Saturation mode transistor

Question

Can somebody explain in a more mathematically way when and how a transistor enters its saturation mode, that's for example the case in a constant current source. I understand the consequences but not why the transistor goes into this mode. As I understand this saturation mode the transistor is able to vary the emitter-collector voltage with its base current, but the base current does not change the base-emitter voltage.

Answer 1

If you are looking for a bright line answer, then the following diagram provides it:

Saturation is anywhere on the left. Active mode is anywhere on the right of the bright yellow line.

In this viewpoint, whenever the BC junction is in any way forward-biased, the bipolar transistor has crossed the line.

Reality is a little more nuanced, as saturation is kind of soft at first. When the BC junction is only slightly forward-biased, it can be largely ignored and the bipolar transistor appears to work as if it were in active mode. You can see this also in the above picture, as the green line shows the \$\beta\$ being quite similar to the active mode \$\beta\$ with only a very gradual decline until the BC junction is forward-biased by \$\ge 450\:\text{mV}\$ and the green line traces out a much more rapid decline in \$\beta\$.

When the BC junction is forward-biased by \$\ge 450\:\text{mV}\$, the collector has, by definition, been pulled about as close to the emitter as possible and the collector starts to look much more like a voltage source and a lot less like a current source.

Some defer the meaning of saturation until this point is reached, rather than saying saturation happens when the BC junction is forward-biased by any amount at all.

It's really just a matter of application space, though. For switch use, this latter "BC junction is forward-biased by \$\ge 450\:\text{mV}\$" definition is better because the goal in switching is to have the collector very close to the emitter and acting like a voltage source. For voltage amplifier use, the "BC junction is forward-biased by \$\ge 0\:\text{mV}\$" definition is better because it is very important to have the collector acting like a current source and to avoid getting into a situation where deeper saturation and switch-like behavior increasingly occurs.

Answer 2

Plain answer

My explanation is not mathematical, but I believe it will be very useful to you because before you can calculate a circuit, you have to understand it intuitively.

Think of the transistor as a variable resistor ("rheostat") that adjusts the current in the voltage-supplied circuit by changing its resistance. In order to obtain a varying voltage, we connect a resistor Rc in series to the collector and take the voltage Vce (Vc) between the collector and the emitter as the output.

As the input voltage increases, the transistor begins to decrease its "resistance" and the current increases. The voltage drop across the collector resistor increases and across the transistor (output voltage) decreases. The transistor is capable of changing the current or, as they say, it is in "active mode".

Eventually, the transistor exhausts its supply of "resistance" and cannot reduce it below zero (only negative resistors can do this magic but the transistor is a "positive resistor"). As they say, the transistor has "saturated".

Here are some quantitative considerations ("math"). The transistor will be saturated when the voltage Vce becomes almost zero, and this will happen when the voltage drop VRc across the collector resistor becomes equal to the supply voltage - VRc = Ic.Rc = Vcc (ie, when all the voltage is lost across the collector resistor and there is nothing left for the transistor.

This will happen when the base current is greater than Ic/β. In the simplest case, the base current is determined by an input voltage Vin and a base resistor Rb, so Ib = Vin/Rb. So β.Rc.Vin/Rb = Vcc is the saturation condition of the transistor.

CircuitLab experiments

Presented in this way, "saturation" is a concept that is not only related to BJT but is much more general. To demonstrate this, I have used three different levels to perform the same CircuitLab experiments - with a variable resistor, a current source and finally a transistor.

Conceptual resistor circuit

Current output: To control the current, let's connect a variable resistor Rce (rheostat) in series to the voltage source. The ammeter Ic serves as a current load. So, when Rc varies...

^{simulate this circuit – Schematic created using CircuitLab}

... the voltage across it is constant...

... and only the current varies. The curve is non-linear because the resistance is in the denominator of the Ohm's law expression Ic = Vcc/Rce.

Voltage output: To get an output voltage, we can insert a resistor Rc in the collector. A "floating" voltage drop VRc occurs across it, but we prefer to take its grounded complement Vce across the transistor.

"Active mode": When we change Rce within certain "reasonable" limits without reaching the final values (zero and infinity), the resistor Rce is "active" (it can change the current and voltage).

^{simulate this circuit}

For example, if we change (sweep) Rce from 100 Ω to 1 kΩ, VRc decreases from 9 V to 5 V, and Vce increases from 1 V to 5 V.

"Saturation": When Rc = 0 Ω, Vce = 0 V and cannot decrease further.

^{simulate this circuit}

We can see this Rce "saturation" in the beginning of the Vce curve.

Negative resistance: For Vce to continue decreasing, Rce must become negative. And here, as a joke, I set Rce = -100 Ω... and it worked! Very interesting, the simulator could work with negative resistances!

^{simulate this circuit}

Now the Vce curve begins from below zero (-1 V), and at 0 V the transistor is not saturated.

Hmmm... let's make sure with some simple example that CircuitLab really works with negative resistances to use it in the future. And indeed, if we connect in series two resistors with the same positive and negative resistance of 100 Ω between the power source and the load, we will see that their resistances (voltages) are mutually neutralized, and the load voltage is equal to the supply voltage. So it works fine!

Current-source circuit

Above Rce was a linear (ohmic) resistor, but the curve turned out to be non-linear. The paradox is that if Rce is a non-linear current-stabilizing resistor (behaving like a current source), the curve becomes linear. Since CircuitLab does not have such an element, we can simulate it with a true current source (producing power).

"Active mode": If we vary (sweep) the current from zero to 9 mA like in Schematic 1.2.1...

^{simulate this circuit}

... we can see that the curves are linear.

"Saturation": At the maximum current of 10 mA, the current source is "saturated", and the voltage across it is zero.

^{simulate this circuit}

True current source: But if this is a true current source producing more than 10 mA, the voltage will go below zero.

^{simulate this circuit}

Transistor circuit

Finally, we repeat the same experiments with a real transistor and see that the results are the same (there is only a difference in the shape of the curves determined by the transfer characteristic of the transistor).

Active mode: The transistor can control the output voltage from 0 V to Vcc.

^{simulate this circuit}

Saturation: The transistor cannot control the output voltage; it is just a "piece of wire".

^{simulate this circuit}

“Negative transistor”: To bring it out of saturation, we can apply the negative resistance trick by inserting an additional negative voltage source Vee in the emitter.

^{simulate this circuit}