What exactly is the saturation state of a transistor

What exactly is the saturation state of a transistor

Next, let's discuss what is the saturation of a transistor, what is the saturation voltage drop, is BE positive bias, BC reverse bias saturation, or when the transistor is in a saturated state, its base current will lose control over the transistor

1: When we adjust the adjustable resistor, R1， When R1=4.3K, we can calculate using Ohm's law that,

Ib=(5-0.7)/4.3K=1mA,

Then we can calculate IC=Hfe * Ib=10mA,

If I adjust R3 to 500 ohms at this time, we can calculate that Vc=5- (10mA * 0.5K)=0V,

At this point, let's take a look at the voltage across the three poles of a transistor, Vb=0.7,Vc=0,Ve=0,

2: When we adjust the adjustable resistor, R1， When R1=4.3K, we can calculate using Ohm's law that,

Ib=(5-0.7)/4.3K=1mA,

Then we can calculate IC=Hfe * Ib=10mA,

If I adjust R3 to 1K ohms at this time, we can calculate Vc=5- (10mA * 1K)=-5V. Will it return to -5V? Of course not, because there is no negative pressure,

So the voltage of Vc will stay at 0V. At this point, let's take a look at what Ic is

Through Ohm's law, we can calculate

Why is Ic=(5-Vc)/1K=(5-0)/1=5mA instead of 10mA,

At this point, let's take a look at the voltage across the three poles of a transistor, Vb=0.7,Vc=0V,Ve=0V,

3: When we adjust the adjustable resistor, R1， When R1=2.15K, we can calculate using Ohm's law that,

Ib=(5-0.7)/2.15K=2mA,

So we can calculate IC=Hfe * Ib=20mA. If I adjust R3 to make R3=1K ohms at this time, we can get Vc=5- (20mA * 1K)=-15V through calculation. Does it return to -15V? Of course not, because there is no negative pressure,

So the voltage of Vc will stay at 0V. At this point, let's take a look at what Ic is

Through Ohm's law, we can calculate

Why is Ic=(5-Vc)/1K=(5-0)/1=5mA, still 5mA instead of 20mA,

At this point, let's take a look at the voltage across the three poles of a transistor, Vb=0.7,Vc=0V,Ve=0V,

I actually think a good analogy is that the process of control is like pushing a gate with our hands and letting water flow through it,

The magnitude of our strength is the base current Ib,

The opening size of the gate is Ib * Hfe,

The water flow passing through the gate is Ic,

So if the collector is connected to a large reservoir and we push it hard, the larger the opening of the collector gate, the greater the water flow will be. If we don't push the collector gate hard, the opening of the collector gate will become smaller, and the water flow will become smaller. At this time, the size of the water flow will be controlled by the size of the gate.

So if the collector is connected to a water pipe and we push it hard, the gate of the collector gate will be opened like a chimney in a thermal power plant, which is useless because the water is only this big, and no matter how big the channel is, it will be useless.

So what we control by the base current is the ability of the collector to flow current, not how much current the collector has. Only when the collector has sufficient capacity, can we control this gate to achieve the goal of controlling the size of the water flow,

So the true saturation should be when the water flow is greater than the maximum flow that our maximum gate can pass through, which is the correct solution. That is, the gate is already at its maximum, and no matter how much water you have, it is useless.

That is to say, the maximum water flow Ic through the gate is limited by the opening size Ib * Hfe of the gate, but the actual water flow cannot be controlled and depends on the water supply equipment. But the maximum will not exceed the tolerance of the gate.