increasing the depth of environmental protection, it is possible to achieve K U noninv=1, i.e. obtaining a non-inverting repeater, the circuit of which is shown in Figure 6.9.

Figure 6.9. Non-inverting op-amp follower

Here, 100% POSN is achieved, so this repeater has the highest input and minimum output impedance and is used, like any repeater, as a matching stage. For a non-inverting follower, you can write:

U osh ≈ U cm + I in sr R g ≈ I in sr R g,

those. The error voltage can reach quite large values.

Based on the considered non-inverting UPT, it is also possible to create an AC amplifier by connecting separating capacitors to the input and output, the ratings of which are determined based on a given frequency distortion factor M n(see subsection 2.5).

In addition to inverting and non-inverting amplifiers based on op-amps, various op-amp options are available, some of which will be discussed below.

6.5. Types of control units on the op amp

difference (differential) amplifier , the diagram of which is shown in Figure 6.10.

Figure 6.10. Op-amp difference amplifier

An op-amp difference amplifier can be considered as a combination of inverting and non-inverting amplifier options. For U out difference amplifier can be written:

U out = K U inv U in 1 +K U noninv U in 2 R 3 /(R 2 + R 3).

Usually, R 1 =R 2 and R 3 =R os, hence, R 3 /R 2 =R os/R 1 =m. Expanding the values ​​of the gain factors, we get:

U out = m(U in 2 – U in 1),

For the special case when R 2 =R 3 we get:

U out = U in 2 – U in 1 .

The last expression clearly explains the origin of the name and purpose of the amplifier in question.

In a difference amplifier based on an op-amp, with the same polarity of input voltages, a common-mode signal occurs, which increases the amplifier error. Therefore, in a difference amplifier it is desirable to use an op-amp with a large CMRR. The disadvantages of the considered difference amplifier include different values ​​of input resistances and difficulty in adjusting the gain. These difficulties are eliminated in devices using several op-amps, for example, in a difference amplifier with two repeaters (Figure 6.11).

Figure 6.11. Repeater difference amplifier

This circuit is symmetrical and is characterized by the same input resistances and low error voltage, but only works for a symmetrical load.

Based on the op-amp it can be performed logarithmic amplifier , the schematic diagram of which is shown in Figure 6.12.

Figure 6.12 Logarithmic op-amp amplifier

The P-n junction of the VD diode is forward biased. Assuming the op-amp is ideal, we can equate the currents I 1 and I 2. Using the expression for the current-voltage characteristic of the p-n junction ( I=I 0 ), it is easy to write:

U in/R= I 0 ·,

from where after transformations we get:

U out = φ T ln( U in/I 0 R) = φ T(ln U in–ln I 0 R),

from which it follows that the output voltage is proportional to the logarithm of the input, and the term ln I 0 R represents the logarithm error. It should be noted that this expression uses voltages normalized to one volt.

When replacing diode VD and resistor R, we get antilog amplifier .

Inverting and non-inverting adders on op-amps, also called summing amplifiers or analog adders. Figure 6.13 shows a schematic diagram of an inverting adder with three inputs. This device is a type of inverting amplifier, many of the properties of which are also manifested in the inverting adder.

Figure 6.13. Op-amp inverting adder

U in 1 /R 1 + U in 2 /R 2 + U in 3 /R 3 = –U out/R os,

From the resulting expression it follows that the output voltage of the device is the sum of the input voltages multiplied by the gain K U inv. At R os=R 1 =R 2 =R 3 K U inv=1 and U out=U in 1 +U in 2 +U in 3 .

When the condition is met R 4 =R os∥R 1 ∥R 2 ∥R 3, the current error is small and can be calculated using the formula U osh=U cm(K U osh+1), where K U osh=R os/(R 1 ∥R 2 ∥R 3) - error signal amplification factor, which has a greater value than K U inv.

Non-inverting adder is implemented in the same way as an inverting adder, but it should use the non-inverting input of the op-amp by analogy with a non-inverting amplifier.

When replacing resistor Roc with capacitor C (Figure 6.14), we obtain a device called analog integrator or just an integrator.

Figure 6.14. Analog integrator on op-amp

With an ideal op-amp, the currents can be equated I 1 and I 2, from which it follows:

The higher the integration accuracy, the greater K u op-amp.

In addition to the considered control units, op amps are used in a number of continuous devices, which will be discussed below.

6.6. Frequency response correction

By correction of frequency characteristics we mean changing the LFC and LPFC to obtain the necessary properties from op-amp devices and, above all, ensuring stable operation. An op-amp is usually used with OOS circuits, however, under certain conditions, due to additional phase shifts in the frequency components of the signal, the OOS can turn into a POS and the amplifier will lose stability. Since the OOS is very deep ( βK U>>1), it is especially important to ensure a phase shift between the input and output signals to ensure that there is no excitation.

Previously, in Figure 6.6, the LFC and LPFC response for the corrected op-amp were shown, in shape equivalent to the LFC and LPFC response of a single amplifier stage, from which it can be seen that the maximum phase shift φ<90° при K u op-amp>1, and the gain decay rate in the HF region is 20 dB/dec. Such an amplifier is stable at any depth of feedback.

If the op-amp consists of several cascades (for example, three), each of which has a decay rate of 20 dB/dec and does not contain correction circuits, then its LFC and LPFC have a more complex shape (Figure 6.15) and contain a region of unstable oscillations.

Figure 6.15. LFC and LPFC of uncorrected op-amp

To ensure stable operation of op-amp devices, internal and external correction circuits are used, with the help of which they achieve a total phase shift with an open feedback loop of less than 135° at the maximum operating frequency. In this case, it automatically turns out that the decline K u op-amp is about 20dB/dec.

It is convenient to use as a criterion for the stability of op-amp devices Bode criterion , formulated as follows: “An amplifier with a feedback circuit is stable if the straight line of its gain in decibels crosses the LFC in a section with a roll-off of 20 dB/dec.” Thus, we can conclude that the frequency correction circuits in the op-amp must provide the decay rate K U inv(K U noninv) at HF ​​about 20 dB/dec.

Frequency correction circuits can be either built into the semiconductor crystal or created by external elements. The simplest frequency correction circuit is carried out by connecting a capacitor C cor of a sufficiently large value to the output of the op-amp. It is necessary that the time constant τ core=R out C cor was greater than 1/2π f in. In this case, high-frequency signals at the output of the op-amp will be shunted C core and the operating frequency band will narrow, most of them quite significantly, which is a significant drawback of this type of correction. The LFC obtained in this case is shown in Figure 6.16.

Figure 6.16. Frequency correction with an external capacitor

Recession K u op-amp here it will not exceed 20 dB/dec, and the op-amp itself will be stable with the introduction of OOS, since φ will never exceed 135°.

Corrective circuits of integrating (lag correction) and differentiating (advanced correction) types are more advanced. In general, an integrating type correction manifests itself similarly to the action of a corrective (load) capacitance. The correcting RC circuit is connected between the op-amp stages (Figure 6.17).

Figure 6.17. Integrating type frequency correction

Resistor R 1 is the input resistance of the op-amp stage, and the correction circuit itself contains R core and C core. The time constant of this circuit must be greater than the time constant of any of the op-amp stages. Since the correction circuit is the simplest single-link RC circuit, its LFC slope is 20 dB/dec, which guarantees stable operation of the amplifier. And in this case, the correction circuit narrows the operating frequency band of the amplifier, but a wide band still does not give anything if the amplifier is unstable.

Stable operation of the op-amp with a relatively wide band is ensured by differential-type correction. The essence of this method of correcting LFC and LPFC is that RF signals pass inside the op-amp, bypassing part of the cascades (or elements) that provide maximum K u Op-amp 0, they are not amplified or delayed in phase. As a result, RF signals will be amplified less, but their small phase shift will not lead to loss of amplifier stability. To implement differential-type correction, a correction capacitor is connected to the special terminals of the op-amp (Figure 6.18).

Figure 6.18. Differential type frequency correction

In addition to the corrective circuits considered, others are known (see, for example). When choosing correction schemes and the values ​​of their elements, you should refer to reference literature (for example,).

The non-inverting amplifier is perhaps one of the three most basic analog electronics circuits, along with the inverting amplifier and voltage follower circuits. It is even simpler than an inverting amplifier, since the circuit does not require bipolar power to operate.

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 you cannot attenuate the signal with such a circuit.

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

The first thing we need to think about is what voltages are present at both inputs of our op amp. Let us recall the first of the rules, which describes the operation of an operational amplifier:

Rule No. 1 - the operational amplifier influences its output on the input through OOS (negative feedback), as a result of which the voltage at both inputs, both inverting (-) and non-inverting (+), is equalized.

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

I = U/R = 3V/10k = 300uA.

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

Rule No. 2 - amplifier inputs do not consume current

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

U = IR = 300uA * 20k = 6V

It turns out that this voltage is the output voltage of the amplifier? No, that's not true. Recall that a 20k resistor has a voltage of 3V at one of its terminals. Notice how the voltages across both resistors are directed.

The current flows in the opposite direction of the arrow, symbolizing the point with 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 resistors R1 and R2 form a simple one. Remember that the sum of the voltages across the individual resistors of the divider must be equal to the voltage supplied to the divider - the voltage cannot disappear without a trace and appear out of nowhere.

Finally, we must check the result obtained with the last rule:

Rule No. 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, it is necessary to check that the voltage we calculated can actually be obtained. Often beginners think that the amplifier works like a “Perpetuum Mobile” and produces voltage out of nothing. But we must remember that the amplifier also needs power to operate.

Classic amplifiers operate on voltages of -15V and +15V. In such a situation, the 9V we calculate is the real voltage, since 9V is in the range of the supply voltage. However, modern amplifiers often operate at voltages as low as 5V or lower. In such a situation, there is no chance that the amplifier will output 9V.

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

The non-inverting amplifier is a basic op-amp circuit. It looks painfully simple:

In this circuit, the signal is applied to the non-inverting input of the op-amp.

So, in order to understand how this circuit works, remember the most important rule that is used to analyze op-amp circuits: the output voltage of the op-amp tends to ensure that the voltage difference between its inputs is equal to zero.

Principle of operation

So, let's denote the inverting input with the letter A:

Following the main rule of the op-amp, we find that the voltage at the inverting input is equal to the input voltage: U A =U in. U A is removed from , which is formed by resistors R1 and R2. Hence:

U A = U out R1/(R1+R2)

Because U A =U in, we get that U in = U out R1/(R1+R2).

The voltage gain is calculated as K U = U out / U input.

We substitute the previously obtained values ​​here and get that K U = 1+R2/R1.

Checking work in Proteus

This can also be easily checked using the Proteus program. The diagram will look like this:

Let's calculate the gain K U. K U = 1+R2/R1=1+90k/10k=10. This means that our amplifier must increase the input signal exactly 10 times. Let's check if this is true. We apply a sinusoid with a frequency of 1 kHz to the non-inverting input and see what we have at the output. For this we need a virtual oscilloscope:

The input signal is a yellow waveform and the output signal is a pink waveform:

As you can see, the input signal has been amplified exactly 10 times. The phase of the output signal remains the same. Therefore, such an amplifier is called NOT inverting.

But, as they say, there is one “BUT”. In fact, real op-amps have design flaws. Since Proteus tries to emulate components close to real ones, let's look at the amplitude-frequency response (AFC), as well as the phase-frequency response (PFC) of our LM358 op-amp.

Frequency response and phase response of a non-inverting amplifier on LM358

In practice, in order to remove the frequency response, we need to apply a frequency from 0 Hertz to some final value to the input of our amplifier, and at the output at this time monitor the change in the amplitude of the signal. In Proteus, this is all done using the Frequency Response function:

On the Y axis we have gain, and on the X axis we have frequency. As you may have noticed, the gain remained almost unchanged up to a frequency of 10 kHz, then began to rapidly drop with increasing frequency. At a frequency of 1 MegaHertz, the gain was equal to unity. This parameter in the op-amp is called unity gain frequency and is denoted as f 1. That is, in essence, the amplifier does not amplify the signal at this frequency. What is given at the input is what comes out.

When designing amplifiers, an important parameter is cutoff frequency f gr. In order to calculate it, we need to know the gain at frequency K gr:

K gr = K Uo / √2 or = K Uo x 0.707, where K Uo is the gain at a frequency of 0 Hertz (direct current).

If we look at the frequency response, we will see that at zero frequency (at direct current) our gain is 10. We calculate K gr.

K gr = 10 x 0.707 = 7.07

Now we draw a horizontal line at level 7.07 and look at the intersection with the chart. I got about 104 kHz. Build an amplifier with a cutoff frequency greater than f gr does not make sense, since in this case the output signal of the amplifier will be greatly attenuated.

It is also very easy to determine the cutoff frequency if you plot a graph in . The cutoff frequency will be at the level K Uo -3dB. That is, in our case, at a level of 17dB. As you can see, in this case we also got a cutoff frequency of 104 kHz.

Okay, we seem to have sorted out the cutoff frequency. Now such a parameter as phase response is important to us. In our case, we seem to have obtained a NON-inverting amplifier. That is, the phase shift between the input and output signal must be zero. But how will the amplifier behave at high frequencies (HF)?

We take the same frequency range from 0 to 100 MHz and look at the phase response:

As you can see, up to 1 kHz the non-inverting amplifier really works as it should. That is, the input and output signals move in phase. But after a frequency of 1 kHz, we see that the phase of the output signal begins to lag. At a frequency of 100 kHz it is already behind by about 40 degrees.

For clarity, the frequency response and phase response can be placed on one graph:

Also, in circuits with a non-inverting amplifier, a compensating resistor R K is often introduced.

It is determined by the formula:

and serves to ensure equality of resistance between each of the inputs and ground. We will look at this in more detail in the next article.

With input from Jeer

There are many important topics in an electronics course. Today we will try to understand operational amplifiers.
Start over. An operational amplifier is a “thing” that allows you to operate with analog signals in every possible way. The simplest and most basic are amplification, attenuation, addition, subtraction and many others (for example, differentiation or logarithm). The vast majority of operations on operational amplifiers (hereinafter referred to as op-amps) are performed using positive and negative feedback.
In this article we will consider a certain “ideal” op-amp, because It doesn't make sense to switch to a specific model. By ideal it is meant that the input resistance will tend to infinity (therefore, the input current will tend to zero), and the output resistance, on the contrary, will tend to zero (this means that the load should not affect the output voltage). Also, any ideal op-amp should amplify signals of any frequency. Well, and most importantly, the gain in the absence of feedback should also tend to infinity.

Get to the point

An operational amplifier is often symbolized in diagrams by an equilateral triangle. On the left are the inputs, which are marked "-" and "+", on the right is the output. Voltage can be applied to any of the inputs, one of which changes the polarity of the voltage (that’s why it was called inverting), the other does not (it is logical to assume that it is called non-inverting). The op-amp power supply is most often bipolar. Typically, positive and negative supply voltages have the same value (but different sign!).
In the simplest case, you can connect voltage sources directly to the op-amp inputs. And then the output voltage will be calculated according to the formula:
, where is the voltage at the non-inverting input, is the voltage at the inverting input, is the output voltage, and is the open-loop gain.
Let's look at the ideal op-amp from the Proteus point of view.


I suggest you “play” with him. A voltage of 1V was applied to the non-inverting input. To inverting 3V. We use an “ideal” op-amp. So, we get: . But here we have a limiter, because we will not be able to amplify the signal above our supply voltage. Thus, we will still get -15V at the output. Result:


Let's change the gain (so you believe me). Let the Voltage Gain parameter become equal to two. The same problem is clearly solved.

Real-life application of op-amps using the example of inverting and non-inverting amplifiers

There are two of these main rules:
I. The op amp output tends to cause the differential voltage (the difference between the voltage at the inverting and non-inverting inputs) to be zero.
II. The op amp inputs do not consume any current.
The first rule is implemented through feedback. Those. the voltage is transferred from the output to the input in such a way that the potential difference becomes zero.
These are, so to speak, the “sacred canons” in the OU topic.
And now, more specifically. Inverting amplifier looks exactly like this (pay attention to how the inputs are located):


Based on the first “canon” we obtain the proportion:
, and after “doing a little magic” with the formula, we derive the value for the gain of the inverting op-amp:

The above screenshot does not need any comments. Just plug everything in and check it yourself.

Next stage - non-inverting amplifier.
Everything is also simple here. The voltage is applied directly to the non-inverting input. Feedback is supplied to the inverting input. The voltage at the inverting input will be:
, but applying the first rule, we can say that

And again, “grandiose” knowledge in the field of higher mathematics allows us to move on to the formula:
I’ll give you a comprehensive screenshot that you can double-check if you want:

Finally, I’ll give you a couple of interesting circuits so that you don’t get the impression that operational amplifiers can only amplify voltage.

Voltage follower (buffer amplifier). The operating principle is the same as that of a transistor repeater. Used in heavy load circuits. Also, it can be used to solve the problem of impedance matching if the circuit contains unwanted voltage dividers. The scheme is simple to the point of genius:

Summing amplifier. It can be used if you need to add (subtract) several signals. For clarity, here is a diagram (again, pay attention to the location of the inputs):


Also, pay attention to the fact that R1 = R2 = R3 = R4, and R5 = R6. The calculation formula in this case will be: (familiar, right?)
Thus, we see that the voltage values ​​that are supplied to the non-inverting input “acquire” a plus sign. On the inverting one - minus.

Conclusion

Operational amplifier circuits are extremely diverse. In more complex cases, you may find active filter circuits, ADC and storage sampling devices, power amplifiers, current-to-voltage converters, and many many other circuits.

List of sources

A short list of sources that will help you quickly get used to both op-amps and electronics in general:
Wikipedia
P. Horowitz, W. Hill. "The Art of Circuit Design"
B. Baker. “What a digital developer needs to know about analog electronics”
Lecture notes on electronics (preferably your own)
UPD: Thank you UFO for invitation

A journey of ten thousand miles 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!.. Soldering - so soldering. There is a computer and the Internet is connected. We choose a scheme. And suddenly it turns out that the diagrams for the intended subject are a carriage and a small cart. And everyone is different. No experience, not enough knowledge. Which one to choose? Some of them contain some kind of rectangles and triangles. Amplifiers, and even operational ones... How they work is unclear. Scary!.. What if it burns? We choose what is simpler, using familiar transistors! Selected, 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 one is a gaming one! And for office work, even the simplest is enough. It's the same with transistors. Soldering a circuit on them is not enough. You still need to be able to configure it. There are too many pitfalls and pitfalls. And this often requires experience that is not at the entry level. So why quit an exciting activity? Not at all! Just don’t be afraid of these “triangles-rectangles”. It turns out that working with them, in many cases, is much easier than with individual transistors. IF YOU KNOW - HOW!

This is what we will now deal with: understanding how an operational amplifier (op-amp, or in English OpAmp) works. At the same time, we will consider his work literally “on the fingers”, practically without using any formulas, except perhaps Ohm’s law: “Current through a section of the circuit ( I) is directly proportional to the voltage across it ( U) and is 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 accept as an assumption that it is a “black box” with some kind of filling. At this stage, we will not consider such op-amp parameters as “bias voltage”, “shift voltage”, “temperature drift”, “noise characteristics”, “common mode suppression ratio”, “supply voltage ripple suppression ratio”, “bandwidth” " and so on. All these parameters will be important at the next stage of its study, when the basic principles of its work “settle” in your head because “it was smooth on paper, but they forgot about the ravines”...

For now, we’ll 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, an 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 terminals: positive and negative. These five conclusions are found in almost any op-amp and are fundamentally necessary for its operation.

The op-amp has a huge gain, at least 50000...100000, but in reality it is 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 influenced solely by the potential difference between its inputs, regardless from them absolute meanings and polarities.

The term "constant current" means that the op amp amplifies input signals starting at 0 Hz. The upper frequency range (frequency range) of signals amplified by an op-amp depends on many reasons, 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 question goes beyond the scope of initial acquaintance with his work and will not be considered here.

The op-amp inputs have a very high input resistance, equal to tens/hundreds of MegaOhms, or even GigaOhms (and only in the ever-memorable K140UD1, and even in the K140UD5 it was only 30...50 kOhm). Such a high resistance of the inputs means that they have virtually 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 - first an important rule that is applied when analyzing the operation of an op-amp. Please remember well what it concerns only the op amp itself, but not schemes with its use!

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

Translated from Latin, one of the meanings of the word “inversio” is “turning around”, “turnover”. In other words, inversion is a mirror image ( mirroring) signal relative to the horizontal X axis(time axis). In Fig. Figure 1 shows several of the many possible options for signal inversion, where red indicates the direct (input) signal and blue indicates the inverted (output) signal.

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 further). In other words, polarity The signal is also not a determining factor.

Op-amps are depicted on circuit diagrams in different ways. Abroad, op-amps used to be depicted, and even now they are very often depicted in the form of an isosceles triangle (Fig. 2, A). The inverting input is represented by a minus symbol, and the non-inverting input is represented by a plus symbol inside a triangle. These symbols do not mean at all that the potential at the corresponding inputs should 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 can easily be confused with power pins, which can turn out to be an unexpected “rake”, especially for beginners.

Rice. 2 Options for conditional graphic images (CGO)
operational amplifiers

In the system of domestic conventional graphic images (UGO) before the entry into force of GOST 2.759-82 (ST SEV 3336-81), op-amps were also depicted in the form of a triangle, only the inverting input - with an inversion symbol - 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 op-amps in diagrams, the inverting and non-inverting inputs can be swapped, if it is more convenient, however, traditionally the inverting input is depicted at the top, and the non-inverting input at the bottom. Power pins, as a rule, are always located in 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 supplying part of the amplifier's output voltage 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 feedback is supplied, a distinction is made between negative feedback (NFB), when part of the output signal is supplied to the inverting input (Fig. 3, A) or positive feedback (POF), when part The output signal is supplied, accordingly, to the non-inverting input (Fig. 3, B).

Rice. 3 The principle of feedback generation (FE)

In the first case, since the output signal is the inverse of the input signal, it is subtracted from the input signal. As a result, the overall gain of the stage 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 POS has a positive effect, and OOS is a completely useless idea: why reduce the gain? This is exactly what US patent examiners thought when, in 1928, Harold S. Black tried patent the OOS. However, by 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 (taking into account 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 deliberately, for example, in generators, comparators with hysteresis (more on this later), etc., but in general its influence on the operation of amplifier circuits with op-amps is rather negative and requires a very careful and reasonable analysis its application.

Since the op-amp has two inputs, the following basic types of its activation using OOS are possible (Fig. 4):

Rice. 4 Basic circuits for connecting op-amps

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

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

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

To analyze the operation of these circuits, one should take into account second most important rule, to which the operation of the op-amp is subordinated: The output of the operational amplifier tends to ensure that the voltage difference between its inputs is zero..

However, any formulation must be necessary and sufficient, to limit the entire subset of cases subject to it. The above formulation, for all its “classicity,” does not provide 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, supplying voltage to them from somewhere “from within”.

If you carefully examine the diagrams in Fig. 4, you can see that the OOS (through Rooos) in all cases is started from the output only to the inverting input, which gives us reason to reformulate this rule as follows: Voltage at 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 “master” when any op-amp with OOS is turned on 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”. The virtual object behaves close to the behavior of similar objects of material reality, i.e., for input signals (due to the action of the feedback loop), the inverting input can be considered connected directly to the same potential to which the non-inverting input is connected. However, “virtual zero” is just a special case that occurs only with a bipolar op-amp supply. When using unipolar power supply (which will be discussed below), and in many other switching circuits, there will be no zero at 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 operating principles of the op-amp.

It is from this point of view that we will analyze the diagrams 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 using 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 via Rooc is initiated from the output of the op-amp to its inverting input. The only difference is which of the inputs is supplied with input voltage.

A) Inverting switching on (Fig. 5).

Rice. 5 Operating principle of an 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 (in total their 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, current will flow through it Iinput= 1 V / 1000 Ohm = 0.001 A (1 mA). The direction of flow of this current is shown by the arrow.

Since Rooc and Rin are included by the divider, and according to Rule 1, the inputs of the op-amp do not consume current, then in order for the voltage to be 0 V at the midpoint of this divider, voltage must be applied to the right pin 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 kOhm + 1 kOhm) = 1 mA), i.e. I input = I oos .

If a voltage of negative polarity is applied to the input, the output of the op-amp will be a voltage of positive polarity. 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 ratings 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 circuit is often used if it is necessary to invert a signal obtained using circuits that are fundamentally inverters. For example, logarithmic amplifiers.

Now let's, keeping the Rin value equal to 1 kOhm, increase the resistance Rooc to 2 kOhm with the same input signal +1 V. The total resistance of the Rooc + Rin divider has increased to 3 kOhm. In order for a potential of 0 V to remain at its midpoint (equal to the potential of the non-inverting input), 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 Rooc rating to 10 kOhm. Now the voltage at the output of the op-amp under the same other conditions will be already 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 much I swore not to use formulas, let’s still display this in the form of an equation:
Ku = – Uout / Uin = – Roos / 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!

Now let's increase the resistance Rooc to 20 kOhm and analyze what happens. According to formula (2), with Ku = 20 and an input signal of 1 V, the output should have a voltage of 20 V. But that’s not the case! We previously accepted the assumption that the supply voltage of our op-amp is only ± 15 V. But even 15 V cannot be obtained (why this is so - a little lower). “You can’t jump above your head (supply voltage)!” As a result of such abuse of the circuit ratings, the output voltage of the op-amp “rests” against the supply voltage (the output of the op-amp enters saturation). Balance of current equality through the divider RoocRin ( Iinput = Ioos) is violated, a potential appears at the inverting input that 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 one, with a nominal value of, say, 10 kOhm (Fig. 6).

Rice. 6 Variable gain inverting amplifier circuit

With the right (according to the diagram) position of its slider, the gain will be Rooc / Rin = 10 kOhm / 1 kOhm = 10. By moving the Roos slider to the left (reducing 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 when any denominator value. The output will also be zero for any value and polarity of the input signal. This circuit is often used in audio amplification circuits, for example, in mixers, where the gain has to be adjusted from zero.

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

Rice. 7 Operating principle of an op-amp in a non-inverting connection

The left Rin pin is connected to the middle point (“ground”), and the +1 V input signal 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 will also take the resistances Rooc and Rin equal to each other and components of 1 kOhm. Because at the 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, there must be a voltage of +2 V at the right terminal of the resistor Rooc (op-amp output). Currents Iinput And Ioos, equal to 1 mA, now flow through resistors Rooc and Rin in the opposite direction (shown by arrows). We did it non-inverting amplifier with a gain of 2, since an input signal of +1 V produces an output signal of +2 V.

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

Now we increase the Rooc rating to 2 kOhm. To maintain a 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 = 3 V / 1 V = 3!

If we compare the values ​​of Ku for 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 = Uout / Uin + 1 = (Rooc / Rin) + 1 (3)

Why is this happening? Yes, very simple! The OOS operates in exactly the same way as with 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 connection, you cannot get a gain of 1? Why can't it - it's possible. Let's reduce the Rooc rating, similar to how we analyzed Fig. 6. When its value is zero - 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 = 1 V / 1 V = 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 Circuit diagram for connecting an op-amp as a voltage follower

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

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

At first glance, such a scheme has no practical meaning: why do we need a single and even non-inverse “gain” - what, you can’t just send the signal further??? However, such schemes are used quite often and here's why. According to Rule 1, current does not flow into the op-amp inputs, i.e., input impedance The non-inverting follower is very large - those same tens, hundreds and even thousands of MOhms (the same applies to the circuit in Fig. 7)! But the output resistance is very low (fractions of an Ohm!). The output of the op-amp is “puffing with all its might”, trying, according to Rule 2, to maintain the same potential at the inverting input as at the non-inverting input. The only limitation is the permissible output current of the op-amp.

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

For most widely used op amps, the technical parameters indicate that the load resistance connected to their output should not be less 2 kOhm. More - as much as you like. For a much smaller number it is 1 kOhm (K140UD...). This means that under worst-case conditions: maximum supply voltage (for example, ±16 V or a total of 32 V), a load connected between the output and one of the power rails, and a 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 = 0.015 A (15 mA). Not too little, but not too much either. Fortunately, most common op amps have built-in output current protection - a typical maximum output current of 25 mA. The protection prevents overheating and failure of the op-amp.

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

IN) Differential switching on (Fig. 9).

Rice. 9 Operating principle of op-amp in differential connection

So, let’s assume that with the same ratings Rin and Rooc 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 (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 essence, we actually have a non-inverting follower (compare with Fig. 8). Accordingly, at the output we will get the same voltage 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 Rin pins and current flows through it Iinput= 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.

This is where the malicious “grin” of the additional unit in the formula for the gain of a non-inverting amplifier appeared. It turns out that with this simplified In differential switching, the difference in gain permanently shifts the output signal by the amount of potential at the non-inverting input. A problem with! However, “Even if you are eaten, you still have at least two options.” 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, we will apply the input signal to the non-inverting input not directly, but through the divider Rin2, R1 (Fig. 9, B). Let us also accept their values ​​of 1 kOhm. Now at the non-inverting (and therefore at the inverting too) input of the op-amp there will be a potential of +0.5 V, current will flow through it (and Rooc) Iinput = Ioos= 0.5 mA, to ensure which the output of the op-amp must have a voltage equal to 0 V. Phew! 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), the op-amp output will maintain a zero voltage equal to the difference in the input signals .

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

Of course, if Rin1 and Rooc (respectively, Rin2 and R1) are equal, we will not receive gain. To do this, you need to increase the ratings of Rooc and R1, as was done when analyzing the previous switching on of the op-amp (I will not repeat), and it should strictly the following ratio is observed:

Rooc / Rin1 = R1 / Rin2. (4)

What practical benefits do we get from such inclusion? 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 sent to the output. This makes it possible to amplify very small signals against a background of interference that affects both inputs equally. For example, a signal from a dynamic microphone against the background of interference from an industrial frequency network of 50 Hz.

However, in this barrel of honey, unfortunately, there is a fly in the ointment. Firstly, equality (4) must be observed very strictly (down to tenths and sometimes hundredths of a percent!). Otherwise, an imbalance of currents acting in the circuit will arise, and therefore, in addition to difference (“antiphase”) signals, combined (“in-phase”) signals will also be amplified.

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

Rice. 10 Signal phase shift

The signal phase is a value characterizing the offset of the signal period reference point relative to the time reference point. Since both the origin of time and the origin of the period are chosen arbitrarily, the phase of one periodic The signal has no physical meaning. However, the phase difference between the two periodic 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. The starting point of the period can be taken as a zero value with a positive slope. It's possible - maximum. Everything is in our power.

In 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, then you can see that they are mutually inverse. Thus, “in-phase signals” are signals that coincide with each other at each point, and “anti-phase signals” are inverse each other relative to each other.

At the same time, the concept inversions broader than the concept phases, because the latter applies only to regularly repeating, periodic signals. And the concept inversions applicable to any signals, including non-periodic ones, such as a sound signal, digital sequence, or constant voltage. To phase was a consistent quantity, 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 a 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 rating Rin1, then the non-inverting input is determined by the ratings sequentially turned on Rin2 and R1 (have you forgotten that the inputs of the op-amp do not consume current?). In the example above, they will be 1 and 2 kOhm, respectively. And if we increase Rooc and R1 to obtain a full-fledged amplifier stage, then the difference will increase even more significantly: with Ku = 10 - to, respectively, the same 1 kOhm and as much as 11 kOhm!

Unfortunately, in practice they usually set the ratings Rin1 = Rin2 and Rooc = R1. However, this is only acceptable if the signal sources for both inputs are very low output impedance. Otherwise, it forms a divider with the input resistance of a given amplifier stage, and since the division coefficient 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 will perform this function poorly .

One way to solve this problem may 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 compliance with 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), the resistance of which is 20% of the desired 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. The option with the inclusion of a tuning resistor (R5) in series with Rooc (Fig. 11, B) is much more stable, since Rooc does not take part in the formation of the input resistance of the inverting input. The main thing is to maintain the ratio of their denominations, similar to option “A” (Rooc / Rin1 = R1 / Rin2).

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

Rice. 12 Switchable inverting/non-inverting follower circuit

The input signal is applied simultaneously to both inputs of the circuit (inverting and non-inverting). The values ​​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 can be connected to a common bus using the SA switch.

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

The principle of operation of this circuit can be expressed in a slightly different way. When the switch SA is closed, it works as an inverting amplifier with a gain equal to minus 1, and when open - simultaneously(!) both as an inverting amplifier with a gain of minus 1, and as a non-inverting amplifier with a gain of +2, whence: Ku = +2 + (–1) = +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 to which there is no access before setting up the device). If you use a transistor (for example, a field-effect transistor) as a key, controlled from the input signal using comparator(we will discuss it below), we get synchronous detector(synchronous rectifier). The specific implementation of such a scheme, of course, goes beyond the scope of an initial acquaintance with the operation of the op-amp and we again will not consider it in detail here.

Now let’s look at the principle of summing input signals (Fig. 13, A), and at the same time let’s figure out what the values ​​of resistors Rin and Rooc should be in reality.

Rice. 13 Operating principle 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. For now, for “training” purposes, we accept the resistance of all resistors, including Rooc, as equal to 1 kOhm. We apply 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 directed from left to right). To maintain the same potential at the inverting input as at the non-inverting input (0 V), a current must flow through the resistor Rooc equal to the sum of the input currents (1 mA + 1 mA = 2 mA), 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 a voltage of +2 V is applied to the input of the inverting amplifier (Fig. 5), or the Rin rating is halved, i.e. up to 500 Ohm. Let's 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 desired. 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 be mutually compensated 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 arises 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 there was still something left for the load connected to the output of the op-amp. But if the input signal voltages are increased to the maximum permissible (close to the supply voltages), then it turns out that the entire output current will go into Rooc. There will be nothing left for the load. And who needs an amplifier stage that works “for itself”? In addition, the values ​​of the input resistors, equal to only 1 kOhm (accordingly, determining the input resistance of the inverting amplifier stage), require excessively large currents to flow through them, heavily loading the signal source. Therefore, in real circuits, the resistance Rin is chosen to be no less than 10 kOhm, but preferably no more than 100 kOhm, so that for a given gain, Rooc is not set to too high a value. Although these values ​​are not absolute, but only approximate, as they say, “as a first approximation” - everything depends on the specific 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 Rooc.

Summation signals can also be supplied to a 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 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 "rakes" that should be taken into account.

Since Rule 2 only applies to the inverting input, which is subject to 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 present 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, it is necessary to use additional circuit design tricks.

Until now, we have considered circuits based on op-amps 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 the English word compare- compare). Its output will be a voltage approaching one of the supply voltages, depending on which signal is greater than the other. Typically, the input signal is applied to one of the inputs, and the other is a constant voltage with which it is compared (the so-called “reference voltage”). It can be anything, including equal to zero potential (Fig. 14, B).

Rice. 14 Circuit diagram for connecting an op-amp as a comparator

However, not everything is so good “in the kingdom of Denmark”... 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 comparator inputs.

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

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

Let's analyze the operation of this scheme. Its supply voltage is ±10 V (for good measure). Resistance Rin is 1 kOhm, and Rpos is 10 kOhm. The midpoint potential is selected as the reference voltage supplied to the inverting input. The red curve shows the input signal arriving at the left pin Rin (input scheme comparator), blue - potential at the non-inverting input of the op-amp and green - output signal.

While the input signal has a negative polarity, the output has a negative voltage, which, through Rpos, is summed with the input voltage in inverse proportion to the values ​​of the corresponding resistors. As a result, the potential of the non-inverting input in the entire range of negative values ​​is 1 V (in 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 polarity to positive. The total potential at the non-inverting input will begin avalanche-like become even more positive, supporting the process of such switching. As a result, the comparator simply “will not notice” minor noise fluctuations in 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 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 switching the comparator output, equal in our case to a total of 2 V, is called hysteresis. The greater the resistance Rpos in relation to Rin (the smaller the depth of the POS), the lower the switching hysteresis. So, at Rpos = 100 kOhm it will be only 0.2 V, and at Rpos = 1 Mohm - 0.02 V (20 mV). The hysteresis (the depth of the PIC) is selected based on the actual operating conditions of the comparator in a specific circuit. In some cases there will be a lot of 10 mV, and in some cases 2 V is not enough.

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 wide application” (LM393/LM339/K554CA3/K597CA3). Their main difference from op-amps is the special design of the output stage, which is made on an open-collector transistor (Fig. 16).

Rice. 16 Output stage of widely used comparators
and its connection to the load resistor

This requires the mandatory use of 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 at 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 circuit design, but quad LM339 (LM139/LM239), the emitter of the output stage transistor is connected to the negative power terminal, which somewhat limits their scope of application. In this regard, I would like to draw attention to the comparator LM31 (LM111/LM211), an analogue of which is the domestic 521/554CA3, in which both the collector and emitter of the output transistor are separately connected, which can be connected to voltages other than the supply voltage of the comparator itself. Its only and relative drawback is that there is only one in an 8-pin (sometimes 14-pin) package.

So far, we have considered circuits in which the input signal was supplied 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 exactly the same as the inverting amplifier discussed above, only the output resistance of the signal source (Rout) will serve as Rin, and we will get converter input current V day off voltage(Fig. 17).

Rice. 17 Circuit 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 input (in this case equal to “virtual zero”), the entire input current ( Iinput) 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 based on it, which have virtually no effect on the current flowing through the measured circuit. Perhaps the only limitation is the permissible range of op-amp input voltages, 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 examined the basic principles of operation of an op-amp in various circuits 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 terminals 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 the operation of their internal circuitry. In some circuits, a non-inverting input was connected to the middle point, 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 nutrition,” if we so stubbornly and with enviable consistency depicted it in the drawings?

It turns out it's simple 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 Output current flow through the load for different op-amp power options

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 kind of input sinusoidal signal is supplied to the input (black sinusoid on the graphs) and the output produces 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 middle point of connection of the power supplies (GB1 and GB2) - Fig. 18, A, the current through the load flows symmetrically relative to the midpoint (red and blue half-waves, respectively), and its amplitude is maximum and the voltage amplitude at Rload. is 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, Rload. and the power supply (red and blue lines showing current flow in the corresponding direction).

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

When powering the 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 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 this is definitely bad for it. However, there are many applications where connecting a load between the op-amp output and one of the power rails (usually negative polarity) is not only acceptable, but also the only possible one.

If you still need to ensure the symmetry of the flow of current through the load with a unipolar supply, then you have to galvanically isolate it from the output of the op-amp using capacitor C1 (Fig. 18, B).

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

Rice. 19 Connecting an input signal source for various op-amp power options

Now we will consider options for connecting input signal sources, excluding load connection from consideration.

Connecting the inverting and non-inverting inputs to the middle point of connection of the power supplies (Fig. 19, A) was considered when analyzing the previously presented circuits. If the non-inverting input does not consume current and simply accepts the midpoint potential, then current flows through the signal source (G) and Rin connected in series, 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 has virtually no effect on the supply voltage.

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

But it is extremely undesirable to connect the signal source (G) to this artificial midpoint because of the same input current. Let's figure it out. Even with divider ratings R1R2 = 10 kOhm and Rin = 10...100 kOhm, the input current Iinput 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 DC signals are amplified with this connection, a separate artificial midpoint potential should be organized for the signal source, formed by resistors R3R4 (Fig. 19, B), or, if the AC signal is amplified, the signal source should be galvanically isolated from the inverting input with capacitor C2 (Fig. 19, B).

It should be noted that in the above circuits (Fig. 18, 19) we have made the default assumption that the output signal must be symmetrical about either the midpoint of the power supplies or an 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 at all necessary that the positive and negative polarities of the power supply be equal in absolute value. One of them may be significantly smaller in absolute value than the other - only such as to ensure the normal functioning of the op-amp.

A natural question arises: “Which one exactly?” To answer this, let’s briefly consider the permissible voltage ranges of the op-amp’s input and output 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 at a load resistance of 10 kOhm is 2 V less than the positive supply rail voltage and 0.5 V less than the negative supply rail voltage. The width of these output voltage “dead zones” that the op amp output cannot reach depends on the series factors such as output stage circuit design, load resistance, etc.). There are op amps that have minimal dead zones, for example, 50 mV before the power rail voltage at a load of 10 kOhm (for OPA340), this feature of the op amp is called “rail-to-rail” (R2R).

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

Let's finally look at and touch these “spider bugs.” You can even sniff and lick it. I allow it. Let's consider their most common options available to beginning radio amateurs. Moreover, if you have to desolder op-amps from old equipment.

Older op-amp designs that necessarily required external circuits for frequency correction in order to prevent self-excitation were characterized by the presence of additional pins. Because of this, some op-amps did not even “fit” into the 8-pin case (Fig. 20, A) and were manufactured in 12-pin round metal-glass ones, for example, K140UD1, K140UD2, K140UD5 (Fig. 20, B) or 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 is translated as “double-pin 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 packages that are quite “exotic” at the moment: a 15-pin rectangular metal-glass one for the hybrid K284UD1 (Fig. 20, D), in which the key is the additional 15th pin from the case, and others. True, I personally have not seen planar 14-pin packages (Fig. 20, D) for placing op-amps in them. They were used for digital microcircuits.

Rice. 20 Cases of domestic operational amplifiers

Modern op-amps mostly contain correction circuits directly on the chip, which makes it possible to make do with a minimum number of pins (for example, the 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 pins) op-amps manufactured on one chip in one package.

Rice. 21 Double-row plastic housings 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 is translated as “single-sided package”.

Rice. 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, surface-mount packages (SMD - Surface Mounting Device) by soldering directly to the board traces had become widespread (Fig. 23). However, for beginners their use presents significant difficulties.

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

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

Rice. 24 Options for placing the same chip in different housings

The pins of all microcircuits are numbered sequentially, counted from the so-called. “key” indicating the location of pin number 1. (Fig. 25). IN any case, if the housing is positioned with leads Push, their numbering is in ascending order against clockwise!

Rice. 25 Operational Amplifier Pinouts
in various housings (pinout), top view;
numbering direction is shown by arrows

In round metal-glass cases, the key has the appearance of a side protrusion (Fig. 25, A, B). With 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 packages, both domestic (302.12) and imported, the key is located between the first and 12th conclusions.

Typically, the inverting input, both in round metal-glass and DIP cases, is connected to the 2nd pin, non-inverting - to the 3rd, output - to the 6th, minus power - to the 4th and plus power - to 7th However, there are exceptions (another possible “rake”!) in the pinout of the OU K140UD8, K574UD1. In them, the pin numbering is shifted by one counterclockwise compared to what is 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).

In recent years, most “domestic use” op amps began to be placed in plastic cases (Fig. 21, 25, B-D). In these cases, the key is either a recess (point) 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, in the middle). The numbering of the pins in these cases is also against clockwise when viewed from above (with conclusions from yourself).

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

In relation to the domestic analogue of LM324 (K1401UD2), there is another “rake”: if in LM324 the plus of the power supply is connected to the 4th pin, and the minus - to the 11th, then in the K1401UD2 it is the other way around: the plus of the power supply is connected to the 11th pin, 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, D), you just need to turn the case 180 degrees so that the 1st pin takes the place of the 8th. That's all.

A few words regarding the labeling 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 the quality group with the first digit of the digital code. For example, op-amps LM158/LM258/LM358, comparators LM193/LM293/LM393, adjustable three-terminal stabilizers TL117/TL217/TL317, 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). Moreover, the price for them may be completely inconsistent with such gradation, or may differ very slightly ( mysterious ways of pricing!). So you can buy them with any marking that is affordable for a beginner, without particularly chasing the first “three”.

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 “ABC” of op-amp operation, covering a little comparators. Next, you need to learn to put words, sentences and entire meaningful “essays” (workable schemes) from these “letters”.

Unfortunately, “It is impossible to embrace the immensity.” If the material presented in this article helped to understand how these “black boxes” work, then further delving into the analysis of their “filling”, the influence of input, output and transient characteristics, is the task of more advanced study. Information about this is presented 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. You just need to not be lazy and read it.

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

Therefore, allow me to take my leave, with respect, etc., author Alexey Sokolyuk ()