Calculation of amplifiers with feedback. Encyclopedia of radio electronics and electrical engineering

Encyclopedia of radio electronics and electrical engineering / Beginner radio amateur

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Feedback (FB) is widely used in amplifiers. The OS allows you to significantly improve their parameters, and in some cases create new devices based on amplifiers - triggers, generators, etc. The generalized circuit of the amplifier with OS is shown in fig. 55.

The input signal Uc and the OS signal Uoc are fed to the adder A1 and then to the amplifier A2 with a transfer coefficient Ko (usually Kc>>1). The signal from the output of the amplifier Uo passes through the feedback circuit with a gain p (usually p<<1), forming a feedback signal Uoc. Let us first assume that neither the amplifier nor the feedback circuit introduces phase shifts. Then, for the case of signal summation in A1, we can write Uo = (Uc + UoC)Ko. At the same time, Uoc = βUo. Substituting, we find the gain of the entire device K:

Uo = UC.Ko(1-Koβ),

K = Uo/Uc = Ko/(1-Koβ).

We see that the gain increases and, at Koβ = 1, goes to infinity. And this means self-excitation - the amplifier becomes a generator. OS of this type is called positive (POS), it is often used to create generators, regenerators, and similar devices. In audio frequency amplifiers (UZCH), it almost never occurs.

Now let's do not summation, but subtraction of signals in node A1. The calculations remain the same, but the signs will change in the formulas:

K = Uo/Uc = Ko/(1+Koβ).

The feedback has become negative (NF) and now reduces the gain. It would seem that this is her major drawback. However, it fully pays off with other useful qualities of the OOS, and obtaining a large initial gain (Ko) in modern transistor devices is not a big problem.

The first useful property of the OOS is the reduction of non-linear distortion. The task of the amplifier is to reproduce at the output an exact copy of the input signal, but with a large voltage and / or power. The distorted output signal can be represented as the sum of the undistorted signal and the distortion products. The latter are not in the input signal, but they go from output to input through a feedback circuit. And since it is negative, the distortion products coming from the input, as it were, compensate themselves, and their share in the output signal is sharply reduced.

Another useful quality of OOS is the equalization and expansion of the frequency response of the amplifier. At those frequencies where the gain is greater, the influence of the CNF, which reduces this gain peak, also becomes greater. If Koβ>>1, then, as can be seen from the formula, K - 1/β.

Having completed the OOS circuit in the form of a frequency-independent divider of two resistors, we get a flat frequency response in a wide frequency range.

There are other advantages: if the OOS signal is removed from the amplifier output in parallel and fed to the input in series with the input signal (in antiphase with it, so that subtraction is carried out), then the output impedance of the amplifier decreases, and the input resistance increases.

This is the most primitive theory of the OS, as you probably already guess, little corresponding to reality. It turns out that there is no purely negative or purely positive feedback in any wide frequency range. Moreover, the NOS at some frequency can turn into a POS. This will happen if the amplifier introduces a phase shift approaching 180 °, and the feedback signal is in phase with the input. If there is enough gain, at that frequency the amplifier will self-excite, and the old amateur radio adage will come true: "when you make an amplifier, you get an oscillator."

The expressions that we have given remain valid, but with a small, albeit very significant, caveat - it is necessary to substitute in them the complex functions of the gains of the amplifier itself Ko(jω) and the OS circuit β(jω). Then the result will be correct. The last formula will now be written as follows;

K(jω)=Ko(jω)/[1+β(jω)Ko(jω)].

Let us explain what has been said with a simple example. Let there be a transistor amplifying stage with a gain of 100 (Fig. 56).

Bias chains are not shown for simplicity, although the existing OS chain can be used for bias as well. The complex gain of the amplifier is determined by the RC chain, where R is formed by the parallel connection of the load resistance R1 and the resistance of the OS divider R2 + R3:

R = R1 (R2 + R3)/(R1 + R2 + R3),

and the capacitance C \u1d CXNUMX is the sum of the output capacitance of the transistor, the mounting capacitance and the capacitance of the output shielded cable (if any). The total gain of the cascaded amplifier and RC circuit is found as their product:

Ko(jω) = 100-1/(1 + jωRC).

We see that, starting from a certain frequency ωc = 1/RC, the modulus of the gain decreases, and the rate of its decrease is 2 times for a twofold increase in frequency, or 6 dB per octave. The frequency response (dependence of the gain modulus on frequency) of our amplifier is shown on a logarithmic scale in fig. 57 thin line.

Let's remove the OS signal from the output of the amplifier in parallel (see Fig. 56) and, having weakened it with a divider with a frequency-independent gain β=R3/(R2+R3)=0,09, feed it to the input in series with the input signal. The OS is negative because the transistor stage inverts the signal. With this inclusion, the OOS will lower the output and increase the input impedance of the amplifier by 1 + βKo, i.e., 10 times. We find the complex gain of the amplifier with OOS

K(jω) = Ko(jω)/[1+β(jω)Ko(jω)] = 100/(1 + jωRC)[ 1+9/(1 + jωRC)] = 10/(1 + jωRC*) ,

where C* = C/10.

What do we see? The gain fell by 10 times and became equal to 10. But the cutoff frequency of the frequency response increased by 10 times, which means the same expansion of the amplifier bandwidth. View of the module chart | K(jω) | remained the same, it is shown by the thickened line in Fig. 57. No undesirable phenomena (self-excitation, peaks in the frequency response) are observed in this simple amplifier with OOS.

Another thing is when the OOS covers several cascades. An example of a practical three-transistor amplifier circuit with a direct connection between the stages is shown in fig. 58.

The first two transistors operate in the so-called "barrier" mode, when the base voltage is equal to the collector voltage and is 0,5 ... 0,6 V. This mode is quite suitable for amplifying small signals. The output stage (VT3) operates normally with a collector voltage equal to half the supply voltage.

Stabilization of the mode of all three cascades is achieved by applying the feedback from the output to the input of the amplifier through the resistor R4.

It also creates the necessary bias current to the base of the transistor VT1. NFB is applied in parallel with the input signal, so the input impedance of the amplifier is low.

Often in such an amplifier self-excitation is observed at high frequencies. Attempts to eliminate it by adding capacitances C1, C2, C3, as a rule, are unsuccessful - the excitation becomes even stronger, although the generation frequency decreases. The reason lies precisely in these capacitances, and the interelectrode capacitances of the transistors are sufficient for excitation. The matter is also aggravated by the input capacitance C4. Let's assume that all four chains R1C1-R4C4 have the same time constant. Then, at the cutoff frequency, they shift the phase by 45° each, and in total by 180°.

Thus, FOS at the cutoff frequency turns into PIC! The attenuation of the signal by chains at the cutoff frequency is only 0.74 = 0,25, the divider formed by the resistor R4 and the input resistance of the cascade on the transistor VT1 makes a rather large attenuation, but the gain can be tens of thousands. Even if the gain is not sufficient for self-excitation, a completely unnecessary peak appears on the frequency response of an amplifier with feedback at higher frequencies, as shown in Fig. 59.

Such a peak will remain even at different time constants of all RC circuits (an accurate calculation must be carried out taking into account the parallel connection of the input resistances of transistors VT2, VT3 and resistors R1, R2). It will be at the frequency where the total phase shift over the entire amplifier loop - OS circuit approaches 180 °.

How to get rid of this unpleasant effect? There is only one way - to make the loop gain (Cor product) less than unity at those frequencies where the OOS turns into a POS. For this, it is possible, for example, to significantly increase the capacitance of C4. thus lowering the cutoff frequency of the R4C4 chain, and, consequently, its transmission coefficient at high frequencies. If shunting the input with a significant capacitance is undesirable, a resistor with a resistance of several kiloohms can be connected in series with C4 (resistance R4 is usually measured in megaohms).

In some cases, the low output impedance of the signal source can serve as such a resistor; in this case, the capacitor C4 is separating. The amplifier will be stable when a signal source is connected, but will self-excite when it is turned off. It is even better to make a resistor R4 of two connected in series, and connect a large capacitor between the point of their connection and the common wire.

There are also more sophisticated methods of frequency correction, for example, using proportionally integrating links (Fig. 60). The resistance of the resistor R2 (Fig. 60, a) is chosen several times less than the resistance R1, then the transfer coefficient equal to unity at low frequencies decreases to the value R2 / (R1 + R2) at high frequencies. The phase shift first increases with increasing frequency, then decreases and approaches zero at sufficiently high frequencies. Another link has similar characteristics (Fig. 60,b), but its input impedance is capacitive in nature and decreases at high frequencies.

In conclusion, let's see how stability issues are solved in operational amplifiers (op-amps), because they must allow operation with 100% OOS (β = 1), and their own gain Ko reaches tens and hundreds of thousands. As a rule, they try to make all stages of the op-amp very broadband, only one stage (usually it also gives maximum gain) is performed with a low cutoff frequency, sometimes even using external corrective capacitors (pay attention to capacitor C1 in the op-amp circuit of the previous chapter). In this case, the frequency response of the amplifier in a very wide frequency range has a slope of 6 dB per octave (see Fig. 57), and the phase shift does not exceed 90 °.

We have considered only amplifiers with a direct connection between stages, amplifying signals of arbitrarily low frequencies, starting from direct current. In amplifiers with coupling capacitors, which also have a lower passband frequency, with the introduction of the feedback, peaks in the frequency response in the low-frequency region can be observed. Self-excitation in this case manifests itself in the form of "motor noise", "dripping", etc. In this case, it is necessary to calculate the phase shift introduced by RC circuits consisting of coupling capacitors and input resistances of subsequent stages. In any case, it is undesirable that there would be more than one such chain inside the OS loop.

So, let's formulate the main conclusion of the above: amplifiers with feedback should be designed so that the loop gain is less than unity at those frequencies where the phase shift in the loop exceeds 90 and approaches 180 °. In more detail, and at a much higher level, the issues discussed are discussed in the article by S. Ageev "Design Considerations for Common Feedback Amplifiers"in "Radio", 2003, No. 4, pp. 16-19. There are also links to primary sources.

Author: V.Polyakov, Moscow

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