Calculation of tube amplifiers. Encyclopedia of radio electronics and electrical engineering

Encyclopedia of radio electronics and electrical engineering / Beginner radio amateur

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An amplifier is one of the most common elements of electronic devices, but why do we start its calculation with an outdated tube amplifier? There are several reasons, and the main one is that interest in tube technology is reviving again, especially among lovers of high-quality sound. Tube amplifiers are unpretentious, reliable, and although overvoltages can cause short-term breakdowns between the electrodes, after them most often the lamp remains operational. Overcurrent causes the electrodes to heat up, but there is enough time to see the hot anode and take the time to turn off the power. Transistors, on the other hand, fail even with short-term overloads, instantly, "silently" and forever. We add, in addition, that the calculations of amplifiers on lamps and field-effect transistors, for example, are very similar.

The calculation of any amplifier begins with determining its parameters based on the purpose of the amplifier: amplified frequency band, output voltage, current or power, load resistance, input voltage and input resistance. For UHF of a home radio complex, for example, the output power can be 5 W at a load resistance (dynamic head) of 4 ohms, the frequency band is 70 Hz ... 12,5 kHz, the input voltage is 20 mV ... 1 V with an input resistance of not less than 500 kOhm. The specified input voltage range will allow you to connect the amplifier to many program sources: a radio receiver, a player with a piezoelectric pickup, line outputs of other devices.

It is advisable to divide such an amplifier into two parts: a voltage pre-amplifier, which will necessarily include a volume control (gain) and, possibly, tone controls (frequency response shapes) and a final power amplifier. The latter is calculated on a constant input signal level corresponding to the output signal of the preamplifier.

So, we calculate the amplifier on the lamps. A diagram of the simplest aperiodic triode amplifier is shown in fig. 48.

For the calculation, some reference data will be needed: the voltage and current of the lamp filament (filament circuits are not shown in the diagram), the recommended bias voltage, anode voltage and current, the slope of the characteristic S and the internal resistance of the lamp RI or its gain μ. The last three parameters are related by a simple relationship: μ = SRI

The lamp cascade is good because at low frequencies it practically does not consume power from the signal source - the anode current is controlled by the voltage on the grid. Nevertheless, the grid leakage resistor R1 with a resistance of 0,5 ... 4,7 MΩ is still necessary so that the rare electrons that have settled on the grid do not charge it negatively, but return through this resistor to the cathode. The same resistor is convenient to use as a volume control.

Capacitor C1 is needed so that the constant component of the input signal (if any) does not fall on the grid and does not change the lamp mode. Its capacitance is calculated by the formula for the cutoff frequency of the HPF, which must be less than the lowest passband frequency fn:

For there to be no grid current, the grid voltage must always be negative with respect to the cathode, so some bias voltage is needed. It is practically inconvenient to use a separate source of negative voltage, therefore, an automatic bias resistor R2 is most often included in the cathode circuit. The anode current of the lamp ia creates a voltage drop Uc on it, applied by the plus to the cathode, and by the minus to the control grid. The formula to calculate it is simple:

It remains to calculate the load resistance, given that about half the voltage of the anode power source Ea will drop on it:

Among the widely used double triodes, the 100N6P lamp with parameters S - 2 mA / V, Ri = 2 kOhm, Uc = -50 V, Ua = 1,5 V, ia = 120 mA has the highest gain μ \u1d 250 (the last two differ from the 1,8 V and 240 mA given in the reference books, but we chose them according to the characteristics of the lamp for reasons of economy.Assuming Ea = 2 V, we find R1,5 = 3 kOhm, R120 = XNUMX kOhm. The gain of the cascade on the triode is calculated as follows:

The gain is not too high, and with an input signal of 20 mV, the output voltage will be only 1,4 V, which may not be enough to completely “build up” the UMZCH output lamp. You will have to use either two cascades on triodes (then the gain will be redundant and it will have to be reduced, for example, using the OOS), or one cascade in another lamp that gives more gain - the pentode (Fig. 49).

It differs only in the power supply circuit of the shielding grid R3C3. The resistance of the quenching resistor R3 is determined by the formula

where Ug2 and ig2 are the screen grid voltage and current.

The internal resistance of the pentode is large, so the gain is calculated using a simpler formula

We will choose the 6Zh1P pentode as the most economical one. Its parameters Ua = = Ug2 = 120 V, S = 5 mA/V, ia = 7 mA and ig2 = = 3 mA at Uc = - 1,5 V, which gives R2 = = 150 Ohm. R3 = 40 kOhm, R4 = 17 kOhm and Kμ = 85. In practice, modes with such a large anode current are not used in preliminary stages. It is advantageous to increase the resistance of all resistors by several times, significantly reducing the anode current. And although the slope of the characteristic in this mode will decrease, the gain will increase and amount to 150 ... 200. To calculate new parameters at a lower anode current of the lamp, you should use its characteristics. However, the lamps are not very sensitive to mode changes and it is easy to choose it experimentally.

Now let's move on to UMZCH. For them, special powerful output beam tetrodes and pentodes are produced. In our example, a 6P14P tetrode with parameters Ua = Ug2 = 250 V, S = 11,5 mA / V, ia = 50 mA and ig2 = 5 mA at Uc = - 6 V is suitable. Our output stage will be single-ended, operating in class A This means that the quiescent current of the lamp will be equal to the nominal, 50 mA, and when the voltage on the control grid changes, it will vary from zero (the lamp is closed) to twice the nominal 100 mA (the lamp is open).

Let's find the required AF voltage on the grid using the formula Δia = SΔUBX:

ΔUBx = Δia/S = 50/11,5 = 4,35 V (peak value).

The resistance of the auto-bias resistor in the cathode circuit should be

If the pentode preamplifier calculated above provides Kμ = 150, then to obtain an amplitude of 4,35 V on the grid of the output stage, the input signal must be equal to 4,35 / 150 = 0,029 V (peak value), or about 20 mV (effective value) that meets the specified requirements.

The circuit design of the UZCH is completed, we can draw its schematic diagram (Fig. 50). The resistances of the resistors are calculated, it remains to choose the capacitances of the capacitors. They are calculated in the same way as the capacitance C1 (see above) for the lowest passband frequency, which must be taken with a margin, below 70 Hz.

Of course, the resistance of the corresponding resistor must be substituted into the formula. For example, if an R1C1 string has a cutoff frequency of 16Hz with a capacitance of 0,01uF, then a R2C2 string will have the same cutoff frequency with a capacitance of 10uF. It is also useful to check the upper frequency of the preamplifier bandwidth by taking the sum of the output capacitance of the VL1 lamp, the input capacitance of the VL2 lamp (taken from reference books) and the mounting capacitance С∑ equal to 3 + 13,5 + 20 - 40 pF:

As you can see, it is higher than required.

A few words must be said about the purpose of the decoupling chain R5C5. Significant fluctuations in the output tube current will inevitably lead to changes in the anode supply voltage, because tube amplifiers are usually powered from unstabilized sources. So that they do not affect the operation of the preliminary cascade (and we absolutely do not need this), and a chain is installed. Capacitor C5 simply does not have time to recharge in time with changes in the anode voltage. In addition, the circuit additionally filters the AC background in case of insufficient smoothing of ripples in the rectifier filter.

Consider now the anode circuit of the output stage. The lamp will give maximum power if current changes from 0 to 100 mA are accompanied by the maximum possible voltage changes at the anode, and the maximum current will correspond to the minimum voltage, which should be at least 20 ... 30 V (otherwise there will be distortion at the peaks signal). Let's take into account another 10 volts of voltage drop across the active resistance of the primary winding of the output transformer and get the amplitude of the AC voltage at the anode 250 - 10 - 30 = 210 V. The AC voltage is added to the DC supply voltage. Please note that when the anode current decreases to zero (on the negative half-wave of the input signal), the instantaneous anode voltage will increase to 250 + 210 = 460 V. As already mentioned, lamps easily tolerate such voltages.

The oscillatory power of the AF signal in the anode circuit will be

P \u2d Um im / 210 \u0,05d 2 5,25 / XNUMX \uXNUMXd XNUMX W.

Taking into account small losses in the output transformer, we fulfilled the set condition (provided 5 W in the load). Let's find the required resistance of the primary winding for the AF currents RH:

RH \u210d Um / im \u50d 4,2/XNUMX \uXNUMXd XNUMX kOhm.

Knowing RH and head resistance Rg, it is now possible to find the transformation ratio of the output transformer T1, taking into account the following: if the transformer lowers the voltage by n times, then it increases the current in the secondary winding circuit by the same amount, then the resistance is transformed into n² once:

At higher frequencies of the audio spectrum, the UMZCH gain increases, since the inductive resistance of the voice coil of the head, converted into the primary winding, and the leakage inductance resistance of the primary winding of the transformer T1 are added to the active load resistance RH. To compensate for the rise, a capacitor C7 is connected in parallel with the primary winding, the capacitance of which is difficult to calculate due to the uncertainty of the named parameters and therefore is selected experimentally, according to the desired shape of the frequency response.

Question for self-test. Maybe you are already tired of theoretical calculations? If not, then calculate the amplifier based on the requirements you set yourself, and if so, then find, for example, an unnecessary tube TV and disassemble it. A good acoustic system is obtained from a wooden case, if the front panel is cut out of chipboard and covered with a cloth. Place a head on the panel, preferably not in the center and preferably two or more, connected in series or in parallel, depending on their resistance. Assemble an amplifier like the one described and enjoy the "tube" sound. All the details necessary for the implementation of the project can be found in the old TV.

Author: V.Polyakov, Moscow

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