How exactly is the transistor collector current generated?

in the manufacture of transistors.

, consciously make the majority of the carrier concentration in the emission area greater than the base area, and at the same time the base area is very thin, and the impurity content should be strictly controlled, so that once the power supply is connected, due to the positive deviation of the emission junction.

Most of the carriers (electrons) in the emission region and most of the carriers (holes) in the base region easily diffuse to each other across the emission junction, but because the concentration base of the former is greater than that of the latter, the current through the emission junction is basically an electron flow, and this electron flow is called the emitter currentn.

Due to the thinness of the base region and the reverse bias of the collector junction, most of the electrons injected into the base region cross the collector junction and enter the collector region to form a collector current IC, leaving only a few (1-10%) electrons to recombine in the holes in the base region, and the recombined base holes are by the base.

The power supply EB is re-replenished, resulting in a base current IBO.

1. Let's put it this way. You can understand that the collector is a water pipe, which has a limited amount of water, but it cannot control the flow of water in it. The flow of water is controlled by a switch (paranoid resistor rc).

When the water flow is within the volume, the pipe has no effect on the water flow, if the water volume is too large, then the pipe can only be discharged so much, and it will be saturated... As for why the IC continues, it is because the collector is used to receive the electrons in the transistor that are ejected by the transmitter... The presence of voltage causes the electrons to be controlled by voltage, so that the electrons continue to flow from the direction of the triode collector to the power supply.

That is, the current is constantly flowing from the power supply to the collector. (The voltage forces the electrons to separate from the holes).

2. Have you ever played checkers? If you think of the beads as electrons, the holes as holes, and then put a row of beads to leave the first empty space. Put the first bead in the first empty space, and the first bead will be vacant.

One more vacant space than a bead). Then place the second bead in the original position of the first bead ......Repeat until the last one is finished. Then you will send the empty slots from the first to the last, which is equivalent to the empty slots flowing from the first to the last...

If you don't understand. Find a checkers and try it yourself...

3. Because only when saturated, the transistor UCE is the smallest, which is equivalent to the smallest voltage division of the triode. But in fact, as long as UBE>=, it can be a triode conduction, but the transistor will divide the voltage due to the existence of UCE, if it is regarded as a resistor divider, then if the UBE is larger, then the triode resistance is smaller. It can be understood in this way.

4. Do you understand the PN junction? If the PN is connected to the forward voltage, the diode will become a resistance of about 1K, which is linear. If it is less than 0,7V, then the resistance of the diode will increase infinitely.

One characteristic of linear resistance is that the voltage and current increase linearly. So if the voltage is larger, the current is larger, and the electron flow will be stronger, so it can be analyzed from this point that the ability to emit electrons actually increases as the voltage increases.

I don't understand what that means, but you have to know that the general low-frequency amplification is a common emitter or a common collector. In other words, the signal is flowing in from the base, so the size of the UB is changing. If you set the operating point to a very small IB, then when the sine wave signal is introduced, the positive half axis can be amplified normally, and the negative half axis will not be (cut-off distortion).

For the sake of so many servings, add a little more ...

Recently, the triode is also being studied, and there are different opinions under the common **, which is welcome to be pointed out.

There is a problem with your description, the IC is due to the current formed by the emitter diffusing to the electrons at the base, reaching the C pole through a drift motion. There are two reasons why this place can drift, one is the processing of the collector, which has a large area for easy penetration, and the other is that the collector applies a reverse voltage. As long as there are electrons from the emitter to the base, this current is constantly flowing.

IB is the current formed by the recombination of electrons and holes.

Such as the second article.

When used as a switch, the smaller the voltage drop is required (the specific reason can refer to the RDS selection when the MOS transistor is used as a switch), the deeper the saturation of the transistor, the smaller the voltage drop. (The minimum can be reached.)

It has a lot to do with the fact that the so-called emitter positive polarizer is also positive bias, and the way to saturate the triode is to increase the IB, so that the IC cannot follow the change with a fixed magnification.

If you say that, it won't help you.

1）i1 = i2 + ib；un = i2*r2 = ib*r3 + ube；i1*r1 + un = ucc；

Solve ib, then uc = ucc - ib* *r4;

2) If < uc < ucc, the transistor is in the amplification region, and the uc < is in the saturation region;

Assuming that the transistor is already in a saturated state, then ignoring the set-emitter saturation voltage drop, then ic=, if the magnification of the triode is assumed to be = 50, then the base current can make the transistor enter the saturation region as long as it satisfies ib=, then, we calculate the actual base current according to the parameters given by the circuit to know the real working state of the triode, according to the parameters listed in the circuit ib=(, the calculation shows that the actual base current has greatly exceeded the required saturation conditions, and the transistor is already in a deep saturation state.

The emitting and base voltages are the signal voltages to be amplified, the conduction of the collector junction is triggered, and the emitter voltage is the signal voltage to be amplified.

The collector voltage of the triode is transmitted through the line and then through the resistor, which triggers the flux and changes the magnitude of the flux, which in turn changes the magnitude of the emitter voltage.

Take the negative ground of a triode as an example: the collector plate has the highest potential, followed by the base, and the lowest is the emitter. In addition to the amplification function, the transistor can also be used as a switching circuit and other logic circuits, and the quality of the two PN junctions can be measured with a multimeter, and its magnification can also be simply measured.

Similar to the macrodispersion characteristics of the two-chamber extinct parallel polar transistor:

3D animation of the working principle of the transistor and MOSFET, and intuitively understand the direction of their current.

The transistor is composed of nonlinear PN sections, so the relationship between voltage and current is not a pure resistive linear relationship such as U=IR. First of all, the base current IB is very small, and the transistor is regarded as a three-terminal component according to the current law IC=IB+IE, so IC is about equal to IE. Silicon tube ube = , germanium tube ube is only about that.

As long as a greater voltage is added to the BE interval, it will be turned on, and the BE inter is a pn section, so the voltage drop will always = regardless of temperature and other factors, which is why a base resistor is added to limit the current, also called voltage division, in the calculation of ib=(vcc-ube) rb. It is precisely because of the nonlinearity of the PN segment, IC=(1+)IB, that there is no definite relationship with the magnitude of UCE. There is indeed a physical resistance inside the triode, but there is also a carrier pn section, which is why the UIR inside the transistor is not a pure resistance relationship.

As for why the constant voltage drop of the PN section, this is a physical phenomenon, you treat it as a component, the current will break down and burn off, and the normal current is constant voltage work. The resistor is another component, U=IR, which presents a linear relationship.

My understanding: the base and emitter are a loop, the emitter and collector are output loops, and the emitter is shared by the common level. Therefore, the current in the emitter is the sum of the base and collector currents, and the voltage drop of the emitter resistance is about equal to IC*RE, and this voltage drop is in series with the base loop and plays a role in weakening the base voltage, which is the original negative feedback.

The reason for the base emitter voltage volt is the forward voltage drop of the p-n junction, which is equivalent to the forward characteristic curve of the diode. You can directly understand that the voltage drop of the diode forward conduction is volts.

When the emission junction (not the pole) is forward-biased, ube = silicon tube), which is due to the p-n junction characteristic.

The emitter current is (+1) times the base current, and the feedback resistor is connected between the emitter and the ground, which does not affect the voltage between the emitter and the base, but only the voltage between the emitter and the ground, and of course, the voltage between the base and the ground. You might think of UBE as UB, the former being the voltage between the two poles, and the latter being the voltage from the base to the ground.

That's the dead zone voltage of the p.n. junction.

The essence is the same, but the connection is different.

Generally, the voltage is taken from the collector, but in fact, the collector current is also amplified.

Emitter current = collector current + base current,

The transistor itself is amplified by current.

Through an external resistor, according to Ohm's law, the amplified current can be converted into voltage.

The current in the emission area, including the base current and the collector current

Emitter current = collector current + base current,

The base current is controlled by the voltage to the base.

The collector current, which is controlled by the base current: IC = Beta * IB.

The collector current, which is then on the RC, forms the output voltage.

In the end, it is the base voltage, which controls the collector voltage, which is voltage amplification.

You just need to know the characteristics of the transistor, and you don't want to make the transistor.