“The components of the amplifying circuit are triodes, so it is necessary to have a certain understanding of the triodes. There are many types of amplifier circuits composed of triodes, and we will use the commonly used ones to explain.

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The components of the amplifying circuit are triodes, so it is necessary to have a certain understanding of the triodes. There are many types of amplifier circuits composed of triodes. Figure 1 is the basic amplifying circuit of co-injection. Generally, what do we need to know about the amplifying circuit?

(1) Analyze the function of each component in the circuit;

(2) The amplification principle of liberating large circuits;

(3) Can analyze and calculate the static operating point of the circuit;

(4) Understand the purpose and method of setting the static working point.

Of the above four items, one is more important.

Example analysis of the basic amplifier circuit of common emission

In Figure 1, C1 and C2 are coupling capacitors, and the coupling is to transmit the signal. The capacitor can couple the signal signal from the front stage to the rear stage because the voltage across the capacitor cannot be abruptly changed. After an AC signal is input at the input end, since the voltage at both ends cannot change abruptly, the voltage at the output end will change with the AC signal input at the input end, thereby coupling the signal from the input end to the output end. But one thing to note is that the voltage across the capacitor cannot change suddenly, but it is not impossible.

R1 and R2 are the DC bias resistors of the transistor V1. What is DC bias? Simply put, work requires food. To require the transistor to work, it is necessary to provide certain working conditions. The Electronic components must be required to have Power Supply, otherwise it will not be called a circuit.

In the operating requirements of the circuit, the condition is that it is required to be stable. Therefore, the power supply must be a DC power supply, so it is called a DC bias. Why is it powered by a resistor? A resistor is like a faucet in a water supply system that adjusts the amount of current. Therefore, the three working states of the triode: “load stop, saturation, and amplification” are determined by the DC bias. In Figure 1, it is determined by R1 and R2.

First of all, we need to know how to distinguish the three working states of the triode. To put it simply, the working state can be judged according to the size of Uce, which is close to the power supply voltage VCC. Then the triode works in the load-stop state, which means that the triode basically does not work, and the Ic current is small (about zero), so since there is no current flowing through R2, the voltage is close to 0V, so Uce is close to the power supply voltage VCC .

If Uce is close to 0V, the transistor works in a saturated state. What is a saturated state? That is to say: Ic current reaches the value, even if Ib increases, it cannot increase any more.

The above two states are generally referred to as switching states. In addition to these two, the third state is the amplification state, generally measured Uce is close to half of the power supply voltage. If the measured Uce is biased to VCC, the triode tends to be in a load-stop state, and if the measured Uce is biased to 0V, the triode tends to be saturated.

Amplifying circuit: It is to amplify the input signal and output it (generally, there are several types of voltage amplification, current amplification and power amplification, which are not included in this discussion). Let’s talk about the signal we want to amplify first, taking a sinusoidal AC signal as an example. In the analysis process, you can only consider whether the signal size change is positive or negative, and the others are not mentioned. It is mentioned above that in the amplifier circuit in Figure 1, the static operating point is set to Uce close to half of the power supply voltage, why?

This is to make the signal positive and negative have a symmetrical change space, when there is no signal input, that is, the signal input is 0. Assuming that Uce is half of the power supply voltage, we take it as a horizontal line as a reference point. When the input signal increases, the Ib increases and the Ic current increases, the voltage of the resistor R2 U2=Ic×R2 will increase accordingly, and Uce=VCC-U2 will decrease. U2 can theoretically reach equal to VCC, then Uce will reach 0V. That is to say, when the input signal increases, the change of Uce is from 1/2 of VCC to 0V.

Similarly, when the input signal decreases, the Ib decreases and the Ic current decreases. Then the voltage of resistor R2 U2=Ic×R2 will decrease accordingly, and Uce=VCC-U2 will increase. When the input signal decreases, Uce changes from 1/2 of VCC to VCC. In this way, when the positive and negative changes occur within a certain range of the input signal, Uce has a symmetrical positive and negative change range based on 1/2VCC, so generally the static operating point of Figure 1 is set to Uce close to half of the power supply voltage.

Designing the Uce to be close to half the supply voltage is our goal, but how can we design the Uce to be close to half the supply voltage? It’s up to us to look at our means.

Here are a few things to know first, the first is Ic, Ib we often say, they are the collector current and base current of the triode, and they have a relationship Ic=β×Ib. But when we were first learning, the teacher obviously did not tell us, what is the appropriate size of Ic and Ib? This question is more difficult to answer because there are so many things involved. But in general, for low-power tubes, Ic is generally set at a few tenths of milliamps to several milliamps, and the power tubes are at a few milliamps to tens of milliamps, and the high-power tubes are at tens of milliamps to a few milliamps.

In Fig. 1, set Ic to be 2mA, then the resistance value of resistor R2 can be calculated by R=U/I. VCC is 12V, then 1/2VCC is 6V, and the resistance value of R2 is 6V/2mA, which is 3KΩ. Ic is set to 2 mA, then Ib can be derived from Ib=Ic/β, and the key is the value of β. The general theoretical 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 far more than 100, between 150 and 400, or higher. Therefore, if the above calculation is done, the circuit may be in a saturated state.

So sometimes we don’t understand that the calculation is correct, but it doesn’t actually work. This is because there is a little less practical guidance to point out the difference between theory and practice. This kind of circuit is greatly affected by the β value. When everyone calculates the same, the results are not necessarily the same. That is to say, the stability of this kind of circuit is poor, and there are few practical applications. However, if it is changed to the voltage-dividing bias circuit shown in Figure 2, the analysis and calculation of the circuit will be closer to the actual circuit measurement.

In the voltage divider bias circuit of Fig. 2, we also assume that Ic is 2mA, and Uce is designed to be 1/2VCC as 6V. Then how should R1, R2, R3, and R4 take the values. The calculation formula is as follows: Because Uce is designed to be 1/2VCC as 6V, then Ic×(R3+R4)=6V; Ic≈Ie. It can be calculated that R3+R4=3KΩ, so what are R3 and R4?

Generally, R4 is 100Ω, and R3 is 2.9KΩ. In fact, we usually directly use 2.7KΩ for R3, because there is no 2.9KΩ in the E24 series resistors. There is no big difference between 2.7KΩ and 2.9KΩ. Because the voltage across R2 is equal to Ube+UR4, that is, 0.7V+100Ω×2mA=0.9V.

We set Ic as 2mA, and β generally takes the theoretical value of 100, then Ib=2mA/100=20#A. There is a current to be estimated here, that is, the current flowing through R1. Generally, the value is about 10 times that 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. In this way, the values of R1, R2, R3, and R4 are respectively 56KΩ, 4.7KΩ, 2.7KΩ, and 100Ω, and Uce is 6.4V.

In the above analysis and calculation, assumptions are made many times, which is necessary in practical applications, and a reference value is often needed to calculate for us. But it is often not. One is that we are not familiar with various devices, and the other is that we have forgotten one thing. We are the people who use the circuit. Some data can be set by ourselves, so that we can avoid detours.

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