Sample Analysis! Tell us about the skills of triode amplifier circuitry.

The main component of amplifier circuit is triode, so it is necessary to have some understanding of triode. There are many types of amplifying circuits made up of triodes, let's use most commonly used ones to explain (as shown in Figure 1). The basic common-emitter amplifier circuit is shown in Figure 1. In general, what do we need to know about amplifier circuit?

(1) Analyze function of each component in circuit;(2) Master principle of amplifying a large circuit;(3) Be able to analyze and calculate static operating point of circuit; < br>(4) Understand purpose and method of setting a static operating point. Among four points listed above, last one is more important.

Analysis of basic circuit of a common-emitter amplifier

(Picture 1)

In figure 1, coupling capacitors C1 and C2 are coupling capacitors. Communication is transmission of signals. Capacitors can carry signal signals from front stage to back stage because voltage across capacitor cannot suddenly change. After an AC signal is applied to input terminal, since voltage at both ends cannot change abruptly, voltage at output terminal will change along with AC signal applied to input terminal, thereby coupling signal from input terminal to output. Terminal. But it should be noted that voltage across capacitor cannot change suddenly, but it is not impossible to change either. R1 and R2 are DC bias resistors of triode V1. What is DC bias? Simply put, work requires food. In order for a triode to work, certain operating conditions must first be provided, and electronic components must require power, otherwise it is not called a circuit. Among operational requirements of a circuit, first condition must be stable. Therefore, power supply must be a constant current source, which is why it is called DC bias. Why is power supplied through a resistor? The resistor is like a faucet in a water supply system, used to regulate current. Therefore, three operating states of triode: "load stop, saturation, boost" are determined by DC bias. In Figure 1, it is determined by resistors R1 and R2. First of all, we need to know how to distinguish between three operating states of a triode. Simply put, it can be judged by value of Uce, which is close to supply voltage VCC. Then triode works in "load-stop" state. The "load-stop" state means that triode is basically not working, and current Ic is small (near zero), so current R2 does not flow, and voltage is close to 0 V. , so Uce is close to power supply voltage VCC. If Uce is close to 0V, then triode is operating in a saturated state, what is a saturated state? That is: current Ic has reached its maximum value, even if Ib increases, it cannot increase any more. These two states are commonly referred to as switching states. In addition to these two, third state is magnified state, and overall measurement of Uce is close to half power supply voltage. If measured value of Uce is biased towards VCC, triode is usually in a load state, and if measured value of Uce is biased towards 0 V, triode is usually in a saturation state.

Understand purpose and method of setting a static operating point

Amplification circuit: designed to amplify input signal and output it (there are usually several types of voltage amplification, current amplification and power amplification, which are not discussed here). Let's first talk about signal we want to amplify, using an AC sine wave as an example. In analysis process, you can only take into accountwhether change in signal size is positive or negative, and nothing else. As mentioned above, in gain circuit shown in Figure 1, static operating point is set to Uce, close to half power supply voltage. Why?This is necessary in order for positive and negative signals to have a symmetrical change space When input, input signal is 0. Assuming Uce is half power supply voltage, we consider it as a horizontal line as a reference point. When input signal increases, Ib increases and current Ic increases, then voltage U2=Ic×R2 of resistor R2 increases accordingly, and Uce=VCC-U2 decreases. Theoretically, maximum value of U2 can be equal to VCC, and minimum value of Uce can reach 0V. That is, when input signal increases, maximum change in Uce is from 1/2 VCC to 0 V. Similarly, when input signal decreases, Ib decreases and current Ic decreases. Then voltage U2=Ic×R2 of resistor R2 will decrease accordingly, and Uce=VCC-U2 will increase. When input signal decreases, maximum change in Uce is from 1/2 VCC to VCC. Thus, when positive and negative changes occur within a certain input signal range, if Uce is based on 1/2VCC, there will be a symmetrical range of positive and negative changes. Therefore, static operating point in Figure 1 is usually set so that Uce is close to half power supply voltage. Our goal is to design Uce to be close to half power supply voltage, but how can we design Uce to be close to half power supply voltage? Here's how we look at it. Here are a few things you need to know first of all, first is what we often call Ic, Ib, these are collector current and base current of triode, there is a relationship between them: Ic=β× Ib. But when we first started learning, teacher obviously didn't tell us what size is right for Ic and Ib? This question is harder to answer because there are many things involved. But generally speaking, for low power lamps, Ic is usually set from a few milliamps to a few milliamps, for medium power lamps from a few milliamps to tens of milliamps, and for high power lamps from tens of milliamps to a few amps. In Figure 1, if Ic is 2 mA, resistance value of R2 can be calculated using formula R=U/I. VCC is 12V, then 1/2VCC is 6V, and R2 is 6V/2mA, which is 3kΩ. Ic is set to 2 mA, then Ib can be derived from Ib=Ic/β and key is value of β. The total 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 reality, β value of low power lamps is much greater than 100, between 150 and 400 or higher. Therefore, if above calculationt is executed, circuit may become saturated.

Therefore, sometimes we do not understand, calculation is correct, but it cannot be used in practice. This is because there is still a small practical guide that points out difference between theory and practice. This type of circuit is heavily influenced by value of β. When everyone calculates same way, results will not necessarily be same. That is, stability of this scheme is poor, and there are few practical applications. But if it is replaced by voltage divider bias circuit in Fig. 2, circuit analysis and calculation will be closer to actual circuit measurement.

In voltage divider bias circuit in Figure 2, we also assume that Ic is 2mA and Uce is designed so that 1/2VCC is 6V. How should R1, R2, R3 and R4 be set? The calculation formula is as follows: Since Uce is designed so that 1/2VCC is 6V, then Ic×(R3+R4)=6V, Ic≈Ie. It can be calculated that R3 + R4 = 3 kΩ. So how much are R3 and R4?Typically, R4 is 100 ohms and R3 is 2.9k ohms. In fact, we usually take 2.7 kΩ for R3. , because there are no 2.9 kΩ resistors in E24 series, there is no big difference between value of 2.7 kΩ and 2.9 kΩ. Since voltage across resistor R2 is Ube+UR4, that is 0.7 V + 100 Ω × 2 mA = 0.9 V.

We set Ic to 2mA, β usually takes theoretical value of 100, then Ib=2mA/100=20#A, it is necessary to estimate current, which is current flowing through R1, and total value is 10 times more, take IR1200#A.

Then R1=11.1V/200#A≈56kΩ R2=0.9V(/200-20)#A=5kΩ

Given that actual value of β can be much greater than 100, actual value of R2 is 4.7 kΩ. Thus, values ​​of R1, R2, R3, and R4 are respectively 56kΩ, 4.7kΩ, 2.7kΩ, and 100Ω, and Uce is 6.4V. In above analysis and calculation, many once assumptions have been made, which is necessary in practical applications. In many cases, we need a reference value for calculation. But often this is not case. Firstly, we are not familiar with various devices, and secondly, we forgot one thing. We are those who use scheme. Some data can be set by ourselves to avoid workarounds.