Transistor saturation problem summary:
1. In practical work, Ib*β=V/R is commonly used as a condition for judging critical saturation. According to Ib*β=V/R calculated Ib value, just make the transistor enter the initial saturation state, in fact it should take several times more than this value, in order to achieve true saturation; the greater the multiple, the deeper the degree of saturation.
2. The larger the collector resistance, the easier it is to saturate.
3. The phenomenon in the saturated zone is that the two PN junctions are all positive and the IC is not controlled by IB.
Question: When the base current reaches the transistor saturation?
Answer: This value should not be fixed. It is related to collector load and β value. The estimation is as follows: Assume that the load resistance is 1K, VCC is 5V, and the maximum resistance current at saturation is 5mA, divided by the tube. The β value (assuming β = 100) 5/100 = 0.05mA = 50μA, then the base current can be saturated with more than 50μA.
For 9013, 9012, when Vce is less than 0.6V at saturation, Vbe is less than 1.2V. The following is a list of 9013 features:
Question: How to judge saturation?
When the saturation is judged, the base-level maximum saturation current IBS should be found, and then the current base-level current is determined according to the actual circuit. If the current base-level current is greater than the base-level maximum saturation current, the circuit can be judged to be in a saturated state at this time.
Saturation conditions: 1. There is a resistance between the collector and the power supply and the greater the easier it is to saturate the tube; 2. The base current is so large that the resistance of the collector draws the collector's power very low, resulting in b Higher voltage than c.
Factors affecting the saturation: 1. The greater the collector resistance is, the easier it is to saturate; 2. The magnification of the tube is larger, the easier it is to saturate; 3. The magnitude of the base set current;
The phenomenon after saturation: 1. The voltage of the base is greater than the voltage of the collector; 2. The voltage of the collector is about 0.3, and the base is about 0.7 (assuming the e-pole is grounded)
Talk about saturation can not mention the load resistance. Assuming that the load resistance of the collector-emitter circuit of the transistor (including the total resistance in the collector and emitter circuits) is R, the collector-emitter voltage Vce=VCC-Ib*hFE*R, as Ib increases, Vce When Vce<0.6V, the BC junction enters the positive deviation, and Ice is difficult to continue to increase. It can be considered that it has entered the saturation state. Of course, if Ib continues to increase, it will cause Vce to decrease again, for example, to 0.3V or even lower, that is, deep saturation. The above is in terms of an NPN silicon tube.
Another problem that should be noted is that when the Ic increases, hFE will decrease, so we should let the transistor into deep saturation Ib>>Ic(max)/hFE, Ic(max) is assumed to be e, c pole The Ic limit in the case of a short circuit, of course, is at the expense of the turn-off speed.
Note: Vb>Vc at saturation, but Vb>Vc is not necessarily saturated. The direct basis for judging saturation is usually the amplification factor, and some tubes can maintain fairly high magnification when Vb>Vc. For example, some pipes define Ic/Ib<10 as saturated and Ic/Ib<1 should be deep saturated.
Looking at the saturation problem from the transistor characteristic curve: I said before: talking about saturation can not mention the load resistance. Let us now explain in more detail.
Take the output characteristic curve of a transistor as an example. Since the original Vce was only drawn up to 2.0V, I extended it to the right by 4.0V for convenience.
If the supply voltage is V and the load resistance is R, Vce and Ic are bound by the following relation: Ic = (V-Vce)/R
On the output characteristic curve diagram of the transistor, the above relation is a diagonal line, the slope is -1/R, the intercept on the X axis is the power supply voltage V, and the intercept on the Y axis is V/R (that is, the front NE5532 According to the second post, "Ic(max) refers to the Ic limit in the case where e and c poles are short-circuited."). This slash is called "static load line" (hereinafter referred to as load line). The intersection of the curve of each base current Ib value and the load line is the operating point of the transistor at different base currents. See below:
The figure assumes that the power supply voltage is 4V, the green diagonal line is the load line with a load resistance of 80 ohms, V/R = 50MA, and the operating point where Ib is equal to 0.1, 0.2, 0.3, 0.4, 0.6, 1.0 mA is marked in the figure A, B, C, D, E, F. Based on this, Ic and Ib are plotted on the right side. According to this curve, the meaning of "saturation" is more clearly seen. The green segment of the curve is a linearly enlarged region, and Ic rises almost linearly with Ib. It can be seen that the β value is about 200. The blue segment begins to bend and the slope becomes smaller. The red segment is almost horizontal, which is "saturation." In fact, saturation is a gradual process, and the blue segment can also be considered as the initial saturated zone. In actual work, Ib*β=V/R is commonly used as a condition for judging critical saturation. In the figure, the hypothetical green segment continues to extend upward and intersects the horizontal line of Ic=50MA. The Ib value at the intersection point is the critical saturated Ib value. The figure shows that the value is about 0.25mA.
It can be seen from the figure that the Ib value calculated from Ib*β=V/R only causes the transistor to enter the initial saturation state. In fact, this value should be taken several times or more to achieve true saturation; the greater the multiple, the saturation level will be. The deeper.
The figure also shows the load line when the load resistance is 200 ohms. It can be seen that when the load resistance is 80 ohms corresponding to Ib=0.1 mA, the transistor is in the linear amplification region, and when the load resistance is 200 ohms, it is already close to the saturation region. The load resistance changes from large to small, and the load line is fan-shaped with Vce = 4.0 as the center. The smaller the load resistance, the larger the Ib value required to enter the saturated state, and the greater the CE voltage drop in the saturated state. In a circuit with a particularly small load resistance, such as a high-frequency resonant amplifier, the collector load is an inductor, the DC resistance is close to zero, and the load line extends almost 90 degrees upwards (red load line in the figure). In such a circuit, the transistor does not enter saturation until it is burned. The above-mentioned "load line" refers to a static load line of DC; "saturation" refers to static saturation of DC.
Problems to consider with triodes:
1) The pressure is not enough
2) The load current is not enough
3) Not fast enough (sometimes slow)
4) B pole control current enough
5) Power considerations may sometimes be considered
6) Sometimes consider the problem of leakage current (can be "completely" cut-off).
7) Generally do not consider how to gain (My application has not required this parameter to be very high)
Actual use, the transistor pays attention to four elements: -0.1 ~ -0.3V oscillator circuit, 0.65-0.7V amplifier circuit, 0.8V or more for the switch circuit, β value in the release, high release is 30-40, low-level 60- 80, switch 100-120 or more on the line, do not have to study the other, study its covalent bond, electron, hole is useless
Vce=VCC (supply voltage)−Vc (collector voltage)=VCC−Ic (collector current) Rc (collector resistance).
As you can see, this is a straight line with a slope of -Rc, called the "load line." When Ic=0, Vce=Vcc. When Vce = 0 (actually Vce cannot be equal to 0 during normal operation, which is determined by its characteristics), Ic = Vcc/Rc. In other words, Ic cannot be greater than this value. The corresponding base current Ib=Ic/β=Vcc/βRc, which is the formula for the calculation of the saturation base current.
Saturation is critical saturation and supersaturation. When Ib = Vcc/βRc, the transistor is essentially in a critical saturation state.
When the base current is greater than twice this value, the transistor will basically go into deep saturation. The deep saturation of the transistor and the critical saturation Vce are very different. The critical saturation pressure drop is large, but it is easy to exit saturation; deep saturation pressure drop is small but it is not easy to exit saturation. Therefore, the base current for different applications is different.
Also, the saturation voltage drop is directly related to the collector current. The smaller the collector resistance, the greater the saturated collector current and the greater the saturation voltage drop. On the other hand, the reverse is true (the greater the collector resistance, the smaller the saturated collector current and the lower the saturation voltage drop). If the triode is saturated with a collector current of 5 mA, the saturation voltage drops, such as 9013 and 9012, will generally not exceed 0.6 volts. When the base current exceeds twice Vcc/βRc, the saturation voltage drop is generally as small as about 0.3V.
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