Analysis of output power and loss of push-pull class A amplifier

Class A amplifiers do not have crossover distortion, and audio signals can be transmitted completely. Class A amplifiers are the goal pursued by audiophiles. What is the output power of a Class A amplifier? What is the power loss? These are the preliminary theoretical calculations for the production of Class A amplifiers. Class A amplifiers mostly use a push-pull working mode of NPN and PNP pairing.

The push-pull Class A power amplifier circuit can be seen as consisting of two single-tube Class A emitters.

The NPN tube of the positive power supply and the PNP tube of the negative power supply work in Class A state respectively to amplify the entire audio signal and output it to the speaker.

Before assembling and debugging a push-pull Class A amplifier, you must know how much power it can produce? How much static current is required? What is the supply current? What is the loss? Information on this aspect is hard to find. Is the power indicated on the Class A amplifier by some manufacturers really that high? Buyers want to verify. How to achieve the above goals?

This requires a theoretical analysis of the push-pull Class A amplifier.

Figure 1 is a Class A push-pull power amplifier output circuit, which can be decomposed into Figure 2.

Figure 1 Class A push-pull amplifier output circuit

Figure 2 Output circuit breakdown diagram

As can be seen from Figure 2, the current obtained by the speaker is provided by the NPN and PNP transistors respectively. The polarity of the audio signal input by the NPN power amplifier tube and the PNP power amplifier tube is the same.

Class A working state means that the triode has current at any time when it is working. Regardless of the drive, in recent years, many people sell C-band receiving devices installed with Zhongxing 6B at low prices. However, generally within two years, these pots will disintegrate due to severe rust, so that no matter whether the signal is positive or negative, the final tube has current flowing through it. The waveform of the collector current of a single-tube Class A working tube is shown in Figure 3. Taking the sine wave as an example, the static current is the peak value of the sine wave, that is, Io=lf, and the maximum current is twice the peak value, that is, Imax=21f=2I. Such a static current setting can ensure that the triode has current flowing through it during the entire signal cycle. To require the output power of the power amplifier, the output current effective value must be calculated. The current effective value is shown in Figure 3.

The equivalent value of the sum of the shaded areas of the output current waveform is:

The Class A output power of each tube is P甲1=I02Z (Z is the output impedance). The total Class A output power of the two final tubes NPN and PNP is P甲2=2P甲1=2I02Z. The general speaker impedance is Z=8Ω. The formula is simplified to P甲2=2I02Z8=16I02z. The output power of a Class A amplifier can be calculated from the above formula. The quiescent current of the amplifier is 1.6A, P甲2=16I02Z=16X1.62=40.96(W). The Class A output power of 40W/8Ω mentioned in the article is credible.

Figure 3 Single-tube Class A working collector current waveform

The Class A output power and quiescent current of 8Ω load are shown in Table 1.

Table 1 8Ω load Class A output power and quiescent current

With Io as the variable, the curve of P甲2 is shown in Figure 4, which is a parabola.

Figure 4 P甲2 curve

Class A amplifier transformer configuration: maximum current Imax=2I0, power supply power is Psued=2VImax=4VIo, transformer capacity S should be greater than Psued, generally S=1.2Psued is sufficient. Calculated from the above example, V=32V, 10=1.6A, Psued=4VI=4x32x1.6=204.8(W). S1=1.2Psued=1.2x204.8=245.76(W). This is the transformer capacity of one channel. If two channels share one transformer, the transformer capacity S2=2S1=2x245.76=491.52W. It is reasonable to use a 500w transformer in this article.

Determination of the power supply voltage of Class A amplifier: The power supply voltage is related to the power required by the load. P=U2/2, U=√PZ, the impedance of the speaker is 8Ω, and the formula is simplified to V1=√8P=2.83√P. When the impedance of a 40W Class A amplifier is 8Ω, the voltage across the two ends is V2=2.83√40=17.9(V). The maximum voltage is calculated based on the maximum current being 2 times the static current, which should be 17.9x2=36V. Therefore, the output voltage of a 40W Class A amplifier is preferably positive or negative 36V. It is slightly lower to use ±32 power supply. The corresponding values ​​of output power and voltage of different Class A amplifiers are shown in the table below. This voltage is the best value. Too large a voltage will cause large losses, and too small a voltage will affect the dynamics.

Class A amplifier efficiency analysis: effective power P1=IoIoZ. The power output power of the power supply is P2=2VIo, and the efficiency η=P1/P2=IoI.8/2VIo=IoIo.8/2VIo=4lo/V. From the above formula, we can see that the higher the voltage, the lower the efficiency. The voltage of the Class A amplifier cannot be too low or too high. Too low affects the dynamic effect, and too high affects the efficiency. Taking 1.6A 32V, 40W Class A output as an example, the calculated efficiency 11=4×1.6/32×100%=20%. Therefore, the efficiency of the Class A amplifier is very low. The efficiency of various Class A amplifiers varies due to different output powers, see Table 2. The Class A amplifier with the highest efficiency is the 40W power amplifier.

Table 2 The efficiency of various Class A amplifiers varies due to different output powers