The output power can be used to calculate the emitter current; when selecting a transistor, you need to pay attention to the frequency characteristics. 2) Determine the power supply voltage. In the first figure, we observe that the maximum output voltage amplitude is 5V. The output voltage amplitude of the transistor is determined by the Vc pole voltage, and the voltage at the Vc terminal should be set to about 1/2 of the power supply voltage. Here we set the power supply voltage to 15V (in order to make the positive and negative signals have 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 treat it as a horizontal line as a reference point. When the input signal increases, Ib increases, and the Ic current increases, then the voltage of the resistor R2 U2=Ic×R2 will increase accordingly, Uce=VCC-U2, and will become smaller. Theoretically, U2 can reach a maximum value equal to VCC, and Uce will reach a minimum value of 0V. This means that when the input signal increases, the maximum change of Uce is from 1/2 of VCC to 0V. Similarly, when the input signal decreases, Ib decreases, Ic current decreases, and the voltage U2=Ic×R2 of resistor R2 will decrease accordingly, and Uce=VCC-U2 will increase. When the input signal decreases, the maximum change of Uce is from 1/2 VCC to VCC. In this way, when the input signal changes positively or negatively within a certain range, Uce has a symmetrical positive and negative change range based on 1/2 VCC. ) 3) Issues to consider when choosing a transistor: 1) Is the withstand voltage enough? 2) Is the load current large enough? 3) Is the speed fast enough (sometimes it needs to be slow) 4) Is the B-pole control current enough? 5) Sometimes the power problem may be considered 6) Sometimes the leakage current problem needs to be considered (whether it can be "completely" cut off). 7) Generally, the gain is not considered (my application has not yet required this parameter very much) 4) Determine the emitter current Ie based on the frequency characteristics of the emitter and the relationship between the frequency characteristics of the emitter. The emitter current of the small signal common emitter is 0.1 to several milliamperes. 5) Determine the value of Rc and Re. Usually Vce is set to half of VCC, Vce=Ic*(Rc+Re), Rc and Re are related to the amplification factor. 6) Determine the value of base bias circuit R1 and R2. We know the value of Ic, and from Ic=β*Ib (β is generally 100), we can estimate the current flowing through R1, which is generally about 10 times Ib. Calculate R1 and R2. R1 and R2 are the DC bias resistors of transistor V1. What is DC bias? Simply put, you have to work to eat. To require the transistor to work, certain working conditions must be provided first. Electronic components must be required to have power supply, otherwise it is not called a circuit. Among the working requirements of the circuit, the first condition is that it must be stable, so the power supply must be a DC power supply, so it is called DC bias. Why is the power supply through resistors? Resistors are like faucets in a water supply system, used to adjust the current. Therefore, the three working states of the transistor: stop, saturation, and amplification are determined by the DC bias. In Figure 1, it is determined by R1 and R2. 7) Determine the coupling capacitors C1 and C2. C1 and the input impedance, and C2 and the load resistor connected to the output end form high-pass filters respectively. The values of C1 and C2 must be obtained by calculating the center frequency. C1 and C2 are coupling capacitors. Coupling is the transmission of signals. The capacitor can couple the signal from the front stage to the back stage because the voltage at both ends of the capacitor cannot suddenly change. After the AC signal is input to the input end, the voltage at the output end will change with the AC signal input at the input end because the voltage at both ends cannot suddenly change, thereby coupling the signal from the input end to the output end. But one thing to note is that the voltage at both ends of the capacitor cannot suddenly change, but it is not impossible to change.
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