Operational amplifier: switching circuits, operating principle. Non-inverting operational amplifier amplifier circuit. Operational amplifier DC voltage amplifier circuit. Operational amplifier

Hence, .

Since U out = U d · K and U d =U out / K, with K → ∞ and U d ≈ 0, we can write that
. Solving the equation, we obtain an expression for the closed-loop gain K os
,(15.3)

which is valid under the condition K » K ​​os.

In the scheme voltage follower at the op-amp ( Fig. 15.4) U out feedback comes from the amplifier output to the inverting input. Since the voltage difference at the inputs of the op-amp - U d - increases, you can see that the voltage at the output of the amplifier is U out = U d · K.

Fig. 15.4. Voltage follower on op-amp

The output voltage of the op-amp is U out = U in + U d. Since U out = U d · K, we obtain that U d = U out / K. Hence,
. Since K is large (K → ∞), then U out / K tends to zero, and as a result we obtain the equality U in = U out.

The input voltage is connected to ground only through the amplifier's input impedance, which is very high, so the repeater can serve as a good matching stage.

Amplifier with differential input has two inputs, with the inverting and non-inverting inputs being at the same voltage, in in this case equal to Uoc, since the voltage difference between the inverting and non-inverting inputs is very small (usually less than 1 mV).

Rice. 15.5. Amplifier with differential input

If you set U 1 equal to zero and apply an input signal to the input U 2, then the amplifier will act as a non-inverting amplifier, in which the input voltage is removed from the divider formed by resistors R 2 and R΄ os. If both voltages U 1 and U 2 are supplied to the corresponding inputs simultaneously, then the signal at the inverting input will cause such a change in the output voltage that the voltage at the connection point of resistors R 1 and R oc becomes equal to U os, where
.

Due to the fact that the amplifier has a very high input impedance,

we have

.

Solving the resulting equation for U out, we have:

Substituting the expression for Uoc, we get:

If we put R 1 = R 2 and R oc = R´ oc (the situation that most often occurs), we get
. The polarity of the output voltage is determined by the larger of the voltages U 1 and U 2 .

Obviously, if U 2 in Fig. 15.5 is equal to zero, then the amplifier will act in relation to U 1 as an inverting amplifier.

Op-amp circuit input impedance can be defined as follows. A voltage is applied to the differential input resistance of the op-amp r d. U d. Thanks to the availability feedback this voltage is small.

U d = U out /K U = U 1 /(1+K U b), (15.6)

where b = R 1 /(R 1 +R 2) is the transmission coefficient of the divider in the feedback circuit. Thus, only a current equal to U 1 /r d (1+K U b) flows through this resistance. Therefore, the differential input resistance, due to the action of feedback, is multiplied by a factor of 1+K U b. According to Fig. 12, for the resulting input resistance of the circuit we have:

R in = r d (1+K U b)||r in

This value even for operational amplifiers with bipolar transistors at the inputs exceeds 10 9 Ohms. It should be remembered, however, that we are talking exclusively about differential value; this means that changes in the input current are small, while the average value of the input current can take on incomparably larger values.

Rice. 15.6. Circuit of a non-inverting amplifier taking into account the op-amp's own resistances.

Op-amp output impedance an operational amplifier not covered by feedback is given by:

(15.7)

When a load is connected, a slight decrease in the output voltage of the circuit occurs, caused by a voltage drop across rout, which is transmitted to the input of the amplifier through a voltage divider R 1, R 2. The resulting increase in differential voltage compensates for the change in output voltage.

In general, the output resistance can have a fairly high value (in some cases from 100 to 1000 Ohms. Connecting the OS circuit allows you to reduce the output resistance

For an amplifier covered by feedback, this formula takes the form:

(15.8)

In this case, the value of U d does not remain constant, but changes by the amount

dU d = - dU n = -bdU out

For an amplifier with a linear transfer characteristic, the change in output voltage is

dU out =K U dU d - r out dI out

The magnitude of the current branching into the feedback voltage divider in this case can be neglected. Substituting the value dU d into the last expression, we obtain the desired result:

(15.9)

If, for example, b =0.1, which corresponds to amplification of the input signal by 10 times, and K U =10 5, then the output impedance of the amplifier will decrease from 1 kOhm to 0.1 Ohm. The above is generally true within the amplifier bandwidth f p, Hz. For more high frequencies the output impedance of the op-amp with feedback will increase, because value |K U | with increasing frequency it will decrease at a rate of 20 dB per decade (see Fig. 3). At the same time, it acquires an inductive character and at frequencies above f t it becomes equal to the value of the output impedance of the amplifier without feedback.

Dynamic parameters of the op-amp, characterizing the performance of the op-amp can be divided into parameters for small and large signals. The first group of dynamic parameters includes the bandwidth f p, the unity gain frequency f t and the settling time t y. These parameters are called small-signal, because they are measured in linear mode operation of op-amp cascades (DU out<1В). Ко второй группе относятся скорость нарастания выходного напряжения r и мощностная полоса пропускания f р. Эти параметры измеряются при большом дифференциальном входном сигнале ОУ (более 50 мВ). Некоторые из этих парамеров рассмотрены выше. Время установления отсчитывается от момента подачи на вход ОУ ступеньки входного напряжения до момента, когда в последний раз станет справедливым равенство |U вых.уст - U вых(t) | = d, где U вых.уст - установившееся значение выходного напряжения, d - допустимая ошибка.

Operating frequency band or bandwidth The op-amp is determined by the type of amplitude-frequency characteristic taken at the maximum possible amplitude of the undistorted output signal. First, at low frequencies, the amplitude of the signal from the harmonic oscillation generator is set such that the amplitude of the output signal U out.max slightly does not reach the saturation limits of the amplifier. Then the frequency of the input signal is increased. The power passband f p corresponds to a value of U out.max equal to 0.707 of the original value. The magnitude of the power passband decreases as the capacitance of the correction capacitor increases.

Operating Parameters The op-amp determines the permissible operating modes of its input and output circuits and the requirements for power supplies, as well as the temperature range of the amplifier. Limitations on operational parameters are determined by the finite values ​​of breakdown voltages and permissible currents through the op-amp transistors. The main operational parameters include: the nominal value of the supply voltage U p; permissible range of supply voltages; current consumed from source I pot; maximum output current I output max; maximum output voltage values ​​at rated power supply; maximum permissible values ​​of common-mode and differential input voltages

Amplitude-frequency response The operational amplifier is an important factor on which the stability of operation of real circuits with such an amplifier depends. In most operational amplifiers, individual stages are connected to each other via DC galvanic couplings, so these amplifiers do not have a gain roll-off in the low-frequency region and it is necessary to analyze the gain roll-off with increasing frequency.

Fig.15.7 . Operational amplifier frequency response

In Fig. 15.7. shows a typical op-amp frequency response.

Rice. 15.8. Simplified equivalent op-amp circuit

As the frequency increases, the capacitance decreases, which leads to a decrease in the time constant τ = R n* C. Obviously, there must be a frequency above which the voltage at the output U out will be less than KU d.

Expression for gain K at any frequency

looks like
, where K is the gain without feedback on low frequencies;f – operating frequency; f 1 – cut-off frequency or frequency at 3 dB, i.e. the frequency at which K(f) is 3 dB lower than K, or equal to 0.707 A.

If, as is usually the case, Rn » Rout, then
.

Usually the amplitude-frequency response is given in general form. How:

. (15.10)

where f is the frequency we are interested in, while f 1 is a fixed frequency, which is called boundary frequency and is a characteristic of a particular amplifier. As the frequency increases, the voltage gain decreases. In addition, from the expression for θ it is clear that when the frequency changes, the phase of the output signal shifts relative to the phase of the input; - the output signal lags in phase from the input.

Adding negative feedback, such as is done in inverting or non-inverting amplifiers, increases the effective bandwidth of the op-amp.

To see this, consider the expression for the open-loop gain of an amplifier with a roll-off of 6 dB/octave (at twice the frequency):

, where K(f) is the open-loop gain at frequency f; A – gain without feedback at low frequencies; f 1 – coupling frequency. Substituting this relationship into the expression for the gain in the presence of feedback
, we get

. (15.11)

This expression can be rewritten as
, wheref 1 oc = f 1 (1+Aβ); K 1 – gain with closed feedback at low frequencies; f 1oc – cut-off frequency in the presence of feedback.

The cutoff frequency with feedback is equal to the cutoff frequency without feedback multiplied by (1+Kβ)>1, so the effective bandwidth actually increases when feedback is used. This phenomenon is shown in Fig. 8, where f 1oc > f 1 for an amplifier with a gain of 40 dB.

If the amplifier roll-off rate is 6 dB/octave, the product of the gain and the bandwidth is constant: Kf 1 = const. To see this, let's multiply the ideal gain at low frequencies by the upper cutoff frequency of the same amplifier in the presence of feedback.

Then we get the product of gain and bandwidth:

, where K is the open-loop gain at low frequencies.

Whereas previously it was shown that to increase bandwidth using feedback, the gain must be reduced, now the derived relationship makes it possible to find out how much of the gain must be sacrificed to obtain the desired bandwidth.

Operational amplifier equivalent circuit allows you to take into account the influence of amplifier imperfections on the characteristics of the circuit. To do this, it is convenient to imagine the amplifier as a complete equivalent circuit containing significant elements of non-ideality. A complete op-amp equivalent circuit for small slow signal changes is shown in Fig. 15.9.

Rice. 15.9.. Operational amplifier equivalent circuit for small signals

For operational amplifiers with bipolar transistors at the input, the input resistance for the differential signal r d is several megaohms, and the input resistance for the common-mode signal r in is several giga ohms. The input currents determined by these resistances are on the order of several nanoamperes. Direct currents flowing through the inputs of the operational amplifier and determined by the bias of the differential stage transistors have significantly larger values. For universal op-amps, input currents range from 10 nA to 2 μA, and for amplifiers with field-effect transistor input stages, they amount to fractions of nanoamps.

Op amp amplifiers use negative feedback (NFB), so there are several simple rules, which determine the behavior of such an amplifier. There are three simplifying assumptions to make about op-amp properties: the op-amp's open-loop gain and input impedance are infinitely large, and the output impedance is zero.

When analyzing, it should be remembered that the large voltage gain of the op-amp leads to the fact that a change in the voltage between the inputs of a few fractions of a millivolt causes a change in the output voltage within its full range. The first rule follows from this: the op-amp amplifies the voltage difference between the inputs and, due to the external OOS circuit, transfers the voltage from the output to the input in such a way that the voltage difference between the inputs is practically equal to zero.

Input impedance various types Op-amps range from megaohms to thousands of megaohms, input currents range from fractions of nanoamps to picoamps. This gives grounds to formulate the second rule: the inputs of the operational amplifier do not consume current. These rules provide a sufficient basis for the analysis of op-amp circuits. The circuit of the inverting amplifier using an op-amp is shown in Fig.

Rice. Op amp inverting amplifier

Analyzing this circuit taking into account the rules formulated above, it can be shown that when the non-inverting input of the op-amp is grounded, the voltage at the inverting input is also zero. This means that the voltage drop across resistor ROC is equal to UOUT, and the voltage drop across resistor R1 is equal to UIN. If the input currents of the op-amp are equal to zero, then UOUT / ROC = -UIN / R1, voltage gain KU = UOUT / UIN = -ROC / R1. The minus sign indicates that the output signal is inverted relative to the input signal (shifted by 180º).

This circuit is an amplifier DC In this circuit, parallel voltage feedback is implemented, since the feedback signal is not connected in series with the input signal, but is supplied in parallel with it to the same input.

As is known, parallel OOS reduces the input impedance of the amplifier. In the circuit, the potential of the connection point R1 and ROC is always zero, and this point is called “virtual zero” (imaginary ground). Therefore, the input resistance of the circuit is RВХ = R1. The output impedance of the circuit is small and equal to fractions of an ohm. Thus, the disadvantage of the circuit is the low input resistance, especially for amplifiers with high voltage gain, in which resistor R1, as a rule, is small. The advantage of the circuit is the low value of the common-mode voltage, almost equal to zero. The fact that the gain is determined simply by the ratio of two resistances makes the use of an inverting amplifier very flexible.

The practical use of op-amp amplifiers has a number of features. The op-amp must be in active mode, its inputs and outputs should not be overloaded. For example, if you apply too much signal to the input of an amplifier, this will cause the output signal to become equal to the saturation voltage (usually its value is 2 V less than the supply voltage).

The op-amp circuit must have a DC feedback circuit, otherwise the op-amp will definitely go into saturation mode. Many op amps have a fairly small maximum permissible differential input voltage. 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, large input currents will arise.

Due to the presence of an input offset voltage, at zero voltage at the input, the output voltage is equal to UOUT=KUUCCM. For an amplifier with a gain of 100 and an input offset voltage of 2 mV, the output offset voltage can reach ±0.2 V. To solve this problem, you need to use external zero correction circuits (using an op-amp with such capabilities), select an op-amp with a small offset value. If DC amplification is not needed, then separating capacitors can be used series circuit transmission of input and output signals.

If in an inverting amplifier one of the inputs is grounded, then even if perfect setting(UСМ = 0), the amplifier output will have a non-zero output voltage. This is because the input bias current IB creates a voltage drop across the resistors, which is then amplified by the amplifier circuit. In this circuit, the resistance on the inverting input side is determined by resistors R1║ROC, but the bias current is perceived as an input signal similar to the current flowing through R1, and therefore it generates an output bias UCM = ICMROC. To reduce errors caused by the input bias current, turn-on is used an additional resistor between the non-inverting input and the common wire. The value of this resistor should be equal to R2 = R1║ROС. For the example given, R1 = 10 kOhm, ROC = 100 kOhm, R2 = 9.1 kOhm.

Rice. Op amp amplifier with compensation resistor

In order to reduce bias currents and their temperature drifts in practical schemes input impedances typically range from 1 to 100 kΩ.

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 consistent OOS in terms of voltage. With an ideal op-amp ( K d = To oc sf = ¥, R BX= ¥ and R OUT = 0) R OUT F= 0 (negative connection and voltage), R VX. F= ¥ (sequential OOS).

, (5.6)

and according to Fig. 5.4,

Substituting (5.7) into (5.6), we obtain

. (5.8)

The NU gain does not depend on the signal source resistance R C, since the input impedance of the NU is equal to ¥, and the current through R C does not leak, 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 completely repeats the input voltage (only for more high level power). Hence the name - voltage repeater.

Unity transfer coefficient, infinitely large input impedance and zero output impedance makes the repeater an ideal buffer stage (impedance transformer).

The resistive balancing method for this circuit depends on the circumstances. If R C= 0, then the balancing resistor R SM connected in series with the non-inverting input (Fig. 5.5).

In this case D u OUT is described by expression (5.5). Non-zero, but known and fixed internal resistance R C could be balanced only with OS resistors, provided that R 1 R 2 /(R 1 +R 2)=R C. However, 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 current balancing of the circuit should be ensured RCM, connected in series with the inverting input (Fig. 5.6). For this circuit

. (5.9)

If it has an uncertain and unstable value, then it is better to use an op-amp with an input stage (differential) using field-effect transistors.

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 study..

If you need additional material on this topic, or you did not find what you were looking for, we recommend using the search in our database of works:

What will we do with the received material:

If this material was useful to you, you can save it to your page on social networks:

All topics in this section:

Purpose, parameters
Comparators are the simplest analog-to-digital converters (ADCs), i.e. devices that convert continuous signal into discrete ones. They are designed to compare the input sig

Features of the use of semiconductor comparators
The most widely used comparators can be divided into four groups: general use (K521CA2, K521CA5), precision (K521CA3, K597CA3), high-speed (K597CA1, K597CA2) and

Specialized comparators on operational amplifiers
When comparing low frequency signals with high accuracy (tens of microvolts) with minimal power consumption, the use of comparators based on op-amps is often more preferable,

Practical application of operational amplifiers. Part one.

Part one.

Hi all.
In this article we will discuss some aspects practical application operational amplifiers in the daily life of a radio amateur.
Without my thoughts spreading over the tree and without going into the dense theoretical foundations operation of the above-mentioned amplifier, let us nevertheless identify some basic terms and concepts that we will encounter in the future.
So - an operational amplifier. From now on we will call it an op-amp, otherwise it’s too lazy to write it in full every time.
On circuit diagrams, most often it is designated as follows:

The figure shows the three most important outputs of the op-amp - two inputs and an output. Of course, there are also power pins and sometimes frequency correction pins, although the latter is becoming less common - most modern op-amps have it built-in. The two inputs of the op-amp - Inverting and Non-Inverting - are named after their inherent properties. If we apply a signal to the Inverting input, then at the output we will receive an inverted signal, that is, shifted in phase by 180 degrees - mirrored; If we apply a signal to the Non-inverting input, then at the output we will receive a phase-unchanged signal.

As well as the main conclusions, there are also three main properties of the op-amp - you can call them TriO (or LLC - as you like): Very high input resistance, Very high gain (10,000 or more), Very low output resistance. Another very important parameter The op-amp is called the rate of voltage rise at the output (slew rate in Burzhuinsky). It actually means the performance of a given op-amp - how quickly it can change the output voltage when it changes at the input.
This parameter is measured in volts per second (V/sec).
This parameter is important primarily for comrades constructing ultrasonic amplifiers, since if the op-amp is not fast enough, then it will not keep up with the input voltage at high frequencies and significant problems will arise. nonlinear distortion. Most modern op-amps general purpose signal slew rate from 10V/µsec and higher. For high-speed op-amps this parameter can reach a value of 1000V/µsec.
You can evaluate whether a particular op-amp is suitable for your purposes based on the signal slew rate using the formula:

where fmax is the frequency of the sinusoidal signal, Vmax is the rate of rise of the signal, Uout is the maximum output voltage.
Well, let's not pull the cat by the tail anymore - let's get down to the main task of this opus - where, in fact, these cool things can be stuck and what can be obtained from it.

The first circuit for switching on the op-amp is inverting amplifier.

The most popular and frequently encountered op-amp amplifier circuit. The input signal is applied to the inverting input, and the non-inverting input is connected to the common wire.
The gain is determined by the ratio of resistors R1 and R2 and is calculated using the formula:

Why "minus"? Because, as we remember, in an inverting amplifier the phase of the output signal is “mirror” to the phase of the input.
The input resistance is determined by resistor R1. If its resistance is, for example, 100 kOhm, then the input impedance of the amplifier will be 100 kOhm.

The following diagram is inverting amplifier with increased input impedance.
The previous circuit is good for everyone, with the exception of one nuance - the ratio of input impedance and gain may not be suitable for the implementation of any specific project. After all, what happens is - let's say we need an amplifier with K=100. Then, based on the fact that the resistor values ​​should be within reasonable limits, we take R2 = 1 MOhm, and R1 = 10 kOhm. That is, the input impedance of the amplifier will be 10 kOhm, which in some cases is not enough.
In these very cases, you can apply the following scheme:

In this case, the gain is calculated using the following formula:

That is, with the same gain, the resistance R1 can be increased, and therefore the input impedance of the amplifier can be increased.

The gain is determined as follows:

In this case, as you can see, there are no disadvantages - the phase of the signal at the input and output is the same.
The main difference from the inverting amplifier is the increased input resistance, which can reach 10 MΩ and higher.
If, when implementing this circuit in practical designs, it is necessary to provide decoupling from the previous stages by direct current - install a separating capacitor, then you need to connect a resistor with a resistance of about 100 kOhm between the op-amp input and the common wire, as shown in the figure.

Power amplifiers. Linear circuits based on op-amps.

Op-amps are widely used in analog electronics devices. It is convenient to consider the functions implemented by an op-amp with OOS if we represent the op-amp in the form ideal model, which has:

1. The input resistance of the operational amplifier is infinity, the currents of the input electrodes are zero (Rin > ∞, i+ = i- = 0).
2. The output impedance of the operational amplifier is zero, i.e. The operational amplifier on the input side is ideal source voltage (Rout = 0).
3. The voltage gain (differential voltage gain) is infinity, and the differential signal in amplification mode is zero (do not short-circuit the op-amp leads).
4. In saturation mode, the output voltage is equal in magnitude to the supply voltage, and the sign is determined by the polarity of the input voltage. It is useful to note that in saturation mode the differential signal cannot always be assumed to be zero.
5. The common-mode signal has no effect on the op-amp.
6. The zero offset voltage is zero.

Op amp inverting amplifier

The circuit of an inverting amplifier covered by parallel voltage feedback is shown in the figures:

OOS is realized by connecting the amplifier output to the input with resistor R2.

At the inverting input of the op-amp, the currents are summed. Since the op amp input current i- = 0, then i1 = i2. Since i1 = Uin /R1, and i2 = -Uout /R2, then . Ku = = -R2/R1. The "-" sign indicates that the sign of the input voltage is inverted.

In Figure (b), resistor R3 is included in the non-inverting input circuit to reduce the influence of op-amp input currents, the resistance of which is determined from the expression:

The input impedance of the amplifier at low frequencies is approximately equal to Rin.os = ≈ R1

Output resistance Rout.os = significantly less than Rout of the op-amp itself.

Non-inverting op-amp amplifier

The circuit of a non-inverting amplifier covered by series voltage feedback is shown in the figure:

OOS is implemented using resistors R1, R2.

Using the previously accepted assumptions for the ideal model, we obtain

Input resistance: Rin.os → ∞

Output resistance: Rout.os = → 0

The disadvantage of amplification is the presence at the inputs of a common-mode signal equal to Uin.

Voltage follower on op-amp

The repeater circuit obtained from the non-verting amplifier circuit, with R1 → ∞, R2 → 0, is shown in the figure:

Coefficient β = 1, Ku.oc = K/1+K ≈ 1, i.e. the voltage at the input and output of the op-amp are equal: Uin = Uout.

Op-amp voltage adder (inverting adder)

Inverting amplifier circuit with additional input circuits shown in the figure:

Considering that i+ = i- = 0, ioc = - Uout /Roс = Uin1 /R1 + Uin2 /R2 + ... + Uin /Rn, we obtain: Uout = -Roс (Uin1 /R1 + Uin2 /R2 + .. . + Uin / Rn)

If Roс = R1 = R2 = ... = Rn, then Uout = - (Uin1 + Uin2 + ... + Uinn ).

The op-amp operates in linear mode.

To reduce the influence of op-amp input currents, a resistor Re (shown as a dotted line in the figure) with resistance: Re = R1//R2//…//Rn//Roc is included in the non-inverting input circuit.

Op-amp subtractive amplifier

The amplifier circuit with a differential input is shown in the figure:

The amplifier is a combination of inverting and non-inverting amplifiers. In the case under consideration, the output voltage is determined from the expression:

Uout = Uin2 R3/(R3+R4) (1+R2/R1) - Uin1 R2/R1

When R1 = R2 = R3 = R4: Uout = Uin2 - Uin1 – i.e. depends on the difference input signals.

Op amp integrating amplifier

The integrator circuit, in which a capacitor is installed in the OOS circuit, is shown in the figure:

Let it be served at the entrance square pulse Uin. In the interval t1...t2, the amplitude Uin is equal to U. Since the input current of the op-amp is zero, then |iin | = |-ic |, iin = Uin /R1, ic = C dUout /dt.

Uin /R1 = C dUout /dt or

where Uout (0) is the voltage at the output (capacitor C) at the start of integration (at time t1).

τ = R1 · C – integration time constant, i.e. time during which Uout will change by the amount ΔUout = U.

Thus, the output voltage in the interval t1...t2 changes according to a linear law and represents the integral of the input voltage. The time constant must be such that until the end of integration Uout< Eпит .

Differentiation amplifier

By swapping R1 and C1 in the integral, we obtain a differentiating amplifier circuit:

By analogy with the integrating amplifier, we write:

Ic = C dUin /dt, IR2 = -Uout /R

Because |Ic | = |-IR2 |, then Uout = - CR dUin /dt

τ = CR – differentiation constant.

The use of op-amps is far from limited to the above circuits.

Active filters

In electronics, a device is widely used to isolate a useful signal from a number of input signals while simultaneously attenuating interfering signals through the use of filters.

Filters are divided into non-passive ones, made on the basis of capacitors, inductors and resistors, and active ones, based on transistors and operational amplifiers.

Typically used in information electronics active filters. The term “active” is explained by the inclusion in the RLC circuit of an active element filter (from a transistor or op-amp) to compensate for losses on passive elements.

A filter is a device that passes signals in the passband and delays them in the rest of the frequency range.

By appearance Frequency response filters are divided into low-pass filters (LPF), and filters treble(HPF), bandpass filters and notch filters.

The diagram of the simplest low-pass filter and its frequency response are shown in the figure:

In the passband 0 - fc, the useful signal passes through the low-pass filter without distortion.

Fс – fз – transition strip,
fз - ∞ – stop band,
fс – cutoff frequency,
fз – delay frequency.

The high-pass filter allows high-frequency signals to pass through and blocks low-frequency signals.

A bandpass filter passes signals from one frequency band located in some inner part of the frequency axis.

The filter circuit is called the Wien bridge. At frequency f0 =

The Wien bridge has a transmission coefficient β = 1/3. With R1 = R2 = R and C1 = C2 = C

A notch filter does not allow signals within a certain frequency band to pass through and allows signals at other frequencies to pass through.

The filter circuit is called an unbalanced double T-bridge.

Where R1 = R2 = R3 = R, C1 = C2 = C3 = C.

As an example, consider a two-pole (according to the number of capacitors) active low-pass filter.

The op-amp operates in linear mode. When calculating, fс is specified. The gain in the passband must satisfy the condition: K0 ≤ 3.

If we take C1 = C2 = C, R1 = R2 = R, then C = 10/fc, where fc is in Hz, C is in µF,

To get more quick change the gain factor away from the passband, similar circuits are sequentially switched on.

By swapping resistors R1, R2 and capacitors C1, C2, we get a high-pass filter.

Selective amplifiers

Selective amplifiers allow you to amplify signals in a limited range of frequencies, highlighting the useful signals and attenuating all others. This is achieved by using special filters in the feedback circuit of the amplifier. The circuit of a selective amplifier with a double T-shaped bridge in a negative feedback circuit is shown in the figure:

The filter transmission coefficient (curve 3) decreases from 0 to 1. The frequency response of the amplifier is illustrated by curve 1. At the quasi-resonant frequency, the filter transmission coefficient in the negative feedback circuit is zero, Uout is maximum. At frequencies to the left and right of f0, the filter transmission coefficient tends to unity and Uout = Uin. Thus, the filter allocates the passband Δf, and the amplifier performs the analog amplification operation.

Harmonic generators

Control systems use signal generators various types. A harmonic oscillation generator is a device that creates an alternating sinusoidal voltage.

The block diagram of such a generator is shown in the figure:

There is no input signal. Uout = K · Uos .

For the occurrence sinusoidal oscillations The self-excitation condition must be satisfied for only one frequency:
K γ = 1 – amplitude balance,
φ + ψ = 2πn – phase balance,
where K is the coefficient amplifier gain,
γ – positive feedback link transmission coefficient,
φ – phase shift for the amplifier,
ψ – phase shift for the feedback circuit,
n = 0, 1, ...

The main generators of sinusoidal signals are filters, such as the Wien bridge. An op-amp-based generator containing a Wien bridge is shown in the figure:

The generator produces a sinusoidal signal with a frequency of .

At frequency f0, the filter transmission coefficient is β = 1/3. The amplifier must have a gain K ≥ 3, which is set by resistors R1 and R2. An important problem is the stabilization of the amplitude Uout, which is ensured by resistor R3 and zener diodes VD1 and VD2. At low Uout, the voltage on VD1 and VD2 is less than the stabilization voltage and R3 is not shunted by zener diodes. In this case, K > 3 and Uout increases. When the voltage on the zener diodes reaches equal to the stabilization voltage, one or another zener diode opens and a pair of zener diodes shunts resistance R3. The gain becomes equal and the voltage Uout begins to decrease, the gain again becomes greater than 3 and Uout will again decrease, but in the opposite direction. This way the zener diodes prevent saturation.

When using this generator, it is advisable to connect the load through a buffer cascade.