Calculation of RC filters. Encyclopedia of radio electronics and electrical engineering

Encyclopedia of radio electronics and electrical engineering / Beginner radio amateur

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Consider frequency-selective or selective circuits that have a filtering effect, i.e., signals with some frequencies pass better, with others - worse. Sometimes this property of circuits is harmful, for example, in high-quality audio frequency amplifiers, where they strive to get the widest possible bandwidth. And sometimes it is useful, say, in radio receivers, when from the mass of signals from radio stations operating at different frequencies, you need to select the signal of a single one broadcasting at a frequency you know.

Filter circuits (filters) must necessarily contain reactive elements - capacitances and / or inductances, since the active resistance of resistors does not depend on frequency (in the ideal case). In reality, there are always parasitic capacitances and inductances (wiring, leads, p-n junctions, etc.), so almost any circuit turns out to be a filter to one degree or another, that is, its parameters depend on frequency. First, consider the simplest RC chains.

On fig. 28a shows a diagram of a simple low-pass filter (LPF) that passes low frequencies and attenuates high frequencies.

The transfer coefficient is the ratio K = Uout / Uin (more precisely, this is the module, or the absolute value of the transfer coefficient). We calculate it using the information we already know about AC circuits. The current in the circuit is:

and the output voltage is equal to the voltage drop across the capacitor C:

Substituting the current, we find

The transmission coefficient turned out to be complex. This means that the output voltage of the filter is out of phase with respect to the input. To emphasize the complex nature of K, it is often denoted as K(jω). Let's find the module (absolute value) and argument (phase) K

Both the modulus and the phase of the gain are frequency dependent, or are said to be functions of frequency. The negative sign of the argument indicates that the phase of the output signal is lagging behind the phase of the input signal. If you build their graphs, you get the amplitude-frequency and phase-frequency characteristics of the filter (AFC and PFC), shown in Fig. 28,6 and in respectively.

The filter works as follows. At the lowest frequencies, the capacitance of the capacitor is large and the signal is transmitted from the input to the output through the resistance R almost without attenuation. As the frequency increases, the capacitance drops and the circuit works as a voltage divider. At the cutoff frequency ωc, the capacitance is equal to the active resistance, and ωcRC = 1. However, the module K is not equal to 1/2, as it would be in the case of active resistances, but is 1/V2 = 0,7, as can be seen from the vector voltage diagram (Fig. 28d). The phase shift introduced by the chain at the cutoff frequency is 45°, which is how much the phase of the output signal lags behind the phase of the input signal. With a further increase in frequency, the gain modulus falls in proportion to the frequency, and the phase shift tends to -90°.

Often, to simplify calculations, the notation RC = τ is introduced. (chain time constant), ωRC = ω/ωc = x (generalized frequency). The transmission coefficient in these notations is written quite simply:

It is advisable to return to the previous notation only after the completion of all calculations.

In our analysis, we tacitly assumed that the circuit is powered by a generator with a very low internal resistance, and its output is not loaded by anything. In reality, the signal source always has some internal resistance R1, and if it is active, you just need to add it to R. Similarly, if the load has a capacitance Cn, you just need to add it to C. If the load has an active resistance RH, then the module K is already at the lowest frequencies, where the effect of capacitance can be neglected, will be less than unity and will be (we simply consider according to Ohm's law) RH / (R + RH). The cutoff frequency will also shift higher and amount, as it is easy to calculate in the manner described above, no longer

where R' is the resistance obtained by connecting R and Rn in parallel.

Here is an example of the practical application of the information presented. The video amplifier of the TV should pass through a frequency band of 6 MHz, and it works on a capacitive load consisting of the output capacitance of the transistor C, the mounting capacitance C and the interelectrode capacitance of the control grid of the kinescope Sk (Fig. 29, a). Their sum can be estimated by any capacitance meter (of course, with the TV turned off!) Or by reference data. Let it be 25 pF - this will be the capacity of the considered RC chain. The resistance R of the circuit is obtained by parallel connection of the internal resistance of the transistor (signal generator) and the load resistance Rn. The first can be found from the collector characteristics of the transistor by taking a small increment ΔUk near the operating collector voltage Uk and finding the corresponding current increment ΔIk

Usually, the internal resistance is much greater than the load resistance, then we can assume R = Rn.

We find the permissible load resistance based on the blockage of the frequency response up to 0,7 (by 3 dB) at a frequency of 6 MHz. The angular frequency of the cutoff will be

(round up). Since RC = 1 /ωс,

Naturally, we would like to choose a larger load resistance, which will increase the gain and reduce the current consumed by the transistor, but this cannot be done due to the blockage of the upper frequencies of the video spectrum, which will lead to loss of image clarity.

For the sake of interest, let's continue the calculation. Let a signal with an amplitude of up to 50 V be applied to the kinescope grid, then the transistor current should be 50 mA. 50 V will also drop on the load resistance, the power supply voltage must be at least 100 V, and 50 V - 50 mA \u2,5d 29 W will be released on the load resistor. The same power will be dissipated in the transistor. The load characteristic for this case is shown in fig. XNUMX,b, along with voltage and current diagrams (which, it should be noted, are rarely sinusoidal in television). Now it should be clear why the output stage of the video amplifier is performed on a powerful transistor, and a powerful resistor is placed in the load, although the kinescope does not consume any power through the control electrode (grid) circuit.

To somehow improve the situation, many ways have been invented. One of them consists in correcting the frequency response by turning on a coil with a small inductance in series with the load (Fig. 29, a), selected so that it resonates with the total capacitance C somewhere at the cutoff frequency or slightly higher. The resulting oscillatory circuit with a very low quality factor (no more than 1...1.5) contributes to the rise of the frequency response near the cutoff frequency. On fig. 29, the solid line shows the frequency response of the amplifier before correction, corresponding to the frequency response of a simple RC circuit, and the dashed line shows after the inductance is turned on. In this way, the bandwidth of the transmitted frequencies is expanded by 1,5 ... 2 times, or the amplification and efficiency of the cascade are increased by the same amount.

The described narrowing of the bandwidth from above occurs in each amplifying stage, which must be taken into account when designing multistage amplifiers. For example, in the case of two identical cascades, the blockage of the frequency response in each should be no more than 0,84 (0,842 = 0,7), in the case of three - no more than 0,89. Sometimes, especially in video amplifiers, "little tricks" are used: the preliminary stage, in which both the interelectrode capacitances and the output voltage swing are smaller, are designed as broadband, with a rise in the frequency response at high frequencies, compensating for the blockage of the frequency response in the output stage.

The described chain (see Fig. 28, a) is called low-pass filter when considering its frequency characteristics, and it is also called integrating when considering the passage of a pulsed signal. Let a voltage drop with a short edge act at the input of the circuit (Fig. 30). The output voltage will not rise immediately, because the capacitor needs time to be charged by the current limited by the resistor R.

Only at the first moment of time after the impact of the drop, the current will be equal to UBX / R, then it will decrease as the voltage across the capacitor increases. By compiling a differential equation for the output voltage and solving it, we can establish that

where e is the base of natural logarithms. During the time τ = RC, the output voltage increases to approximately 0,63 of the input value and then asymptotically approaches it. Thus, the integrating chain "overwhelms" the steep edges of the signal, which, by the way, explains the decrease in the clarity of the television image.

Let's move on to high-pass filters (HPF), the simplest of which (differentiating RC-chain) is shown in fig. 31, a. The transfer coefficient is now expressed as:

The frequency response of the chain is shown in fig. 31b. The formula for the cutoff frequency remains the same. The phase response is also the same, but the sign of f changes - the phase of the output signal is ahead of the phase of the input. It is close to 90° at the lowest frequencies and approaches zero at high frequencies (it is enough to shift the graph of Fig. 28c upwards along the φ axis by 90°). Actually, all expressions for the HPF are obtained from the formulas for the HPF when the generalized frequency x is replaced by -1/x', which is very often used when calculating any filters.

The impulse response of the chain is shown in fig. 32. It is, as it were, the opposite of the previous one - the output voltage increases abruptly, but then falls according to an exponential law in accordance with the view. Over a time equal to the time constant of the chain t, it decreases to 0,37 input, over the next interval t - again to 0,37 and so on (by the way, this is a good rule for plotting exponentials - for each horizontal division, the vertical coordinate of the curve should increase or decrease by the same percentage).

Almost every interstage dividing RC chain is a described HPF. Even if there is no explicit resistance R, it is the input resistance of the cascade connected behind the coupling capacitor. If we also take into account that the parasitic capacitance at the output of the cascade forms a high-pass filter, then it becomes clear that any amplifying cascade limits the bandwidth of the transmitted frequencies both from below and from above, that is, it is a band-pass filter. For rectangular pulses passing through the amplifying stage, steep fronts are smoothed out (LPF action) and the top collapses (HPF action).

To increase the filtering effect of RC circuits, several of them are turned on, one after another, and in order to exclude shunting of the next circuits, they are separated by intermediate amplification stages on transistors. Sometimes, for the same purpose, subsequent chains are chosen with great resistance. However, in any case, the frequency response of the filters in the region of the cutoff frequency is very flat.

Active filters allow correcting the situation, in which the amplifying element (transistor) itself serves as a filter element. On fig. 33 is a diagram of an active low-pass filter (Sallena-Key). The active element in it must have unity gain and not invert the signal. Additionally, high input and low output impedances are required. These requirements are met by a emitter (source) follower on a transistor or (better) an operational amplifier, the inverting input of which is connected to the output. Resistors are usually selected with the same resistance, and the capacitance of the capacitor C2 is 2 ... 2,5 times less than the capacitance of C1. Filter cutoff frequency

The filter works like this. At frequencies below the cutoff frequency of the RC circuits, the output voltage practically repeats the input voltage and the capacitor C1 is turned off from work, since both of its plates have the same potential. The signal is transmitted without attenuation. As the frequency increases, the RC2 circuit comes into action and the output voltage decreases. Then the RC1 circuit also comes into play, further attenuating the output signal. As a result, a steep drop in the frequency response above the cutoff frequency is formed.

By changing the ratio of capacitances C1 and C2, you can get a smooth and monotonically falling frequency response within the passband (Butterworth filter), and you can even form some rise before the cutoff frequency (Chebyshev filter).

Having formed such a rise (curve 1 in Fig. 34), it is advisable to add another passive link (curve 2), which will compensate for the rise and make the slope of the frequency response behind the cutoff frequency even steeper (curve 3) - |K| will decrease by a factor of 8 when the frequency is doubled. The result is a third-order filter with a slope of 18 dB per octave. As an example, in fig. 35 shows a diagram of such a low-pass filter with a cutoff frequency of 3 kHz. It is easy to tune the filter to other frequencies by changing the values ​​of all capacitances inversely proportional to the frequency. An HPF with similar characteristics is obtained by interchanging the resistors and capacitors and changing their ratings accordingly.

About the order of filters: it is determined by the number of reactive filter elements, and the slope of the frequency response slope depends on the order. So, the first-order links (Fig. 28, a and 31, a) give a signal attenuation by 2 times with a double change in frequency (6 dB / oct.), the second-order filter (Fig. 33) - by 4 times (12 dB / oct.). oct.), third-order filter (Fig. 35) - 8 times (18 dB / oct.).

Question for self-test. Some high-quality (band 20 Hz ... 20 kHz) 3H amplifier has an input impedance of 100 kOhm, the signal source has the same output impedance. They are connected by a shielded cable with a linear capacitance of 100 pF/m. The cable length is 3,2 m. In addition, a 0,01 μF isolation capacitor is included at the amplifier input. Has everything been done correctly, what will the frequency band actually be and what should be done to correct the situation?

Response. Let's draw an equivalent circuit (Fig. 63) containing a signal source G1 with an internal resistance r, a cable with a capacitance C1, a coupling capacitor C2 and the input impedance of the amplifier R1.

The upper frequencies are attenuated by the capacitance of the cable, in parallel with which the input resistance R1 and the internal resistance of the signal source r are connected. The isolation capacitor C2 at high frequencies has a negligible resistance and can be ignored. Parallel connection of two 100 kΩ resistors gives a total value of 50 kΩ. The capacitance of cable C1 is 100 pF / m x 3,2 m = = 320 pF. Using the formula fc= 1/2πRC, we determine the upper frequency of the bandwidth:

f B = 1/6,28 320 10^-12-50 10³ = 104 Hz = 10 kHz.

To increase it to 20 kHz, you must either shorten the cable by half, or choose a cable with half the linear capacity, or lower the output impedance of the signal source to about 30 kΩ so that the total resistance connected in parallel to the cable is not 50, but 25 kΩ .

The latter method is preferable, since it also increases the voltage at the input of the amplifier. Indeed, if the resistances of the signal source and the amplifier are equal, it is half the EMF of the source, and when the resistance of the signal source decreases to 30 kOhm, it will reach 75% of the EMF of the source.

For this reason, cathode, emitter or source followers with low output impedance are often installed at the output of signal sources operating on long connecting cables.

Let us now calculate the lower cutoff frequency of the passband. It is determined by the isolation capacitor C2 (0,01 μF) and the total resistance of the signal source and amplifier input connected in series (r + R1 = 100 + 100 = 200 kOhm). Using the same formula, we calculate the cutoff frequency of this RC chain (HPF): fH = 1/2πRC = 1/6,28 2 10⁵10^-8 = 80 Hz. To lower the cutoff frequency to 20 Hz, the capacitance of the isolation capacitor must be increased at least 4 times. The closest standard capacitance value is 0,047uF.

If, in accordance with the above recommendation, the output impedance of the signal source r is reduced to 30 kΩ, then the total resistance of the HPF chain will be r + R1 = 30 + 100 = 130 kΩ, and the required capacitance of the isolation capacitor will be equal to:

C \u1d 2 / 1πf HR \u6,28d 20 / 1,3 10 XNUMX-XNUMX⁵= 0,07 uF.

Author: V.Polyakov, Moscow

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