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The figure here presents the output characteristics of a common emitter BJT.

My question is that why the region mentioned as saturated region is known as the saturation region?

As far as I know is that saturation is the term used when there is no further increases but in this region current (collector current, I(c) is increasing as the V(CE) increases but it's named a saturated region.

On the other hand collector current remains unchanged for an increase in V(CE) and it's named as active region.

So my question is why it's called saturated region?

Note the image is taken from this link.

enter image description here

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6 Answers 6

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"Saturation" is confusing. It has at least three different meanings in electronics.

  1. "Saturation current" is the theoretical diode current under reverse bias in the diode equation.
  2. When a field effect transistor is operated in the region where the drain current versus voltage is nearly constant, the transistor is said to be "saturated".
  3. When a bipolar junction transistor is operated in the region where the collector voltage is near zero, the transistor is said to be "saturated".

Note that the FET and BJT definitions are almost opposites.

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Jos and Simon's answers do make sense intuitively, but for what it's worth, a related but ultimately different explanation is given for the name "saturation" in Chenming Hu's Modern Semiconductor Devices for Integrated Circuits:

So far, we have avoided examining the part of the \$I–V\$ curves in Fig. 8–12 that is close to \$V_\text{CE}\$ = 0. This portion of the \$I–V\$ curves is known as the saturation region because the base is saturated with minority carriers injected from both the emitter and the collector. (Unfortunately the MOSFET saturation region is named in exactly the opposite manner.) The rest of the BJT operation region is known as the active region or the linear region because that is where BJT operates in active circuits such as the linear amplifiers.

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  • \$\begingroup\$ BJT term saturation actually makes sense physically, whereas MOSFET doesn't really. That's why the terms "active region" and linear one (low VCE/VDS) are much clearer. \$\endgroup\$
    – edmz
    Feb 22 at 16:25
  • \$\begingroup\$ @edmz I think one can argue either way. In a MOSFET's saturation region, the current has "saturated" as it does not increase much with increasing \$|V_\text{DS}|\$. The linear (triode) region is also aptly named because current increases approximately linearly with \$|V_\text{DS}|\$, resembling triode characteristics. \$\endgroup\$
    – Puk
    Feb 22 at 16:42
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This graph is obtained by explicitly varying \$V_{CE}\$ using a voltage source connected directly between collector and emitter. It must not be confused with the common-emitter setup, with collector resistor, which I address later.

This is the arrangement used to produce the graph you provided:

schematic

simulate this circuit – Schematic created using CircuitLab

The independent variable in that graph is \$V_{CE}\$. This is the one being varied, while the dependent variable, the resulting collector current \$I_C\$, is plotted. This process is repeated for different values of constant base current \$I_B\$:

enter image description here

One definition of saturation is: the condition where an increase in base current does not result in an increase of collector current.

In this scenario, this can be interpreted as the region of the graph where, if you jump from one plot (say \$I_B = 3mA\$, orange) to another (say \$I_B = 5mA\$, brown), you do not see a significant change in current \$I_C\$. This behaviour is visible in the left-most region, where the plots for various \$I_B\$ all overlap, sharing the same \$I_C\$.

Confusion about this graph, and its representation of "saturation" may stem from better familiarity with the common emitter arrangement:

schematic

simulate this circuit

This time, it is base current \$I_B\$ that is the independent variable, which I will sweep from 0A upwards. If I plot a graph of the resulting collector current \$I_C\$, it looks like this:

enter image description here

It's very similar to the first graph, but that similarity is only due to a sudden change of behaviour when collector current is no longer able to rise any further. That occurs, of course, when collector potential is as low as it can get, or in other words, \$V_{CE}\$ is close to zero.

The same definition of saturation is true: the condition wherein an increase in base current does not result in an increase of collector current. However, in this scenario the "saturation" region is everything to the right of the green marker.

Everything to the left of that marker is the "active", or "linear" region, where collector current is approximately proportional to base current.

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Today we use "saturated" for MOSFETs when they are in the region of relatively high drain-source voltage where the drain current does not increase very much when the drain voltage is further increased.

Originally the term was introduced for BJTs with almost the completely opposite meaning. That also is a form of saturation if a fixed load is present: the collector current cannot increase further if you increase the base drive, because the load would otherwise make the collector voltage drop to zero.

The reason why people chose to use the term saturation in a different way for a MOSFET than for a BJT is probably lost in time (none of the discussions we see about it explains it). But since the BJT came first, it would be better to ask whay saturation for a MOSFET is defined as it is, not questioning the BJT's definition. After all it would have been reasonable to just duplicate the way it was already used.

Interestingly, what is called saturation region for a MOSFET was called "pentode region" for a BJT, so the BJT did actually try to honor its predecessor by not messing up the terms!

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Basic idea

"Saturation" is a concept that in both cases has a meaning of "current limitation".

You can think if this arrangement as a network of two resistors R1 and R2 in series supplied by a voltage source V where, for some reasons, the current I does not change - I = V/(R1 + R2) = const. There are two possibilities:

Exhausted Rce

Vcc/(Rc + 0) = const. Think of the transistor as a "variable resistor" whose "resistance" Rce is controlled by the input base-emitter voltage Vbe. When Vbe increases, Rce gradually decreases, and finally becomes zero. Then the collector current is determined only by Rc (Ic = Vcc/Rc) and stops changing.

Dynamic Rds

Vdd↑/(Rd + Rds↑) = const. When Vdd increases, at some point Rds begins increasing with the same rate of change. As a result of this dynamic resistance, the drain current stops changing.

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As far as I can see, the actual reason why the collector current reaches a maximum value and therefore cannot increase any further has not been clearly explained up to now.

And this has to do with the definition of the saturation state: Vce < Vbe or Vb>Vc or Vbc>0 .

So if the voltage drop across the collector resistor Rc is so large (due to a sufficiently increased Vbe voltage) that the base-collector junction is no longer operated in the reverse direction (Vc<Vb), the normally almost constant relationship Ic=B*Ib no longer applies, as now - in addition to the "normal" base current - a second current component flows through the forward-biased base-collector junction.

This means that the significantly increased base current - in contrast to normal amplifier operation - is not the cause, but rather the result (and a clear indication) of the saturation state.

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