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From my knowledge,

In active mode of operation of a NPN transistor, Base-Emitter Junction is in Forward Bias, and Collector-Base Junction is in Reverse Bias. Width of Base is very low.

During Saturation mode, both Base-Emitter Junction and Collector-Base Junction are in Forward Bias.

Vbc = -0.4V (approx.)

There, as usual, The electrons from the emitter make it into the base with a lot of kinetic energy. Some recombine in the base, and most of them want to move towards the collector, but since it is in forward bias and the collector is at a lower potential wrt base, it cannot "pull" that many electrons towards it.

As a result, the base current increases, and the collector current decreases right?

What I do not understand:

Then why is it mentioned everywhere, that during saturation mode, the transistor acts as a short-circuit or a switch, if practically, almost no current passes through to the collector?

Added a picture for added reference: enter image description here

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  • \$\begingroup\$ Saturation is a gradual process. As the BC junction begins to be forward biased, only a very tiny modification occurs and it may still look active. It's when the BJT goes into deeper saturation that it reaches the point where it kind of looks like switch. The easy way to say it is that the collector looks like a current source/sink when in active mode and that when the BC junction is largely forward-biased (the collector is near the emitter voltage) then the collector looks like a voltage source that is close to the emitter voltage. It's the collector load that forces saturation. \$\endgroup\$ Commented Jun 12, 2023 at 23:02

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if practically, almost no current passes through to the collector

Because this statement is incorrect.

As the collector voltage goes below the base voltage, the collector does indeed go into forward bias. I think where your misunderstanding lies is that while the proportion of electrons that go from the emitter to the collector decreases, the bulk amount of electrons can still be quite high.

It's a rule of thumb that when you run a transistor into saturation you plan on the \$H_{FE}\$ in saturation to be about one tenth as much as the figure out of saturation, for the same collector current.

So -- drive enough current into the base, and you'll see the transistor go into saturation and switch plenty of current -- as long as you do it right.

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  • \$\begingroup\$ Alright let's say a bulk of the electrons, do make it to the collector, wouldn't they be moving back again, towards the emitter? Because the collector is still forward biased, and so wouldn't there be a current component in the reverse direction, which would be also kind of significant in value for a constant supply of electrons from the emitter? \$\endgroup\$
    – SubbSE
    Commented Jun 12, 2023 at 23:29
  • \$\begingroup\$ @SubbSE In the non-linear (large-scale, not the linearized small scale) Ebers-Moll model, there is a saturated mode generator current (called \$I_{_\text{CT}}\$ in some papers) internally that represents the difference between the forward-biased BE junction diode current and the forward-biased BC junction diode current, moving from base, to internal collector, then to internal emitter. This generator current is not directly seen at the collector pin. See Figure 2.4 here. \$\endgroup\$ Commented Jun 12, 2023 at 23:55
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In saturation, the collector-base junction is in forward bias because it's being starved of current by the external circuit. That is, the collector current is controlled by the external circuit.

If you increase the available current supply to the collector, it will happily sink that current, as long as it stays below HFE * Ib.

Once the supplied current increases above that, the transistor will come out of saturation, collector voltage will rise, and the transistor will control its collector current.

That's why the behaviour in saturation is described as 'being a switch'. An on switch has its current controlled by the external circuit.

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A few charts may help a little, starting with one similar to yours

enter image description here

The value of \$R_{_\text{C}}\$ doesn't impact the near-zero (left) side of the results. (It only affects where each line ends on the right side.)

The small negative value region for \$V_{_\text{BC}}\$ is the forward-biased BC junction area. The BC junction cannot be passively driven forward-biased more than the BE junction. So the collector cannot go beyond the emitter voltage. That's why the left side doesn't exceed about a diode drop below zero on the curves there.

But that's a poor curve for what you want to see, I think. Let's look at it against an applied voltage, \$V_1\$, where the effect of the resistor is made more visible:

enter image description here

Now, you can more easily see that the collector current does in fact go towards zero. But only when the applied voltage goes to zero, also. Since the base junction is being driven with a current source, it has no choice about its current. But the collector current is going to obviously go to zero when the applied voltage there also goes to zero.

The slope on the left side is exactly what's required for a \$200\:\Omega\$ resistor. The slope is about \$5\:\text{mhos}\$, which is what you'd expect for the collector resistor. And the current doesn't actually reach zero until and unless the applied collector voltage also reaches zero.

The active mode is everything to the right of the left-side slope. The slope is just showing you that once it enters saturation, then the collector acts like a voltage source fixed hard and close to the emitter voltage and that the collector current is then entirely determined by the collector load resistor and the applied voltage. (Which is what would happen if you just grounded the collector resistor.)

It's not terribly complicated. For a while, when the collector current allowed by the recombination current at the base isn't sufficient to cause a large voltage drop across the passive collector load, then the collector "floats" and tracks applied voltage less the collector resistor drop and the collector current tracks the base current by a ratio. But when the base current is sufficiently large and allows more current than can be supported through the collector resistor (the applied voltage minus the collector resistor voltage drop would, if allowed, cause the collector voltage to go below the emitter voltage), then the collector just "gets stuck" close to the emitter and stays there. Then the collector current is just a result of the applied voltage across the collector resistor and is no longer a function of the base current -- saturation.

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  • \$\begingroup\$ That diagram I provided there, was for the Output Characteristics of Common Base Configuration of the NPN transistor... I believe all the graphs of the different configurations mostly imply the same characteristics, it's just the input characteristics that change right? \$\endgroup\$
    – SubbSE
    Commented Jun 13, 2023 at 8:50
  • \$\begingroup\$ @SubbSE The BJT itself has no idea what it is connected into. It's just a thing. I only tried to help highlight a detail that I thought may help out. In common base and in common emitter there is a collector load. In common collector, the load isn't present but it changes very little. The first diagram I show could just as well have used a 200 milliOhm resistor load as the 200 Ohm load I showed. Curves would have been almost identical except for the tails on the right. (If you want, I can add that curve set just to make the point.) \$\endgroup\$ Commented Jun 13, 2023 at 10:39

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