Transistor depth Saturation analysis

Transistor Saturation Problem Summary:

1. In practical work, the common ib*β=v/r is used as the condition for judging the critical saturation. According to the IB value calculated by IB*Β=V/R, only the transistor enters the initial saturation, in fact should take the value of more than a few times to achieve true saturation, the larger the multiplier, the deeper the saturation .

2. The larger the collector resistance, the more easily saturated;

3. Saturation zone phenomenon is: Two PN junction is positive, IC is not controlled by IB

question: What is the transistor saturation when the base current is reached?

Answer: This value should be not fixed, it is related to the collector load, β value, the estimate is this: assuming that the load resistance is 1K,VCC is 5V, saturation when the resistor through the current maximum is 5mA, divided by the tube's β value (assuming β=100) 5/100=0.05ma=50μa, Then the base current is greater than 50μa to saturate.

For 9013, 9012, the saturation of VCE less than 0.6v,vbe less than 1.2V. The following is a 9013 feature table:

Question: How to judge saturation?

When judging saturation, the maximum saturation current IBS should be calculated, then the current base level current is calculated according to the actual circuit, if the current base level current is greater than the maximum saturation current of the base level, the circuit can be judged to be saturated at this time.

Saturation conditions: 1. There is a resistor between the collector and the power supply, and the larger the pipe saturation, the more easily the tube is saturated; 2. The base set current is larger so that the resistor of the collector pulls the collector's power supply low, resulting in a case where B is higher than the C voltage.

Factors that affect saturation: 1. The larger the collector resistance, the more easily saturated; 2. The larger the magnification of the tube, the greater the saturation; 3. The size of the base set current;

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 e-pole grounding)

Talk about saturation without lifting the load resistor. Assuming that the transistor set-emitter circuit of the load resistance (including the collector and the emitter circuit in the total resistance) is R, the set-emitter voltage vce=vcc-ib*hfe*r, with the increase of the Ib, VCE reduced, when vce<0.6v, the b-c knot is to enter the positive bias, ice is difficult to continue to increase, Can be thought to have reached saturation. Of course, if IB continues to grow, it will reduce VCE a little bit, for example, to 0.3V or lower, that is, depth saturation. The above is for NPN type silicon tube.

Another problem that should be noted is that when the Ic is enlarged, the HFE will be reduced, so we should let the transistor enter the depth saturation ib>>ic (max)/hfe,ic (max) is the IC limit in the case of a short circuit with the assumption that E, C pole, Of course, this is at the expense of shut-off speed.

Note: Saturation is VB>VC, but VB>VC is not necessarily saturated. The direct basis of the general judgment saturation is the magnification, and some of the pipe VB>VC can maintain a very high magnification. For example: Some tubes define ic/ib<10 as saturated and ic/ib<1 should be deep saturated.

Saturation problem from transistor characteristic curve: As I said before: talking about saturation cannot be without the load resistor. Now let me give you a little more explanation.

Take the output characteristic curve of a transistor as an example. As the original VCE was only drawn to 2.0V, I extended right to 4.0V for illustrative purposes.

If the supply voltage is V and the load resistance is R, then Vce and IC are constrained by the following relationship: IC = (v-vce)/R

On the output characteristic curve of the transistor, the above relation is a diagonal line, the slope is the intercept on the -1/r,x axis is the Intercept on the v,y axis of the supply voltage is v/r (that is, the "IC (max") referred to in the preceding NE5532 2nd refers to the IC limit under the assumption that E, C pole short-circuit. This slash is called a "static load Line" (hereinafter referred to as a load line). The intersection of the curve of each base current IB value with the load line is the operating point of the transistor at different base currents. See:

The figure assumes that the supply voltage is 4V, the green slash is a load line with a load resistance of 80 ohm, v/r=50ma, the figure of IB is equal to 0.1, 0.2, 0.3, 0.4, 0.6, 1.0mA of the operating point A, B, C, D, E, F. The relationship curve between IC and IB is made on the right. According to this curve, it is more clear that the meaning of "saturation". the green segment of the curve is a linear amplification area, and the IC increases almost linearly with the increase of the IB, which shows that the β value is about 200. The blue section begins to bend and the slope becomes smaller. The red segment becomes almost horizontal, which is "saturated". In fact, saturation is a gradual process, and the blue segment can also be thought of as the initial entry saturation. In practical work, the common ib*β=v/r is used as the condition to judge the critical saturation. In the picture, the hypothetical green segment continues to extend upward, intersecting the ic=50ma horizontal line, and the IB value corresponding to the intersection point is the IB value for critical saturation. The figure shows that the value is approximately 0.25mA.

The figure shows that, according to the IB value calculated by IB*Β=V/R, only the transistor into the initial saturation, in fact, should take the value of more than a few times to achieve true saturation; the larger the multiplier, the deeper the saturation .

A load line with a load resistance of 200 ohms is also shown. As can be seen, for Ib=0.1ma, when the load resistance is 80 ohms, the transistor is in a linear amplification area, while the load resistance is 200 ohms, it is close to entering the saturation zone. The load resistance varies from large to small, and the load line is fan-shaped upward as the center of the vce=4.0. The smaller the load resistance, the greater the IB value required to enter saturation, and the greater the C-E pressure drop under saturation. in circuits with particularly small load resistance, such as high-frequency resonant amplifiers, the collector load is an inductive coil, the DC resistance is close to 0, and the load line is stretched almost 90 degrees upward (in the red Load line). In such a circuit, the transistor will not enter saturation until it burns down. The above mentioned "load line" refers to the static load line of the DC, "saturation" refers to DC static saturation.

Questions to consider with transistors:

1) Withstand pressure enough

2) load current is not large enough

3) fast enough (sometimes slow)

4) B-pole control current enough

5) Sometimes power issues may be considered

6) Sometimes consider the problem of leakage current (can be "complete" cutoff).

7) The gain is not generally considered (my application has not yet asked for this parameter very high)

Actual use, the transistor note four elements on the line: -0.1~-0.3v oscillation circuit, 0.65-0.7v amplifier circuit, 0.8V above for switching circuit, beta value, HLW 30-40, low 60-80, switch 100-120 above the line, do not study other, study its covalent bond , electrons, holes useless

VCE=VCC (supply voltage)-VC (collector voltage) =vcc-ic (collector current) Rc (collector resistor).
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 (VCE cannot be equal to 0 when it is actually working, this is the characteristic of it), IC=VCC/RC. In other words, the IC cannot be larger than this value. The corresponding base current IB=IC/Β=VCC/ΒRC, which is the formula for calculating the current of the saturated base electrode.
Saturation is the critical saturation and over saturation of the two states. When IB=VCC/ΒRC, the transistor is basically in a critical saturation.
When the base current is greater than twice times this value, the transistor is essentially in depth saturation. Transistor depth saturation and critical saturation of VCE difference is very large. The critical saturation pressure drop is large, but the exit saturation is easy; the depth saturation pressure drop is small but not easy to exit saturation. Therefore, the base current chosen for different uses is not the same.
Also, there is a direct relationship between the saturation pressure drop and the collector current. The smaller the collector resistance, the greater the current of the saturated collector, and the greater the saturation pressure drop. Conversely (the larger the collector resistance, the smaller the current of the saturated collector, the smaller the saturation pressure drop). If the collector current 5 mah transistor saturation, 9013, 9012 and the like saturation pressure drop generally not more than 0.6 volts. When the base current exceeds twice times VCC/ΒRC, the general saturation pressure drop is about 0.3V.

Turn: This is my experience when teaching electronic technology, talk about transistors, beginners are difficult to understand, in order to talk about the thorough, I gave students a metaphor for the image: Transistor is a capitalist (full classroom lowered), such as a production of mobile phone capitalists, production of a mobile phone, raw materials 100 yuan, the price of 400 yuan, Profit margin of 400%, relative to the magnification of the transistor is 4, the original production of 100, profit tens of thousands of, the capitalists think this business is good, want to expand profits, improve production capacity, into a day to produce 200, that is, the transistor's input current increased, then the capitalists found, profits multiplied, good ah! It was then converted into a one-day production of 300, which was later converted into a one-day production of 400, 500 ... Until 1000, but the capitalists soon found that when the production capacity of more than 800 units, the profit will not rise in proportion, but slowly rise, more than 1000 units, the profit does not rise at all, maintained as is, because the output is too large, the market saturated, the price drop and so on, then the transistor into the saturation, The input current is increased and the output current does not increase. As a result of economic crisis, product sales do not go out, the capitalist had to stop production, a day does not produce, this is equivalent to the transistor into the cut-off state, but the factory always to maintain, so that every day selling raw materials, waste equipment, waste materials, or organize workers to clean, clean up warehouses and workshops, selling the junk, At least every day can be a little profit, this gain is the transistor cutoff state leakage current. That is, the input does not have a bit of current, the output is still some micro-current. From this process, we can find that the capitalist just magnified the profits, the raw materials into the finished product, which consumes a lot of manpower, brain power and electricity. Transistor and this analogy, transistor current amplification is actually amplified transistor input signal, output is amplified signal, the middle to consume a large amount of electrical energy, which must be direct current, such as batteries or rectifier after the alternating current. As the capitalists maintain their factories, human, mental and electrical power must remain stable and not be changed every day. Of course, for the power amplifier transistor, the rationale is basically the same, but the amplification is the signal of the current and voltage, of course, the input of human, mental and electrical power is still essential. Three levels, the base is the procurement, the collector is the processing workshop, the launch pole is sales.

Transistor depth Saturation analysis