A non-inverting amplifier (NA) is an amplifier that has a stable gain with zero phase difference between the input and output signals.
In the NU (Fig. 5.3), there is a sequential OOS for voltage. For an ideal OA ( K d = K oc sf = ¥, R IN= ¥ and R OUT = 0) R OUT. F= 0 (connection is negative and voltage), R IN. F= ¥ (sequential OOS).
, (5.6)
and according to fig. 5.4,
Substituting (5.7) into (5.6), we obtain
. (5.8)
The gain of the NU does not depend on the impedance of the signal source R C, since the input resistance of the NU is ¥, and the current through R C does not flow, then there is no voltage drop across this resistance and 
... At R 2 = 0, R 1 = ¥ K e F= 1. This means that the output voltage is exactly the same as the input voltage (only at a higher power level). Hence the name - voltage follower.
The unit gain, infinitely high input impedance and zero output impedance make the repeater an ideal buffer stage (impedance transformer).
The method of resistive balancing of this circuit depends on the circumstances. If R C= 0, then the balun resistor R CM turns on in series with a non-inverting input (Fig.5.5).
Moreover, D u OUT is described by expression (5.5). Non-zero but known and fixed internal resistance R C could only be balanced with OS resistors, provided that R 1 R 2 /(R 1 +R 2)=R C. However, in this case, the gain of the circuit (5.8) will also change. Simpler resistors R 1 and R 2 should be selected based on the required gain, and the current balancing of the circuit should be ensured R CM connected in series with the inverting input (fig. 5.6). For this scheme
. (5.9)
If it has an undefined and unstable value, then it is better to use an op-amp with an input stage (differential) on field-effect transistors.
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To reduce the potential component of the output static error D u OUT it is necessary either to use the appropriate outputs of the op-amp, or, if they are absent, to balance the circuit at the input (Fig. 5.7). Setting zero in this circuit lowers its gain slightly.
End of work -
This topic belongs to the section:
Analog electronic devices
Analog electronic devices. Part II. Lecture notes for students of the specialty "Radio Engineering" of all forms of education ..
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Purpose, parameters
Comparators are the simplest analog-to-digital converters (ADCs), i.e. devices that convert a continuous signal into a discrete signal; they are designed to compare the input signal
Features of the use of semiconductor comparators
The most widely used comparators can be divided into four groups: general use (K521SA2, K521SA5), precision (K521SA3, K597SA3), high-speed (K597SA1, K597SA2) and
Specialized comparators based on operational amplifiers
When comparing low-frequency signals with high accuracy (tens of microvolts) with minimal power consumption, the use of op-amp-based comparators is often preferable.
Something often began to ask me questions about analog electronics. Did the session take students for eggs? ;) Okay, it's high time to move a small educational program. In particular, on the operation of operational amplifiers. What is it, what is it eaten with and how to calculate it.
What is it
An operational amplifier is an amplifier with two inputs, nevier ... hhm ... a large signal gain and one output. Those. we have U out = K * U in and K is ideally equal to infinity. In practice, of course, there are more modest numbers. Let's say 1,000,000. But even such numbers blow your mind when you try to apply them directly. Therefore, as in kindergarten, one Christmas tree, two, three, many Christmas trees - we have a lot of reinforcement here;) And that's it.
And there are two entrances. And one of them is straight and the other is inverse.
Moreover, the inputs are high resistance. Those. their input impedance is equal to infinity in the ideal case and VERY much in real life. The account there goes to hundreds of MegaOhms, and even to gigaohms. Those. it measures the voltage at the input, but is minimally affected. And we can assume that the current does not flow in the op-amp.
In this case, the output voltage is calculated as:
U out = (U 2 -U 1) * K
Obviously, if the voltage at the direct input is greater than at the inverse one, then at the output plus infinity. Otherwise, it will be minus infinity.
Of course, in a real circuit, there will be no plus and minus infinity, and they will be replaced by the highest and lowest possible supply voltage of the amplifier. And we will get:
Comparator
A device that allows you to compare two analog signals and make a verdict - which of the signals is greater. Already interesting. You can think of a lot of applications for it. By the way, the same comparator is built into most microcontrollers, and I showed how to use it using the AVR example in articles and about creation. Also the comparator is great for creating.
But the matter is not limited to one comparator, because if you introduce feedback, then a lot can be done from the op-amp.
Feedback
If we take the signal from the output and send it straight to the input, then there will be a feedback.
Positive feedback
Let's take and drive into a direct input the signal immediately from the output.
- U1 voltage is greater than zero - at the output -15 volts
- U1 voltage is less than zero - at the output +15 volts
What happens if the voltage is zero? In theory, the output should be zero. But in reality, the voltage will NEVER be zero. After all, even if by one electron the charge of the right one outweighs the charge of the left one, then this is already enough to roll the potential into the output at an infinite gain. And at the output, a shaped hell will begin - signal jumps here and there with the speed of random perturbations induced at the inputs of the comparator.
To solve this problem, hysteresis is introduced. Those. a kind of gap between switching from one state to another. For this, a positive feedback is introduced, like this:
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We consider that at the inverse input at this moment there is +10 volts. At the output from the op-amp minus 15 volts. At the direct input, it is no longer zero, but a small part of the output voltage from the divider. About -1.4 volts Now, until the voltage at the inverse input drops below -1.4 volts, the op-amp output will not change its voltage. And as soon as the voltage drops below -1.4, the output of the op-amp will sharply jump to +15 and there will already be an offset of +1.4 volts at the direct input.
And in order to change the voltage at the output of the comparator, the signal U1 will need to increase by as much as 2.8 volts to get to the upper bar of +1.4.
There is a kind of gap where there is no sensitivity, between 1.4 and -1.4 volts. The gap width is controlled by the ratio of the resistors in R1 and R2. The threshold voltage is calculated as Uout / (R1 + R2) * R1 Let's say 1 to 100 will give +/- 0.14 volts.
Still, op amps are more often used in negative feedback mode.
Negative feedback
Okay, let's stick it differently:
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In the case of negative feedback, the op-amp has an interesting property. It will always try to adjust its output voltage so that the input voltages are equal, resulting in zero difference.
Until I read it in the great book from comrades Horowitz and Hill, I could not fit into the work of the OU. But everything turned out to be simple.
Repeater
And we got a repeater. Those. at the input U 1, at the inverse input U out = U 1. Well, it turns out that U out = U 1.
The question is, what for is such happiness to us? It was possible to directly throw the wire and no op-amp would be needed!
It is possible, but not always. Imagine this situation, there is a sensor made in the form of a resistive divider:
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The lower resistance changes its value, the distribution of the output voltages from the divider changes. And we need to take readings from it with a voltmeter. But the voltmeter has its own internal resistance, albeit large, but it will change the readings from the sensor. Moreover, if we do not want a voltmeter, but want the light bulb to change brightness? There is no way to connect a light bulb here already! Therefore, the output is buffered by an operational amplifier. Its input impedance is huge and it will have a minimal effect, and the output can provide quite tangible current (tens of milliamperes, or even hundreds), which is quite enough for the light bulb to work.
In general, you can find applications for a repeater. Especially in precision analog circuits. Or where the circuitry of one stage can influence the operation of another to separate them.
Amplifier
And now let's make a feint with our ears - take our feedback and put it on the ground through the voltage divider:
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The inverse input now supplies half of the output voltage. And the amplifier still needs to equalize the voltages at its inputs. What will he have to do? That's right - to raise the voltage at its output twice as high as before in order to compensate for the resulting divider.
Now there will be U 1 on the straight line. On inverse U out / 2 = U 1 or U out = 2 * U 1.
Let's put a divisor with a different ratio - the situation will change in the same way. In order not to twirl the voltage divider formula in your mind, I will give it right away:
U out = U 1 * (1 + R 1 / R 2)
It is mnemonically remembered what is divided into very simply:
In this case, it turns out that the input signal goes through the chain of resistors R 2, R 1 to U out. In this case, the direct input of the amplifier is set to zero. We recall the behavior of the op-amp - he will try, by hook or by crook, to make sure that a voltage equal to the direct input is formed at its inverse input. Those. zero. The only way to do this is to lower the output voltage below zero so that zero appears at point 1.
So. Let's imagine that U out = 0. So far it is equal to zero. And the voltage at the input is, for example, 10 volts relative to U out. The divisor of R 1 and R 2 will halve it. So there are five volts at point 1.
Five volts is not zero and the op-amp lowers its output until there is zero at point 1. For this, the output should be (-10) volts. In this case, the difference relative to the input will be 20 volts, and the divider will provide us exactly 0 at point 1. We got an inverter.
But you can also choose other resistors so that our divider gives out other coefficients!
In general, the gain formula for such an amp will be as follows:
U out = - U in * R 1 / R 2
Well, and a mnemonic picture for quickly memorizing xy from xy.
Let's say U 2 and U 1 will be 10 volts each. Then at the 2nd point there will be 5 volts. And the output will have to be such that at the 1st point there will also be 5 volts. That is, zero. So it turns out that 10 volts minus 10 volts is equal to zero. That's right :)
If U 1 becomes 20 volts, then the output should drop to -10 volts.
Calculate for yourself - the difference between U 1 and U out will be 30 volts. The current through the resistor R4 will be (U 1 -U out) / (R 3 + R 4) = 30/20000 = 0.0015A, and the voltage drop across the resistor R 4 will be R 4 * I 4 = 10000 * 0.0015 = 15 volts ... Subtract the 15 volt drop from the input 20 to get 5 volts.
Thus, our op-amp solved the arithmetic problem from 10, subtracted 20, getting -10 volts.
Moreover, the problem has coefficients determined by resistors. It's just that for me, for simplicity, the resistors are of the same value and therefore all the coefficients are equal to one. But in fact, if we take arbitrary resistors, then the dependence of the output on the input will be as follows:
U out = U 2 * K 2 - U 1 * K 1
K 2 = ((R 3 + R 4) * R 6) / (R 6 + R 5) * R 4
K 1 = R 3 / R 4
The mnemonic technique for memorizing the formula for calculating the coefficients is as follows:
Directly according to the scheme. The numerator of the fraction is at the top, so we add the upper resistors in the current flow circuit and multiply by the lower one. The denominator is at the bottom, so we add the lower resistors and multiply by the upper one.
Everything is simple here. Because point 1 is constantly reduced to 0, then we can assume that the currents flowing into it are always equal to U / R, and the currents entering the node number 1 are summed up. The ratio of the input resistor to the feedback resistor determines the weight of the input current.
There can be as many branches as you like, but I drew only two.
U out = -1 (R 3 * U 1 / R 1 + R 3 * U 2 / R 2)
The resistors at the input (R 1, R 2) determine the amount of current, which means the total weight of the input signal. If we make all the resistors equal, like mine, then the weight will be the same, and the multiplication factor of each term will be 1. And U out = -1 (U 1 + U 2)
Non-inverting adder
Everything is a bit more complicated here, but it looks like.
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Uout = U 1 * K 1 + U 2 * K 2
K 1 = R 5 / R 1
K 2 = R 5 / R 2
Moreover, the resistors in the feedback should be such that the equation R 3 / R 4 = K 1 + K 2
In general, on operational amplifiers you can do any math, add, multiply, divide, calculate derivatives and integrals. And almost instantly. Analog computers are made at the op-amp. I even saw one of these on the fifth floor of SUSU - a fool the size of a room. Several metal cabinets. The program is typed by connecting different blocks with wires :)
24. Inverting amplifier for op.
25. Non-inverting amplifier on op.
A schematic diagram of a non-inverting amplifier is shown in Fig. 9.6. The expression for the voltage gain for this circuit is obtained, in the same way as for the previous one, from the equations drawn up according to Kirchhoff's law
Taking into account (9.13), the expression for the gain will have the form
It follows that the voltage gain in the non-inverting amplifier circuit is always greater than 1. In contrast to the inverting amplifier circuit in this circuit, the op-amp is covered by a voltage feedback circuit sequential at the input. Therefore, the input impedance of this circuit is much higher than the input impedance of an op-amp without an OS: 
The output impedance is determined, as for an inverting amplifier, according to (9.16).
26. Scheme of the adder on the op.
Summing circuits include adders and subtraction circuits. These circuits are used to solve algebraic equations and in analog signal processing devices. An adder is a device at the output of which the signals supplied to its inputs are summed. The adders are built using inverting and non-inverting amplifiers.
Inverting adder
An inverting adder circuit with three input signals is shown in Fig. 11.10. For simplicity of reasoning, we assume that R1 = R2 = R3 = Roc. 
Since the ideal op-amp K U → ∞, Rvx → ∞, and the bias current is very small compared to the feedback current, then according to Kirchhoff's law I1 + I2 + I3 = Ios. (11.19) Due to the fact that the inverting input has practically zero potential, there is no mutual influence of the input signals in it. Expression (11.19) can be represented as Consequently, the output is the inverted sum of the input voltages. If R1 ≠ R2 ≠ R3, then the output is the inverted sum of the input voltages (11.20) with different scale factors. The inverting adder combines the functions of an adder and an amplifier while maintaining circuit simplicity. Resistor R is used to compensate for the offset zero at the op-amp output caused by the time and temperature fluctuations in the input current. Resistance R is chosen current value so that the equivalent resistances connected to the inputs of the op-amp are the same: R = Roc || R1 || R2 || R3. 
Non-inverting adder
The circuit of a non-inverting adder, which is built on the basis of a non-inverting amplifier, is shown in Fig. 11.11. Since at U0 = 0 the voltages at the inverting and non-inverting inputs are equal, then
Considering that RinxОУ at the non-inverting input is very large, the input current is 0. According to Kirchhoff's law, you can write
If in the circuit (Fig. 11.11) signals are still supplied to the inverting inputs, then the circuit performs an addition-subtraction operation. For the adder to work correctly, it is necessary to balance the inverting and non-inverting gain, i.e. ensure the equality of the sums of the gains of the inverting and non-inverting parts of the circuit.
27. Differentiating amplifier on op.
The differentiating amplifier (differentiator) is designed to obtain an output signal proportional to the rate of change of the input. When differentiating the signal, the op-amp must pass only the AC component of the input voltage, and the gain of the differentiating link must increase with an increase in the rate of change of the input voltage. The circuit of the differentiator, at the input of which the capacitor C is connected, and the resistor in the OS circuit, is shown in Fig. 11.13. Assuming that the op-amp is ideal, the current through the feedback resistor can be considered equal to the current through the capacitor Iс + Ir = 0,
, then
The considered differentiator is rarely used due to the following disadvantages:
1. Low input impedance at high frequencies, determined by the capacitance C;
2. Relatively high output noise due to high gain at high frequencies;
3. Propensity to self-excitement. (this circuit can be unstable in the frequency range where the frequency response of the differentiator (curve 1 in Figure 11.14), which has a rise of 20 dB / dec, intersects with the frequency response of the corrected op-amp, which has a roll-off of −20 dB / dec (curve 2 in Figure 11.14) . The amplitude-frequency characteristic of an open-loop system in some part of the frequency range has 
the decay of –40 dB / dec, which is determined by the difference between the slopes of curves 1 and 2, and the phase shift ϕ = –180 °, which indicates the possibility of self-excitation.)
To avoid the manifestation of these shortcomings of the differentiator, the following circuit solutions are taken:
1. The feedback resistor is shunted by a capacitor, the capacitance of which is chosen such that the frequency response of the op-amp with a drop of -20 dB / dec starts at a frequency higher than the maximum frequency of the useful differential signal. This results in a reduction in high frequency noise components in the output signal. This segment begins at the frequency f = 1 / (2πRocCoc).
2. A resistor is connected in series with the input capacitor C, which limits the gain at high frequencies of the differentiator. This provides dynamic stability and reduces the input capacitive current from the signal source.
3. The use of op amps with low bias voltage and low input currents, as well as capacitors with low leakage currents and low noise resistors.
A practical diagram of the differentiator and its frequency response are shown in
rice. 11.15. The introduction of the resistor R leads to the appearance on the frequency response (curve 1 in Fig. 11.15, b) of a horizontal section, where there is no differentiation at frequencies exceeding the frequency
In a non-inverting amplifier, the input signal is fed to the non-inverting input of the op-amp (+), this is the main difference between the non-inverting amplifier on the op-amp from. In this case, the signal source "sees" the infinite input impedance of the op-amp. The zero offset voltage is zero, and therefore the inverting input of the op amp must be at the same potential as the non-inverting input. The current from the op-amp output creates a voltage drop across the resistor R G, which must be equal to the input voltage V IN.
Rice. 1. Non-inverting op-amp
To calculate the output voltage V OUT and the gain, the voltage divider calculation rule will be used:
After transformation, an expression for the gain is obtained in the following form:
It is important to note that expression (2) contains only the denominations of passive elements.
If the resistance of the resistor R G is chosen much more than R F, then the ratio (R F / R G) tends to zero, and at zero resistance R F, expression (2) is converted into
In this case, the non-inverting amplifier turns into a buffer (signal follower) with a unity gain, with infinite input and zero output resistances. Resistor R G in this case can also be excluded from the circuit. In practice, some op amps may “burn out” when turned on without resistor R F. For this reason, this resistor is present in many buffer designs. Its function is to protect the inverting input from voltage surges by limiting the current to a safe level. The commonly used value for this resistor is 20 kΩ. In drain feedback amplifier circuits, the resistor R F determines stability and is always required. However, do not be lazy and look through the datasheet for the op-amp. If the inclusion is described there as in Fig. 2 - feel free to use!
It was shown that when using an operational amplifier in various switching schemes, the gain of a stage on a single operational amplifier (op-amp) depends only on the depth of the feedback. Therefore, the formulas for determining the gain of a particular circuit do not use the gain of the "bare" op-amp itself, so to speak. That is, just that huge coefficient, which is indicated in reference books.
Then it is quite appropriate to ask the question: "If the final result (gain) does not depend on this huge" reference "coefficient, then what is the difference between an op-amp with a gain of several thousand times, and with the same op-amp, but with a gain of several hundred thousand and even millions? "
The answer is simple enough. In either case, the result will be the same, the amplification of the cascade will be determined by the OOS elements, but in the second case (op-amp with high gain), the circuit works more stably, more accurately, the speed of such circuits is much higher. It is not for nothing that op-amps are divided into general-use op-amps and high-precision, precision ones.
As already mentioned, the name "operational" considered amplifiers received at that distant time, when they were mainly used to perform mathematical operations in analog computers (AVM). These were operations of addition, subtraction, multiplication, division, squaring and many other functions.
These antediluvian op amps were made on electronic tubes, later on discrete transistors and other radio components. Naturally, the dimensions of even transistor op amps were large enough to be used in amateur designs.
And only after, thanks to the achievements of integrated electronics, op-amps became the size of an ordinary low-power transistor, then the use of these parts in household equipment and amateur circuits became justified.
By the way, modern op-amps, even of fairly high quality, are priced slightly higher than two or three transistors. This statement applies to general-purpose op amps. Precision amplifiers can cost a little more.
As for the circuits on the op-amp, it is worth immediately making a remark that they are all designed to be powered from a bipolar power source. This mode is the most "usual" for an op amp, allowing to amplify not only AC voltage signals, such as a sinusoid, but also DC signals or simply voltage.
And yet, quite often, the power supply of circuits at the op-amp is made from a unipolar source. True, in this case, it is not possible to increase the constant voltage. But it often happens that this is simply not necessary. We will talk about circuits with a unipolar power supply later, but for now we will continue about the circuits for switching on an op-amp with a bipolar power supply.
The supply voltage of most op amps is most often within ± 15V. But this does not mean at all that this voltage cannot be made somewhat lower (higher is not recommended). Many op amps operate very stably from ± 3V, and some models even ± 1.5V. This possibility is indicated in the technical documentation (DataSheet).
Voltage follower
It is the simplest device on an op-amp in circuitry, its circuit is shown in Figure 1.
Figure 1. Circuit of a voltage follower on an operational amplifier
It is easy to see that not a single detail was needed to create such a circuit, except for the op-amp itself. True, the figure does not show the power connection, but such an outline of the circuits is found all the time. The only thing I would like to note is that between the power pins of the op-amp (for example, for op-amp KR140UD708, these are pins 7 and 4) and the common wire should be connected with a capacity of 0.01 ... 0.5 mkF.
Their purpose is to make the operation of the op-amp more stable, to get rid of self-excitation of the circuit in the power supply circuits. Capacitors should be connected as close as possible to the power supply pins of the microcircuit. Sometimes one capacitor is connected based on a group of several microcircuits. The same capacitors can be seen on boards with digital microcircuits, their purpose is the same.
The repeater gain is equal to unity, or, to put it another way, there is no gain at all. Then why such a scheme is needed? Here it is quite appropriate to remember that there is a transistor circuit - an emitter follower, the main purpose of which is to match cascades with different input resistances. Such cascades (repeaters) are also called buffer cascades.
The input impedance of the repeater on the op-amp is calculated as the product of the input impedance of the op-amp and its gain. For example, for the mentioned UD708, the input impedance is approximately 0.5 MΩ, the gain is at least 30,000, and maybe even more. If these numbers are multiplied, then the input resistance is 15GΩ, which is comparable to the resistance of not very high-quality insulation, for example, paper. Such a high result is unlikely to be achieved with a conventional emitter follower.
So that the descriptions do not give rise to doubts, below there will be figures showing the operation of all the described circuits in the Multisim simulator program. Of course, all these circuits can be assembled on prototypes, but you can get no worse results on the monitor screen as well.
Actually, it's even a little better here: you don't have to climb somewhere on the shelf to change the resistor or microcircuit. Everything here, even the measuring devices, is in the program, and "gets" with the help of a mouse or keyboard.
Figure 2 shows the repeater circuit made in the Multisim program.
Figure 2.
The study of the circuit is quite simple to carry out. A sinusoidal signal with a frequency of 1KHz and an amplitude of 2V is applied to the input of the follower from the function generator, as shown in Figure 3.
Figure 3.
The signal at the input and output of the repeater is observed by an oscilloscope: the input signal is displayed as a blue beam, the output beam is red.
Figure 4.
And why, the attentive reader will ask, is the output (red) signal twice the input blue? Everything is very simple: with the same sensitivity of the oscilloscope channels, both sinusoids with the same amplitude and phase merge into one, hide behind each other.
In order to see both of them at once, we had to reduce the sensitivity of one of the channels, in this case the input one. As a result, the blue sine wave became exactly half the size on the screen, and stopped hiding behind the red one. Although to achieve a similar result, you can simply shift the beams using the oscilloscope controls, leaving the sensitivity of the channels the same.
Both sinusoids are located symmetrically about the time axis, which means that the constant component of the signal is zero. What happens if you add a small DC component to the input signal? The virtual generator allows you to shift the sine wave along the Y axis. Let's try to shift it upwards by 500mV.
Figure 5.
What came of this is shown in Figure 6.
Figure 6.
It is noticeable that the input and output sinusoids have risen up by half a volt, while not changing at all. This indicates that the repeater also accurately transmitted the DC component of the signal. But most often they try to get rid of this constant component, make it equal to zero, which makes it possible to avoid the use of such circuit elements as interstage decoupling capacitors.
A repeater is, of course, good and even beautiful: not a single additional detail was needed (although there are repeater circuits with insignificant "additions"), but they did not receive any amplification. What kind of amplifier is it then? To get an amplifier, it is enough to add just a few details, how to do this will be described further.
Inverting amplifier
In order to make an inverting amplifier from the op-amp, it is enough to add only two resistors. What came of this is shown in Figure 7.
Figure 7. Schematic of an inverting amplifier
The gain of such an amplifier is calculated by the formula K = - (R2 / R1). The minus sign does not mean that the amplifier turned out to be bad, but only that the output signal will be opposite in phase to the input signal. No wonder the amplifier is called inverting. Here it would be appropriate to recall a transistor connected according to a circuit with an OE. There, too, the output signal on the collector of the transistor is in antiphase with the input signal fed to the base.
This is where it is worth remembering how much effort will have to be made in order to obtain a pure undistorted sinusoid on the collector of the transistor. It is required to appropriately select the bias based on the transistor. This is usually quite difficult, depending on many parameters.
When using an op-amp, it is enough to simply calculate the resistance of the resistors according to the formula and get the specified gain. It turns out that setting up an op-amp circuit is much easier than setting up multiple transistor stages. Therefore, there is no need to be afraid that the scheme will not work, it will not work.
Figure 8.
Here everything is the same as in the previous figures: the input signal is shown in blue, and in red it is after the amplifier. Everything corresponds to the formula K = - (R2 / R1). The output signal is in antiphase with the input signal (which corresponds to the minus sign in the formula), and the amplitude of the output signal is exactly twice the input. This is also true when the ratio (R2 / R1) = (20/10) = 2. To make the gain, for example, 10, it is enough to increase the resistance of the resistor R2 to 100KΩ.
In fact, the inverting amplifier circuit can be somewhat more complicated, this option is shown in Figure 9.
Figure 9.
A new part appeared here - resistor R3 (rather, it simply disappeared from the previous circuit). Its purpose is to compensate for the input currents of a real op-amp in order to reduce the temperature instability of the constant component at the output. The value of this resistor is chosen according to the formula R3 = R1 * R2 / (R1 + R2).
Modern highly stable op-amps allow connecting a non-inverting input to the common wire directly without resistor R3. Although the presence of this element will not do anything bad, but at the current scale of production, when they save on everything, they prefer not to install this resistor.
Formulas for calculating the inverting amplifier are shown in Figure 10. Why in the figure? Yes, just for clarity, in a line of text they would not look so familiar and understandable, they would not be so noticeable.
Figure 10.
The gain was mentioned earlier. Only the input and output resistances of a non-inverting amplifier are worthy of attention here. With the input resistance, everything seems to be clear: it turns out to be equal to the resistance of the resistor R1, but the output resistance will have to be calculated using the formula shown in Figure 11.
The letter K ”denotes the reference coefficient of the DT. Here, please, calculate what the output impedance will be. It will turn out to be a rather small figure, even for an average OA of the UD7 type with its K ”equal to no more than 30,000 , of course, within the limits, you can connect to this cascade.
A separate note should be made about the unit in the denominator of the formula for calculating the output resistance. Suppose that the ratio R2 / R1 will be, for example, 100. This is the ratio that will be obtained in the case of the gain of the inverting amplifier 100. It turns out that if this unit is discarded, then nothing will change much. In fact this is not true.
Let's assume that the resistance of resistor R2 is equal to zero, as is the case with the repeater. Then, without one, the entire denominator turns to zero, and the output resistance will be the same zero. And if then this zero turns out to be somewhere in the denominator of the formula, how do you order to divide by it? Therefore, it is simply impossible to get rid of this seemingly insignificant unit.
In one article, even a fairly large one, you can't write everything. Therefore, you will have to everything that does not fit into the next article. There will be a description of a non-inverting amplifier, a differential amplifier, an amplifier with a unipolar supply. There will also be a description of simple circuits for testing the op-amp.
