How to learn the triode amplifier circuit design? This engineer has a good experience
The core component of the amplifier circuit is a triode, so it is necessary to have a certain understanding of the triode. There are many types of amplifying circuits composed of triodes, and we use several commonly used ones to explain them (see Figure 1). Figure 1 is a basic amplification circuit of a common beam. What do we have to master the amplification path? Here we must know a few things first. The first one is what we often say Ic, Ib, which are the collector current and base current of the triode. They have a relationship of Ic=β×Ib, but when we first started, the teacher Obviously, I didn’t tell us how big Ic and Ib are. This question is more difficult to answer because there are more things involved, but in general, for small power tubes, Ic is usually set at a few milliamps to a few milliseconds. Ann, the middle power tube is in the range of a few milliamps to tens of milliamps, and the high power tube is in the tens of milliamperes to a few amps. Waterproof Connector,Idc Connector,Idc Socket Connector,Horizontal Waterproof Idc Socket Connector Shenzhen Jinyicheng Electronci Technology Co.,Ltd. , https://www.jycconnector.com
(1) analyzing the role of each component in the circuit;
(2) The principle of amplification of the liberation of large circuits;
(3) can analyze the static working point of the calculation circuit;
(4) Understand the purpose and method of setting static working points.
Of the above four items, the last item is more important. 
In Figure 1, C1 and C2 are coupling capacitors. Coupling is the signal transfer function. The capacitor can couple the signal signal from the previous stage to the latter stage because the voltage across the capacitor cannot be abrupt. After inputting the AC signal at the input terminal, The voltage at the terminal cannot be abrupt, and the voltage at the output changes with the AC signal input at the input, thereby coupling the signal from the input to the output. But one thing to note is that the voltage across the capacitor cannot be abrupt, but it cannot be changed.
R1 and R2 are the DC bias resistance of the transistor V1. What is the DC offset? Simply put, it is necessary to eat. When the triode is required to work, it must first provide certain working conditions. The electronic components must require power supply, otherwise it will not be called a circuit.
In the working requirements of the circuit, the first condition is to be stable, so the power supply must be a DC power supply, so it is called DC bias. Why is it powered by a resistor? A resistor is like a faucet in a water supply system that regulates the current. Therefore, the three operating states of the triode ": load, saturation, amplification" are determined by the DC offset, which is determined by R1, R2 in Figure 1.
First of all, we need to know how to determine the three working states of the triode. In simple terms, it can be judged according to the size of Uce, and Uce is close to the power supply voltage VCC, then the triode works in the loading state. The stop state means that the triode is basically not working, and the Ic current is small (about zero), so since R2 has no current flowing, the voltage is close to 0V, so Uce is close to the power supply voltage VCC.
If Uce is close to 0V, the triode works in saturation. What is the saturation state? That is, the Ic current reaches the maximum value, and even if Ib increases, it cannot be increased any more.
The above two states are generally referred to as the switch state. Except for the two states, the third state is the amplification state. Generally, the Uce is close to half of the power supply voltage. If Uce is biased toward VCC, the transistor tends to be in the load-carrying state. If Uce is biased toward 0V, the transistor tends to be saturated.
Understanding the purpose and method of setting the static working point The amplifying circuit is to amplify the input signal and output it (generally there are voltage amplification, current amplification and power amplification, which are not discussed in this discussion). Let's start with the signal we want to amplify, taking the sinusoidal AC signal as an example. In the analysis process, it can be considered only that the signal size change is positive or negative, and others do not say. As mentioned above, in the amplifying circuit of Fig. 1, the static working point is set to Uce close to half of the power supply voltage. Why?
This is to make the signal positive and negative energy have a symmetrical change space. When there is no signal input, the signal input is 0. If Uce is half of the power supply voltage, we use it as a horizontal line as a reference point. When the input signal increases, Ib increases and the Ic current increases. The voltage U2=Ic×R2 of the resistor R2 increases, and Uce=VCC-U2, which becomes smaller. U2 can theoretically be equal to VCC, then Uce will reach 0V minimum. This means that when the input signal increases, the maximum change of Uce is from 1/2 VCC to 0V.
Similarly, when the input signal decreases, Ib decreases, and the Ic current decreases. Then, the voltage U2=Ic×R2 of the resistor R2 decreases, and Uce=VCC-U2, which becomes larger. When the input signal decreases, the maximum change in Uce is from VCC of 1/2 to VCC. Thus, when a positive or negative change occurs within a certain range of the input signal, Uce has a symmetrical positive and negative variation range with 1/2 VCC as the standard, so generally the static operating point of Figure 1 is set to Uce close to half of the power supply voltage.
It is our goal to design Uce to be close to half the supply voltage, but how can we design Uce to be close to half the supply voltage? This is the means. 
In Fig. 1, if Ic is 2 mA, the resistance of the resistor R2 can be calculated by R = U / I, VCC is 12V, 1/2VCC is 6V, and the resistance of R2 is 6V / 2mA, which is 3KΩ. Ic is set to 2 mA, then Ib can be derived from Ib=Ic/β, the key is the value of β, and the general value of β is 100, then Ib=2mA/100=20#A, then R1=(VCC -0.7V)/Ib=11.3V/20#A=56.5KΩ, but in fact, the beta value of the small power tube is much more than 100, between 150 and 400, or higher, so if you do the above calculation, The circuit is likely to be saturated, so sometimes we don't understand, the calculation is correct, but it can't be used. This is because there is still less practical guidance, pointing out the difference between theory and reality. This type of circuit is greatly affected by the value of β. When everyone calculates the same, the results are not necessarily the same. That is to say, the stability of such a circuit is poor and practical applications are few. However, if the voltage-dividing bias circuit of Figure 2 is changed, the analytical calculation of the circuit is closer to the actual circuit measurement. 
In the voltage-divided bias circuit of Figure 2, we assume that Ic is 2mA and Uce is designed to be 1/2VCC to 6V. Then how do R1, R2, R3, and R4 take values? The calculation formula is as follows: Since Uce is designed to have 1/2VCC of 6V, Ic × (R3 + R4) = 6V; Ic ≈ Ie. It can be calculated that R3+R4=3KΩ, so what is the R3 and R4?
Generally, R4 takes 100Ω and R3 is 2.9KΩ. In fact, we generally take 2.7KΩ for R3, because there is no 2.9KΩ in E24 series resistors. The value of 2.7KΩ is not much different from 2.9KΩ. Because the voltage across R2 is equal to Ube+UR4, that is, 0.7V+100Ω×2mA=0.9V, we set Ic to 2mA, β is theoretically 100, then Ib=2mA/100=20#A, there is a current here. Estimated is the current flowing through R1, generally about 10 times the value of Ib, take IR1200#A. Then R1=11.1V/200#A≈56KΩR2=0.9V(/200-20)#A=5KΩ; considering that the actual β value may be much larger than 100, the actual value of R2 is 4.7KΩ. Thus, the values ​​of R1, R2, R3, and R4 are 56KΩ, 4.7KΩ, 2.7KΩ, 100Ω, and Uce is 6.4V.
In the above analysis and calculation, many hypotheses are proposed, which is necessary in practical applications. Many times, a reference value is needed to give us calculations, but often there is no. One of them is that we are not familiar with various devices. Second, I forgot one thing. We ourselves are the people who use the circuit. Some data can be set by ourselves, so that we can take fewer detours.