Class A amplifiers in simple type are widely used in various kinds of ICs with Cout output. Due to its multiple applications, in this article, we’ll introduce how to improve the sound quality of Class A amplifiers. See the circuit (figure 1) of Class A amplifier as below. And then try to take a test for this circuit.
1. Test Condition
Power voltage: 3V
Q1: Respectively use 8050D(モis 225) and 8050C(モis 152).
Speaker impedance: 8з.
Load resistor: 470з for 8050D; 470з or 820з for 8050C.
Test signal: 500Hz sine wave signal written in 4E Target Demo Board (sampling rate is 8KHz).
2. Test Process
Firstly, use 8050D transistor and 470з load resistor to test. See the output waveform of speaker in figure 2.
Through analyzing the waveform of figure 2, you can find that the upper part steps in sine wave are obviously wider than that in the lower part. The lower steps are not enlarged and the upper steps are cut off. Try to adjust the resistance of load resistor, you can also find that the smaller the resistance is, the more cut off region will be. And the bigger the resistance is, the smaller steps of lower part will be. Therefore, compared with other resistance, 470з is more suitable for the cases that some customers want louder sound but common sound quality.
Then use 8050C transistor and 470з load resistor to test. See the output waveform of speaker in figure 3.
Through analyzing the waveform of figure 3, you can find that the upper part of sine wave is obviously wider than the lower part. The upper part is cut off and its linearity is no good. Try to adjust the resistance of load resistor, you can also find that the smaller the resistance is, the more cut off region will be. And the bigger the resistance is, the smaller steps of lower part will be. Therefore, to reduce the cut off region, you can increase the resistance. But inevitably, the lower part will become smaller.
Then use 8050C transistor and 820з load resistor to test. See the output waveform in figure 4.
As shown in figure 4, the steps of both upper and lower part are equal. But there are also cut off regions in both parts, which will cause distortion in both parts. If customers use 8050C transistor and 820з resistor, the volume and current generated by them is the same as that of 8050D transistor and 470з resistor. But using 8050C can get better sound quality than using 8050D.
According to the above two waveforms, you can find that both waveforms have cut off regions in upper parts. Through further analyzing, you will find that the cut off regions are caused by the small current. The voltage on load resistor is lower than the conduction (switch-on) voltage of PN junction, which is not large enough to turn on the transistor. Therefore, there is no waveform output in speaker. Moreover, the smaller resistance of load resistor will lead to the lower voltage and more cut off regions by the same current.
In order to avoid the cut-off region occurring, the solution is to turn on the transistor once it get current. Try to remove the load resistor. Then have a test for it, and you will find that the cut off region disappears. But the lower part gets narrower. This is because the remove of resistor makes the current become much stronger. The too strong current will cause saturation conduction. So the method of removing the resistor still cannot solve the problem.
Try to find another better way. Suppose that we could use some components instead of the load resistor, which have the similar character of transistor to bypass parts of current, and the transistor can be turn on by any input current. Finally we got the ideal way that using the diode of an 8050D (PN junction) and a 20з resistor to replace the load resistor. After having a test for it, we found that the sound quality almost approached perfection. See the figure 5 and figure 6 as below:
As shown in figure 6, there is no cut off region in both upper and lower part. All the steps in this waveform are quite equal. Compared with the above several circuits, this circuit is the best one. It can create the best sound quality with the lowest distortion although the sound volume is relatively low. Hence, this circuit is the best choice.
Nevertheless, considering that this best circuit will cost more, we recommend another method. That is, you could use a diode and a 100з resistor, and an 8050D transistor. See the circuit and waveform in figure 7 and figure 8.
For this circuit, the waveform is still not equal because the V-I character of diode is different from 8050D PN junction. But using diode is better than only using resistor.
In addition, we have also taken some tests for the previous introduced circuits by music, and found that the sound quality of this circuit (figure 7) is also higher. So if customers want the high-quality sound, the circuit in figure 7 is the suitable choice. But if customers want to save cost, it is recommended to use 8050C combined with 820з resistor.
As for the package, if customers need to use SMT instead of DIP on 8050, there will be no problem for this circuit (figure 7). But note that most 8050 with SMT cannot offer the enough power, whose output power is only 9014. Thus, our customers need to check whether the 8050 is real when using.
3. Test Conclusion
According to the above analysis, we draw the conclusion as following: if the output current of IC is larger than the input current of transistor need, only using simple resistor to bypass the current must cause the sound distortion. Therefore, there is no need to add a load resistor but make the character of IC output similar to that of 8050. Or use a large size load resistor to slight adjust current. In this way, the usage will become more convenient, but it requires higher design of IC. In addition, different transistors have different characters. Actually most モ value of 8050C in the market is around 160 (a little larger than 150). In Shenzhen (China), generally 8050C+1kз resistance is suitable for manufacture.
 Improving the Sound Quality of Class A Amplifier.rar
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