How to choose an operational amplifier. What is an operational amplifier and how does it work? Positive feedback, negative feedback
The main active element of modern analog circuitry is an operational amplifier - a complex circuit made in an integrated design (i.e., an integrated microcircuit). The term "operational amplifier" (abbreviated as op-amp) historically goes back to tube analog computers (AVM) - devices that allow some non-electrical processes to be represented by changes in time of electrical quantities (currents, voltages); in other words, “operations” are performed on currents and voltages in the AVM. The main components of the AVM are amplifiers, the transmission coefficients of which can be quickly changed during operation (using jumpers and potentiometers). It was these amplifiers that were first called "operating".
The scope of OA application has now expanded significantly,
the technology of their manufacture has also changed. However, the main advantage remained - the ability to quickly and without large costs change not only the gain of the amplifier, but also generally change the purpose and function of the electronic circuit. As a rule, a common amplifier is used in combination with two or three additional elements: resistances, capacitors, diodes, etc. The nature of the connection of these additional elements, as will be shown in this section, determines the fundamental properties of the resulting electronic circuit. Changing just one element radically changes the function and purpose of the circuit.
On the diagrams, next to the figure denoting the DU, the letters are usually put DA, which corresponds to an analog microcircuit (as opposed to a digital, that is, a "discrete" microcircuit having a letter designation DD). Operational amplifiers (microcircuits) produced by the industry of the Russian Federation are series (140 series, 544 series, etc.); a sign that some microcircuit is an op-amp are the letters UD (less often - UT), for example 140UD8A. A simplified block diagram of such an op-amp is shown in Fig. 5.2. As can be seen from the figure, there are four main units in the circuit: a differential amplifier DU (1), a linear amplifier LU (2), an amplifier-limiter UO (3) and an emitter follower EP (4). The remote control provides amplification of the difference between two signals arriving at the non-inverting and inverting inputs of the op-amp (respectively, 
and 
). The LO consists of several amplifier stages and has a huge overall gain. The presence of the UO allows you to use the OA as a converter of the waveform, expands the scope of their application. The final block of the op-amp - emitter follower - performs the function of a resistance transformer and determines the value of the op-amp output resistance R you x. Usually R you x has the order of units of kilo-ohms, for some types of op-amp - hundreds of ohms. Without ES value R you x would be larger: thus, thanks to the presence of the electric drive, the op-amp is protected from shunting by a low-impedance load.
|
|
The main parameters and characteristics of the op-amp. As with any amplifier, the important parameters of an op-amp are the amplitude (transfer) characteristic, the gain, the amplitude-frequency characteristic (AFC), the phase-frequency characteristic (PFC), as well as the input and output impedances. Obviously, since the op-amp has two inputs, each of the listed parameters, except R output, should be considered separately for the case when the amplified signal arrives at the inverting input (with inverting turn on), and for the case when the non-inverting input is used (with non-inverting turn on). The given set of parameters characterizes the amplifier in a linear mode, that is, with a "small" signal. If, when the signal passes through the op amp, its shape changes due to nonlinear distortions, then you have to use other parameters that describe the output signal as a pulse. These are the slew rate of the output signal, the amplitude of the pulses, the shape of the pulse front, and its duration. Op-amp parameters for "small" and "large" signals are closely related, since they refer to the same amplifier. Let's consider the main parameters and characteristics of the op-amp.
1. Transfer characteristicОУ - dependence of the amplitude of the output signal U out from the amplitude of the input signal.
|
|
point out the difference in measurement methods: in transistor and tube amplifiers, a constant signal, as a rule, is not amplified and the amplitude characteristic is taken at the signal frequency f 0. On the contrary, in the OA, the transfer characteristic tends to be measured at f = 0. Due to the latter consideration, the transfer characteristic is measured at both polarities U in.
The transfer characteristics of the op-amp during normal operation are shown in Fig. 5.3. Here 1 is the transfer characteristic when the input signal is applied to the non-inverting input ( U in x = 
); 2 - when applied to the inverting input ( U in x = 
). Plot - U in x. max<
U in x<
< U at x max corresponds to linear amplification, for | U in x | > U non-linear distortions occur at x max, the signal is limited "from above". We can roughly assume that the restriction levels are + E and - E, but U in x. max = E/TO, where TO Is the amplification factor of the op-amp.
2. Gain OU TO can be determined by the slope of the linear section of the transfer characteristic: it is quantitatively equal to the tangent of the angle α (Fig. 5.3). Note that the transfer characteristics (Fig. 5.3) are qualitative: taking into account the real values of the amplification factors, the transfer characteristics of industrial samples of the op-amp have almost vertical linear sections.
3. Amplitude-frequency characteristic. In operational amplifiers, in the overwhelming majority of samples, the properties are identical for inverting and non-inverting connections (for example, the gains for both connections are equal in magnitude). The identity of the properties of the op-amp at different inclusions makes it possible to consider not two, but one single frequency response (as well as phase response). The frequency response of a typical op-amp is shown in Fig. 5.4.
|
|
||||||||||||||||||||
|
4. Phase response. Although with an inverting op-amp, the phase shift between the input and output signals should be 180 °, and with a non-inverting 0 °, in fact, in real samples 5.
Input and output resistances. Due to the identity of the properties of the op-amp with inverting and non-inverting connections, the values of the input resistances at both inputs of the amplifier (respectively, 6. Large signal slew rate- the parameter is complex, covering immediately both the amplitude of the pulse signal at the output of the op-amp, and the duration of the front. As it comes about a large signal, which in the process of amplification acquires an amplitude close to E(Fig. 5.4), then, denoting the duration of the front through τ fr, for the speed and rise of the signal, we write 2 E/ τ fr. Meaning is closely related to the frequency properties of the OA: this is obvious, since τ fr ~ 1 / f in. gr, where f in. gr - upper cutoff frequency. Disadvantages of operational amplifiers. The main disadvantages of op amps are: - decrease in the gain when connecting a low-impedance load; –Displacement of the transfer characteristic from the origin of coordinates (imbalance); Consider these phenomena and measures to combat them. 1. Reducing the gain when connecting the load. Despite the fact that the op-amp includes an emitter follower and R output is reduced as a result of this, yet it remains quite large: when connecting a load with a resistance of the order of a few tens of ohms, negative phenomena take place: a decrease in the gain and, at the same time, the level of the maximum output signal. These effects are shown graphically in Fig. 5.5: Transfer characteristic 1 corresponds to no-load (load resistance R n ), characteristics 2 and 3 correspond to loads with R n 2> R n 3.
At the same time, we note that if the entire complex electronic circuit is built from stages based on an op-amp, then in this case for each op-amp (except for the op-amp of the final stage) an idle mode is automatically provided for the load: after all, the op-amp load is also an operational amplifier with R in, many times (two to three orders of magnitude) exceeding R out. Thus, the developers of the op-amp took care of combining the circuits based on them. 2. Displacement of the transfer characteristic from the origin (imbalance). The presence of two power supplies, and with not always the same voltages, often causes a shift in the transfer characteristic of the op-amp from the origin. This phenomenon is often referred to as imbalance. There are other possible reasons for the imbalance. The unbalance phenomenon is illustrated by the graph in Fig. 5.6. Here, the unbalance voltage is denoted as Δ U. The shift of the transfer characteristic from the origin leads to the following negative consequences:
- to the appearance of an undesirable "pedestal" when amplifying a small alternating signal; - to the occurrence of nonlinear distortions when amplifying an alternating signal with an amplitude close to E/TO. Other negative consequences of the imbalance are also possible: it is especially dangerous in the adders of constant signals, since an addition error occurs in this case. The fight against imbalance is reduced to voltage compensation Δ U... If the op-amp is turned on in such a way that only one input is used to supply a useful signal, then to compensate for the imbalance, the second input can be disconnected from the ground and a voltage equal in value and opposite in sign to the voltage Δ U. Let's consider this method in more detail. As indicated earlier, the first node of the op-amp is a differential amplifier, the operation of which is described by the formula U you x = TO DU ( Hence it is clear that the "return" to a directly proportional relationship U you x from The compensation voltage is usually supplied from a power source through a potentiometer; another way is using the input current I in of the op-amp itself. In the latter case, a potentiometer (the so-called balancing resistance) is turned on between the input of the op-amp that is not used for supplying a useful signal and the ground, the voltage drop across which, when the input current of the op-amp flows, is Δ U. The schemes that implement the two considered methods of dealing with the imbalance are shown in Fig. 5.7 (since in practice the inverting switching of the op-amp is more often used, these circuits also correspond to the inverting inclusion). It should be noted that the phenomenon of imbalance is not constant, the value of Δ U changes under the influence of many factors, and therefore the op-amp mode must be regularly monitored and the compensation voltage must be promptly changed.
When considering the parameters of the op-amp, it was noted that at high frequencies there is, on the one hand, a decrease in TO, and on the other hand, an increase in the value of Δφ TO... If we assume that the values of γ and Δφ γ do not depend on frequency, and Δφ γ = 0 (this is true for many op-amp-based circuits), then at low and medium frequencies (where Δφ TO = 180 °, with the inverting switching on of the op-amp), the phase balance condition is not met and generation does not occur. With increasing frequency Δφ TO increases and can reach 360 ° and larger values. However, generation occurs only when the amplitude balance condition is satisfied at these frequencies, i.e., at TO> 1 / γ. The circuit implementation of the OA correction is usually as follows: it does not cover
the whole amplifier, and one or more stages - one or more external elements (capacitors, resistors) are connected to the special pins of the microcircuit. The most common are single-pole, double-pole correction, phase advance and phase lag corrections. Single-pole correction consists in connecting in parallel part of the amplifying stages of the op-amp capacitance WITH K (fig.5.9). This capacitance at high frequencies bypasses the amplifier and lowers the op amp gain. The two-pole correction circuit is shown in Fig. 5.10, but: it consists of two capacitors WITH 1 and WITH 2 and resistor R 3, and WITH 2 10 WITH one . The action of the circuit is different at different frequencies: at sufficiently low frequency values f resistance WITH 2 is large and the signal does not pass through the circuit; the circuit has no corrective effect. With increasing frequency, the resistance WITH 1 decreases and the two-pole correction circuit turns into a single-pole correction circuit, and the function WITH K performs equivalent capacity WITH e = WITH 1 WITH 2 /(WITH 1 + WITH 2). Therefore, we can assume that the scheme
Phase advance correction scheme (Fig.5.11, but) is connected in series with the used input of the op-amp and contains a resistor R 1 and capacitor WITH... The essence of this circuit is to introduce an additional differentiating circuit into the amplifier C – R at x OU, where R in x op amp - input impedance of op amp. In this case, there is compensation of the phase shift in the amplifier Δφ TO phase shift in the correction circuit, since Δφ TO and Δφ core have different signs (Figure 5.11, b, where curve 1 is the graph of Δφ TO, 2 - graph of Δφ core, 3 - their sums). The phase-advance correction circuitry, like a differentiating circuit, is a high-pass filter; as a result of this, the gain of the amplifier at low frequencies decreases, which is a disadvantage (so that the gain would not be at the frequency f = 0 is equal to zero, WITH 5 shunt resistor R 1).
Note. The terms "phase advance" and "phase lag" in the names of correction schemes can be explained by comparing Δφ TO in the op-amp without correction and with the connection of one or another correcting circuit. For example, when connecting a differentiating circuit (Fig.5.10, but) the phase shift acquires a positive addition. Chain R 1 – C has a complex resistance with a negative imaginary part, therefore, connecting this chain to the input of the op-amp, in addition to shunting the input resistance at high frequencies, causes a negative addition of the phase shift at the same frequencies. If we now imagine, say, a harmonic signal passing through the op amp as a rotating vector on a complex plane, then the presence of a positive phase addition means that the vector rotates ahead of the vector of the signal with a smaller phase. The vector of a signal with a negative "addition" in phase rotates with a delay. | ||||||||||||||||||||
and 
) are practically the same and range from hundreds of kilo-ohms to units-tens of megohms (an op-amp of the 140UD8A type even has R at x = 10 9 Ohm). Order of values R you x specified earlier: the output resistances of the op-amp are in the range from units of kilo-ohms to hundreds of ohms.
–
). Let's assume that a non-inverting op-amp is used. In this case, the inverting input is connected to ground, 
=
0:
U you x = TO DU 
... If an imbalance occurs, this formula is incorrect and must be replaced with another one: U you x = TO DU ( 
–
Δ U).
it is possible with 
= –
Δ U, i.e. U you x = TO DU ( 
–
Δ U
–
)
=
= TO DU ( 
–
Δ U
+ Δ U)
= TO DU 
.
An operational amplifier (op-amp) is an amplifier direct current with a differential input, the characteristics of which are close to the characteristics of the so-called “ideal amplifier.” The op-amp has a large voltage gain K >> 1 (K = 10 4 - 10 6), a large input (R in = 10-100 Ohm) resistance.
In linear amplifiers, an op-amp is used only with negative feedback circuits (NFB), which reduces the voltage gain K to 1-10 3, but at the same time reduces the dependence of K on temperature, supply voltage, increases R in.us and decreases R out .us. The use of an op-amp in amplifiers without OOS circuits is unacceptable, since the risk of breaking the stability of the op-amp increases and the frequency response correction circuits in a wide frequency band become more complex.
The op-amp (Figure 15.1.) Contains a differential amplifier as the first stage. The differential amplifier has a high gain for the difference between the input signals U 2 - U 1 and a low gain for the common-mode signals, i. E. identical signals applied simultaneously to both inputs. This allows you to reduce the sensitivity to common-mode signals (external noise) and the offset voltage determined by the non-identity of the op-amp arms.
Figure 15.1. The internal structure of the operational amplifier.
The input stage is followed by one or more intermediate ones; they provide the required voltage and current gain.
The complementary output stage must provide a low op-amp output impedance and sufficient current to power the expected loads. The output stage is usually a simple or complementary emitter follower.
To reduce the sensitivity of the circuit to common-mode signals and increase the input resistance, the emitter current of the first differential stage is set using a constant current source.
Basic parameters of operational amplifiers
1. K - own gain of the op-amp (without feedback).
2. U shift - Output shear voltage. A small voltage arising from the asymmetry of the op-amp arms at zero voltage at both inputs. Usually Ushv has a value of 10 - 100 mV.
3. I cm - Input bias current. The current at the inputs of the amplifier required for the operation of the input stage of the operational amplifier.
4. I shift - Input shift current (). The difference in bias currents appears due to inaccurate matching of the input transistors. ...
5. R in - Input resistance. Typically, R in has a value of up to 1-10 megohms.
6. R out - Output resistance. Usually R out does not exceed hundreds of Ohms.
7. Koss - Coefficient of attenuation of the common mode signal. It characterizes the ability to attenuate signals applied to both inputs at the same time.
8. Current consumption. Quiescent current drawn by the operational amplifier.
9. Power consumption. The power dissipated by the operational amplifier.
10. Maximum speed rise of the output voltage (V / μs).
11. U pit. - Supply voltage.
12. Transient response. The signal at the output of the amplifier when a voltage surge is applied to its input.
The op-amp has several options for switching circuits, which differ significantly in their characteristics.
To analyze the operation and calculate the characteristics of various schemes for switching on the OS, it is further necessary to remember that, based on the properties of the control device:
1. The voltage difference between the inputs of the op-amp is very small and can be taken equal to zero.
2. The operational amplifier has a high input impedance, so it consumes very little input current (up to 10 nA).
Basic schemes for switching on the op-amp
IN inverting amplifier(Figure 15.2.), the input and output signals are 180º out of phase. If U in, positive, then the voltage at point A, and hence U d, will also become positive, and U out will decrease, which will lead to a decrease at the inverting input to the value U d = U out / K ≈ 0.
Point A is often called virtual land, because its potential is almost equal to the potential of the earth, since U d, as a rule, is very small
Rice. 15.2. Op-amp inverting amplifier
To obtain an expression for the gain with feedback, we take into account that, since the R input of the amplifier is very large. Since and, then.
Setting U d = 0 (since K → ∞), we obtain. The feedback gain of the circuit under consideration is
The output voltage is inverted, as evidenced by the negative value of K os.
Since, thanks to feedback, approximately zero potential is maintained at point A, the input resistance of the inverting amplifier circuit is equal to R 1 .. Resistance R 1 should be chosen so as not to load the input signal source, and, naturally, R os should be large enough so as not to overload the op amp.
Non-inverting amplifier can also be implemented on an op-amp (Fig. 15.3) with a high input impedance, the voltage gain of which can also be set using the resistances R 1 and R os.
As before, we assume that, since R вх → ∞.
The voltage at the inverting input of the amplifier is, therefore
15.3. Non-inverting op-amp amplifier
Consequently, .
Since U out = U d · K and U d = U out / K, at 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 o.
In the scheme voltage follower on op-amp(Figure 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 is amplified - U d, it can be seen that the voltage at the output of the amplifier U out = U d K.
Figure 15.4. Voltage follower on op-amp
OU output voltage U out = U in + U d. Since U out = U d · K, we obtain that U d = U out / K. Consequently, . 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 input impedance of the amplifier, which is very high, so the follower can serve as a good matching stage.
Amplifier with differential input has two inputs, and the inverting and non-inverting inputs are under the same voltage, in this case equal to U os, 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 U 1 is set to zero and the input signal is applied to the U 2 input, then the amplifier will act as a non-inverting amplifier, from which the input voltage is removed from the divider formed by the resistors R 2 and R? wasp If both voltages U 1 and U 2 are applied to the corresponding inputs at the same time, then the signal at the inverting input will cause such a change in the output voltage that the voltage at the junction point of the resistors R 1 and R OS becomes equal to U OS, where.
Due to the fact that the amplifier has a very high input impedance,
Solving the resulting equation for U out, we have:
Substituting the expression for U os, 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 Figure 15.5 is zero, then the amplifier will act as an inverting amplifier with respect to U 1.
The input impedance of the op-amp circuit can be defined as follows. Voltage is applied to the differential input resistance of the op-amp r d. U d. Due to the presence of 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 transfer ratio 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 impedance, due to the action of the feedback, is multiplied by the factor 1 + K U b.
According to fig. 12, for the resulting input impedance 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. However, it should be remembered that we are talking exclusively about differential value; this means that the changes in the input current are small, while the average value of the input current can take on incomparably large values.
Rice. 15.6. Non-inverting amplifier circuit, taking into account the intrinsic resistance of the op-amp.
Op-amp output impedance operational amplifier, not covered by feedback, is determined by the expression:
When the load is connected, there is a slight decrease in the output voltage of the circuit, caused by a voltage drop across rout, which is transmitted to the input of the amplifier through the voltage divider R 1, R 2. The resulting increase in differential voltage compensates for the change in output voltage.
In general, the output impedance can be quite high (in some cases, from 100 to 1000 ohms. Connecting the OC circuit will reduce the output impedance.
For a feedback amplifier, this formula takes the form:
In this case, the value of U d does not remain constant, but changes by the value
dU d = - dU n = -bdU out
For an amplifier with a linear transfer characteristic, the change in the 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 get the desired result:
If, for example, b = 0.1, which corresponds to the 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 foregoing, generally speaking, is true within the amplifier passband f p, Hz. At higher frequencies, the output impedance of the feedback op-amp will increase as it increases. the quantity | K U | with increasing frequency, it will decrease at a rate of 20 dB per decade (see Fig. 3). In this case, it acquires an inductive character and at frequencies above f t becomes equal to the value of the output impedance of the amplifier without feedback.
The dynamic parameters of the op-amp, characterizing the operating speed of an op-amp can be divided into parameters for small and large signals. The first group of dynamic parameters includes the bandwidth f p, the unit gain frequency f t and the settling time t y. These parameters are called small-signal, because they are measured in the linear mode of operation of the op-amp stages (DU out< 1В).
The second group includes the slew rate of the output voltage r and the power bandwidth f p. These parameters are measured with a large differential input signal of the op amp (more than 50 mV). Some of these parameters are discussed above. The settling time is counted from the moment the input voltage step is applied to the OA input until the last time the equality | U out.set - U out (t) | = d, where U out.set is the steady-state value of the output voltage, d is the permissible error.
Working bandwidth or bandwidth The op-amp is determined by the form of the 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 oscillator is set so that the amplitude of the output signal U out.max does not slightly reach the saturation limits of the amplifier. Then the frequency of the input signal is increased. The power bandwidth f p corresponds to the value of U out.max equal to 0.707 of the initial value. The value of the power bandwidth decreases with increasing capacitance of the correcting capacitor.
Performance 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 of operational parameters are due to the final 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 the source I pot; maximum output current I out max; maximum values of the output voltage at nominal power supply; maximum permissible values of common-mode and differential input voltages
Frequency response an operational amplifier is an important factor on which the stability of the operation of real circuits with such an amplifier depends. In most operational amplifiers, the individual stages are DC coupled to each other by galvanic connections, so these amplifiers do not have a gain rolloff in the low frequency region and it is necessary to analyze the gain rolloff with increasing frequency.
Figure 15.7 . Frequency response of the operational amplifier
Figure 15.7. shows the typical frequency response of an operational amplifier.
Rice. 15.8. Simplified op-amp equivalent circuit
With increasing frequency, the capacitive resistance drops, which leads to a decrease in the time constant τ = R n * C. Obviously, there must be a frequency, when exceeding which the voltage at the output U out will be less than KU d.
Expression for the gain K at any frequency:
has the form, where K is the gain without feedback at low frequencies; f is the operating frequency; f 1 - cutoff 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, R n "R out, then.
Usually the frequency response is given in general terms. how:
where f is the frequency of interest to us, while f 1 is a fixed frequency called cutoff frequency and is a characteristic of a particular amplifier. Voltage gain decreases with increasing frequency. In addition, from the expression for θ, it can be seen that when the frequency changes, the phase of the output signal is shifted relative to the phase of the input; - the output signal is out of phase with the input.
Adding negative feedback, such as in inverting or non-inverting amplifiers, increases the op-amp's effective bandwidth.
To verify this, consider the expression for the open-loop gain of an amplifier with a roll-off of 6dB / octave (when the frequency is doubled):
Where K (f) is the gain without feedback at frequency f; A - gain without feedback at low frequencies; f 1 - coupling frequency. Substituting this ratio into the expression for the gain in the presence of feedback, we obtain
This expression can be rewritten as, where f 1 oc = f 1 (1 + Aβ); K 1 - gain with closed feedback at low frequencies; f 1oc - cutoff frequency in the presence of feedback.
The feedback cutoff frequency is equal to the open loop cutoff frequency multiplied by (1 + Kβ)> 1, so the effective bandwidth does increase 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 roll-off rate of the amplifier is 6dB / octave, the gain-bandwidth product is constant: Kf 1 = const. To verify this, we multiply the ideal low frequency gain by the high 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 earlier it was shown that in order to increase the bandwidth using feedback, the gain must be reduced, now the derived ratio makes it possible to know how much of the gain must be sacrificed to obtain the desired bandwidth.
Operational amplifier equivalent circuit allows you to take into account the effect of imperfection of the amplifier on the characteristics of the circuit. For this, it is convenient to represent the amplifier as a complete equivalent circuit containing essential elements of imperfection. The complete OA equivalent circuit for small, slow signal changes is shown in Fig. 15.9.
Rice. 15.9 .. Equivalent circuit of an operational amplifier for small signals
In operational amplifiers with bipolar transistors at the input, the input impedance for the differential signal r d is several megohms, and the input impedance for the common-mode signal r input is several gigaohms. The input currents determined by these resistances are on the order of a few nanoamperes. Significantly higher values are for direct currents flowing through the inputs of the operational amplifier and determined by the bias of the differential stage transistors. For universal op amps, the input currents are in the range from 10 nA to 2 μA, and for amplifiers with input stages made on field effect transistors, they are fractions of nanoamperes.
Op amp parameters
Since the op-amp is a universal device, a large number of parameters are used to describe its properties.
1. The gain K is equal to the ratio of the output voltage to the differential input signal that caused this increment in the absence of feedback (is 10 3-10 7) and is determined at no-load at the output. TO = U out / U in.
2. The zero bias voltage U cm shows what voltage must be applied to the input of the op-amp in order to obtain U out = 0 at the output (it is 0.5-0.15 mV). This is due to inaccurate emitter-base voltage matching of the input transistors.
3. The input current I in is determined by the normal operating mode of the input differential stage on bipolar transistors. This is the base current of the remote control input transistor. If field effect transistors are used in the differential stage, then these are leakage currents.
When signal sources with different internal resistances are connected to the inputs of the op-amp, different voltage drops across these resistances by bias currents are created. The resulting differential signal changes the input voltage. To reduce it, the resistances of the signal sources must be the same.
4. The difference of the input currents DI in is equal to the difference in the values of the currents flowing through the inputs of the op-amp, at a given value of the output voltage, is 0.1-200 nA.
5. Input resistance R bx (resistance between input terminals) is equal to the ratio of the input voltage increment to the input current increment at a given signal frequency. R bx is defined for the low frequency region. Depending on the nature of the applied signal, the input impedance is differential (for a differential signal) and common-mode (for a common-mode signal).
Differential input impedance is the total input impedance on the side of any input, when the other input is connected to the common terminal, is tens of kOhms - hundreds of megohms. Such a large R bx is obtained due to the input remote control and a stable constant voltage source. Common-mode input resistance is the resistance between the shorted input pins and ground. It is characterized by a change in the average input current when a common-mode signal is applied to the inputs and is several orders of magnitude higher than R in diff.
6. The coefficient of attenuation of the common-mode signal K osl sf is defined as the ratio of the voltage of the common-mode signal applied to both inputs, to the differential input voltage, which causes the same value of the output voltage. Attenuation shows how many times the gain of the differential signal is greater than the gain of the common-mode input signal and is 60-120 dB:
With an increase in the common-mode signal attenuation coefficient, the more accurate it is possible to distinguish the differential input signal against the background of the common-mode noise, the better quality OU. Measurements are carried out in the low frequency range.
7. Output resistance R out is determined by the ratio of the output voltage increment to the increment in the active component of the output current at a given value of the signal frequency and is units to hundreds of Ohms.
8. Thermal offset voltage drift is equal to the ratio of the maximum offset voltage change to the temperature change that caused it and is estimated in μV / deg.
Thermal drifts of bias voltage and input currents cause thermal errors in op amps.
9. The coefficient of influence of the instability of the power supply on the output voltage shows the change in the output voltage when the supply voltages change by 1 V and is estimated in μV / V.
10. The maximum output voltage U out max is determined by the limiting value of the op-amp output voltage at a given load resistance and input signal voltage, ensuring stable operation of the op-amp and distortions not exceeding the specified value. U out max is 1-5 V lower than the supply voltage.
11. The maximum output current I out max is limited by the permissible collector current of the op-amp output stage.
12. Power consumption - the power dissipated by the op-amp when the load is off.
13. The frequency of unity gain f 1 is the frequency of the input signal at which the gain of the op-amp is equal to 1: | K (f 1) | = l. With integrated op amps, the unity gain frequency has a limit value of 1000 MHz. The output voltage at this frequency is about 30 times lower than for direct current.
14. Cutoff frequency f c OA - the frequency at which the gain is reduced by a factor. It estimates the bandwidth of the op-amp and is tens of MHz.
15. The maximum rate of rise of the output voltage V max is determined by the highest rate of change in the output voltage of the op-amp when a rectangular pulse with an amplitude equal to the maximum value of the input voltage acts at the input and lies in the range of 0.1-100 V / μs. Under the influence of the maximum input voltage, the output stage of the op-amp falls into the saturation region in both polarities. This parameter is specified for wideband and pulse op-amp devices and results in output edges with finite durations. V max characterizes the operating speed of the op-amp in the large signal mode.
16. The settling time of the output voltage t yc t (transient decay time) is the time required for the amplifier to return from the saturation state at the output to the linear mode.
The settling time is the time during which, after a jump in the input voltage, the output voltage differs from the steady-state value by the value of the permissible relative error dU out. During the settling time, the output voltage of the op-amp under the influence of the input voltage of the rectangular shape changes from the level of 0.1 to the level of 0.9 of the steady-state value.
17. Noise voltage, reduced to the input, is determined by the effective value of the voltage at the output of the amplifier at zero input signal and zero resistance of the signal source divided by the gain of the op-amp. The noise spectral density is estimated as the square root of the square of the reduced noise voltage divided by the bandwidth in which the noise voltage measurement is made. The dimension of this parameter. In the DU on the op amp, the noise figure (dB) is sometimes specified, which is defined as the ratio of the reduced noise power of an amplifier operating from a source with internal resistance R g, to the noise power of the active resistance
where U w - reduced noise voltage at R g = 0;
4kTR r is the thermal noise spectral density of the resistor.
The requirements for the parameters of the op-amp depend on the functions it performs. It is desirable in all practical cases to reduce the error of the operations performed, to increase the reliability and speed. The simultaneous improvement of all parameters puts forward conflicting requirements for the circuit and its manufacture. All this is explained by a wide variety of op amps, in which only specific parameters are optimized at the expense of degrading others.
For example, the instrumentation uses precision op amps with high gain, high input impedance, low offset voltage, and low noise. And high-speed op amps must have a high slew rate, large bandwidth, and low settling time of the output voltage. Such op-amps have found application in pulsed and broadband amplifying devices and in devices for analog-to-digital converters.
To create comparators that serve to compare the instantaneous values of two voltages, high-speed op amps operating in switching mode are used.
Operational amplifiers are one of the main components in modern analog electronic devices. Due to the simplicity of calculations and excellent parameters, operational amplifiers are easy to use. They are also called differential amplifiers because they can amplify the difference in input voltages.
The use of operational amplifiers in audio technology is especially popular to amplify the sound of music speakers.
Designation on the diagrams
Five pins usually come out of the amplifier case, of which two pins are inputs, one is an output, and the other two are power.
Operating principle
There are two rules to help you understand how an op amp works:
- The output of the operational amplifier tends to zero voltage difference across the inputs.
- The amplifier inputs do not consume current.
The first input is marked "+", it is called non-inverting. The second input is marked with "-" and is considered to be inverting.
The amplifier inputs have a high resistance called impedance. This allows a current consumption of several nanoamperes at the inputs. At the input, the magnitude of the voltages is evaluated. Depending on this estimate, the amplifier outputs an amplified signal.
The gain, which sometimes reaches a million, is of great importance. This means that if at least 1 millivolt is applied to the input, then the output voltage will be equal to the voltage of the amplifier power supply. Therefore, opamp is not used without feedback.
The amplifier inputs operate according to the following principle: if the voltage at the non-inverting input is higher than the voltage of the inverting input, then the output will be the highest positive voltage. In the opposite situation, the output will be the largest negative value.
Negative and positive voltages at the output of the operational amplifier are possible due to the use of a power supply that has a split bipolar voltage.
Operational amplifier power supply
If you take a finger-type battery, then it has two poles: positive and negative. If the negative pole is counted as the zero point, then the positive pole will show +1.5 V. This can be seen from the connected.
Take two elements and connect them in series, you get the following picture.
If the negative pole of the lower battery is taken as the zero point, and the voltage is measured at the positive pole of the upper battery, then the device will show +10 volts.
If we take the midpoint between the batteries as zero, then a bipolar voltage source is obtained, since there is a voltage of positive and negative polarity, equal to +5 volts and -5 volts, respectively.
There are simple split-power block diagrams used in amateur radio designs.
The circuit is powered from the household network. The transformer reduces the current to 30 volts. The secondary winding in the middle has a branch, with the help of which +15 V and -15 V rectified voltage is obtained at the output.
Varieties
There are several different op-amp circuits that are worth considering in detail.
Inverting amplifier
This is the basic scheme. A feature of this circuit is that the opamp is characterized, in addition to amplification, also by a phase change. The letter "k" indicates the gain parameter. The graph shows the effect of the amplifier in this circuit.
The blue color represents the input signal and the red color represents the output signal. The gain in this case is: k = 2. The amplitude of the output signal is 2 times greater than that of the input signal. The amplifier's output is inverted, hence its name. Inverting operational amplifiers have a simple circuit:
These op-amps have become popular due to their simple design. To calculate the gain, use the formula:
This shows that the amplification of the opamp does not depend on the resistance R3, so you can do without it. Here it is used for protection.
Non-inverting operational amplifiers
This circuit is similar to the previous one, the difference is that the signal is not inverted. This means maintaining the phase of the signal. The graph shows the amplified signal.
The gain of the non-inverting amplifier is also: k = 2. A signal in the form of a sinusoid is supplied to the input, only its amplitude has changed at the output.
This circuit is no less simple than the previous one, it has two resistances. At the input, the signal is applied to the positive terminal. To calculate the gain, you need to use the formula:
It can be seen from it that the gain is never less than unity, since the signal is not suppressed.
Subtraction scheme
This circuit makes it possible to create a difference between two signals at the input, which can be amplified. The graph shows the principle of operation of the differential circuit.
This amplifier circuit is also called a subtraction circuit.
It has a more complex design, in contrast to the previously discussed schemes. To calculate the output voltage, use the formula:
The left side of the expression (R3 / R1) defines the gain, and the right side (Ua - Ub) is the voltage difference.
Addition scheme
This is called an integrated amplifier. It is the opposite of the subtraction scheme. Its peculiarity is the ability to process more than two signals. All sound mixers operate on this principle.
This diagram shows the possibility of summing several signals. To calculate the voltage, the formula is applied:
Integrator circuit
If you add a capacitor to the feedback loop, you get an integrator. This is another device that uses operational amplifiers.
The integrator circuit is similar to an inverting amplifier, with the addition of capacitance in the feedback. This leads to a dependence of the system on the frequency of the input signal.
The integrator is characterized by an interesting feature of the transition between signals: first, a rectangular signal is converted into a triangular signal, then it is converted into a sinusoidal one. The gain is calculated using the formula:
In this formula, the variable ω = 2 π f increases with increasing frequency, therefore, the higher the frequency, the lower the gain. Therefore, the integrator can act as an active low pass filter.
Differentiator circuit
In this scheme, the situation is reversed. A capacitance is connected at the input, and a resistance is connected in the feedback.
Judging by the name of the circuit, its principle of operation lies in the difference. The higher the rate of change of the signal, the greater the value of the gain. This parameter allows you to create active filters for high frequency.
The gain for the differentiator is calculated using the formula:
This expression is inverse to that of the integrator. The gain increases in the negative direction with increasing frequency.
Analog comparator
The comparator device compares the two voltage values and converts the signal to a low or high value at the output, depending on the state of the voltage. This system includes digital and analog electronics.
A feature of this system is the lack of feedback in the main version. This means that the loop resistance is very high.
A signal is supplied to the positive input, and the main voltage is applied to the negative input, which is set by a potentiometer. Due to the lack of feedback, the gain tends to infinity.
When the voltage at the input exceeds the value of the main reference voltage, the highest voltage is obtained at the output, which is equal to the positive supply voltage. If the input voltage is less than the reference, then the output value will be a negative voltage equal to the voltage of the power supply.
There is a significant disadvantage in the analog comparator circuit. As the voltage values at the two inputs approach each other, the output voltage may change frequently, which usually leads to gaps and malfunctions in the relay. This could result in equipment malfunction. To solve this problem, a hysteresis circuit is used.
Analog comparator with hysteresis
The figure shows the scheme of action of the scheme with, which is similar to the previous scheme. The difference is that switching off and on does not occur at the same voltage.
The direction of the arrows on the graph indicates the direction of movement of the hysteresis. When looking at the graph from left to right, it can be seen that the transition to a lower level is carried out at voltage Uph, and moving from right to left, the voltage at the output will reach the highest level at voltage Upl.
This principle of operation leads to the fact that at equal values of the input voltages, the state at the output does not change, since the change requires a difference in voltages by a significant amount.
Such operation of the circuit leads to some inertia of the system, but it is safer, in contrast to the circuit without hysteresis. Usually, this principle of operation is used in heating devices with a thermostat: stoves, irons, etc. The figure shows an amplifier circuit with hysteresis.
The stresses are calculated according to the following relationships:
Voltage repeaters
Operational amplifiers are often used in voltage follower circuits. The main feature of these devices is that they do not amplify or attenuate the signal, that is, the gain in this case is equal to one... This feature is due to the fact that the feedback loop has zero resistance.
Such voltage follower systems are most often used as a buffer to increase the load current and device operability. Since the input current is close to zero, and the output current depends on the type of amplifier, that is, the possibility of unloading weak signal sources, for example, some sensors.
Operational amplifier (OA) eng. Operational Amplifier (OpAmp), commonly referred to as an opamp, is a direct current amplifier (DCA) with a very high gain. The phrase "DC amplifier" does not mean that an operational amplifier can only amplify DC current. This means, starting from a frequency of zero Hertz, and this is a direct current.
The term "operational" has become firmly established for a long time, since the first samples of OA were used for various mathematical operations such as integration, differentiation, summation, and so on. The op-amp gain depends on its type, purpose, structure and can exceed 1 million!
Operational amplifier circuit
In the diagrams, the operational amplifier is indicated like this:
or so
Most often, op-amps in the diagrams are indicated without power pins
An input with a plus sign is called a non-inverting input, and an input with a minus sign is called an inverting input. Do not confuse these two signs with the polarity of the power supply! They do NOT say that it is imperative to apply a negative signal to the inverting input, and a positive polarity to the non-inverting input, and then you will understand why.
Operational Amplifier Power
If the power pins are not specified, then it is considered that the op-amp is supplied with bipolar power + E and -E Volts. It is also labeled + U and -U, V CC and V EE, Vc and V E. Most often it is +15 and -15 Volts. Bipolar nutrition is also called bipolar nutrition. How to understand this - bipolar nutrition?
Let's imagine a battery
I think you all know that the battery has a plus and a minus. In this case, the “minus” of the batteries is taken as zero, and the batteries are counted relative to zero. In our case, the battery voltage is 1.5 volts.
Let's take another such battery and connect them in series:
So, the total voltage will be 3 Volts, if we take the minus of the first battery as zero.
But what if we take the minus of the second battery to zero and measure all the voltages relative to it?
This is where we got bipolar power.
Ideal and real model of an operational amplifier
In order to understand the essence of the op-amp, consider it perfect and real models.
1) the ideal op-amp is infinitely large.
In real op amps, the value of the input impedance depends on the purpose of the op amp (universal, video, precision, etc.), the type of transistors used and the circuitry of the input stage and can range from hundreds of ohms to tens of megohms. A typical value for general purpose op amps is a few megohms.
2) The second rule follows from the first rule. Since the input impedance of an ideal op-amp is infinitely large, the input impedance will be zero.
In fact, this assumption is quite true for an op-amp with an input, for which the input currents can be less than picoamperes. But there is also an op-amp with an input. Here, the input current can already be tens of microamperes.
3) The output impedance of an ideal op-amp is zero.
This means that the voltage at the output of the op-amp will not change when the load current changes. In real general-use op amps, it is tens of ohms (usually 50 ohms).
In addition, the output impedance depends on the frequency of the signal.
4) The gain in an ideal op-amp is infinitely large. In reality, it is limited by the internal circuitry of the op amp, and the output voltage is limited by the supply voltage.
5) Since the gain is infinitely large, therefore, the voltage difference between the inputs of an ideal op-amp is zero. Otherwise, even if the potential of one input is greater or less than at least the charge of one electron, then the output will be an infinitely large potential.
6) The gain in an ideal op-amp is independent of the signal frequency and is constant at all frequencies. In real op-amps, this condition is fulfilled only for low frequencies up to some cutoff frequency, which is individual for each op-amp. Typically, the cutoff frequency is taken as a 3 dB gain drop or 0.7 times the gain at zero frequency (DC).
The circuit of the simplest op-amp on transistors looks something like this:
The principle of operation of an operational amplifier
Let's take a look at how op-amp works
The operating principle of the op-amp is very simple. It compares the two voltages and at the output already gives out a negative or positive supply potential. It all depends on which input the potential is greater. If the potential at the non-inverting input U1 is greater than at the inverting U2, then the output will be + Usup, if the potential at the inverting input U2 is greater than that at the non-inverting U1, then the output will be -Upit. That's the whole principle ;-).
Let's take a look at this principle in the Proteus simulator. To do this, we will choose the simplest and most common operational amplifier LM358 (analogs 1040UD1, 1053UD2, 1401UD5) and assemble a primitive circuit showing the principle of operation
We supply 2 Volts to the non-inverting input, and 1 Volt to the inverting input. Since the potential is greater at the non-inverting input, therefore, at the output we must get + Usup. We got 13.5 volts, which is close to this value.
But why not 15 volts? The internal circuitry of the op-amp is to blame for everything. The maximum value of the op-amp may not always be equal to the positive or negative supply voltage. It can deviate from 0.5 to 1.5 volts, depending on the type of op-amp.
But, as they say, the family is not without freaks, and therefore, op-amps have long appeared on the market that can produce an acceptable supply voltage at the output, that is, in our case, these are values close to +15 and -15 Volts. This feature is called Rail-to-Rail, which is literally translated from English. “From rail to rail”, and in the language of electronics “from one power bus to another”.
Let us now apply more potential to the inverting input than to the non-inverting one. We supply 2 Volts to the inverting one, and 1 Volt to the non-inverting one:
As you can see, in this moment the output “lay down” on -Upit, since the potential at the inverting input was greater than at the non-inverting one.
In order not to download the Proteus software package once again, it is possible to simulate the operation of an ideal op-amp online using the Falstad software. To do this, select the Circuits — Op-Amps—> OpAmp tab. As a result, the following diagram will appear on your screen:
On the right control panel, you will see sliders for adding voltage to the inputs of the op-amp and you can already visually see what happens at the output of the op-amp when the voltage at the inputs changes.
So, we have considered the case when the voltage at the inputs can be different. But what if they are equal? What will Proteus show us in this case? Hmm, showed + Upit.
What will Falstad show? Zero Volts.
Whom to believe? Nobody! In real life, this cannot be done in order to drive absolutely equal voltages onto two inputs. Therefore, such a state of the op-amp will be unstable and the output values can take on values of either -E Volts, or + E Volts.
Let's apply a sinusoidal signal with an amplitude of 1 volt and a frequency of 1 kilohertz to the non-inverting input, and put the inverting one on the ground, that is, to zero.
Let's see what we have on the virtual oscilloscope:
What can be said in this case? When the sinusoidal signal is in the negative region, at the output of the op-amp we have -Upit, and when the sinusoidal signal is in the positive region, then at the output we have + Upit. Also pay attention to the fact that the voltage at the output of the op-amp cannot change its value abruptly. Therefore, the op-amp has such a parameter as the slew rate of the output voltage V Uout .
This parameter shows how quickly the output voltage of the op-amp can change when working in pulse circuits. Measured in Volts / sec. Well, how do you understand what more value this parameter, the better the op-amp behaves in pulse circuits. For LM358, this parameter is 0.6 V / μs.
Featuring Jeer
It was shown that when using an operational amplifier in different schemes inclusion, the amplification of the stage on one operational amplifier (OA), 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 both cases, 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-purpose 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.
Regarding 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 supply. This mode is the most "familiar" for an op amp, allowing you 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 on 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 schemes 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 diagrams 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 stages 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.
To ensure 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 prototyping boards, but you can get no worse results on the monitor screen as well.
Actually, it's even 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. 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 by a beam of blue color, 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 with 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 out 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 allows avoiding 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 applied 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 a circuit on an op-amp is much easier than setting up several 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 DC 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 with 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 the non-inverting amplifier deserve 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 equal to. 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 much will change. In fact this is not true.
Let's assume that the resistance of resistor R2 is zero, as is the case with a 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.
