Op-amp microcircuits in the vast majority of cases are used with feedback circuits. Most often, negative feedback is used, but in some cases positive feedback is also used.
2.2.1. Op-amps with negative feedback circuits
As shown in Section 2.1, negative feedback coverage of the amplifier can significantly increase the stability of the gain. This is all the more important, since the open-loop gain of the op-amp can vary over a wide range with temperature. For example, the gain of the op-amp MC1556G (Motorola) at t = 25 ◦ C is equal to at t = -50 ◦ C, its value decreases to 
, and at t = 100 ◦ C, it increases to 
. The use of negative feedback can significantly reduce the instability of the gain. Indeed, if an amplifier with a gain equal to 100 (at t = 25 ◦ C) is needed, then using the MC1556G microcircuit from formula (2.1) we obtain:
.
From here 
.
Substituting into (2.1) the value of K at t = 100 ◦ C, equal to 
, we get:
.
Thus, the gain with the introduction of negative feedback due to a change in the gain without feedback, it changed by only 0.02059%.
A result very close to this can be obtained if we take into account that the instability of the feedback amplifier gain, due to the instability of the open-loop gain, decreases by a factor of the feedback. The relative instability of the open-loop gain can be determined by the formula:
or 
%.
The feedback factor F must be calculated, given that it increased with increasing temperature, but assuming that K os practically does not change and remains equal to 100.
From here 
or 
.
Considering that 
, we get 
V %: 
%.
Compared with the previous result 
%, we note that the error obtained when calculating using approximate formulas using the second method differs from the error obtained using exact formulas by only 0.0087421%. So the instability 
can be calculated using approximate formulas, taking into account that 
.
The large value of the feedback factor obtained by using op amp chips allows us to use a simplified formula for calculating 
. Indeed, if in the formula 
take into account that 
And 
, then we get 
.
Thus, the gain of a negative feedback amplifier is only determined by the feedback circuit. Equity conditions 
becomes obvious if we consider the operational amplifier to be ideal ( 
). It is this circumstance that made it possible to use the “imaginary earth” principle in the calculation of amplifiers based on op amps.
We illustrate the use of this principle on the basis of an amplifier - an inverter, made on an op-amp chip (Fig. 2.5).
Rice. 2.5. The scheme of the amplifier-inverter on the op-amp
In the above diagram, the op-amp is covered by a negative parallel voltage feedback. The connection is negative because if Ug is positive, then Uout is negative, which reduces the voltage at the input of the op-amp. The connection is parallel, since with respect to the inputs of the op-amp, the generator Ug and the feedback signal are connected in parallel. Voltage feedback, as the feedback signal is proportional to the output voltage.
Considering that U OUT = K 
U IN, and U OUT cannot be more than +E and less than -E, its value can be considered finite. Hence for the ideal op amp ( 
), we get 
, i.e. the potential at the inverting input is 0. That is why the word “ground” is used in the name of the principle. (In the vast majority of electronic devices there is a common bus, which is usually really connected to the ground). However, the "ground" at the inverting input of the op-amp is "imaginary" or "virtual". Indeed, if the voltage generator Ug, one terminal of which is grounded, is connected to some kind of resistor with resistance R, and the second terminal of the resistor is connected to the "ground", then a current will flow from the generator through the resistor R to the "ground" 
. In our case (Fig. 2.5), a current I equal to 
, but this current will not flow into the "ground" (or common bus), but into the feedback circuit of the op-amp. From here 
And 
.
It should be noted that the current I or any part of it cannot go to the input of the op-amp, because. given that the ideal op amp has 
, this would create a voltage at the input of the op-amp equal to 
.
It is possible to formulate the conditions under which the principle of “imaginary earth” can be used:
OU is ideal;
OU covered by negative feedback;
The op-amp does not leave the linear mode, i.e. its amplitude characteristic can be considered linear.
Consider other op-amp circuits with negative feedback using the “imaginary earth” principle.
Inverting amplifier-adder.
The scheme of the inverting amplifier-adder on the op-amp is shown in fig. 2.6.
The principle of operation of the circuit is similar to the principle of operation of an inverting amplifier on an op-amp. In this case, the current in the feedback circuit I 0 is the sum of the currents from the input voltage generators U 1 , U 2 ,…, U n:
.
Rice. 2.6. Scheme of an inverting amplifier-adder on an op-amp
In turn, each of these currents, according to the “imaginary earth” principle (Uin = 0), is equal to:
…,
Let's assume that 
. In this case 
.
Current 
, flowing through the resistance 
, creates a voltage Uout:
.
Thus, at 
the circuit not only sums and inverts the signals, but also amplifies them. In this case, the voltages U 1, U 2, ..., U n can be not only positive, but also negative.
non-inverting amplifier.
The diagram of a non-inverting amplifier on an op-amp is shown in fig. 2.7. In the circuit, the op amp is covered by negative serial voltage feedback.
Rice. 2.7. Op-amp non-inverting amplifier circuit
The connection is negative because if the generator voltage is positive and is applied to the non-inverting input, then the output voltage will also be positive, but through the feedback circuit R 2 ̶ R 1 it is applied to the inverting input, reducing the voltage acting between the inputs of the op-amp. The connection is serial, because the voltage generator and the feedback signal are connected to the inputs of the op-amp in series. Voltage feedback, as the feedback signal is proportional to the output voltage.
In the above diagram, the voltage between the inputs of the op-amp, according to the “imaginary earth” principle, must also be equal to zero: Uin = 0. It follows that the voltage at the inverting input, as well as the voltage at the non-inverting input, is equal to Ug. Therefore, according to the resistance 
current flows I = 
This current I flows through the feedback resistance 
and creates a voltage at the output U OUT.
Thus, the current I can be expressed in terms of Ug and Uout:
.
Hence the feedback gain 
equals:
.
If we put in the scheme (Fig. 2.7) 
, A 
, That 
. Such a circuit is called a voltage follower. The scheme is shown in fig. 2.8.
Fig.2.8. Voltage follower circuit
Since everything output voltage applied to the input of the op-amp, the connection is considered 100 percent.
Non-inverting amplifier-adder.
The diagram of a non-inverting amplifier-adder on an op-amp is shown in fig. 2.9.
Rice. 2.9. Diagram of a non-inverting adder amplifier on an op-amp
To determine the voltage at the output of the circuit Uout, it is first necessary to determine the voltage at the non-inverting input U 0. At the same time, we take into account that the current cannot flow into the op-amp through the non-inverting input ( 
). Therefore, according to Kirchhoff's first law, we can write: 
.
Each of the currents from voltage sources U 1, U 2, ..., U n can be expressed by the following formulas:
…,
Thus
.
Let's assume that 
. From here
And 
.
As a result, given that the voltage 
amplified by a non-inverting amplifier in ( 
) time, we get:
.
Differential amplifier.
A differential amplifier is an amplifier that amplifies the difference between input signals.
The scheme of the differential amplifier on the op-amp is shown in fig. 2.10. To determine U OUT, it is advisable to use the overlay method (superposition method).
Rice. 2.10. Op-amp differential amplifier circuit
The method of superpositions (superpositions) is that the voltage or current in any part of a linear circuit containing voltage and current sources can be found by determining the required voltage or current from one of the voltage or current sources. In this case, all other voltage sources are closed, and current sources are excluded from the circuit. This determines the voltage or current from each source. Then the results are summarized.
Find the output voltage in the differential amplifier circuit by first considering the voltage 
and then voltage 
. In the first case, closing 
, we obtain the diagram in Fig. 2.11a. Considering that the input current of an ideal op-amp is zero, we can assume that the voltage 
at the non-inverting input is zero. Therefore, the voltage at the inverting input is also zero. This means that the one shown in Fig. 2.11a, the circuit is equivalent to the circuit of an inverting amplifier, i.e. 
.
In the second case, closing 
, we obtain the circuit shown in Fig. 2.11b. For determining 
it is necessary to determine 
. Considering that the input impedance of the op-amp is equal to 
, we get
.
Voltage 
amplified by a non-inverting amplifier by (1+m) times. From here
Rice. 2.11. Circuits for determining the output voltage of a differential amplifier using the overlay method
Thus, , i.e. at the output of the amplifier, we get the difference of input voltages amplified by m times 
And 
.
Current to voltage converter.
If the current of the current source needs to be converted into voltage, then this current can be passed through the resistance and voltage is obtained (the circuit is shown in Fig. 2.12). However, this technical solution In a number of cases, it turns out to be unacceptable for the following reasons:
Rice. 2.12. Scheme of the simplest current-to-voltage converter
The output impedance of such a voltage source, converted from a current source, turns out to be excessively large, since 
and at small I to obtain the necessary 
you should choose a large R. Therefore, if necessary, a possible further strengthening 
, an amplifier with a very high input impedance is required;
Voltage 
, arising on the resistance R, can interfere with the normal functioning of the current source I, if any sensor is used as such a source. Indeed, when the current I changes, the voltage at the sensor output will change, which can lead to a nonlinear dependence of the sensor current I on some physical parameter.
In order to eliminate these unfavorable factors, you can use the converter circuit made on the OS (Fig. 2.13)
In this case, the op-amp is covered by negative parallel 100% voltage feedback.
Taking into account the “imaginary earth” principle, the output voltage 
will be equal to: 
, i.e. 
turns out to be equal in absolute value to the output voltage in the circuit in Fig. 2.12. However, in the above circuit on the op-amp, the voltage at the output terminals of the current source will always be zero, the output resistance of the voltage source due to negative voltage feedback will also be close to zero.
Rice. 2.13. Scheme of the current-voltage converter, made on the op-amp
In addition, parallel negative feedback reduces the input impedance by a factor of (1+K), where K is the gain of the open-loop amplifier. Indeed, taking into account the gain, we obtain the following equations:
, I OS = 
.
Given that 
, we get
.
Since the input impedance of the op-amp is quite large, then with great confidence we can assume that R BX ∙ (1 + K)> ˃ R, i.e.
.
Voltage to current converter.
It is necessary to carry out such a conversion in which the current would not depend on the load resistance (at simple connection to voltage source 
load resistance current will be equal to 
, i.e. will depend on the load resistance). With the help of an op-amp, you can make sure that the current does not depend on the load resistance. The scheme of the voltage-to-current converter is shown in fig. 2.14.
Rice. 2.14. Voltage to current converter circuit
According to the “imaginary earth” principle, given that the amplifier is covered by negative feedback, 
. From here, the current in the resistance R will be equal to 
. This current cannot flow from the op-amp's inverting input, but will flow from the op-amp's output. At the same time, from the resistance 
it will not be affected if the voltage 
will not go beyond the linear amplitude characteristic of the op-amp.
Voltage stabilizer based on OU.
You can stabilize the voltage using reference diodes (zener diodes). However, the stabilizer circuit on a zener diode and one resistor, shown in Fig. 2.15., has a number of significant drawbacks:
Rice. 2.15. Voltage stabilizer circuit on a zener diode
These shortcomings of the voltage regulator can be corrected by using an operational amplifier. A voltage regulator circuit using an op amp as a regulator amplifier is shown in fig. 2.16.
Rice. 2.16. Voltage stabilizer using
amplifier-regulator on the op-amp
The voltage from the zener diode is applied to the non-inverting input of the amplifier with adjustable using a variable resistor 
gain: 
.
The above scheme has the following advantages:
logarithmic amplifier.
The scheme of the logarithmic amplifier on the op-amp is shown in fig. 2.17.
Rice. 2.17. Op-Amp Logarithm Amplifier Circuit
The operational amplifier is covered by negative feedback and therefore the “imaginary earth” principle can be used, i.e. 
. Hence 
.
Current I flows through the diode, and the p-n junction is forward biased. diode current 
is defined by the following formula:
,
Where 
- current of minority carriers, 
- diode voltage, 
- temperature potential, k - Boltzmann's constant, Т - absolute temperature, q - electron charge.
For t = 20 ◦ C, we can assume that 
. Provided U D >> 
T , i.e. U D >>25mB formula for 
simplified:
or 
.
Logarithmizing, we get 
.
Diode voltage 
equal to the voltage at the output of the op-amp with a minus sign: 
. Thus, we get 
Anti-logarithmic amplifier.
The anti-log amplifier circuit can be obtained from the logarithm amplifier circuit by swapping the resistor and diode. The scheme of the antilogarithmic amplifier is shown in fig. 2.18.
We use the principle of "imaginary earth". We get that 
And 
. As for the logarithmic amplifier, we assume that 
.
Rice. 2.18. Anti-log amplifier circuit
Hence, 
. The diode current, flowing through the feedback resistance R, creates a negative voltage at the output of the op-amp 
, i.e. 
.
Given that U d \u003d U g, we get U OUT \u003d I 0 ∙R∙ 
.
OS integrator.
Op-amp integrator circuit that integrates the input voltage over time 
, shown in fig. 2.19.
Rice. 2.19. Op-amp integrator circuit
In the circuit, the op amp is covered by negative feedback only for alternating current. For this reason, in the real case, i.e. when using any op amp chips with 
at the output of the op-amp, a voltage is set that is close to either 
, or 
. Therefore, in practice, measures must be taken to keep the OS in a linear mode. This can be done either by introducing additional negative DC feedback, for example, by shunting a capacitor C with a high resistance resistor. 
, or by periodically setting the output voltage to zero, for example, periodically shorting the capacitor electronic key Cl.
Consider the work of the integrator, assuming that the op-amp is ideal and operates in a linear mode. In this case, according to the “imaginary earth” principle, the voltage at the inverting input is zero ( 
). For this reason, the output voltage is equal to the voltage across the capacitor. On the other hand, the voltage across the capacitor 
equals the charge on the capacitor divided by the capacitance of the capacitor 
. And the charge on the capacitor is equal to the time integral of the current 
going to charge the capacitor. Thus 
And 
.
Considering that the current 
, we get
U OUT = 
U C = 
.
Provided that 
, we have 
, i.e. voltage that changes linearly with time.
Let us assume that a sinusoidal voltage is applied to the input of the circuit 
. In this case, you can find the output voltage by taking the integral of 
. However, it can be made simpler, assuming that the AC integrator is an amplifier-inverter, in the feedback circuit of which a capacitor with AC resistance is included 
equal to 
. From here 
.
OA Differentiator.
The circuit of the differentiator on the op amp, which produces an output voltage proportional to the time derivative of the input voltage, is shown in fig. 2.20.
Rice. 2.20. Op-amp differentiator circuit
The circuit is covered by 100 percent negative feedback. For this reason, the op-amp in the circuit will always be in linear mode, i.e. when calculating 
you can use the principle of "imaginary earth" ( 
). From here we get U OUT = 
I C ∙R.. It is known that the current 
, going to charge the capacitor, is equal to 
where is the voltage across the capacitor 
equals 
. Therefore, we get
.
When applied to the input of a voltage differentiator 
as well as in the case of an integrator, you can consider the circuit as an amplifier-inverter with a capacitor at the input connected instead of a resistor. Wherein 
.
SelectiveRC- op-amp amplifier.
The selective amplifier is designed to amplify input signal at one frequency and suppressing signals at all other frequencies. The frequency response of the selective amplifier is shown in fig. 2.21.
Rice. 2.21. Frequency response of a selective amplifier on an op-amp
The selective amplifier is characterized by the following parameters: the resonance frequency at which the gain reaches its maximum value - 
, the gain at the resonant frequency - 
, the quality factor, defined as the ratio of the resonant frequency to the frequency difference ∆ω, at which the modulus of the gain at the resonant frequency decreases by 
once.
A diagram of a selective RC amplifier with a frequency-dependent negative feedback circuit is shown in fig. 2.22.
Rice. 2.22. Selective RC amplifier circuit
The circuit has 100% negative DC feedback through a resistor. 
. Therefore, the op-amp will always be in linear mode and the “imaginary earth” principle can be used ( 
). Assume that the input voltage is positive. Then the currents flowing through individual sections of the feedback circuit: 
,
,I 2 have the direction shown in Fig. 2.22. Let's take into account that I=I 1 +I 2 . We denote the resistance of capacitors C 1 and C 2 to a sinusoidal current Z 1 and Z 2, and we assume that Z 1 \u003d Z 2 \u003d Z.
Let's express the currents 
,
And 
through voltage U g, U 1 and U out:
;
;
.
From the equality of currents in the node with voltages U 1, we get:
.
Hence, given that 
we get:
The gain modulus is:
.
Differentiating with respect to ω and equating the derivative to zero,
it can be shown that 
has a maximum at the frequency 
:
.
.
Assuming that at frequencies 
And 
decreases in 
times, we get the equation
.
Solving this equation, we find:
,
..
From here 
And 
. Note that when 
And 
=0.
The main parameters of a selective amplifier can also be determined using simpler formulas known from the theory of active filters. To do this, in the formula K OS we will make the replacement 
to operator p. Get
Where 
,
,
,
.
It is known from filter theory that
,
,
= 
.
From here 
,
,
.
If we use obtained from the operator expression 
formulas, it is obvious that it is much easier to determine the main parameters of a selective amplifier than using the symbolic method.
Most of the citizens of this virtual city came here with the desire to make a good amplifier.
Some will say it's better to do tube amplifier... But this is not the simplest solution. We need rather scarce spare parts - lamps, an output transformer ...
Others will answer them: "Why lamps? Microcircuit or transistor amplifiers are much more compact and powerful! Well, even if their sound is not so good ..."
And everyone will be right. This is a matter of taste and possibilities for everyone.
It is for the second category of citizens that I decided to write this article;)
In this diagram you see the simplest circuit turning on the power amplifier, which is used in the vast majority of modern amplifiers.
The sound is booming, blurry and unpleasant. In particular, when using mass-produced Chinese spare parts.
But I can assure you that even without major modifications, you can make this circuit sound!
I'll start with a small lyrical digression.
I have a friend. Just like me, slightly turned on the sound, though it does not communicate with electronics.
So, more than once he praised the sound of my amplifier. Although it was made at the dawn of my passion for sound. Worked in class B (with all the disadvantages inherent in this class).
The only difference in the circuit was OOS in current. What do not say, once I heard this sound, I could not refuse it!
And once this friend persuaded me to remake his Vega 50U according to the same principle.
As a result, I was terribly pleased, and the owner of this miracle of Soviet engineers was shocked. Neither he nor I expected such a clear and rich sound from this amplifier :) It has been working for 5 years. Safely ate already 2 sets of S90 (he likes more bass) and to this day pleases the ear of the owner
Why am I doing all this? Yes, I just assure you that You should listen to this amp at least once...
And also, this same friend gave me SVEN speakers to use while I remake my amplifier.
Everything would be fine, but their sound did not suit me ...
Therefore, I decided, without any permission, to mock them
The amplifier in them is built according to the standard scheme on two microcircuits.
I looked at the datasheet. A simplified diagram of the inclusion of this MS is given at the beginning of the article.
Refinement. actually the simplest! And at a cost of less than 10 rubles per channel!
Resistor R4 creates a voltage drop that is directly proportional to the current passing through the speaker. This voltage is fed through capacitors C3 and C4 to the inverting input of the amplifier. Capacitors connected in this way create a non-polar capacitor with a capacity of two less, i.e., 110uF. this is necessary in order not to buy expensive non-polar capacitors.
And if you add a switch to this circuit ...
Then you can feel the differences in the sound of the standard circuit and the circuit with current feedback. True, it will be necessary to select the resistor R3 so that the volume in both modes is approximately the same.
In essence, it turns out almost a tube amplifier (in any case, in terms of sound, may lovers of tube amplifiers not curse me!). After all, a tube amplifier that is not covered by an OS loop is a current amplifier (voltage across control grid regulates the cathode current).
Any amplifier can be upgraded in this way. Though transistor, though microcircuit. The only exception will be the bridge - there the circuit will become much more complicated.
All in all, I highly recommend you give it a try.
The comparison is best done on choir recordings. After the alteration, you can separate the singing voices from each other without straining, and not listen to them in porridge, as on a conventional amplifier. Or instrumental music...
For example, Gregorian, Hilary Stagg or, what everyone has, Aria - Careless Angel (introduction, guitar strumming).
(I can send them to good quality. who needs it - knock on ICQ)
Addition :
In order to avoid the same type of questions, I decided to add an addition to this article ...
Applicability:
This scheme can be fully implemented only in non-bridge amplifier with bipolar supply.
The speaker in such amplifiers is connected with one output to the output of the amplifier, and with the other - to a common wire, without coupling capacitors.
Additional resistor power:
The power of the resistor is calculated quite simply:
We know from physics that P=U*I
The voltage across the resistor is approximately equal to Ur \u003d Ud * (Rr / Rd), where Ud is the voltage on the speaker, Rr is the resistance of the resistor, Rd is the speaker resistance.
The current through the resistor and the speaker are equal.
Accordingly, Pr=Pout*(Rr/Rd).
Ideally, I advise you to take a resistor of twice the power Pr \u003d 2 * Pout * (Rr / Rd) in order to achieve maximum reliability (because the resistance of the speaker winding at some frequencies becomes much less than its resistance to direct current).
Accordingly, for an amplifier power of 20 W and a speaker resistance of 4 ohms, the power of the resistor should be 1 W. And for a speaker with a resistance of 8 ohms at the same power, a 0.5 W resistor is enough.
Operational amplifiers are often used to perform various operations: summing signals, differentiating, integrating, inverting, etc. Also, operational amplifiers have been developed as advanced
balanced amplification circuits.
Operational amplifier– universal functional element, widely used in modern circuits for the formation and conversion of information signals for various purposes, both in analog and digital technology. Let's take a look at the types of amplifiers.
Inverting amplifier
Consider a simple inverting amplifier circuit:
a) the voltage drop across the resistor R2 is Uout,
b) the voltage drop across the resistor R1 is Uin.
Uout/R2 = -Uin/R1, or voltage gain = Uout/Uin = R2/R1.
To understand how feedback works, imagine that a certain voltage level, say 1 V, is applied to the input. To be specific, let's say that the resistor R1 has a resistance of 10 kΩ, and the resistor R2 has a resistance of 100 kΩ. Now imagine that the output voltage decides to go out of control and becomes 0 V. What happens? Resistors R1 and R2 form a voltage divider that maintains the potential of the inverting input equal to 0.91 V. The operational amplifier detects the input mismatch, and the voltage at its output begins to decrease. The change continues until the output voltage reaches -10 V, at which point the potentials of the op-amp inputs become the same and equal to the ground potential. Similarly, if the output voltage begins to decrease further and becomes more negative than -10 V, then the potential at the inverting input will become lower than the ground potential, as a result, the output voltage will begin to rise.
The disadvantage of this circuit is that it has a low input impedance, especially for amplifiers with high voltage gain (with a closed feedback loop), in which the resistor R1, as a rule, is small. This shortcoming is eliminated by the circuit shown below in Fig. 4.
non-inverting amplifier. DC amplifier.
Consider the diagram in Fig. 4. Its analysis is extremely simple: UA = Uin. The voltage UA is removed from the voltage divider: UA = Uout R1 / (R1 + R2). If UA = Uin, then gain = Uout / Uin = 1 + R2 / R1. This is a non-inverting amplifier. In the approximation that we will use, the input impedance of this amplifier is infinite (for a 411 type op amp it is 1012 ohms or more, for bipolar transistor op amps it usually exceeds 108 ohms). The output impedance, as in the previous case, is equal to fractions of an ohm. If, as in the case of the inverting amplifier, we carefully examine the behavior of the circuit when the input voltage changes, we will see that it works as promised.
AC amplifier
The circuit above is also an amplifier direct current. If the signal source and amplifier are ac coupled, the input current (which is very small) must be grounded as shown in fig. 5. For the component values shown in the diagram, the voltage gain is 10, and the -3 dB point corresponds to a frequency of 16 Hz.
Amplifier alternating current. If only AC signals are being amplified, then the DC gain can be reduced to unity, especially if the amplifier has a large voltage gain. This makes it possible to reduce the effect of the always existing finite "input-referred shear stress".
For the circuit presented in fig. 6, the -3 dB point corresponds to a frequency of 17 Hz; at this frequency, the capacitor impedance is 2.0 kΩ. Please note that the capacitor must be large. If a high gain non-inverting amplifier is used to build an AC amplifier, the capacitor may be too large. In this case, it is better to do without a capacitor and adjust the offset voltage so that it is equal to zero. You can use another method - increase the resistance of resistors R1 and R2 and use a T-shaped divider circuit.
Despite the high input impedance that designers always strive for, a non-inverting amplifier circuit is not always preferred over an inverting amplifier circuit. As we will see later, the inverting amplifier does not place such high demands on the op-amp and, therefore, has several the best performance. In addition, due to the imaginary ground, it is convenient to combine signals without their mutual influence on each other. And finally, if the circuit in question is connected to the output (stable) of another op-amp, then the value of the input impedance is indifferent to you - it can be 10 kOhm or infinity, since in any case the previous stage will perform its functions in relation to the next.
Repeater
On fig. 7 shows an emitter-like follower based on an operational amplifier.
It is nothing more than a non-inverting amplifier, in which the resistance of the resistor R1 is equal to infinity, and the resistance of the resistor R2 is zero (gain = 1). There are special op amps designed to be used only as repeaters, they have improved performance (mostly faster), an example of such an op amp is a circuit like LM310 or OPA633, as well as simplified circuits, such as circuit type TL068 (it is available in transistor package with three terminals).
A unity gain amplifier is sometimes referred to as a buffer because it has isolating properties (large input impedance and low output).
Basic precautions when working with op amps
1. The rules are valid for any operational amplifier, provided that it is in active mode, i.e. its inputs and outputs are not overloaded.
For example, if you apply too much signal to the input of the amplifier, this will lead to the fact that the output signal will be cut off near the level of UCC or UEE. While the output voltage is fixed at the cutoff voltage, the input voltage cannot remain unchanged. The op-amp's output voltage swing cannot be greater than the supply voltage range (usually 2V less than the supply range, although some op amps limit the output voltage swing to one or the other supply voltage). A similar limitation is imposed on the output stability range of a current source based on an operational amplifier. For example, in a current source with a floating load, the maximum voltage drop across the load in the “normal” current direction (the direction of the current coincides with the direction of the applied voltage) is UCC - Uin, and in the reverse direction of the current (the load in this case can be quite strange, for example, it may contain reversed batteries to receive a direct charge current, or it may be inductive and work with currents that change direction) - Uin - UEE.
2. Feedback must be negative. This means (among other things) that inverting and non-inverting inputs should not be confused.
3. The op-amp circuit must have a DC feedback circuit, otherwise the op-amp will definitely go into saturation.
4. Many op amps have a fairly low differential input voltage rating. The maximum voltage difference between the inverting and non-inverting inputs can be limited to 5 V for either voltage polarity. If this condition is neglected, then large input currents will occur, which will lead to deterioration in performance or even destruction of the operational amplifier.
The concept of "feedback" (FB) is one of the most common, it has long gone beyond the narrow field of technology and is now used in a broad sense. In control systems, feedback is used to compare the output signal with a setpoint and correct accordingly. Anything can act as a "system", for example, the process of controlling a car moving along the road - the output data (the position of the car and its speed) is monitored by the driver, who compares them with the expected values \u200b\u200band corrects the input accordingly (using the steering wheel, speed switch, brakes). In an amplifying circuit, the output signal must be a multiple of the input signal, so in a feedback amplifier, the input signal is compared with a certain part of the output signal.
All about feedback
negative feedback is the process of passing the output signal back to the input, in which part of the input signal is cancelled. It may seem that this is a stupid idea, which will only lead to a decrease in the gain. Harold S. Black, who in 1928 attempted to patent negative feedback, received just such a response. “Our invention was treated the same as a perpetual motion machine” (IEEE Spectrum, December 1977). Indeed, negative feedback reduces the gain, but at the same time it improves other parameters of the circuit, for example, eliminates distortion and non-linearity, smoothes the frequency response (brings it in line with the desired characteristic), and makes the behavior of the circuit predictable. The deeper the negative feedback, the less the external characteristics of the amplifier depend on the characteristics of the amplifier with open feedback (without feedback), and ultimately it turns out that they depend only on the properties of the feedback circuit itself. Op-amps are typically used in deep feedback mode, and the open-loop voltage gain (without feedback) is in the millions in these circuits.
The feedback circuit can be frequency-dependent, then the gain will depend in a certain way on the frequency (an example is a preamplifier audio frequencies in an RIAA-compliant player); if the OS circuit is amplitude-dependent, then the amplifier has a non-linear characteristic (a common example of such a circuit is a logarithmic amplifier, in which the OS circuit uses the logarithmic dependence of the voltage UBE on the current IK in a diode or transistor). Feedback can be used to form a current source (output impedance close to infinity) or a voltage source (output impedance close to zero) and can be used to produce very large or very small input impedance. Generally speaking, the parameter on which feedback is introduced improves with its help. For example, if we use a signal proportional to the output current for feedback, we get a good current source.
Feedback can also be positive; it is used, for example, in generators. Oddly enough, it is not as useful as the negative OS. Rather, it is associated with trouble, since in a circuit with a negative OS on high frequency sufficiently large phase shifts can occur, leading to the appearance of a positive feedback and undesirable self-oscillations. In order for these phenomena to arise, it is not necessary to make great efforts, but to prevent unwanted self-oscillations, correction methods are resorted to.
Operational amplifiers
In most cases, when considering feedback circuits, we will be dealing with operational amplifiers. An operational amplifier (op-amp) is a DC differential amplifier with a very high gain and a single-ended input. The classic differential amplifier with two inputs and a single-ended output can serve as a prototype of the op-amp; however, it should be noted that real op-amps have much higher gains (typically around 105 - 106) and lower output impedances, and also allow the output signal to change almost over the entire supply voltage range (split power supplies of ±15 V are usually used).
The symbols "+" and "-" do not mean that the potential must always be more positive at one input than at the other; these symbols simply indicate the relative phase of the output signal (important if negative feedback is used in the circuit). To avoid confusion, it is better to call the inputs "inverting" and "non-inverting" rather than the input "plus" and the input "minus". The diagrams often do not show the connection of power sources to the op-amp and the output intended for grounding. Operational amplifiers have a colossal voltage gain and are never (with rare exceptions) used without feedback. We can say that operational amplifiers are designed to work with feedback. The gain of the open-loop circuit is so high that in the presence of a closed feedback loop, the characteristics of the amplifier depend only on the feedback circuit. Of course, on closer examination it should turn out that such a generalized conclusion is not always true. We'll start by simply looking at how an op amp works, and then study it more carefully as needed.
The industry produces literally hundreds of types of op-amps that offer various advantages over each other. A very good LF411 (or simply "411") circuit, introduced to the market by National Semiconductor, has become ubiquitous. Like all op amps, it is a tiny element housed in miniature case with a two-row mini-DIP pinout. This circuit is inexpensive and easy to handle; the industry produces an improved version of this circuit (LF411A), as well as an element housed in a miniature package containing two independent operational amplifiers (a circuit of the LF412 type, which is also called a "dual" operational amplifier). We recommend you the LF411 circuit as a good starting point in development electronic circuits.
The Type 411 circuit is a silicon chip containing 24 transistors (21 bipolar transistors, 3 FETs, 11 resistors and 1 capacitor). On fig. 2 shows the connection to the case pins.
The dot on the housing cover and the notch on its end serve to designate the reference point for pin numbering. In most electronic circuit housings, the pin numbering is done in a counter-clockwise direction from the side of the housing cover. The “zero setting” (or “balance”, “adjustment”) pins serve to eliminate the slight asymmetry that is possible in the operational amplifier.
Important Rules
Now we will get acquainted with the most important rules that determine the behavior of an operational amplifier covered by a feedback loop. They are valid for almost all cases of life.
First, the op amp has such a large voltage gain that a few fractions of a millivolt change between inputs causes the output voltage to change over its full range, so let's not consider this small voltage and formulate the I rule:
I. The output of the operational amplifier tends to ensure that the voltage difference between its inputs is zero.
Second, the op amp draws very little input current (an LF411 op amp draws 0.2 nA; an op amp with inputs on field effect transistors- of the order of picoamperes); Without going into deeper details, we formulate Rule II:
II. The inputs of the operational amplifier do not consume current.
A clarification needs to be made here: Rule I does not mean that the op amp actually changes the voltage at its inputs. This is impossible. (This would be inconsistent with Rule II.) The op-amp "evaluates" the state of the inputs and, using an external OS circuitry, passes the voltage from output to input, so that the voltage difference between the inputs becomes zero (if possible).
These rules provide a sufficient basis for considering op-amp circuits.
The journey of ten thousand miles begins with the first step.
(Chinese proverb)
It was in the evening, there was nothing to do ... And so suddenly I wanted to solder something. Sort of ... Electronic! .. Solder - so solder. The computer is available, the Internet is connected. We choose a scheme. And suddenly it turns out that the schemes for the conceived subject are a wagon and a small cart. And everyone is different. No experience, little knowledge. Which one to choose? Some of them contain some kind of rectangles, triangles. Amplifiers, and even operational ones ... How they work is not clear. Stra-a-ashno! .. What if it burns down? We choose what is simpler, on familiar transistors! Chose, soldered, turned on ... HELP !!! Does not work!!! Why?
Yes, because "Simplicity is worse than theft"! It's like a computer: the fastest and most sophisticated - gaming! And for office work, the simplest is enough. It's the same with transistors. Soldering a circuit on them is not enough. You still need to know how to set it up. Too many "pitfalls" and "rake". And this often requires experience that is by no means an entry level. So what, quit an exciting activity? By no means! Just do not be afraid of these "triangles-rectangles". It turns out that in many cases it is much easier to work with them than with individual transistors. IF YOU KNOW - HOW!
Here we are: understanding how an operational amplifier (op-amp, or in English OpAmp) works, we will now deal with. At the same time, we will consider his work literally “on the fingers”, practically without using any formulas, except perhaps, except for Ohm’s grandfather’s law: “Current through a circuit section ( I) is directly proportional to the voltage across it ( U) and inversely proportional to its resistance ( R)»:
I=U/R. (1)
To begin with, in principle, it is not so important how exactly the op-amp is arranged inside. Let's just take as an assumption that it is a "black box" with some stuffing there. On this stage we will not consider such parameters of the op amp as “bias voltage”, “shift voltage”, “temperature drift”, “noise characteristics”, “common mode suppression coefficient”, “supply voltage ripple suppression coefficient”, “bandwidth”, etc. .P. All these parameters will be important at the next stage of its study, when the basic principles of its work “settle down” in the head, because “it was smooth on paper, but forgot about the ravines” ...
For now, let's just assume that the parameters of the op-amp are close to ideal and consider only what signal will be at its output if some signals are applied to its inputs.
So, the operational amplifier (op-amp) is a DC differential amplifier with two inputs (inverting and non-inverting) and one output. In addition to them, the op-amp has power leads: positive and negative. These five conclusions are found in almost any OS and are fundamentally necessary for its operation.
The op-amp has a huge gain, at least 50,000 ... 100,000, but in reality - much more. Therefore, as a first approximation, we can even assume that it is equal to infinity.
The term "differential" ("different" is translated from English as "difference", "difference", "difference") means that the output potential of the op-amp is affected exclusively by the potential difference between its inputs, regardless from them absolute meaning and polarity.
The term "DC" means that the op-amp amplifies the input signals starting from 0 Hz. The upper frequency range (frequency range) of the signals amplified by the op amp depends on many factors, such as frequency characteristics transistors of which it consists, the gain of a circuit built using an op-amp, etc. But this issue is already beyond the scope of the initial acquaintance with his work and will not be considered here.
Op-amp inputs have a very high input impedance equal to tens/hundreds of MegaOhm, or even GigaOhm (and only in the memorable K140UD1, and even in K140UD5 it was only 30...50 kOhm). Such a high impedance of the inputs means that they have almost no effect on the input signal.
Therefore, with a high degree of approximation to the theoretical ideal, we can assume that current does not flow into the inputs of the op-amp . This - first an important rule that is applied in the analysis of the operation of the OS. Please remember well what it concerns only the OU itself, but not schemes with its use!
What do the terms "inverting" and "non-inverting" mean? In relation to what is the inversion determined and, in general, what kind of “animal” is this - signal inversion?
Translated from Latin, one of the meanings of the word "inversio" is "wrapping", "coup". In other words, inversion is mirror reflection (mirroring) signal relative to horizontal axis X(time axis). On Fig. 1 shows a few of the many possible signal inversion options, where the direct (input) signal is marked in red and the inverted (output) signal is in blue.
Rice. 1 Concept of signal inversion
It should be especially noted that to the zero line (as in Fig. 1, A, B) the signal inversion not tied! Signals can be inverse and asymmetrical. For example, both are only in the region of positive values (Fig. 1, B), which is typical for digital signals or with unipolar power supply (this will be discussed later), or both are partially in the positive and partially in the negative regions (Fig. 1, B, D). Other options are also possible. The main condition is their mutual specularity relative to some arbitrarily chosen level (for example, an artificial midpoint, which will also be discussed later). In other words, polarity signal is also not a determining factor.
Depict OU on circuit diagrams in different ways. Abroad, OS were previously depicted, and even now they are very often depicted in the form of an isosceles triangle (Fig. 2, A). The inverting input is marked with a minus symbol, and the non-inverting input is marked with a plus symbol inside a triangle. These symbols do not mean at all that the potential at the respective inputs must be more positive or more negative than at the other. They simply indicate how the output potential reacts to the potentials applied to the inputs. As a result, they are easy to confuse with power leads, which can be an unexpected "rake", especially for beginners.
Rice. 2 Conditional options graphic images(UGO)
operational amplifiers
In the system of domestic conditional graphic images (UGO) before the entry into force of GOST 2.759-82 (ST SEV 3336-81), OUs were also depicted as a triangle, only the inverting input - with an inversion symbol - a circle at the intersection of the output with a triangle (Fig. 2, B), and now - in the form of a rectangle (Fig. 2, C).
When designating the op amp on the diagrams, the inverting and non-inverting inputs can be interchanged if it is more convenient, however, traditionally, the inverting input is shown at the top, and the non-inverting input at the bottom. Power pins are usually always placed 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 applying a portion of the output voltage of an amplifier to its input, where it is algebraically (subject to sign) added to the input voltage. The principle of signal summation will be discussed below. Depending on which input of the op-amp, inverting or non-inverting, the OS is fed, there is a negative feedback (NFB), when part of the output signal is applied to the inverting input (Fig. 3, A) or positive feedback (PIC), when part the output signal is fed, respectively, to the non-inverting input (Fig. 3, B).
Rice. 3 The principle of feedback formation (OS)
In the first case, since the output is the inverse of the input, it is subtracted from the input. As a result, the overall gain of the stage is reduced. In the second case, it is added to the input, the overall gain of the cascade is increased.
At first glance, it may seem that POS has a positive effect, and OOS is a completely useless undertaking: why reduce the gain? This is exactly what U.S. patent examiners thought when, in 1928, Harold S. Black tried patent the OS. However, by sacrificing gain, we significantly improve other important parameters circuits, 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 POS (given its own huge gain of the op-amp) has the opposite effect on the characteristics of the circuit and the most unpleasant thing is that it causes its self-excitation. It, of course, is also used consciously, for example, in generators, comparators with hysteresis (more on this later), etc., but in general view its influence on the operation of amplifier circuits with op amps is rather negative and requires a very thorough and reasonable analysis of its application.
Since the OS has two inputs, the following main types of its inclusion using the OS are possible (Fig. 4):
Rice. 4 Basic schemes for switching on the OS
A) inverting (Fig. 4, A) - the signal is applied to the inverting input, and the non-inverting one is connected directly to the reference potential (not used);
b) non-inverting (Fig. 4, B) - the signal is applied to the non-inverting input, and the inverting one is connected directly to the reference potential (not used);
V) differential (Fig. 4, B) - signals are fed to both inputs, inverting and non-inverting.
To analyze the operation of these schemes, one should take into account second the most important rule, to which the operation of the OS is subject: The output of an op-amp tends to have zero voltage difference between its inputs..
However, any wording must be necessary and sufficient to limit the entire subset of cases that obey it. The above formulation, for all its “classicism”, does not give any information about which of the inputs the output “seeks to influence”. Based on it, it turns out that the op-amp seems to equalize the voltages at its inputs, applying voltage to them from somewhere “from the inside”.
Looking closely at the diagrams in Fig. 4, you can see that the OOC (through Rooc) in all cases is started from the exit only to the inverting input, which gives us reason to reformulate this rule as follows: Voltage on the output of the op-amp, covered by the OOS, tends to ensure that the potential at the inverting input is equal to the potential at the non-inverting input.
Based on this definition, the “leading” at any inclusion of the OA with OOS is the non-inverting input, and the “slave” is the inverting one.
When describing the operation of an op amp, the potential at its inverting input is often referred to as "virtual zero" or "virtual midpoint". The translation of the Latin word "virtus" means "imaginary", "imaginary". A virtual object behaves close to the behavior of similar objects of material reality, i.e., for input signals (due to the action of the FOS), the inverting input can be considered connected directly to the same potential as the non-inverting input. However, "virtual zero" is just a special case that takes place only with bipolar power supply of the op-amp. When using a unipolar power supply (which will be discussed below), and in many other switching circuits, there will be no zero on either the non-inverting or inverting inputs. Therefore, let's agree that we will not use this term, since it interferes with the initial understanding of the principles of operation of the OS.
From this point of view, we will analyze the schemes shown in Fig. 4. At the same time, to simplify the analysis, we will assume that the supply voltages are still bipolar, equal to each other in value (say, ± 15 V), with a midpoint (common bus or “ground”), relative to which we will count the input and output voltages. In addition, the analysis will be carried out in direct current, because. a changing alternating signal at each moment of time can also be represented as a sample of direct current values. In all cases, feedback through Rooc is connected from the output of the op-amp to its inverting input. The difference is only in which of the inputs the input voltage is applied.
A) inverting switching on (Fig. 5).
Rice. 5 The principle of operation of the op-amp in an inverting connection
The potential at the non-inverting input is zero, because it is connected to the midpoint ("ground"). An input signal equal to +1 V relative to the midpoint (from GB) is applied to the left terminal of the input resistor Rin. Let us assume that the resistances Rooc and Rin are equal to each other and amount to 1 kOhm (their total resistance is 2 kOhm).
According to Rule 2, the inverting input must have the same potential as the zeroed non-inverting, i.e., 0 V. Therefore, a voltage of +1 V is applied to Rin. According to Ohm's law, a current will flow through it Iinput= 1 V / 1000 ohms = 0.001 A (1 mA). The direction of flow of this current is shown by an arrow.
Since Rooc and Rin are connected by a divider, and according to Rule 1, the inputs of the op-amp do not consume current, in order for the voltage to be 0 V at the midpoint of this divider, a voltage must be applied to the right output of Rooc minus 1 V, and the current flowing through it Ioos should also be equal to 1 mA. In other words, a voltage of 2 V is applied between the left terminal Rin and the right terminal Rooc, and the current flowing through this divider is 1 mA (2 V / (1 kΩ + 1 kΩ) = 1 mA), i.e. I input = I oos .
If a negative polarity voltage is applied to the input, the output of the op-amp will be a positive polarity voltage. Everything is the same, only the arrows showing the flow of current through Rooc and Rin will be directed in the opposite direction.
Thus, if the values of Rooc and Rin are equal, the voltage at the output of the op-amp will be equal to the voltage at its input in magnitude, but inverse in polarity. And we got inverting repeater . This scheme is often used if you need to invert the signal received using circuits that are fundamentally inverters. For example, logarithmic amplifiers.
Now let's keep Rin equal to 1 kOhm and increase the resistance Rooc to 2 kOhm with the same input signal +1 V. The total divider resistance Rooc+Rin has increased to 3 kOhm. In order for a potential of 0 V (equal to the potential of the non-inverting input) to remain at its midpoint, the same current (1 mA) must flow through Rooc as through Rin. Therefore, the voltage drop across Rooc (voltage at the output of the op-amp) should already be 2 V. At the output of the op-amp, the voltage is minus 2 V.
Let's increase the value of Rooc to 10 kOhm. Now the voltage at the output of the op-amp under the same other conditions will already be 10 V. Wow! Finally we got inverting amplifier
! Its output voltage is greater than the input voltage (in other words, the gain Ku) as many times as the resistance Rooc is greater than the resistance Rin. No matter how I swore not to use formulas, let's still display this as an equation:
Ku \u003d - Uout / Uin \u003d - Rooc / Rin. (2)
The minus sign in front of the fraction on the right side of the equation only means that the output signal is inverse with respect to the input. And nothing more!
And now let's increase the resistance Rooc to 20 kOhm and analyze what happens. According to formula (2), with Ku \u003d 20 and an input signal of 1 V, the output should have been a voltage of 20 V. But it wasn’t there! We previously assumed that the supply voltage of our op-amp is only ± 15 V. But even 15 V cannot be obtained (why so - a little lower). "You can't jump above your head (supply voltage)"! As a result of such abuse of the ratings of the circuit, the output voltage of the op-amp “rests” on the supply voltage (the output of the op-amp enters saturation). The balance of current equality through the divider RoocRin ( Iinput = Ioos) is violated, a potential appears at the inverting input, which is different from the potential at the non-inverting input. Rule 2 no longer applies.
input resistance inverting amplifier is equal to the resistance Rin, since all the current from the input signal source (GB) flows through it.
Now let's replace the constant Rooc with a variable, with a nominal value of, say, 10 kOhm (Fig. 6).
Rice. 6 Variable gain inverting amplifier circuit
With the right (according to the circuit) position of its slider, the gain will be Rooc / Rin = 10 kOhm / 1 kOhm = 10. By moving the Rooc slider to the left (decreasing its resistance), the gain of the circuit will decrease and, finally, at its extreme left position it will become equal to zero, since the numerator in the above formula will become zero at any the value of the denominator. The output will also be zero for any value and polarity of the input signal. Such a scheme is often used in audio signal amplification circuits, for example, in mixers, where you have to adjust the gain from zero.
B) non-inverting switching on (Fig. 7).
Rice. 7 The principle of operation of the op-amp in a non-inverting inclusion
The left pin of Rin is connected to the midpoint ("ground"), and the input signal equal to +1 V is applied directly to the non-inverting input. Since the nuances of the analysis are “chewed” above, here we will pay attention only to significant differences.
At the first stage of the analysis, we also take the resistances Rooc and Rin equal to each other and equal to 1 kOhm. Because at the non-inverting input, the potential is +1 V, then according to Rule 2, the same potential (+1 V) must be at the inverting input (shown in the figure). To do this, there must be a voltage of +2 V on the right terminal of the Rooc resistor (output of the op-amp). Currents Iinput And Ioos, equal to 1 mA, now flow through the resistors Rooc and Rin in the opposite direction (shown by arrows). We got it non-inverting amplifier with a gain of 2, since an input of +1V produces an output of +2V.
Strange, isn't it? The ratings are the same as in the inverting connection (the only difference is that the signal is applied to another input), and the gain is obvious. We'll look into this a little later.
Now we increase the value of Rooc to 2 kOhm. To keep the balance of currents Iinput = Ioos and the potential of the inverting input is +1 V, the output of the op-amp should already be +3 V. Ku \u003d 3 V / 1 V \u003d 3!
If we compare the values of Ku with a non-inverting connection with an inverting one, with the same ratings Rooc and Rin, it turns out that the gain in all cases is greater by one. We derive the formula:
Ku \u003d Uout / Uin + 1 \u003d (Rooc / Rin) + 1 (3)
Why is this happening? Yes, very easy! NFB works exactly the same as in an inverting connection, but according to Rule 2, the potential of the non-inverting input is always added to the potential of the inverting input in a non-inverting connection.
So, with a non-inverting inclusion, it is impossible to obtain a gain equal to 1? Why not, why not. Let's reduce the value of Rooc, similar to how we analyzed Fig. 6. With its zero value - by short-circuiting the output with the inverting input (Fig. 8, A), according to Rule 2, the output will have such a voltage that the potential of the inverting input is equal to the potential of the non-inverting input, i.e., +1 V. We get: Ku \u003d 1 V / 1 V \u003d 1 (!) Well, since the inverting input does not consume current and there is no potential difference between it and the output, then no current flows in this circuit.
Rice. 8 Scheme of switching on the op-amp as a voltage follower
Rin becomes generally superfluous, because it is connected in parallel with the load on which the output of the op-amp should work, and its output current will flow through it in vain. And what happens if you leave Rooc, but remove Rin (Fig. 8, B)? Then in the gain formula Ku = Roos / Rin + 1, the resistance Rin theoretically becomes close to infinity (in reality, of course, not, because there are leaks on the board, and the input current of the op-amp, although negligible, is still zero is still not equal), and the ratio Rooc / Rin is equated to zero. Only one remains in the formula: Ku \u003d + 1. Can the gain be less than one for this circuit? No, less will not work under any circumstances. You can’t go around the “extra” unit in the gain formula on a crooked goat ...
After we removed all the "extra" resistors, we get a circuit non-inverting repeater shown in Fig. 8, V.
At first glance, such a scheme does not make practical sense: why do we need a single, and even non-inverse "amplification" - what, you can’t just send a signal further ??? However, such schemes are used quite often and here's why. According to Rule 1, current does not flow into the inputs of the op-amp, i.e., input impedance the non-inverting follower is very large - the same tens, hundreds and even thousands of MΩ (the same applies to the circuit according to Fig. 7)! But the output resistance is very small (fractions of Ohm!). The output of the op-amp “chuffs with all its might”, trying, according to Rule 2, to maintain the same potential at the inverting input as at the non-inverting one. The only limitation is the permissible output current of the op-amp.
But from this place we will wag a little to the side and consider the issue of the output currents of the op-amp in a little more detail.
For most general purpose op amps, the technical specifications state that the resistance of the load connected to their output should not be less 2 kOhm More - as much as you want. For a much smaller number, it is 1 kOhm (K140UD ...). This means that under the worst-case conditions: maximum supply voltage (e.g. ±16 V or 32 V in total), a load connected between the output and one of the supply rails, and maximum output voltage of opposite polarity, a voltage of about 30 V will be applied to the load. In this case, the current through it will be: 30 V / 2000 Ohm = 0.015 A (15 mA). Not so little, but not too much either. Fortunately, most general purpose op amps have built-in overcurrent protection - typical maximum output current is 25 mA. Protection prevents overheating and failure of the op-amp.
If the supply voltages are not the maximum allowable, then the minimum load resistance can be proportionally reduced. Say, with a power supply of 7.5 ... 8 V (total 15 ... 16 V), it can be 1 kOhm.
IN) differential switching on (Fig. 9).
Rice. 9 The principle of operation of the op-amp in a differential connection
So, let's assume that with the same ratings Rin and Rooc equal to 1 kOhm, the same voltages equal to +1 V are applied to both inputs of the circuit (Fig. 9, A). Since the potentials on both sides of the resistor Rin are equal to each other (the voltage across the resistor is 0), no current flows through it. This means that the current through the resistor Rooc is also zero. That is, these two resistors do not perform any function. In fact, we actually got a non-inverting follower (compare with Fig. 8). Accordingly, we will get the same voltage at the output as at the non-inverting input, i.e., +1 V. Let's change the polarity of the input signal at the inverting input of the circuit (turn GB1 over) and apply minus 1 V (Fig. 9, B). Now a voltage of 2 V is applied between the terminals Rin and a current flows through it Iin\u003d 2 mA (I hope that it is no longer necessary to describe in detail why this is so?). In order to compensate for this current, a current of 2 mA must also flow through Rooc. And for this, the output of the op-amp must have a voltage of +3 V.
That's where the malicious "grin" of an additional one appeared in the formula for the gain of a non-inverting amplifier. It turns out that with such simplified In differential switching, the difference in gain constantly shifts the output signal by the potential at the non-inverting input. A problem with! However, "Even if you were eaten, you still have at least two exits." This means that we somehow need to equalize the gains of the inverting and non-inverting inclusions in order to “neutralize” this extra one.
To do this, let's apply the input signal to the non-inverting input not directly, but through the divider Rin2, R1 (Fig. 9, B). Let's take their denominations also for 1 kOhm. Now, at the non-inverting (and therefore also at the inverting) input of the op-amp, there will be a potential of +0.5 V, a current will flow through it (and Rooc) Iin = Ioos\u003d 0.5 mA, to ensure which the output of the op-amp must have a voltage equal to 0 V. Phew! We got what we wanted! With equal magnitude and polarity signals at both inputs of the circuit (in this case +1 V, but the same will be true for minus 1 V and for any other digital values), the output of the op-amp will maintain zero voltage equal to the difference in input signals .
Let's check this reasoning by applying a signal of negative polarity minus 1 V to the inverting input (Fig. 9, D). Wherein Iin = Ioos= 2 mA, for which the output should be +2 V. Everything was confirmed! The output level corresponds to the difference between the inputs.
Of course, if Rin1 and Rooc are equal (respectively, Rin2 and R1), we will not get amplification. To do this, you need to increase the values of Rooc and R1, as was done when analyzing previous inclusions of the op-amp (I will not repeat it), and it should strictly respect the ratio:
Rooc / Rin1 = R1 / Rin2. (4)
What useful do we get from such an inclusion in practice? And we get a remarkable property: the output voltage does not depend on the absolute values of the input signals, if they are equal to each other in magnitude and polarity. Only the difference (differential) signal is output. This makes it possible to amplify very small signals against the background of noise acting equally on both inputs. For example, a signal from a dynamic microphone against the background of a 50 Hz industrial frequency mains pickup.
However, in this barrel of honey, unfortunately, there is a fly in the ointment. First, equality (4) must be observed very strictly (up to tenths and sometimes hundredths of a percent!). Otherwise, there will be an unbalance of the currents acting in the circuit, and therefore, in addition to the difference ("anti-phase") signals, the combined ("common-mode") signals will also be amplified.
Let's understand the essence of these terms (Fig. 10).
Rice. 10 Signal phase shift
The phase of the signal is a value that characterizes the offset of the origin of the signal period relative to the origin of time. Since both the origin of time and the origin of the period are chosen arbitrarily, the phase of one periodical signal has no physical meaning. However, the phase difference between the two periodical signals is a quantity that has physical meaning, it reflects the delay of one of the signals relative to the other. What is considered the beginning of the period does not matter. For the point of the beginning of the period, you can take a zero value with a positive slope. It is possible - maximum. Everything is in our power.
On Fig. 9, red indicates the original signal, green - shifted by ¼ period relative to the original, and blue - by ½ period. If we compare the red and blue curves with the curves in Fig. 2, B, it can be seen that they are mutually inverse. Thus, “in-phase signals” are signals that coincide with each other at each of their points, and “anti-phase signals” are inverse relative to each other.
At the same time, the concept inversions broader than the concept phases, because the latter applies only to regularly repeated, periodic signals. And the concept inversions applicable to any signals, including non-periodic ones, such as sound signal, digital sequence, or constant voltage. To phase is a consistent value, the signal must be periodic at least over a certain interval. Otherwise, both phase and period turn into mathematical abstractions.
Secondly, the inverting and non-inverting inputs in the differential connection, with equal ratings Rooc = R1 and Rin1 = Rin2, will have different input resistances. If the input resistance of the inverting input is determined only by the value Rin1, then the non-inverting input is determined by the values successively included Rin2 and R1 (have not forgotten that the op-amp inputs do not consume current?). In the example above, they will be 1 and 2 kΩ, respectively. And if we increase Rooc and R1 to obtain a full-fledged amplifying stage, then the difference will increase even more significantly: with Ku \u003d 10 - up to, respectively, all the same 1 kOhm and as much as 11 kOhm!
Unfortunately, in practice, the ratings Rin1 = Rin2 and Rooc = R1 are usually set. However, this is only acceptable if the signal sources for both inputs are of very low output impedance. Otherwise, it forms a divider with the input impedance of this amplifying stage, and since the division factor of such “dividers” will be different, the result is obvious: a differential amplifier with such resistor values will not perform its function of suppressing common-mode (combined) signals, or perform this function poorly .
One of the ways to solve this problem can be the inequality of the values of the resistors connected to the inverting and non-inverting inputs of the op-amp. Namely, so that Rin2 + R1 = Rin1. Another important point is to achieve exact observance of equality (4). As a rule, this is achieved by splitting R1 into two resistors - a constant, usually 90% of the desired value, and a variable (R2), whose resistance is 20% of the required value (Fig. 11, A).
Rice. 11 Differential amplifier balancing options
The path is generally accepted, but again, with this method of balancing, albeit slightly, the input impedance of the non-inverting input changes. A much more stable option with the inclusion of a tuning resistor (R5) in series with Rooc (Fig. 11, B), since Rooc does not participate in the formation of the input resistance of the inverting input. The main thing is to keep the ratio of their denominations, similar to option "A" (Rooc / Rin1 = R1 / Rin2).
Since we talked about differential switching and mentioned repeaters, I would like to describe one interesting circuit (Fig. 12).
Rice. 12 Switched inverting/non-inverting follower circuit
The input signal is applied simultaneously to both inputs of the circuit (inverting and non-inverting). The ratings of all resistors (Rin1, Rin2 and Rooc) are equal to each other (in this case, let's take their real values: 10 ... 100 kOhm). The non-inverting input of the op-amp with the SA key can be closed to a common bus.
In the closed position of the key (Fig. 12, A), the resistor Rin2 does not participate in the operation of the circuit (only current “uselessly” flows through it Ivx2 from the signal source to the common bus). We get inverting follower with a gain equal to minus 1 (see Fig. 6). But with the SA key in the open position (Fig. 12, B), we get non-inverting follower with gain equal to +1.
The principle of operation of this scheme can be expressed in a slightly different way. When the SA key is closed, it works as an inverting amplifier with a gain equal to minus 1, and when it is open - simultaneously(!) And as an inverting amplifier with a gain, minus 1, and as a non-inverting amplifier with a gain of +2, from where: Ku = +2 + (–1) = +1.
In this form, this circuit can be used if, for example, the polarity of the input signal is unknown at the design stage (say, from a sensor that is not accessible until the device is set up). If, however, a transistor (for example, a field-effect transistor) is used as a key, controlled from the input signal using comparator(which will be discussed below), we get synchronous detector(synchronous rectifier). The specific implementation of such a scheme, of course, goes beyond the initial acquaintance with the operation of the OS, and we will not consider it in detail here again.
And now let's consider the principle of summing the input signals (Fig. 13, A), and at the same time we will figure out what values of the resistors Rin and Rooc should be in reality.
Rice. 13 The principle of operation of the inverting adder
We take as a basis the inverting amplifier already discussed above (Fig. 5), only we connect not one, but two input resistors Rin1 and Rin2 to the input of the op-amp. So far, for "educational" purposes, we accept the resistance of all resistors, including Rooc, equal to 1 kOhm. We supply input signals equal to +1 V to the left terminals Rin1 and Rin2. Currents equal to 1 mA flow through these resistors (shown by arrows pointing from left to right). To maintain the same potential at the inverting input as at the non-inverting one (0 V), a current equal to the sum of the input currents (1 mA + 1 mA = 2 mA) must flow through the Rooc resistor, shown by an arrow pointing in the opposite direction (from right to left ), for which the output of the op-amp must have a voltage of minus 2 V.
The same result (output voltage minus 2 V) can be obtained if +2 V is applied to the input of the inverting amplifier (Fig. 5), or the value of Rin is halved, i.e. up to 500 Ohm. Let's increase the voltage applied to the resistor Rin2 up to +2 V (Fig. 13, B). At the output we get a voltage of minus 3 V, which is equal to the sum of the input voltages.
There can be not two inputs, but as many as you like. The principle of operation of this circuit will not change from this: the output voltage in any case will be directly proportional to the algebraic sum (taking into account the sign!) of the currents passing through the resistors connected to the inverting input of the op-amp (inversely proportional to their ratings), regardless of their number.
If, on the other hand, signals equal to +1 V and minus 1 V are applied to the inputs of the inverting adder (Fig. 13, B), then the currents flowing through them will be in different directions, they will cancel each other out and the output will be 0 V. Through the resistor Rooc in this case no current will flow. In other words, the current flowing through Rooc is algebraically summed with input currents.
An important point also follows from this: while we were operating with small input voltages (1 ... 3 V), the output of a widely used op-amp could well provide such a current (1 ... 3 mA) for Rooc and something else remained for the load connected to the output of the op-amp. But if the voltages of the input signals are increased to the maximum allowable (close to the supply voltages), then it turns out that the entire output current will go to Rooc. Nothing left to load. And who needs an amplifying stage that works "for itself"? In addition, input resistor values of only 1 kΩ (respectively, determining the input resistance of the inverting amplifier stage) require excessively high currents to flow through them, heavily loading the signal source. Therefore, in real circuits, the resistance Rin is chosen not less than 10 kOhm, but it is also desirable not more than 100 kOhm, so that at a given gain, Rooc is not set too high. Although these values \u200b\u200bare not absolute, but only estimates, as they say, "in the first approximation" - it all depends on the specific circuit. In any case, it is undesirable that a current flowing through Rooc exceeds 5 ... 10% of the maximum output current of this particular op-amp.
The summed signals can also be applied to the non-inverting input. It turns out non-inverting adder. In principle, such a circuit will work in exactly the same way as an inverting adder, the output of which will be a signal that is directly proportional to the input voltages and inversely proportional to the values of the input resistors. However, in practice it is used much less frequently, because. contains a "rake" that should be taken into account.
Since Rule 2 is valid only for the inverting input, which has a “virtual zero potential”, then the non-inverting input will have a potential equal to the algebraic sum of the input voltages. Therefore, the input voltage available at one of the inputs will affect the voltage supplied to the other inputs. There is no “virtual potential” at the non-inverting input! As a result, additional circuitry tricks have to be applied.
So far, we have considered circuits based on OS with OOS. What happens if feedback is removed altogether? In this case, we get comparator(Fig. 14), i.e., a device that compares the absolute value of two potentials at its inputs (from the English word compare- compare). At its output, there will be a voltage approaching one of the supply voltages, depending on which of the signals is greater than the other. Typically, the input signal is applied to one of the inputs, and to the other - a constant voltage with which it is compared (the so-called "reference voltage"). It can be anything, including zero potential (Fig. 14, B).
Rice. 14 Scheme of switching on the op-amp as a comparator
However, not everything is so good "in the kingdom of Denmark" ... And what happens if the voltage between the inputs is zero? In theory, the output should also be zero, but in reality - never. If the potential at one of the inputs even slightly outweighs the potential of the other, then this will already be enough for chaotic voltage surges to occur at the output due to random disturbances induced at the inputs of the comparator.
In reality, any signal is "noisy", because ideal cannot be by definition. And in the area close to the point of equality of the potentials of the inputs, a burst of output signals will appear at the output of the comparator instead of one clear switching. To combat this phenomenon, the comparator circuit is often introduced hysteresis by creating a weak positive PIC from the output to the non-inverting input (Figure 15).
Rice. 15 The principle of operation of the hysteresis in the comparator due to the POS
Let's analyze the operation of this scheme. Its supply voltage is ± 10 V (for an even account). The resistance Rin is 1 kOhm, and Rpos is 10 kOhm. The midpoint potential is chosen as the reference voltage applied to the inverting input. The red curve shows the input signal coming to the left pin Rin (input scheme comparator), blue - the potential at the non-inverting input of the op-amp and green - the output signal.
While the input signal has a negative polarity, the output is a negative voltage, which, through Rpos, is added to the input voltage in inverse proportion to the values of the corresponding resistors. As a result, the potential of the non-inverting input 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 to positive polarity. The total potential at the non-inverting input will start like an avalanche become even more positive, supporting the process of such a switch. As a result, the comparator simply “will not notice” insignificant noise fluctuations of the input and reference signals, since they will be many orders of magnitude smaller in amplitude than the described “step” of the potential at the non-inverting input when switching.
When the input signal decreases, the reverse switching of the comparator output signal will occur at an input voltage of minus 1 V. This difference between the input signal levels leading to the switching of the comparator output, which in our case is equal to a total of 2 V, is called hysteresis. The greater the resistance Rpos with respect to Rin (the smaller the POS depth), the smaller the switching hysteresis. So, with Rpos \u003d 100 kOhm, it will be only 0.2 V, and with Rpos \u003d 1 MΩ, it will be 0.02 V (20 mV). The hysteresis (PIC depth) is selected based on the actual operating conditions of the comparator in a particular circuit. In which 10 mV will be a lot, and in which - and 2 V will be small.
Unfortunately, not every op amp and not in all cases can be used as a comparator. Specialized comparator ICs are available for matching between analog and digital signals. Some of them are specialized for connecting to digital TTL microcircuits (597CA2), some - to digital ESL microcircuits (597CA1), but most are so-called. "comparators for general use" (LM393/LM339/K554CA3/K597CA3). Their main difference from the op amps lies in the special device of the output stage, which is made on an open collector transistor (Fig. 16).
Rice. 16 Comparator output stage for general applications
and its connection to the load resistor
This requires the mandatory use of an external load resistor(R1), without which the output signal is simply physically unable to form a high (positive) output level. The voltage +U2 to which the load resistor is connected may be different from the supply voltage +U1 of the comparator chip itself. This allows simple means to provide the desired output level - be it TTL or CMOS.
Note In most comparators, an example of which can be dual LM393 (LM193 / LM293) or exactly the same in circuitry, but quad LM339 (LM139 / LM239), the emitter of the output stage transistor is connected to the negative power terminal, which somewhat limits their scope. In this regard, I would like to draw attention to the comparator LM31 (LM111 / LM211), the analogue of which is the domestic 521/554CA3, in which both the collector and the emitter of the output transistor are separately output, which can be connected to other voltages than the supply voltage of the comparator itself. Its only and relative disadvantage is that it is only one in an 8-pin (sometimes 14-pin) package. |
So far, we have considered circuits in which the input signal was fed to the input(s) through Rin, i.e. they were all converters input voltage in day off voltage same. In this case, the input current flowed through Rin. What happens if its resistance is taken equal to zero? The circuit will work in exactly the same way as the inverting amplifier discussed above, only the output impedance of the signal source (Rout) will serve as Rin, and we get converter input current V day off voltage(Fig. 17).
Rice. 17 Scheme of the current-to-voltage converter at the op-amp
Since the potential at the inverting input is the same as at the non-inverting one (in this case it is "virtual zero"), the entire input current ( Iin) will flow through Rooc between the output of the signal source (G) and the output of the op-amp. The input resistance of such a circuit is close to zero, which makes it possible to build micro/milliammeters on its basis, which practically do not affect the current flowing through the measured circuit. Perhaps the only limitation is the permissible input voltage range of the op-amp, which should not be exceeded. It can also be used to build, for example, a linear photodiode current-to-voltage converter and many other circuits.
We have considered the basic principles of operation of the OS in various schemes for its inclusion. One important question remains: nutrition.
As mentioned above, an op amp typically has only 5 pins: two inputs, an output, and two power pins, positive and negative. In the general case, bipolar power is used, that is, the power supply has three outputs with potentials: + U; 0; -U.
Once again, carefully consider all the above figures and see that a separate output of the midpoint in the op-amp NO ! It is simply not needed for their internal circuitry to work. In some circuits, a non-inverting input was connected to the midpoint, however, this is not the rule.
Hence, overwhelming majority modern op amps are designed to power UNIPOLAR tension! A logical question arises: “Why then do we need bipolar power,” if we depicted it so stubbornly and with enviable constancy in the drawings?
It turns out it's just very comfortably for practical purposes for the following reasons:
A) To ensure sufficient current and output voltage swing through the load (Fig. 18).
Rice. 18 Output current flow through the load at various options OS power supply
For now, we will not consider the input (and OOS) circuits of the circuits shown in the figure (“black box”). Let's take it for granted that some input sinusoidal signal is applied to the input (black sinusoid on the graphs) and the output is the same sinusoidal signal, amplified with respect to the input colored sinusoid on the graphs).
When connecting the load Rload. between the output of the op-amp and the midpoint of the connection of the power supplies (GB1 and GB2) - Fig. 18, A, the current flows through the load symmetrically about the midpoint (respectively, the red and blue half-waves), and its amplitude is maximum and the voltage amplitude at Rload. also the maximum possible - it can reach almost supply voltages. The current from the power source of the corresponding polarity is closed through the OS, Rload. and a power source (red and blue lines showing current flow in the corresponding direction).
Since the internal resistance of the op-amp power supplies is very low, the current through the load is only limited by its resistance and the op-amp's maximum output current, which is typically 25 mA.
When the op-amp is powered by a unipolar voltage as common bus the negative (negative) pole of the power source is usually selected, to which the second output of the load is connected (Fig. 18, B). Now the current through the load can only flow in one direction (shown by the red line), the second direction simply has nowhere to come from. In other words, the current through the load becomes asymmetrical (pulsating).
It is impossible to say unequivocally that this option is bad. If the load is, say, a dynamic head, then for it it is bad unambiguously. However, there are many applications where connecting a load between the output of the op-amp and one of the power rails (usually negative polarity) is not only acceptable, but also the only possible one.
If, nevertheless, it is necessary to ensure the symmetry of the current flow through the load with a unipolar supply, then it is necessary to galvanically decouple it from the output of the op-amp with a galvanic capacitor C1 (Fig. 18, B).
B) To ensure the required current of the inverting input, as well as bindings input signals to some arbitrarily selected level accepted for the reference (zero) - setting the operation mode of the OS for direct current (Fig. 19).
Rice. 19 Connecting the input signal source with various options for supplying the op-amp
Now consider the options for connecting input signal sources, excluding from consideration the connection of the load.
Connecting the inverting and non-inverting inputs to the midpoint of the power supply connection (Fig. 19, A) was considered when analyzing the previously given diagrams. If the non-inverting input draws no current and simply accepts the mid-point potential, then through the signal source (G) and Rin connected in series, the current flows, closing through the corresponding power source! And since their internal resistances are negligible compared to the input current (many orders of magnitude less than Rin), it practically does not affect the supply voltage.
Thus, with a unipolar supply of the op-amp, you can quite easily form the potential supplied to its non-inverting input using the divider R1R2 (Fig. 19, B, C). Typical resistor values of this divider are 10 ... 100 kOhm, and it is highly desirable to shunt the lower one (connected to a common negative bus) with a 10 ... 22 microfarad capacitor in order to significantly reduce the effect of supply voltage ripples on the potential of such artificial middle point.
But it is extremely undesirable to connect the signal source (G) to this artificial midpoint because of the same input current. Let's guess. Even with the ratings of the divider R1R2 = 10 kOhm and Rin = 10…100 kOhm, the input current Iin will be at best 1/10, and at worst - up to 100% of the current passing through the divider. Consequently, the potential at the non-inverting input will “float” by the same amount in combination (in phase) with the input signal.
To eliminate the mutual influence of the inputs on each other when amplifying DC signals with such a connection, for the signal source it is necessary to organize a separate potential of the artificial midpoint, formed by resistors R3R4 (Fig. 19, B), or, if the AC signal is amplified, galvanically isolate the signal source from the inverting input by capacitor C2 (Fig. 19, B).
It should be noted that in the above diagrams (Fig. 18, 19) we assumed by default that the output signal should be symmetrical about either the midpoint of the power supplies or the artificial midpoint. In reality, this is not always necessary. Quite often, you want the output signal to have predominantly either positive or negative polarity. Therefore, it is not at all necessary that the positive and negative polarities of the power supply be equal in absolute value. One of them can be much smaller in absolute value than the other - only in such a way as to ensure the normal functioning of the OS.
A logical question arises: “Which one exactly?” To answer it, let's briefly consider the allowable voltage ranges of the input and output signals of the op-amp.
For any op amp, the output potential cannot be higher than the potential of the positive power rail and lower than the potential of the negative power rail. In other words, the output voltage cannot go beyond the limits of the supply voltages. For example, for an OPA277 op amp, the output voltage at a load resistance of 10 kΩ is less than the voltage of the positive power rail by 2 V and the negative power rail by 0.5 V. The width of these "dead zones" of the output voltage, which the op amp output cannot reach, depends on the series factors such as output stage circuitry, load resistance, etc.). There are op amps that have minimal dead zones, for example, 50 mV to the supply rail voltage at a load of 10 kΩ (for OPA340), this feature of the op amp is called "rail-to-rail" (R2R).
On the other hand, for general-purpose op-amps, the input signals should also not exceed the supply voltage, and for some, be less than 1.5 ... 2 V. However, there are op-amps with specific input stage circuitry (for example, the same LM358 / LM324) , which can work not only from the negative power level, but even “negative” by 0.3 V, which greatly facilitates their use with unipolar power supply with a common negative bus.
Let's finally look at and feel these "spider bugs". You can even sniff and lick. I allow. Consider their most common options available to novice radio amateurs. Especially if you have to solder the op amp from the old equipment.
For op-amps of old designs, which necessarily require external circuits for frequency correction, in order to prevent self-excitation, it was typical to have additional conclusions. Because of this, some op amps did not even “fit” into an 8-pin package (Fig. 20, A) and were made in 12-pin round metal-glass, for example, K140UD1, K140UD2, K140UD5 (Fig. 20, B) or in 14-pin DIP packages, for example, K140UD20, K157UD2 (Fig. 20, B). The abbreviation DIP is an abbreviation of the English expression "Dual In line Package" and translates as "double-sided package".
The round metal-glass case (Fig. 20, A, B) was used as the main one for imported op-amps until about the mid-70s, and for domestic op-amps - until the mid-80s and is now used for the so-called. "military" applications ("5th acceptance").
Sometimes domestic op-amps were placed in currently rather “exotic” cases: a 15-pin rectangular metal-glass for the hybrid K284UD1 (Fig. 20, D), in which the key is an additional 15th pin from the case, and others. True, I personally have not met planar 14-pin packages (Fig. 20, E) for placing an op-amp in them. They were used for digital circuits.
Rice. 20 Cases of domestic operational amplifiers
Modern op amps, for the most part, contain correction circuits right on the chip, which made it possible to get by with a minimum number of pins (as an example, a 5-pin SOT23-5 for a single op amp - Fig. 23). This made it possible to place two to four completely independent (except for common power outputs) op-amps made on a single chip in one case.
Rice. 21 Two-row plastic cases of modern op amps for output mounting (DIP)
Sometimes you can find op-amps placed in single-row 8-pin (Fig. 22) or 9-pin packages (SIP) - K1005UD1. The abbreviation SIP is an abbreviation of the English expression "Single In line Package" and translates as "housing with one-way pinout."
Rice. 22 Single-row plastic case of double op-amps for through-hole mounting (SIP-8)
They were designed to minimize the space occupied on the board, but, unfortunately, they were “late”: by this time, surface mount packages (SMD - Surface Mounting Device) had become widespread by soldering directly to the board tracks (Fig. 23). However, for beginners, their use presents significant difficulties.
Rice. 23 Cases of modern imported op amps for surface mounting (SMD)
Very often, the same microcircuit can be "packed" by the manufacturer in different packages (Fig. 24).
Rice. 24 Placement options for the same chip in different packages
The conclusions of all microcircuits have a sequential numbering, counted from the so-called. "key", indicating the location of the output at number 1. (Fig. 25). IN any if the body is positioned with terminals Push, their numbering goes in ascending order against clockwise!
Rice. 25 Pin assignment of operational amplifiers
in various cases (pinout), top view;
numbering direction shown by arrows
In round metal-glass cases, the key has the form of a side protrusion (Fig. 25, A, B). Here, from the location of this key, huge “rakes” are possible! In domestic 8-pin cases (302.8), the key is located opposite the first pin (Fig. 25, A), and in imported TO-5 - opposite the eighth pin (Fig. 25, B). In 12-pin cases, both domestic (302.12) and imported, the key is located between the first and 12th conclusions.
Typically, the inverting input, both in round glass-metal and DIP packages, is connected to the 2nd pin, the non-inverting input to the 3rd pin, the output to the 6th pin, the power minus to the 4th pin, and the power plus to the pin 4. 7th. However, there are exceptions (another possible "rake"!) In the pinout of the OU K140UD8, K574UD1. In them, the numbering of the conclusions is shifted by one counterclockwise compared to the generally accepted for most other types, i.e. they are connected to the terminals, as in imported cases (Fig. 25, B), and the numbering corresponds to domestic ones (Fig. 25, A).
IN last years most of the OS "domestic purposes" began to be placed in plastic cases (Fig. 21, 25, C-D). In these cases, the key is either a recess (dot) opposite the first pin, or a cutout in the end of the case between the first and 8th (DIP-8) or 14th (DIP-14) pins, or a chamfer along the first half of the pins (Fig. 21, middle). The pin numbering in these cases also goes against clockwise when viewed from above (with conclusions away from you).
As mentioned above, internally corrected op amps have a total of five outputs, of which only three (two inputs and an output) belong to each individual op amp. This made it possible to place two completely independent (with the exception of plus and minus power, which require two more pins) op amps on one chip in one 8-pin package (Fig. 25, D), and even four in a 14-pin package (Fig. 25, D). As a result, at present, most op-amps are produced at least dual, for example, TL062, TL072, TL082, cheap and simple LM358, etc. Exactly the same in internal structure, but quad - respectively, TL064, TL074, TL084 and LM324.
With regard to the domestic analogue of the LM324 (K1401UD2), there is one more “rake”: if in the LM324 the plus of the power supply is connected to the 4th pin, and the minus to the 11th, then in K1401UD2 it is the other way around: the plus of the power is brought 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, E), you just need to turn the case 180 degrees so that the 1st pin takes the place of the 8th. Yes, that's all.
A few words about the labeling of imported OUs (and not only OUs). For a number of developments of the first 300 digital designations, it was customary to designate the quality group with the first digit of the digital code. For example, LM158/LM258/LM358 op amps, LM193/LM293/LM393 comparators, TL117/TL217/TL317 adjustable three-pin stabilizers, etc. are completely identical in internal structure, but differ in temperature operating range. For LM158 (TL117) the operating temperature range is from minus 55 to +125 ... 150 degrees Celsius (the so-called "combat" or military range), for LM258 (TL217) - from minus 40 to +85 degrees ("industrial" range) and for LM358 (TL317) - from 0 to +70 degrees ("household" range). At the same time, the price for them may be completely inappropriate for such a gradation, or differ very slightly ( inscrutable ways of pricing!). So you can buy them with any marking available “for the pocket” of a beginner, without particularly chasing the first “troika”.
After the first three hundred digital markings were exhausted, reliability groups began to be marked with letters, the meaning of which is deciphered in datasheets (Datasheet literally translates as “data table”) for these components.
Conclusion
So we studied the "alphabet" of the operation of the op-amp, capturing a little and comparators. Next, you need to learn how to add words, sentences and whole meaningful “compositions” (workable schemes) from these “letters”.
Unfortunately, "It is impossible to grasp the immensity." If the material presented in this article helped to understand how these "black boxes" work, then further deepening into the analysis of their "stuffing", the influence of input, output and transient response, is a task for more advanced study. Information about this is described in detail and thoroughly in a variety of existing literature. As grandfather William of Ockham used to say: "Entities should not be multiplied beyond what is necessary." There is no need to repeat what has already been well described. All you need to do is not be lazy and read it.
11. http://www.texnic.ru/tools/lekcii/electronika/l6/lek_6.html
Therefore, let me take my leave, with respect, etc., the author Alexey Sokolyuk ()
Feedback(OS) The phenomenon of transferring part of the energy of amplified oscillations from the output circuit of the amplifier to its input circuit is called.
The reasons that contribute to the transfer of energy from the output to the input of the amplifier may be:
A) physical properties and design features applied transistors (presence of capacitances and inductances of outputs, capacitances R-P transitions, etc.). The resulting OS is called internal feedback;
c) special circuits introduced by the designer to transmit vibrations from the output of the amplifier to its input in order to give the device the desired properties. This kind of feedback is called external feedback.
Of the listed types of operating systems, the first two are undesirable, so the designer is forced to take additional measures to eliminate them.
The circuit through which energy is transferred from the output of the amplifier to its input is called feedback loop.
Usually the OS circuit is some linear passive quadripole with a transfer coefficient g, the input of which is connected to the output of the amplifier, and the output - to the input of the amplifier (Figure 2.9). In the general case, the OS quadripole can be linear or non-linear, with a frequency-dependent or frequency-independent transmission coefficient.
Figure 2.9 - Amplifier with feedback circuit
The feedback loop can be general covering all or several stages of the amplifier (Figure 2.10, A, b), or local covering individual cascades (Figure 2.10, b, OS circuit with transfer coefficient g 1).
A
b
Figure 2.10 - Types of feedback
When the oscillations of the signal source are added to the oscillations coming from the output of the amplifier through the feedback circuit, the resulting oscillation is formed at the input of the amplifier. The resulting fluctuation is sum two vibrations, if both of these vibrations are added in phase, or differences two vibrations if they add up out of phase. In the first case, there is positive feedback (POS), in the second - negative feedback (OOS).
Practical coincidence or opposition of phases is possible only in a limited range of amplified frequencies, since the phase shifts inherent in amplifiers change with frequency. This can cause feedback that is negative for some frequencies to become positive for others. Therefore, it is customary to attribute feedback to negative or positive by the fact that what sign does it have in the main part of the amplified frequency range(that is, within the bandwidth of the amplifier).
External feedback created with the help of a special feedback circuit can always be attributed to one form or another, knowing how this circuit is connected to the amplifier.
There are the following four main types of feedback in the amplifier (the first part of the name determines the method of connecting the output of the feedback circuit to the input of the amplifier, and the second - the method of connecting the input of the feedback circuit to the output of the amplifier):
- serial voltage feedback;
- parallel voltage feedback;
- serial feedback over current;
- parallel current feedback.
If the input signal source is connected in series with the input of the amplifier and the output of the feedback circuit, then the feedback is called consistent (Figure 2.11, A). In this case, the feedback signal u sv applied to the input of the amplifier in series with the input signal and in.
Parallel feedback occurs when the feedback circuit is connected in parallel with the input signal source (Figure 2.11, b). With parallel feedback at the input of the amplifier, algebraic addition (taking into account polarity or initial phase) of currents occurs, and not voltages, as in the case of series feedback.
Thus, in series negative feedback, the voltage is used as the feedback signal, which is subtracted from the voltage of the signal source, and in parallel negative feedback, the current is used as the feedback signal, which is subtracted from the current of the external signal source.
a b
Figure 2.11 - Sequential ( A) and parallel ( b) OS
According to the way the feedback is turned on at the output of the amplifier, voltage and current feedback are distinguished. With voltage feedback, the output of the amplifier, the load and the feedback circuit are connected in parallel to each other (Figure 2.12, A). In this case, the feedback signal is proportional to the output voltage of the amplifier. If the amplifier output, load and feedback circuit are connected in series (Figure 2.12, b), then there is current feedback, in which the feedback signal is proportional to the current through the load.
a b
Figure 2.12 - voltage OS ( A) and current ( b)
To determine which OOS is taking place, in terms of current or voltage, the following must be considered. In load short circuit mode (when R N= 0), the voltage feedback disappears, and the current feedback remains. At idle (ie. R N® ¥), voltage feedback is preserved, and current feedback disappears.
Influence of negative feedback on the main parameters and characteristics of amplifiers
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Effect of NFB on Amplifier Gains.
An amplifier covered by feedback (Figure 2.13) can be represented as an amplifier itself (without feedback) with a gain K U, at the input of which the voltage U, and a four-terminal feedback network with a transfer coefficient g.
Figure 2.13 - Amplifier with series CNF circuit
Consider the case when there is a sequential FOS on the input. Then the voltage U in coming from the output of the signal source to the input of the amplifier is opposite in phase to the feedback voltage U sv. In this case, one can write
. (2.24)
We divide the left and right sides of equation (2.24) by U out:
. (2.25)
In equation (2.25) - the voltage gain of the amplifier without feedback. Attitude 
represents the voltage gain of the amplifier covered by the CNF circuit, and is the transfer coefficient of the four-terminal circuit of the CFO. Then equality (2.25) can be rewritten as
,
. (2.26)
Thus, from the obtained expression it can be seen that with a sequential FOS at the input, the voltage gain of the amplifier covered by feedback K U OOS, less than its own gain K U(that is, the voltage gain of the same amplifier, but without the CFO circuit). Moreover, the expression is true, regardless of what type of OOS output is sequential by current or serial by voltage. Product g K U called loop amplification , and the value F= 1 + g K U - depth OOC. For a positive feedback, the feedback depth is determined by the expression: F= 1 - g K U.
The feedback depth shows how many times the gain of the amplifier will change when the feedback circuit is introduced. If the condition g K U>> 1, then the amplifier is said to be covered by deep (one hundred percent) feedback. In this case, the gain of the feedback amplifier is independent of its own gain and is determined only by the feedback gain g. Indeed, under the condition g K U >> 1
. (2.27)
At consistent feedback current amplification factor does not change, since in this case the current amplification factor is equal to
, (2.28)
i.e. does not differ from the current gain of an open-loop amplifier K I. This is explained as follows. With the parameters of the signal source and amplifier load unchanged, negative feedback reduces the signal voltage at the amplifier output by F times and the same number of times the output current decreases. But since serial feedback increases the input impedance of the amplifier also in F times (will be shown later), the input current decreases and the current amplification factor does not change.
At parallel negative feedback (for both current and voltage, Figure 2.14), the voltage gain does not change, that is, in this case, you can write
. (2.29)
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Figure 2.14 - Amplifier with parallel CAB circuit
Let us derive a relation for determining the current amplification factor in the amplifier in the presence of parallel input feedback.
Intrinsic Amplifier Current Gain K I equals:
. (2.30)
Given that , we get
. (2.31)
It can be shown that the resulting expression is valid, no matter what kind of negative output feedback is parallel by current or parallel by voltage.
Influence of NFB on the input and output impedance of the amplifier.
Feedback has a significant effect on the input and output impedances of the amplifier.
Input impedance amplifier with feedback depends on how the feedback circuit is connected to the input of the amplifier and does not depend on how it is connected to the output. output impedance on the contrary, depends on the way the feedback circuit is connected to the output of the amplifier and does not depend on the way it is connected to the input of this amplifier.
Let's see how the influence manifests itself various kinds OOS on amplifier input impedance.
To determine the impact consistent feedback on the input impedance of the amplifier, we use the circuit shown in Figure 2.13. Analysis of the circuit shows that the expression for determining input impedance amplifier with serial feedback will have the form
(2.32)
Where R in- input impedance of the amplifier without OOS;
K U- voltage gain of the amplifier without feedback within the bandwidth (in the mid-range).
From the last expression, it follows that with serial feedback, the input impedance of the amplifier increases by (1 + g K U) once.
However, the input impedance of an amplifier is usually of a complex nature, therefore, in order to fully assess the effect of the NOS on the input impedance, the latter must be written in a complex form
. (2.33)
To determine the impact parallel OOS on the input impedance of the amplifier, we use the circuit shown in Figure 2.14. Circuit analysis shows that parallel FOS reduces the input impedance of the amplifier, since with this type of OOS to the input impedance of the amplifier R in as if a resistance is connected in parallel R sv.
To quantify the impact parallel OOS on the input impedance of the amplifier use the expression:
, (2.34)
or, in general, the expression
. (2.35)
Thus, the OOS allows you to control the value of the input impedance of the amplifier and provide both sufficiently high (hundreds of kΩ - tens of MΩ) - with serial OOS, and sufficiently low (tenths - thousandths of an ohm) - with parallel OOS input resistances.
The output impedance of the amplifier is highly dependent on how the feedback signal is taken. If it is removed by voltage, then the output resistance decreases, and if by current, it increases.
To quantify the effect of voltage feedback on the output impedance of the amplifier, the following expression is used:
, (2.36)
Where R out- output impedance of the amplifier without OOS.
To calculate the output impedance of an amplifier in the frequency range outside the passband, use the expression:
. (2.37)
From the last expression it follows that the introduction of voltage feedback into the amplifier reduces its output impedance F once.
The physical meaning of the voltage feedback action is as follows. Any OOS strives to maintain the value of the parameter that is used to receive feedback unchanged. Therefore, voltage feedback under the action of external disturbances, in particular, when the output current changes, tends to maintain the value of the output voltage of the amplifier unchanged. This is equivalent to reducing its output impedance.
Environmental impact assessment by current on the output impedance of the electronic amplifier is carried out on the basis of the expression
or, respectively,
From (2.39) it follows that with current feedback, the output impedance of the amplifier increases.
Thus, the introduction of feedback can be used to purposefully change the output impedance of the amplifier and allows you to implement an amplifier with a very small (hundredths of Ohms) or very large (hundreds of kOhm - tens of MΩ) output impedance. When using voltage feedback, the amplifier approaches an ideal voltage source, the output signal of which changes little at various load resistances. Current feedback stabilizes the load current, bringing the amplifier closer to an ideal current source.
Influence of NFB on nonlinear distortions and amplitude characteristic of the amplifier.
Previously, it was found that series OOS reduces the voltage gain, and, consequently, reduces the slope of the amplitude characteristic (Figure 2.15). It can be seen from the figure that the introduction of serial feedback into the amplifier leads to expansion its dynamic range (since 
) and to decrease the magnitude of non-linear distortions.
Figure 2.15 - Change in the amplitude characteristic of the amplifier in the presence of the feedback circuit
If the voltage U out 2 (Figure 2.15) - the maximum voltage at the output of the amplifier, at which it can still be considered a linear device - to be taken the same for an amplifier without OOS and an amplifier with OOS (this is acceptable, since the value U out 2 mainly depends on the parameters of the active element used and the voltage of the power supply), then we can write
,
According to (2.12), nonlinear distortions in an open-loop amplifier can be estimated using the formula
,
where is the equivalent total voltage of higher harmonics.
The introduction of a series OOS circuit into the amplifier leads to a decrease in the output voltage of the amplifier, equal to 
, and, consequently, each harmonic of this voltage, in F times, that is, you can write
From (2.41) it follows that in order to maintain the output voltage in an amplifier with CNF at the same level as in an amplifier without CNF, it is necessary to increase the input voltage by F once. But at the same time, the amplitude of the first harmonic in the output voltage, at a constant voltage , will also increase in F once. Then one can write
. (2.42)
So the introduction to the amplifier consistent OOS allows expand its dynamic range and decrease harmonic distortion (reduce non-linear distortion) by about 1 + g K U once.
The effect of environmental feedback on the frequency and phase characteristics of the amplifier.
Previously, when analyzing the effect of feedback on various amplifier parameters, we proceeded from the fact that the gain of the amplifier K U and the NFB circuit gain g are real (i.e., the influence of the NFB was evaluated at frequencies within the passband). However, as shown in § 2.1.3.2, the gain is complex outside the passband.
The transmission coefficient of the CNF circuit in the general case can also be complex. And this means that a real amplifier always introduces additional phase shifts into the amplified signal, the values of which depend on the parameters of the components, the amplifier circuit and the range of amplified frequencies. These phase shifts are due to the presence of reactive elements in the amplifier circuits and the inertial properties of active devices (for example, transistors).
Taking into account the above reasons, expression (2.26) should be written in the form:
, (2.43)
Where 
(j To- phase angle between the output and input voltages amplifier);
(j g - phase angle between the voltages at the output and input of the feedback circuit).
Usually, the complex nature is taken into account at frequencies and less than the changes
For any frequency, the loop gain is a real negative value (phase balance);
The value of the loop gain at this frequency is greater than or equal to unity (amplitude balance).
In single-stage amplifiers, it is most often possible to use a fairly deep feedback without fear that at the edges of the frequency range it can cause self-excitation in the amplifier. At the same time, in multistage amplifiers (which are in most cases used in practice), additional measures must be taken to prevent self-excitation. This is especially important in broadband amplifiers.
Figure 2.17 shows an example of the frequency response of a single-stage amplifier without feedback ( K U(w)) and the same amplifier covered by the OOS circuit ( K Uooc(w)). It can be seen from the figure that when the cascade is covered by the OOS circuit, simultaneously with a decrease in the voltage gain, the bandwidth of the amplifier expands. The cutoff frequencies of the passband of a single-stage amplifier with OOS are determined from the expressions
, (2.45)
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Figure 2.17 - Illustration of the influence of CNF on the bandwidth of the amplifier
Summing up the above, we note that the introduction of frequency-independent feedback improves the frequency characteristics of the amplifier, helps to expand the bandwidth and reduce frequency distortion within a given frequency range. In addition, voltage feedback provides stabilization of the output voltage and voltage gain of the amplifier, and current feedback provides stabilization of the output current.
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