Transistor in saturation mode

Question

I am having great trouble in understanding the operation of transistor in saturation region. This is how I understand the working of a transistor:

In a bipolar junction transistor the width of base is very small, therefore a very small number of carriers undergo recombination inside the base, whereas, the rest of the carriers get conducted through the collector because of the polarity of voltage applied across the collector-base junction.

In this case, electrons flow towards collector because collector is at higher potential than the base. However, when the transistor is being operated in saturation region, the collector-base junction is forward biased i.e. collector is at lower potential than the base. If this is the case, how do the electron flow towards the collector?

Please help me understand this concept as I am experiencing great frustration not being able to understand it.

The original question is here: https://www.physicsforums.com/threads/transistor-in-saturation-region.401308/

Answer 1

In an ideal transistor (e.g. ultra low Vce(sat) parts from Diodes Inc are close to ideal but more $$)) Vbe=Vcb and both are saturated as charges flow between CE with a reduced hFe of about 10% to 25%..

• i.e. Ic:Ib ratio of 50:1 or 10:1 are typically used in many specs for these types, otherwise 10:1 is typical.

• The example below has hFE of 300~900, but Vce(sat) is given for 10:1 even though many use designs with 50:1 at reduced current.

  Then Rce is just a linear bulk R value with very little offset (<0.1) between Vbe-Vbe and Rce values of 10~1000mOhms are possible. N.B. see diodes inc search tool for range choosing any parameter on discrete bipolar.

Otherwise all others have an offset + Rce effective voltage rise on Vce

let me find some examples from Diodes Inc

200mOhms =Rce

Note Fig 7 Rce when operating at moderate to high current, Rce is fairly constant.

Answer 2

I'm not sure what to make of Tony's answer. But I'm not sure it answers your question, as you asked it.

Take a look at the following lateral planar BJT layout:

(Keep in mind that saturation isn't something that isn't determined by the device physics itself, but is determined by the surrounding circuit, as well. And also keep in mind that BJTs operate based upon both minority and majority currents, despite the fact that I'm going to not over-complicate this by including both. I'm going to describe this more like a vacuum tube than a BJT, to get the point across more easily.)

Normally, the NPN collector is much more positive than the emitter and the base is only a little more positive. So I think you can see that electrons emitted from the emitter would be accelerated towards the collector region and less so to the base. But as the collector is driven to increasingly less positive voltages, this accelerating force also diminishes as well. At some point, the base is actually more positive than the collector (and the transistor now becomes saturated.) But accelerated electrons will still reach the collector. But with the base far more positive, you also can see that the base would pull more electrons towards it than before.

It's a gross simplification. And I apologize for that. But it may get across one way of keeping in mind why the collector still collects most of the electrons. Also, the base tends to be lightly doped (higher resistance, lower mean drift velocities), as well. Regardless, perhaps this helps. But also keep in mind that saturation is caused by the surrounding circuit and that it's not solely internal device physics that defines it. It is when the collector is driven so that the base-collector junction is forward biased, however that may be achieved.