Electronic – Common emitter characteristics in the saturation region

In general, it is the states of the PN junctions inside the transistor which will determine what operation region it is in. However, after gathering some experience, one can deduce the states of the above junctions by inspecting the circuit itself without actually measuring the voltages at the terminals.

An example:

Lets analyze the circuit you've referenced.

Once the switch is closed a current of approximately \$1mA\$ will flow into the base, which will cause:

$$V_{BE} \approx 2V$$

Since this is higher than the minimum of \$0.6V-0.7V\$ for being out of cut-off - the transistor is in one of its operational modes. In reality, the Base-to-Emitter voltage will not rise much beyond \$0.6V-0.7V\$ (due to presence of protection resistor R1), which means that the Base current will be a bit higher than \$1mA\$.

Knowing that the motor is \$12V, 100mA\$, and that the transistor is capable of handling \$100mA\$ Collector-to-Emitter current, we can deduce that:

$$I_C = I_{Motor} \approx 100mA$$

Given that we know (from motor's specs) that the motor will consume \$100 mA\$ at \$12V\$, the voltage on the motor:

$$V_{Motor} \approx 12V$$

Which leads to:

$$V_C \approx 0V$$

But this means that Collector-to-Base junction is forward biased which implies that the transistor in saturation.

The above analysis is quite general for this configuration (full voltage rated motor switched by matching BJT), therefore, in majority of circuits like this one, the transistor will be in saturation.

Experienced engineers perform the analysis above at a glance, knowing that the transistor in saturation a second after they see the schematics.

When electrons flow through a forward-biased diode junction, such as the base-emitter junction of a transistor, it actually takes a non-zero amount of time for them to recombine with holes on the P side and be neutralized.

In an NPN transistor, the P-type base region is constructed so as to be so narrow that most of the electrons actually pass all of the way through it before this recombination occurs. Once they reach the depletion region of the reverse-biased base-collector junction, which has a strong electrical field across it, they are quickly swept away from the base region altogether, creating the collector current.

The total current through the base-emitter junction is controlled by the base-emitter voltage, which is independent of the collector voltage. This is described by the famous Ebers-Moll equation. If the collector is open-circuit, all of this current flows out the base connection. But as long as there's at least a small positive bias on the collector-base junction, most of the current is diverted to the collector and only a small fraction remains to flow out of the base.

In a high-gain transistor, fewer than 1% of the electrons actually recombine in the base region, where they remain as the base-emitter current, which means that the collector current can be 100× or more the base current. This process is optimized through careful control of both the geometry of the three regions and the specific doping levels used in each of them.

As long as the transistor is biased in this mode of operation, a tiny change in base-emitter voltage (and a correspondingly small change in base-emitter current) causes a much larger change in collector-emitter current. Depending on the external impedance connected to the collector, this can also cause a large change in collector voltage. The overall circuit exhibits power gain because the output power (ΔV_C × ΔI_C) is much greater than the input power (ΔV_B × ΔI_B). Depending on the specific circuit configuration, this power gain can be realized as either voltage gain, current gain, or a combination of both.

Essentially the same thing happens in a PNP transistor, but now you have to think of the holes (the absence of an electron) as being the carrier of a positive charge that drifts all of the way through the N-type base to the collector.