Teaching of fault analysis of voltage-dividing common-emitter amplifier

Students majoring in electronics (electronic technology, electronic information, electrical engineering, mechatronics, computer) must learn the course "Analog Electronic Technology". It is a very practical subject, requiring students to be able to combine theory with practice, analyze and eliminate faults in specific electronic circuits after learning. Amplifiers are one of the key learning points in the teaching of analog electronic technology, especially voltage divider common emitter amplifiers, which are very common in actual electronic products. This article is the author's specific introduction to the troubleshooting method of this amplifier.

The voltage-dividing common-emitter amplifier is widely used in many electronic products because it can stabilize the static operating point well. The main purpose of this article is to make a detailed analysis of the various possible situations of the DC voltage of the circuit after the various components (transistors, various bias resistors, and related welding points) in the voltage-dividing common-emitter amplifier circuit fail. (Focusing on the analysis of the DC state, the AC analysis is relatively simple for the fault and will not be described in detail.) When we inspect the amplifier, we first use a multimeter to detect the DC voltage (potential) of the three electrodes of the transistor to the ground to determine whether the transistor is working in the amplification state (according to the conditions for the transistor to be in the amplification state: the emitter is forward biased and the collector is reverse biased at the same time). As can be seen from the right figure, the DC voltage of the three electrodes of the core component transistor Q1 in this voltage-dividing common-emitter amplifier circuit is normal, and the transistor is in the amplification state (emitter forward biased and collector reverse biased at the same time). Most NPN transistors are made of silicon. The emitter dead zone voltage is about 0.5V, and the emitter forward biased voltage is about 0.7V. In the amplification state, the tube voltage drop UCE>1V and is less than the DC power supply voltage VCC.

1 Fault Analysis

The author used the YB02-2 analog electronic experiment box produced by the Jiangsu Green Yang Group Yangzhong Green Yang Electronics Factory to conduct the following experiment. There is no AC signal source at the input end of the amplifier, so the amplifier is in a pure DC state. After theoretical calculation, the DC potential of each pole of the amplifier in the circuit of Figure 1 when it is working normally is marked in Figure 1. It is easy to see that the emitter of the transistor Q1 is forward biased and the collector is reverse biased. At the same time, the tube voltage drop UCE>1V, and the transistor is in a normal amplification state. The author actually records the data of DC fault phenomena in the circuit under various conditions as follows, and then conducts qualitative or quantitative theoretical analysis of different fault phenomena.

1.1 Fault phenomenon 1

UC=12V, UB=0V, UE=0V.

Analysis of the cause of the fault: From the DC voltage of the three electrodes, it can be seen that the transistor is in the cut-off state. Because UB=0V, there is no current flowing through the Rb2 resistor. The current in the Rb2 resistor comes from the Rb1 resistor, so this situation can only be caused by an open circuit in the Rb1 resistor or the corresponding solder joint. It is easy to know that IB=0 at this time. According to the transistor principle, there is no IC and IE (controlled source) without IB. Therefore, IC=0, IE=0.

1.2 Fault phenomenon 2

UC=6.4V, UB=7.0V, UE=6.3V.

Analysis of the cause of the fault: Both PN junctions of the transistor are in a forward biased state, and the tube voltage drop UCE=0.1V, and the transistor is in a deep saturation state. For transistors, their IC and IE are affected by IB. At this time, the saturated state IB is bound to be much larger than that in the amplified state. For the transistor base, the KCL law is satisfied: IB1=IB2+IB. IB becomes very large, which is likely to be caused by IB2 becoming very small, that is, Rb2 or the related solder joint is open. IB becomes very large, and the corresponding IC and IE also become very large (see Figure 2), which in turn makes URC larger, so UC is smaller than in the amplified state; URE becomes larger, so UE is larger than in the amplified state. The corresponding tube voltage drop UCE becomes much smaller than in the amplified state.

1.3 Fault phenomenon 3

UC=1.1V, UB=1.8V, UE=1.1V.

Analysis of the cause of the fault: Both PN junctions of the transistor are in a forward biased state, and the tube voltage drop UCE=0V, and the transistor seems to be in a deep saturation state. But if this is true, it will appear: the corresponding IC and IE also become very large, which in turn makes URC larger, and the corresponding UC should be smaller than in the amplified state; URE should also increase, and the corresponding UE should be larger than in the amplified state. But the facts are different: UE is much smaller than in the amplified state. It can be seen that IE has actually become smaller. According to KCL: IE=IB+IC. It is easy to see that IB has not become larger (if IB becomes larger, IE will also become larger), IC will only become smaller, that is, Rv becomes larger when open.

1.4 Fault phenomenon 4

UC=12V, UB=4.0V, UE=3.6V.

Analysis of the cause of the fault: The emitter voltage UBE of the transistor is 0.4V, which is less than the dead zone voltage. Therefore, the transistor is in the cut-off state, so IB=0, IC=0, and IE should also be very small. So UC=12V. (Fault 1 has been analyzed) The corresponding UE has become larger. This can only be because RE is open and larger.

1.5 Fault phenomenon 5

UC=12V, UB=1.32V, UE=1,32V.

Analysis of the cause of the fault: The emitter voltage USE of the transistor is 0V, which is less than the dead zone voltage. Obviously, the emitter of the transistor has been broken down, and the transistor is in the cut-off state, so IC=0. So UC=12V. (Fault 1 has been analyzed) At this time, IC should also be very small, making UB smaller than when it is in the amplified state.

1.6 Fault phenomenon six

UC=5.8V, UB=5.8V, UE=5.2V.

Analysis of the cause of the fault: The emitter voltage UBE=0.6V of the transistor is greater than the dead zone voltage. The transistor is in the on state. However, its collector voltage UBC=0V, and the collector should have been broken down. At this time, IC becomes very large. According to KCL: IE=IB+IC, the corresponding IE also becomes very large. Therefore, UE is larger than the amplified state.

1.7 Fault phenomenon seven

UC=6V, UB=4V, UE=6V.

Analysis of the cause of the fault: Both PN junctions of the transistor are in the reverse bias state, and the transistor is in the cut-off state. If the transistor itself is not damaged, then there will be IC=0, IE=0. And it will appear: UC=12V, UE=0V. But the fact is not the case, UC=UE=6V. This fact can only be that the transistor itself is broken down, causing the C pole and the E pole to be short-circuited. At this time, IC=IE is much larger than the amplified state. Therefore, the actual UE is much larger than that in the enlarged state.

1.8 Brief analysis of AC fault phenomenon

If the coupling capacitors C1 and C2 are in an open circuit state, the signal will not be able to be transmitted to the next stage. An oscilloscope can easily check the cause of such AC faults (shorting the two ends of the suspected capacitor with a wire can also identify the cause of the fault). When the emitter bypass capacitor CE is working normally, the AC voltage amplification factor is about several dozen times: when the emitter bypass capacitor CE is open circuit, the AC voltage amplification factor is generally less than two times (an oscilloscope can easily check the cause of such AC faults); when the emitter bypass capacitor CE is short-circuited and broken down, UE=0V, UB=0.7V, UC=0.06V, at this time the emitter current is obviously much larger than normal, which eventually puts the transistor in a deep saturation state, and the tube voltage drop UE