Feedback. Negative feedback - Amateur Radio ABC

First of all, let's consider the effect of serial voltage feedback on the voltage gain. For an amplifier covered by feedback,

to uoc = U out /(U in ±U os) = U out /

but the voltage gain of the amplifier without feedback to u = U out / U in, therefore, after carrying out the transformation for OOS, you can write:

Kios =k and/ (1+χk and).

For PIC, the minus sign should be used in the denominator on the right side.

Let's introduce the concept depth of feedbackF. For OOS F = 1+χк u. It follows that the depth of the environmental protection increases with increasing χ And K and, With very deep OOS F = χк u, so in this case we can write

to uoc = 1/χ = (R1+R2)/R2

Conclusion: with deep OOS ( F>10) it is possible to almost completely eliminate the influence of the parameters of the transistor and the entire amplifier on its To iOS. N factors such as temperature changes, radiation exposure, parameter scatter, aging, etc. will not be affected. Thus, it can be argued that the introduction of deep serial voltage feedback ensures the stability of the voltage gain.

Improving gain stability using OOS is widely used to expand the frequency response of an amplifier. With deviation in the LF or HF region, it decreases K and, but the depth of environmental protection also decreases, i.e. 1+ χ K i. As a result To iOS changes slightly and an frequency response with a wide bandwidth is realized.

Input impedance of amplifier with OOS R in.os determined by the way feedback signals are supplied to the input circuit. With sequential voltage feedback R in.os can be represented as R input.os = U input (1+χк u)/I input = R input F.

It follows that the series voltage feedback increases the input impedance of the amplifier by F once.

The output impedance of an amplifier with feedback is determined by the method of removing the feedback signal from the output of the device. With sequential voltage feedback U VCH The amplifier is less dependent on the load current, which corresponds to a decrease in its output resistance. For the type of environmental protection under consideration, we can write

R out.oc = R out /F

It follows that the series voltage feedback reduces the output resistance by F once. Thus, the deeper the OOS, the less R out.os. The above allows us to conclude that serial voltage feedback reduces and stabilizes the voltage gain, reduces both linear and nonlinear distortions, increases the input impedance and reduces the output impedance of the amplifier.

Series amplifier with voltage feedback 100% serial voltage feedback

Serial current feedback

With serial current feedback, a special resistor is switched on in the output circuit of the amplifier,

voltage drop across which

proportional to the output current.

In the amplifier input circuit, this is algebraically added to the input voltage.

.

With deep current OOS, this formula can be converted to the following form:

Consistent feedback in current, as well as in voltage, reduces frequency distortion (expands the frequency response bandwidth) and nonlinear distortion of the amplifier. The introduction of OOS also reduces the influence of interference and interference penetrating the amplifier.

The input impedance of an OOS amplifier is determined by the way signals are supplied to the input circuit

The most significant difference between sequential OOS in voltage and current is manifested through the value R exitOS. The output impedance of an amplifier with feedback is determined by the method of removing the feedback signal from the output of the device. In this case, the method of supplying the OOS signal to the input circuit does not play any role. For R exitOS amplifier covered by current feedback, we can write the following expression:

from which it follows that the output resistance increases. Thus, the considered environmental protection leads to an increase R vxOS, and to a greater extent, the deeper the feedback.

The above allows us to conclude that sequential current feedback stabilizes the voltage gain at a constant load, reduces distortion, and increases the input and output resistance of the amplifier.

Parallel current feedback

With parallel current feedback, a special resistor is switched on in the output circuit of the amplifier R, the voltage drop across which is proportional to the output current. This voltage forms a feedback current in the input circuit, flowing through a special additional resistor R os. Algebraic addition occurs in the input circuit of the amplifier I os and input signal current. The figure shows a block diagram of an amplifier with parallel current feedback. Here, and the current feedback coefficient Depth of current protection

Current gain

where is the current gain without OOS. With deep parallel current feedback

We also note that the introduction of parallel current feedback reduces both linear and nonlinear distortions of current signals.

Since the input impedance of the amplifier in the OOS is determined only by the way the feedback signal is supplied to the input circuit, then for parallel OOS we can write:

Here, in the input circuit of the amplifier, the currents are added algebraically. Thus, parallel OOS reduces R input, and the value R input is inversely proportional to the depth of current feedback.

As shown above, current feedback helps to increase the output impedance of the amplifier. For parallel current feedback R exitOS can be calculated using the following approximate formula:

The above allows us to conclude that parallel current feedback reduces and stabilizes the current gain, reduces distortion of current signals, reduces the input resistance and increases the output resistance of the amplifier.

Parallel OS voltage

With parallel voltage feedback, the output voltage is removed from the load resistance, which in the input circuit forms a feedback current flowing through a special resistor . The figure shows a block diagram of an amplifier with parallel voltage feedback. Although the currents in the input circuit of the amplifier are algebraically added, when analyzing an amplifier with parallel voltage feedback, the voltage feedback coefficient is often used. In this case, it is necessary to take into account the shunting effect of the amplifier input circuit, since in this case Rin. Therefore, it can be represented in the following form:

.

Voltage is released in the input circuit of the amplifier through resistances.

Due to small R in on the internal resistance of the signal source R g a significant share will be lost Eg As a result, a voltage is applied to the amplifier input

.

Voltage gain at deep parallel voltage feedback:

With parallel voltage feedback To iOS stable at Thus, with deep parallel OOS in voltage, it is possible to exclude the influence of external factors on the value K i0 s, reduce linear and nonlinear distortions. However, such an amplifier is completely unsuitable in its properties for the input stage of a multistage amplifier, in particular, due to its high sensitivity to changes R g. Amplifiers with parallel voltage feedback are recommended to be used as intermediate and output stages.

Conclusion: Parallel voltage feedback stabilizes the voltage gain at a constant signal source resistance, reduces distortion, and reduces the input and output resistance of the amplifier.

Operational amplifier: purpose, device, characteristics, types. Circuits of electronic devices based on operational amplifiers: inverting and non-inverting amplifiers, adding and subtracting devices, differentiating and integrating devices, analog signal comparator.

Operational amplifier (op-amp)- amplifiers with galvanic (capacitor-free) couplings, which have a differential input, one output and operate in the presence of a deep OS, which almost completely determines the parameters and characteristics of the devices assembled on them.

Designation:

"-" - inverting input

“+” - non-inverting input

Full designation: In accordance with GOST 2759-82, the designation of analogue equipment elements is based on a rectangle.

Not all op-amps have ground terminals; if it is not needed, then it is not drawn.

F c – terminals for connecting frequency correction circuits.

N c – terminals for connecting initial displacement correction circuits.

Type of op amp.

K140UD1, UD2, UD5, UD7, UD9, UD10, UD11, UD12, UD13, UD14, UD17, UD18, UD20;

K153UD1, UD2, UD3, UD4, UD5, UD6;

K154UD1, UD2, UD3, UD4;

K157UD1, UD2;

554UD1, UD2;

551UD1, UD2;

574UD1, UD2, UD3;

740UD1, UD3, UD4, UD5;

K1401UD1, UD2;

K1407UD1, UD2, UD3, UD4;

The operational amplifier consists of 3 main stages: 1) the differential stage acts as a common-mode signal attenuator; 2) a cascade with a common emitter with a current source in the collector circuit - the main amplifier voltage stage Ku=10 3 ..10 5; 3) push-pull emitter follower in class B mode - designed to match the high input resistance of the current source with a low load resistance, in addition, it provides amplification of the output signal power. In addition, the op-amp may contain an output protection circuit against short circuit, an input overvoltage protection circuit.

According to the types of input stages, op-amps are divided into:

On a BPT - a wide range of applications, good balancing, high input impedance, larger shift and drift;

On the PT – high input impedance, large shift and zero drift compared to the BPT;

On a BPT with ultra-high gain (super β transistors) - they provide an input impedance comparable to a cascade on a PT, the magnitude of shifts and drifts are the same as in conventional BPTs;

With galvanic isolation of the input from the output - modulation or optical methods are used, used in medicine and high voltage technology;

On a varicap - they have a very small input bias current, they are used to amplify the current on photomultipliers.

Op-amp characteristics:

Input voltage

Max diff. input voltage

Max common mode input voltage

Input bias current

Max weekend U and I

Offset Options

- drift (temperature and time)

- frequency

Dynamic

Output voltage slew rate

The most important characteristics of the op amp are the amplitude (transfer) U out =f(U in) and amplitude-frequency (AFC) to U(f). Amplitude-frequency response has the form of a frequency response of a direct current amplifier with the exception of special frequency-dependent devices (selective amplifier, etc.). The transfer characteristics have a linear section for which to U = =const, and nonlinear - to U¢<к U . When implementing specific devices, linear and nonlinear sections are used. Let's look at examples of constructing devices based on op-amps.

Frequency response:

1MHz bandwidth means that

to u f = const.

f gr = 10 6 Hz

Op-amp parameters:

Input

Weekend

Amplifiers

Energy

Drift

Frequency

Express

Op-amp input parameters are input resistance, input bias currents, difference and drift of input bias currents, maximum, input and differential voltages. The presence of input bias currents is determined by the final value of the input resistance of the differential stage, and their difference is determined by the spread of transistor parameters. The input impedance of an op-amp is considered in relation to the input signal. For an ideal op-amp, but in practice it ranges from 300KOhm to 10MOhm, if the differential stage is made on a BPT, and if on a PT, then Mom.

The input voltage supplied to the inputs of the op-amp is limited by the maximum differential input voltage, therefore, to avoid damage to the transistors of the differential stage between the inputs of the op-amp, two cascades or zener diodes are connected back-to-back.

Output parameters of the op-amp are output impedance, maximum output voltage and current. The op-amp must have a low output resistance to ensure high output voltages at low load resistances. The range of real output resistance values ​​ranges from a few to several hundred ohms. The minimum load resistance value is given in the data sheet.

The maximum output voltage is close to the supply voltage .

The maximum output current is limited by the permissible collector current from both power supplies and, accordingly, the total power consumption.

Dynamic parameters of the op-amp are the rate of rise of the output voltage and the settling time of the output voltage. They are determined by the effect of a voltage surge at the input in the area where the output voltage changes from 0.1 to .

Energy parameters of the op-amp are estimated by the maximum current consumption from both power sources and, accordingly, the total power consumption.

Inverting amplifier:

If you use a simple voltage divider in the feedback circuit, you get a basic inverting amplifier circuit.

Inverting input potential U- =0. Since the op amp is in linear mode, then U- - U + = U out /K 0. For example, when U out =5 V, K 0 = 2 10 5 we get U A =25μV. Such a low voltage (it is comparable to thermo-emf at ∆Т=1ºС) cannot even be measured with a conventional digital voltmeter. It follows that the potentials at the outputs of the op-amp can be considered equal with good accuracy. If one of the op-amp inputs is grounded, the second input will also be maintained at zero potential, although the op-amp inputs are not directly galvanically connected. This effect is called imaginary grounding. Thus, from U+=0 should U - =0, Uin = U R 5(voltage drop by R5); Uout = U R 19(voltage drop by R19). Since the input current of the op-amp is very small, it can be neglected, then we get I5 = Uin/R5= -Uout/R19. This means that for an inverting amplifier Ku = Uout/Uin = -R19/R5.

Gain

.

Non-inverting amplifier:

Because U + ≈U -, That Uin = U - = U R 8(voltage drop by R8); Uout = U R 8 + U R 20(voltage drop by R20 And R8). Since the input current of the op-amp is very small, it can be neglected, then we get Ioc = Uin/R8= Uout/(R20+R8). This means that for a non-inverting amplifier Ku = Uout/Uin = 1+R20/R8.

Integrator implements the operation

,

Where t=R 1 C o.s- time constant.

Can serve as a first order low pass filter

Differentiator: performs an operation

.

U out = -I os R os

I os = C dU s /dt

U c = U in

U out = -R os C dU in /dt

Used to isolate the leading and trailing edges of the signal, as well as as a first-order high-pass filter.

Inverting and non-inverting adders:

The action of this scheme exactly corresponds to its name. An inverting adder forms an algebraic sum of several voltages and reverses its sign.

If individual input voltages need to be given different weights, then a summation circuit with scale factors is used. Used for summing signals for a digital-to-analog converter. There is no mutual influence of signal sources in the adder.

For an inverting adder, the output voltage is determined by the formula

With equal input resistances R 1 =R 2 =R

U out =- (U in.1 +U in.2 +...+U in.n)- for an inverting adder;

For a non-inverting adder.

In the adder circuit, the variable parameter is the feedback resistance R os, which determines the gain. Formulas are given for constant quantities (numerical adder) U in.1, U in.2 etc.

Subtractor:

The conditions that must be met for the correct operation of this circuit are that the sum of the gains of the inverting part of the circuit is equal to the sum of the gains of its non-inverting part . In other words, the inverting and non-inverting gains must be balanced.

For the circuit shown in the figure, the output voltage is proportional to the difference in voltage at the inputs Uin1 and Uin2.

With R9=R11=R10=R21, we get

Used in differential measuring circuits.

Comparator– device for comparing two signals. The comparator jumps the output level when the continuously time-varying output signal goes above or below a certain level.

Comparators are digital and analog (compares voltages)

Diodes are used to protect the op-amp inputs from voltage overload. At U = 100V the diodes do not open.

Often at one input of the comparator there is a fixed Uin. The comparator compares the input voltages and amplifies their difference with K and = 10 4 -10 5 . Those. at the slightest excess of one signal over another, we get a max signal of positive or negative polarity at the output. Due to its high gain, the circuit switches at a very small voltage difference, making it suitable for comparing two voltages with high accuracy.

The operation of the comparator when comparing two voltages is illustrated by the diagram:

In order to increase performance, additional forcing Re chains are introduced into specialized comparators (CA), which can lead to nonlinearity in the operation of the op-amp, which is not significant for the comparator. Those. The op-amp can work as a comparator.

Disadvantage of the comparator: insufficiently clear response with slowly changing and protected input signals.

Operational amplifiers are one of the main components in modern analog electronic devices. Thanks to the simplicity of calculations and excellent parameters, operational amplifiers are easy to use. They are also called differential amplifiers because they are capable of amplifying the difference in input voltages.

The use of operational amplifiers in audio technology is especially popular to enhance the sound of music speakers.

Designation on diagrams

There are usually five pins coming out of the amplifier case, of which two pins are inputs, one is output, and the remaining two are power supply.

Operating principle

There are two rules to help you understand the operating principle of an operational amplifier:

1. The output of the operational amplifier tends to zero voltage difference across the inputs.
2. The amplifier inputs do not consume current.

The first input is designated “+” and is called non-inverting. The second input is marked with a “–” sign and is considered inverting.

The amplifier's inputs have a high resistance called impedance. This allows current consumption at the inputs of several nanoamps. At the input, the voltage value is assessed. Depending on this assessment, the amplifier outputs an amplified signal.

The gain factor is of great importance, sometimes reaching a million. This means that if at least 1 millivolt is applied to the input, then the output voltage will be equal to the voltage of the amplifier's power supply. Therefore, opamps are not used without feedback.

The amplifier inputs operate on the following principle: if the voltage at the non-inverting input is higher than the voltage at the inverting input, then the output will have the highest positive voltage. In the opposite situation, the output will have the largest negative value.

Negative and positive voltage at the output of the operational amplifier is possible due to the use of a power supply that has a split bipolar voltage.

Op amp power

If you take a AA battery, it has two poles: positive and negative. If the negative pole is considered the zero reference point, then the positive pole will show +1.5 V. This can be seen from the connected one.

Take two elements and connect them in series, you get the following picture.

If the negative pole of the lower battery is taken as the zero point, and the voltage is measured at the positive pole of the upper battery, the device will show +10 volts.

If we take the midpoint between the batteries as zero, then we get a bipolar voltage source, since there is a voltage of positive and negative polarity, equal to +5 volts and -5 volts, respectively.

There are simple split-power supply circuits used in amateur radio designs.

Power to the circuit is supplied from a household network. The transformer reduces the current to 30 volts. The secondary winding in the middle has a tap, with the help of which the output is +15 V and -15 V rectified voltage.

Varieties

There are several different op amp designs that are worth looking at in detail.

Inverting amplifier

This is the main scheme. The peculiarity of this circuit is that the opamps are characterized, in addition to amplification, by a phase change. The letter "k" represents the gain parameter. The graph shows the effect of the amplifier in this circuit.

Blue color represents the input signal and red color represents the output signal. The gain in this case is equal to: k = 2. The amplitude of the output signal is 2 times greater than the input signal. The amplifier's output is inverted, hence its name. Inverting op-amps have a simple circuit:

These op amps have become popular due to their simple design. To calculate the gain, use the formula:

This shows that the gain of the op-amp does not depend on the resistance R3, so you can do without it. Here it is used for protection.

Non-inverting operational amplifiers

This circuit is similar to the previous one, the difference is the absence of inversion (inversion) of the signal. This means maintaining the phase of the signal. The graph shows an amplified signal.

The gain of the non-inverting amplifier is also equal to: k = 2. A signal in the form of a sinusoid is supplied to the input; only its amplitude has changed at the output.

This circuit is no less simple than the previous one; it has two resistances. At the input, the signal is applied to the positive terminal. To calculate the gain you need to use the formula:

It shows that the gain is never less than unity, since the signal is not suppressed.

Subtraction circuit

This circuit makes it possible to create a difference between two input signals, which can be amplified. The graph shows the operating principle of a differential circuit.

This amplifier circuit is also called a subtraction circuit.

It has a more complex design, in contrast to the previously discussed schemes. To calculate the output voltage use the formula:

The left side of the expression (R3/R1) determines the gain, and the right side (Ua – Ub) is the voltage difference.

Addition circuit

This circuit is called an integrated amplifier. It is the opposite of the subtraction scheme. Its special feature is the ability to process more than two signals. All sound mixers operate on this principle.

This diagram shows the ability to sum multiple signals. To calculate the voltage, the formula is used:

Integrator circuit

If you add a feedback capacitor to the circuit, you get an integrator. This is another device that uses operational amplifiers.

The integrator circuit is similar to an inverting amplifier, with capacitance added to the feedback. This leads to the dependence of the system operation on the frequency of the input signal.

The integrator is characterized by an interesting feature of the transition between signals: first, the rectangular signal is converted into a triangular one, then it turns into a sinusoidal one. The gain is calculated using the formula:

In this formula the variable ω = 2 π f increases with increasing frequency, therefore, the higher the frequency, the lower the gain. Therefore, the integrator can act as an active low-pass filter.

Differentiator circuit

In this scheme the opposite situation occurs. A capacitance is connected at the input, and a resistance is connected in the feedback.

Judging by the name of the circuit, its operating principle lies in the difference. The faster the signal changes, the higher the gain. This option allows you to create active filters for high frequencies.

The gain for the differentiator is calculated using the formula:

This expression is the inverse of the integrator expression. The gain increases in a negative direction as the frequency increases.

Analog comparator

A comparator device compares two voltage values ​​and drives the signal to a low or high output value, depending on the state of the voltage. This system includes digital and analog electronics.

A special feature of this system is the absence of feedback in the main version. This means that the loop resistance is very high.

A signal is supplied to the positive input, and the main voltage, which is set by a potentiometer, is supplied to the negative input. Due to the lack of feedback, the gain tends to infinity.

When the voltage at the input exceeds the value of the main reference voltage, the output receives the highest voltage, which is equal to the positive supply voltage. If the input voltage is less than the reference voltage, then the output value will be a negative voltage equal to the voltage of the power source.

There is a significant flaw in the analog comparator circuit. When the voltage values ​​at the two inputs approach each other, the output voltage may change frequently, which usually leads to skips and malfunctions in the relay. This may cause equipment malfunction. To solve this problem, a circuit with hysteresis is used.

Analog comparator with hysteresis

The figure shows the operation diagram of circuit c, which is similar to the previous circuit. The difference is that turning off and turning on does not occur at the same voltage.

The direction of the arrows on the graph indicates the direction in which the hysteresis moves. When examining the graph from left to right, it can be seen that the transition to a lower level occurs at voltage Uph, and moving from right to left, the output voltage will reach the highest level at voltage Upl.

This principle of operation leads to the fact that at equal values ​​of input voltages, the state at the output does not change, since a change requires a voltage difference of a significant amount.

This operation of the circuit leads to some inertia of the system, but it is safer, in contrast to a circuit without hysteresis. Typically, this principle of operation is used in heating devices with a thermostat: stoves, irons, etc. The figure shows an amplifier circuit with hysteresis.

Voltages are calculated according to the following dependencies:

Voltage repeaters

Operational amplifiers are often used in voltage follower circuits. The main feature of these devices is that they do not amplify or attenuate the signal, that is, the gain in this case is equal to unity. This feature is due to the fact that the feedback loop has a resistance equal to zero.

Such voltage follower systems are most often used as a buffer to increase the load current and device performance. Since the input current is close to zero, and the output current depends on the type of amplifier, it is possible to unload weak signal sources, for example, some sensors.

Feedback can be positive (POS) or negative (NOS). Positive feedback increases and turns it into Schmitt (see Volume I). reduces the gain and transfers the op-amp to linear mode - into an analog signal.

Rice. 1.27. Amplifiers with OOS: a - non-inverting; b - inverting; c - with isolating capacitors; d - connection of a low-impedance load to an amplifier with bipolar power supply; d - repeater; e - with high input resistance; g, h - powerful op-amp with sequential connection of powerful transistors

Rice. 1.27. Amplifiers with OOS:

and - a powerful op-amp with parallel connection of powerful transistors; k - powerful op-amp with high-voltage power supply; l - these resistors (usually R1 = R2 = 10...1000 kOhm) the voltage at the midpoint is equal to half the supply voltage. C1 is a filter, it is needed to reduce the output divider (i.e., to improve the operation of the amplifier). Its capacity is determined from the expression Cl x R = 0.5...2, where C1 is the capacitance of capacitor C1 in microfarads, and R is one of the resistors R1 or R2 (R1 = R2) in megaohms.

The advantage of an op-amp repeater is its huge input, i.e. it consumes virtually no current from the signal source. The input output of op-amp repeaters with bipolar transistors at the input reaches several to hundreds of gigaohms, i.e., approximately the same as with poor insulation. The input at the input is almost impossible to measure - it achieves good isolation and is thousands to millions of times greater than that of a bipolar transistor op amp. The disadvantage of repeaters based on widespread and cheap op-amps is their significant input capacitance. It usually does not exceed 5 pF, but at a frequency of 100 kHz (and, accordingly, the input follower) of such a capacitor is 400 kOhm. In high-frequency and some precision, as well as special buffer and measuring op-amps, the input capacitance is thousands of times smaller.

Resistor R balances the op-amp - with its help, at zero voltage at the input, the pointer of device P1 is set to the zero position. When using precision op-amps in a circuit, this is optional.

A supply voltage divider is assembled on elements Rl, R2, C1; the voltage at the connection point of these elements is equal to half the supply voltage. C1 is optional, it has virtually no effect on operation. But it is not advisable to remove it - self-excitation may occur at high frequencies.

When the input voltage changes, the current flowing through the R fl0I] -P1 chain changes, because of this the voltage at the midpoint of the voltage divider changes. But since the signal (voltage) to the input is supplied precisely relative to this very middle point (one of the inputs is connected to it), this will not affect the accuracy of voltage measurement: for example, with an input voltage of 1.0 V, the voltage on the circuit R non -Pl will be equal to 1.0 V, regardless of the voltage at the midpoint. But this is only if the supply voltage is controlled not from the supply voltage, i.e. it must be powered from its own battery (battery).

Let's now apply a positive half-wave of the input signal to the op-amp input. A positive voltage will appear at the output of the op-amp - it will decrease to zero only when the voltage at the inverse input of the op-amp becomes equal in value to the voltage at the forward one (this is the principle of operation of the op-amp). In order for the output voltage to increase, the inside of the op-amp (see Fig. 1.25, a) must open, connected to the “+U” terminal (VT1 in Fig. 1.25, a). That is, in this case, pins 2 and 6 of the op-amp are closed, and the current flows through the series-connected R4 and R6, and the voltage drop across resistor R4 increases (and across resistor R5 it decreases; but not more than 2...3 times). At some point in time, the voltage drop across resistor R4 increases to a value at which VT1 begins to open - it seems to “help” the op-amp to increase the voltage at the output of the amplifier, while a current of approximately b 2| e times more than through an op-amp - that is, this ensures that the output current of a relatively low-power op-amp is amplified by powerful transistors.

As soon as the voltage at both inputs of the op-amp, due to the OOS, is equal, the voltage at the output of the amplifier (at the load) will be fixed and stop changing. In this case, some current will flow through the load and, accordingly, a current will also flow through R4 and R6, approximately h 2 l 3 times less than the output. VT1 will be slightly open, and with the slightest increase/decrease in the voltage at the input, the current flowing through it will also increase/decrease. If the input voltage sharply decreases (which is typical for an audio signal), the output voltage may not keep up with the input - in this case, the voltage drop across resistor R5 will increase (since the op-amp always tries its best to equalize the voltages on the direct and inverse inputs) and VT2 will “help” the output voltage decrease.

The current gain of transistors VT1 and VT2 in this circuit does not exceed 5... 10 times - due to the shunting of the base junction with resistors R4 and R5. If a higher gain is needed, it is advisable to replace VT1 and VT2 with composite ones (as in Fig. 1.27, h), but at the same time it is impossible to remove or even change the value of resistors R4 and R5 (why, see above). In this case, even when using low-power op-amps, you can get a significant current at the output.

At one time I made a lot of amplifiers according to the circuit in Fig. 1.27, and therefore I want to give some useful tips:

Resistor R6 can be short-circuited - especially if single power ones are used (I prefer compound ones). This will improve the performance of the amplifier, but will increase the heating of the case - a radiator is required. If you do not have a special radiator, then you can simply glue several metal plates on top of the microcircuit - the larger their area, the better.

The amplifier develops power up to 70 W. You can check whether your op-amp can operate at such a high voltage as follows: the bases of both transistors, as well as the load, are not connected to the circuit, everything else is assembled according to Fig. 1.27, and the supply voltage is turned on (via R4 and R5!) and the voltage drop across R4, R5 is measured. If it does not exceed 0.3 V, everything is fine.

At a supply voltage of ±28 V or more, all op-amps burned out. If you need the op-amp to operate at such a high voltage, power must be supplied to it through (Fig. 1.27, k; the correction and feedback circuits are connected in the same way as in Fig. 1.27, i). The stabilization voltage for both zener diodes must be the same and such that the supply voltage of the op-amp does not exceed 25 V (for example, U ni)T = ±32 V, U CT = 32 – 25 = 7 V).

As you know, the voltage drop across a zener diode is very weak due to the current flowing through it. It is precisely because of this effect that they are suitable for use in such an amplifier: thanks to them, the supply voltage of the op-amp is limited to a level that is safe for it, and those connected according to the circuit are fully open even when the voltage drop across the base resistor is only 1...2 V - this is much less voltage op-amp supply.

Disadvantages of such an amplifier:

1. All of them, even the best ones, are very noisy (that is, their stabilization voltage, with a constant flowing current, fluctuates chaotically around some “average” level), therefore the assembled one makes more noise than the same one, but without zener diodes (and with lower supply voltage). To combat noise, you can turn on electrolytic capacitances of several units...tens in parallel with the zener diodes, but because of these capacitors, immediately after turning on the supply voltage, the voltage at the power supply terminals of the op-amp will increase to dangerous values ​​(a discharged capacitor is close to zero) and the op-amp may fail .

2. Since the voltage at the collectors of the transistors may be greater than the voltage at the output of the op-amp, the latter may fail (current flows through R6). You can insure yourself against this problem if the terminal of resistor R6, which is right according to the diagram, is connected not to the collectors of the transistors, but to the common wire. In this case, the resistance is about 100 ohms.

3. The stabilization voltage of the zener diodes should not exceed 10 V: the higher it is, the greater the chances that at one point your amplifier will spontaneously fail.

The amplitude of the output voltage in this case does not depend on the supply voltage of the op-amp and, just like in Fig. 1.27, and can reach U nHT – 0.7 V (modulo). When using composite transistors, it is approximately 0.5 V less.

In these circuits (Fig. 1.27, i; 1.27, j) it can also be used with an insulated gate (VT1 - p-channel, VT2 - η-channel). Such an amplifier does not have any noticeable advantages compared to an amplifier based on bipolar transistors. It is more difficult to configure, so I will not provide its diagram here. If you do not have sufficient experience with field-effect transistors and amplifiers based on them, do not try to draw its circuit yourself.

Capacitor C2 is charged through series-connected R3, R4. As soon as the voltages at both inputs are equal, the voltage at the op-amp output will decrease to the voltage at the direct input (i.e., to half the supply voltage) and, in the absence of an input signal, will be maintained at this level. When a high-frequency signal is applied to the input of the amplifier, the output voltage will also change; but, since the capacitance of capacitor C2 is quite large, the voltage on its plates will not have time to change significantly during one half-cycle of the amplified signal, so we can assume that the voltage at the left terminal of resistor R3 according to the diagram is constant and equal to half the supply voltage. In this case, the voltage gain of the amplifier is equal to the ratio of resistor R4 to the resistance of resistor R3.

When using a capacitor C2 of too small a capacity, the gain of the amplifier at low frequencies will be less than at high frequencies, and in extreme cases (the capacitance of the capacitor C2 is zero, that is, there is no capacitor at all), the gain is equal to one (this turns into, similar to that shown in Fig. 1.27, d). This is due to the fact that in this case, the voltage at the terminals of the capacitor will fluctuate within significant limits when the output signal changes, which is why the low-frequency component of the signal will be “smoothed out”.

For example in Fig. Figure 1.31 shows graphs of the input (Fig. 1.31, a) and output (Fig. 1.31,6) signal of such an “amplifier”. In Fig. 1.31, but the low-frequency component of the input signal is clearly visible, and when such a signal is amplified by the “correct” amplifier, it will be just as clearly audible. But if the amplifier according to the circuit in Fig. 1.27, l capacity of capacitor C2 is too

Rice. 1.31. Explanations for Fig. 1.27, l. Input signal (a) and output signal (b) with too small capacitance C2. If the capacitance C2, as well as C1 and S3, is significant, the shape of the output signal repeats the shape of the input signal, then the low-frequency component (the “bass” so beloved by today’s music lovers) will be weakened so much that it will become completely unnoticeable, and it will be very difficult to restore it. In Fig. 1.31, and the dotted line conventionally shows the change in voltage across the capacitor. As you can see, the larger its capacity, the better. But you shouldn’t strive for the ideal, and a capacitance of 47 μF, with an OOS resistor resistance of 100 kOhm, is quite sufficient.

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 an 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 accept 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.

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 packages (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 packages, 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.

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