“In fact, as long as you understand the characteristics of the triode, it will be easier for you to use the microcontroller. Everyone actually knows that the triode has an amplifying effect, but how to truly understand it is the key to whether you will use most of the Electronic circuits and 1C in the future.
In fact, as long as you understand the characteristics of the triode, it will be easier for you to use the microcontroller. Everyone actually knows that the triode has an amplifying effect, but how to truly understand it is the key to whether you will use most of the electronic circuits and 1C in the future.
What we generally call an ordinary triode is a device with a current amplification effect. Other transistors also extend their functions on the basis of this principle. The symbol of the triode is shown on the left side of the figure below. Let’s take the NPN type triode as an example to explain its working principle. Since the triode is evolved from a diode, so everyone remembers that the PN junction always points to N, so that PNP or XPN-it is very clear.
It is a device that uses b (base) current lb to drive the current Ic flowing through CE, and its working principle is much like a controllable valve.
In the thin pipe on the left, the small water flow impulses the lever to open the valve of the large water pipe, allowing the larger red water flow to pass through the valve. The larger the blue water flow, the greater the red water flow in the big pipe. If the magnification is 100, then when the small blue water flow is 1 kg/h, then 100 kg/h of water is allowed to flow through the big pipe. The principle of the triode is the same. When the magnification is 100, when the lb (base current) is 1M, a current of 100mA is allowed to pass through Ice. Can you understand me when I say that?
Everyone may know this principle, but if it is used in a circuit and can be understood, there is a major obstacle to the use of the single-chip microcomputer. The most commonly used connections are shown below.
Let’s analyze this circuit. If its magnification is 100, we don’t count the base voltage. The base current is 10V+10K=lmA, and the collector current should be 100mA. According to Ohm’s law, the voltage on Rc is 0.1AX50=5V. Then the remaining 5V is eaten on the C and E poles of the transistor. good! Now if we let Rb be 1K, then the base current is 10V+lK=10mA, so if the magnification is 100, Ic is 1000mA or 1A? If it is really 1A, then the voltage on Rc is 1AX50Q=50V. Ah? 50V! Over the power supply voltage, are the triodes turned into generators? This is not the case. See below:
We still use the flowing water in the water pipe to compare the current. When the control current is 10mA, the valve on the main water pipe is opened to allow 1A current to flow, but isn’t it possible to have 1A current flow? No, because there is a resistor on it, it is equivalent to a valve with a fixed opening. It is stringed on top of the main water pipe. When the opening of the lower controllable valve is greater than the opening of the upper fixed resistor, The water flow will not increase any more but is equal to the water flow passing through the fixed valve opening above. Therefore, it is useless to open the lower triode to a large opening. Therefore, we can calculate the maximum current of the fixed resistance 10V+50Q=0.2A, which is 200mA. That is to say, in the circuit, the base current of the triode increases and the collector current also increases. When the base current lb increases to 2mA, the collector current increases to 200mA. When the base current increases again, the collector current will no longer increase, and it won’t move at 200mA. At this time, the upper resistor also acts as a current limiter.
Common emitter circuit NPN tube, when ib becomes larger, it essentially injects holes into the base area. If this is the case, the injected holes will neutralize more electrons from the emitter, and theoretically ic will become It’s right to be smaller, why does ic still magnify by the corresponding multiple of P?
The drawing in the picture is the current flow inside the transistor[NPN type tube, the arrow points to represent the direction of the current]and now the base current increases to 2, indicating that more electrons in the base area are recombined by the holes in the base area. It stands to reason. In other words, the collector current should be reduced[because more electrons are recombined in the base region, fewer electrons flow to the collector region]but the reality is that the collector current is amplified to 6. Obviously, when I increase the base current, the emitter current is also increasing, and the base current has doubled, and the emitter current has also doubled. Why?
In other words, if I double the base current, the electrons arriving from the emitter area to the base area will be recombined by twice as many holes. However, what causes the emitter to emit at this moment? There are twice as many electrons as before. For example, in the picture on the right, one more unit of electrons is recombined by holes in the base area, but at the same time, 4 more units of electrons are emitted from the emitter area. I artificially increased the base current by 1 unit, but the emitter emitted 4 more electrons, which increased the current by 4 units, why?
Do not use the formula ie=ib-ic=(1+3)ib to explain, please analyze and explain from the microscopic movement of carriers inside the transistor,
1 The emitter area emits electrons to the base area
Because the emission junction is in a forward bias, the diffusion movement of carriers is strengthened, and the carriers (electrons) in the emission region diffuse to the base region (called emission), and the majority carriers in the same base region (empty Holes) also diffuse to the emitter, but since the electron concentration in the emitter is much higher than the hole concentration in the base region, the diffusion of holes from the base region to the emitter is negligible in comparison between the two. Since the negative poles of the two power supplies Eb and Ec are connected to the emitter, the electrons from the emitter region to the base region from the emitter region to the base region can be supplemented from the power source, thus forming the emitter current le.
2 The diffusion and recombination of electrons in the base area
After the electrons emitted from the emitter region to the base region reach the base region, the electron concentration near the emitter junction is higher than the electron concentration near the collector junction, so these electrons will continue to diffuse toward the collector junction. During the diffusion process, a small part of the electrons will recombine with the holes in the base area. Since the positive electrode of the power supply Eb is connected to the base, these recombined holes can be supplemented by Eb, thus forming the base current lb. Because the base area is made very thin, the time for electrons to pass through the base area in the diffusion process is very short, and the hole concentration in the base area is very low, so the electrons emitted from the emitter area to the base area continue to the collector junction in the base area In the process of nearby diffusion, there is little chance of recombination with holes in the base area, so the current of the base electrode is very small, and most of the electrons can pass through the base area to reach the vicinity of the collector junction, so the collector current is very large.