The amplifier circuit working in Class B, although the tube consumption is small, is conducive to improving efficiency, but there is severe distortion, so that half the waveform of the input signal is eliminated. How to solve the above contradiction?
(A) Basic complementary symmetric circuit (b) Emitter output device composed of NPN tubes (c) Emitter output device composed of PNP tubes
Figure XX_01 Let's examine the complementary symmetric circuit shown in Figure XX_01a. T1 and T2 are NPN tube and PNP tube respectively. The base and emitter of the two tubes are connected to each other. The signal is input from the base and output from the emitter. RL is the load. Since this circuit has no base bias, vBE1 = vBE2 = vi. When vi = 0, T1 and T2 are in the cut-off state, so this circuit is a Class B amplifier circuit. This circuit can be seen as a combination of two emitter output stages shown in figures XX_01b and c.
Considering that the BJT transmitting junction is forward biased, it conducts electricity, so when the signal is in the positive half cycle, vBE1 = vBE2> 0, then T2 is cut off, T1 undertakes the amplification task, and there is current through the load RL; and when the signal is in the negative half cycle , VBE1 = vBE2 <0, then T1 ends, T2 undertakes the amplification task, and there is still current through the load RL; in this way, one works in the positive half cycle and the other works in the negative half cycle, the two tubes complement each other ’s deficiencies, so that the load Get a complete waveform, called complementary circuit.
The complementary circuit solves the contradiction between efficiency and distortion in the Class B amplifier circuit. In order to make the positive and negative half cycles of the waveform obtained on the load the same, it is also required that the characteristics of the two tubes must be completely consistent, that is, the working performance is symmetric. Therefore, the circuit shown in Figure XX_01a is usually called a Class B complementary symmetric circuit.
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