Does Ic reverse its current flow when BJT in common-emitter configuration goes from linear to saturated mode?

Question

I don't understand how both junctions of BJT are forward biased in saturation mode. Does \$I_c\$ reverses? Does the saturation happen when \$V_{BE}\$ > \$V_{CE}\$ ?

Please explain the phenomenon using common emitter configuration.

P.S:

Consider a NPN transistor. During saturation \$V_{CE}\$ is approx 0.2V . Since \$V_{BE}\$ is always around 0.7V for NPN. This means $$VC=0.2V+VE$$$$VB=0.7V+VE.$$ Subtracting we get \$V_B−V_C=0.5V\$ , hence both junction are forward biased.

I get this. What I don't understand is since \$V_{B}\$ > \$V_{C}\$ now, why doesn't current flow from \$B\$ to \$C\$ now? If this happens how is there even something called saturation region. Why doesn't the whole processes that happened while in the working region stop?

Answer 1

EDIT #2: In your edited question, you have (rightfully so) shown your continued confusion, as your question wasn't really answered yet (I have added emphasis on the word "working"):

What I don't understand is since Vb>Vc now, why doesn't current flow from B to C now? If this happens how is there even something called saturation region. Why doesn't the whole processes that happened while in the working region stop?

FIRST, let me clear it up for you: BJT transistor NEVER STOPS WORKING as long as its base-emitter junction "diode" is forward-biased.
Your confusion stems partially from your incorrect view and from the slightly misleading naming of the transistor operating modes as active and saturation - a transistor is ACTIVE/WORKING (amplifying its base current) throughout both its LINEAR and SATURATION region!
In your mind, you may be thinking active and non-active. Our subconscious minds play such simple tricks on us sometimes, especially when our picture, knowledge or understanding of something is incomplete.

To help demonstrate my point, I have taken one of my examples from below to the most extreme case which never happens in reality but it helps you understand.
Let's assume the collector-emitter channel is so deep in saturation that it has ZERO resistance and the collector and emitter are practically shorted and at exactly the same, GROUND potential.
This would make the base-emitter junction diode and base-collector junction diode have EXACTLY the same voltages since both PN junctions will be connected in parallel.

^{simulate this circuit – Schematic created using CircuitLab}

Even in this unrealistic worst case scenario, the base-emitter diode conducts as much current as does the base-collector diode, in other words their currents are equal and the "amplification" of the transistor becomes 1, meaning no current amplification.
But this is practically impossible because the basic theory of "transistor-action" means that there has to be base-emitter current for the collector-emitter current to flow, this base-emitter current can't be greater than the collector-emitter current, AND because the collector-emitter channel ALWAYS has some resistance.
Finally, a BJT is not just two diodes sandwiched back-to-back, but it has an additional collector-emitter channel represented as a resistor connected between collector and emitter, and its resistance depends upon the base current.
It is almost like the base-emitter is a diode whose current controls the collector-emitter channel resistance, as below.

^{simulate this circuit}
The above schematic would be a more complete representation of a BJT.

For all of the above reasons, you will always have at least some voltage drop (at least a few mV) across the collector-emitter channel and the base-emitter voltage will never be less than the sum of the base-collector and collector-emitter voltages, even when the base-collector voltage is lower than the base-emitter voltage or even when the base-collector voltage is zero; its voltage relative to ground is still no less than the base-emitter voltage.
If the base-emitter voltage ever started becoming less than the Vbc+Vce, the base current would start going lower, which would start increasing the collector-emitter channel resistance, so this sort of a negative feedback prevents the transistor from ever being so saturated that it drops less voltage via the combination of BC and CE voltages than across the BE alone.

EDIT: Your question in the comment below has helped me figure out a better answer.

Let's start with an unsaturated transistor working in its linear region, as below.
The base-emitter junction is forward-biased with its typical forward diode voltage drop.
The base-collector junction is reverse biased.
Notice one thing here:
the Vbe voltage or the voltage across the base-emitter diode is equal to the sum of the Vbc and Vce voltages, in other words to the sum of the base-collector diode and collector-emitter channel voltage drops!

First an actual transistor schematic:

^{simulate this circuit}

Now a diode representation "equivalent":

^{simulate this circuit}

Now we look at the saturated transistor situation:
Again, first the transistor schematic:

^{simulate this circuit}
And now the diode "equivalent":

^{simulate this circuit}
Notice that we have exactly the same thing here with a saturated transistor as we did above with the non-saturated transistor:
the Vbe voltage or the voltage across the base-emitter diode is equal to the sum of the Vbc and Vce voltages, in other words to the sum of the base-collector diode and collector-emitter channel voltage drops!
So, apparently a bipolar junction transistor becomes saturated when its base-collector junction becomes forward biased and its voltage equals the base-emitter junction voltage.
Additionally, driving such a transistor deeper into saturation allows for a lower voltage drop across the collector-emitter channel, and this voltage adds to the base-collector diode voltage drop to equal the base-emitter voltage drop.
One more thing to notice is that the collector-emitter channel in all cases acts like a variable resistor whose resistance depends on the amount of the base current being injected into a part of the channel.

Just for an exercise and a further clarification, let's make the values of resistors such that the base-emitter diode and collector-emitter channel voltages are equal.
This is the point at which a transistor starts to enter into a saturation.
Notice that the voltage between base and collector or on the base-collector diode is actually ZERO!
What does this mean?
Does it mean that the base-collector diode/junction is shorted?
NO! Such thing can't happen inside a functional transistor, although it would seem to be the case based on zero voltage between base and collector.
What is ACTUALLY happening here is that the base current value has changed the collector-emitter resistance to the value which in that particular circuit (in combination with a particular load) develops a voltage drop equal to the base-emitter voltage at that moment.
The base and the collector electrodes are not shorted in this case, even if zero voltage between would make you believe they are.
This is another lesson to "take home": zero volts between two points or equal potential on two different points does not automatically mean they are shorted or connected to each other; it may mean that they are simply at the same potential at the moment.

^{simulate this circuit}

Again, a diode "equivalent" schematic which should make what's going on a little more obvious:

^{simulate this circuit}

OLD ANSWER BELOW:

No, the Ic does NOT reverse its direction. It still keeps flowing from collector to emitter (assuming positive/conventional current flow here).

^{simulate this circuit}

As you can see in the above circuits, the one on the left is unsaturated, and the collector potential is above the base potential.
The one on the right is saturated, and the collector potential is lower than the base potential.
In both cases, the emitter current is equal to the sum of the base and collector currents.
Feel free to click on "simulate this circuit" to see it better and experiment with it.

And yes, the saturation happens when Vbe is larger than Vce, as the Vbe can't go below a certain voltage (typically around 0.6V for silicon transistor) if you want to keep the transistor switched ON, while the Vce saturation voltage can go close to 0V.
In that case, both PN junctions are forward biased.

Answer 2

Under some (unusual) conditions, current can indeed flow out of an NPN transistor's collector.

^{simulate this circuit – Schematic created using CircuitLab}

However, under normal conditions V1 is much higher than the open-circuit saturation voltage of the transistor, so external current flows into the collector as usual in saturation.

Answer 3

Consider a NPN transistor.
During saturation \$V_\text{CE}\$ is approx \$0.2\text{V}\$. Since \$V_\text{BE}\$ is always around \$0.7\text{V}\$ for NPN. This means $$ \begin{align} V_\text{C} = 0.2\text{V} + V_\text{E},\\ V_\text{B} = 0.7\text{V} + V_\text{E}. \end{align} $$ Subtracting we get \$V_\text{B} - V_\text{C} = 0.5\text{V}\$, hence both junction are forward biased.

Answer 4

Because the B-C junction is forward biased, there is a component of current flowing (assuming an npn transistor) into the base and out of the collector, yes.

There is also a component flowing, in most applications, into the collector and out of the emitter.

The total current at the collector is the sum of these two components, which can be directed into or out of the collector, or even be zero if they balance exactly.

There's a component into the base and out the emitter as well, but this doesn't (directly, anyway) impact the collector current.

Answer 5

What I don't understand is since 𝑉𝐵 > 𝑉𝐶 now, why doesn't current flow from 𝐵 to 𝐶 now?

Saturation in a BJT is truly something that challenges naive assumptions about electricity.

In a forward biased PN junction diode, we would expect to see free electrons move from the negative N region to the positive P region, and/or to see holes move from the positive P region to the negative N region. But in a saturated NPN transistor, where the base collector PN junction is forward biased, we see electrons flowing in the opposite direction, i.e. from the (more) positive P region to the (more) negative N region.

^{simulate this circuit – Schematic created using CircuitLab}

In an NPN transistor in saturation, the base more positive than the collector, yet electrons flow from the base into the collector, that is from the base region, not the base terminal. This challenges the naive intuition that electrons always flow from a more negative region to less negative (i.e. more positive region), except when there is an electromotive force, such as a battery, to make them move against the gradient.

Thinking in terms of conventional current, as opposed to electron flow, in a saturated NPN transistor, current flows from the less positive collector, through the more positive base.

To understand this, we must understand that conventional current does NOT always flow from more positive to less positive, and electrons do NOT always flow from more negative to less negative.

A gradient in the electric potential causes charges to accelerate. Electrons accelerate in the direction of a more positive potential. Interactions (called collisions) between electrons and their environment result in the acceleration being short lived, but after a collision an electron will begin to accelerate again. This process imparts a net velocity component to the average free electron. This is called drift.

The average drift velocity of electrons in a typical circuit is quite small in comparison to the individual velocities of electrons due to thermal motion. This is similar to how individual molecules in the air might be traveling at 500 m/s while a breeze might be blowing a mass of air at only a few cm/s. However, in the case of electrons, the difference in velocities is much more dramatic. For example, in copper at room temperature, the average speed of an individual electron due to thermal motion is about 100 km/s. On the other hand, the drift velocity might be only 1 mm/s!

So, an electric field causes electron drift velocity, but this drift is only a very small portion of the total motion of an individual electron.

There is another effect which causes net movement of charges, and that is diffusion. Because electrons are in motion, free electrons will tend to diffuse from a region of higher concentration to one of lower concentration. (Holes will do something similar).

In a saturated NPN transistor, the collector is only very lightly doped. So unless electrons are injected into the collector from the base, the concentration of free electrons will be very small. On the other hand, the emitter is much more heavily doped than the base, and so when the the base-emitter junction is forward biased, a high concentration of free electrons is injected into the base from the emitter. The higher concentration of free electrons in the base than in the collector means that free electrons will diffuse to the collector, even when the electric field would potentially cause them to drift in the opposite direction. As long as the external circuit continuously removes those electrons from the collector (which will happen as long as the collector is more positive than the emitter), if the base has a high concentration of free electrons, they will flow into the collector even though the base is more positive than the collector.

Thus we have electrons flowing from the more positive base to the less positive collector in a saturated NPN transistor!

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Comment from @LvW (in chat)

Only now I have detected a severe error in the answer from @MathKeepsMeBusy. He wrote (saturation case): " electrons are flowing..... and then from the more positive P region of the base to the less positive N region of the collector. They are flowing against the gradient of the electric potential." This is, of course , wrong. For the B-C junction also applies the same rule as for the B-E juncton: Electrons move in a direction towards the more positive potential - that means: From C to B.

@LvW think about the situation again. In an NPN transistor in saturation, conventional current is flowing from collector to base. That is, from the collector region into the base region, not out through the base terminal. Physically electrons are flowing from base (region) to collector (region). But the base-collector junction is forward biased, That means the base region is more positive than the collector region. So electrons are flowing from the more positive base region to the more negative collector region. That is what is so counter-intuitive about charge flow in a saturated (NPN transistor).

Answer 6

It depends on the collector voltage.

If (assuming a grounded emitter) you take the collector voltage below ground, then current DOES flow out of the collector, sourced from the base, as the BC 'diode' is now forward biassed.

This allows you to use a BJT transistor as a shunt switch in bipolar analogue applications. It's not however a very good one. For low distortion, a shunt FET is a far better, that is a FET is a far more linear shunt switch.

I have used just such a bipolar shunt switch as a crude phase detector, for a lock-in amplifier type application.

Answer 7

The phenomenon of saturation (without going into the physical-mathematical descriptions) in the common emitter configuration of a BJT is clear when the collector characteristic curves of the BJT and the static load line are drawn. The operating point Q must be chosen at the center of the linear region so as to have equal variations of the voltage Vce, i.e. increases and decreases. When the operating point Q moves on the load line in Q', following an increase in the base current, the transistor is on and is in saturation. By further increasing the base current, the operating point is in Q'' and the more the base current increases, the more the saturation increases. In any case moving Q in the linear region, the base-collector junction is reverse biased, while the base-emitter junction is forward biased. On the contrary in the saturation region both junctions are forward polarized See photo:

Answer 8

When both junctions are forward biased (as a result of a large Ic which causes a voltage drop across Rc large enough to make Vbe>Vce) the base curent Ib increases correspondingly due to an additional portion Ibc through the B-C junction. Nevertheless, the classical equation Ie=Ic+Ib is still valid.

However, due to the additional base current Ibc the net collector current Ic (through the node C) is somewhat smaller as anticipted by the relation Ic=B*Ib.

Therefore, Ic does not "reverse" its direction but there is a small additional portion in opposite direction.

Comment (Edit): The collector current Ice will flow still in the "normal" direction from C to E (npn case) because there is still a driving voltage Vce>0. However, The current Ic through the node C is not identical to the (largest) portion of Ic (here called Ice) which arrives at the emitter node (Ice=Ic+Ibc). Here are the equations:

Ie=Ib+Ic with Ib=Ibc+Ibe and Ic=Ice-Ibc, which leads to

Ie=Ibe+Ice