# Transistor in saturation mode

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/

• I'm not really sure what to do with this. You have simply copied a six year old question verbatim from another forum. What is it you don't understand from the answers given there? It's still a good question though, and maybe it deserves its place on EE.SE as well. – pipe Sep 25 '16 at 11:41
• @pipe - not only is it a verbatum copy of the entire question, a quick skim suggests there is a complete answer at physicsforum.com. So I don't understand why it would be replicated here. So unless you know physicsforum.com is going to be closed down, this should just be closed. Or am I missing something? – gbulmer Sep 25 '16 at 12:02
• @gbulmer The only reason I can think of is if the answers over there are not complete. I don't know enough about the subject to judge, but they sure seem complete and easy to understand to me. – pipe Sep 25 '16 at 12:06
• @pipe - okay. I'll give the OP some time to explain. – gbulmer Sep 25 '16 at 12:14
• I think the OP is specifically confused about this: at heavy saturation, the collector-base region becomes forward biased therefore, electrons from the emitter should much prefer to flow into the base rather than take the more arduous path into the collector. – Andy aka Sep 25 '16 at 12:39

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.

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.

• Thank u for your answer.I copied the question here because I could not find a satisfactory answer and thought it would be a time waste to type the same question.Sir I am still not getting it, for ex-In CB configuration NPN transistor,when it enters saturation mode the EB junction becomes forward biased hence the electrons flow into the base fro the emitter terminal simultaneously the CB junction is also forward biased hence electrons flow from the collector to base so base current should increase,I may be wrong but according to my consciousness reasoning this seems to be correct. – Ashutosh Yadav Sep 27 '16 at 19:32
• @AshutoshYadav Yes, the base current would increase. Is that a problem? I was trying to explain why the collector would still collect current, which it also does do, and to explain why the collector current can still be somewhat more than the base current. – jonk Sep 27 '16 at 19:51
• But sir,in saturation the base current is fixed,and sir if i have to forward bias the collector junction shouldn't I apply negative voltage in an NPN transistor – Ashutosh Yadav Sep 28 '16 at 15:34
• @AshutoshYadav Why is the base current fixed? – jonk Sep 28 '16 at 16:47
• Sir could you please explain me npn transistor in saturation mode in CB & CE config. in detail once again,it would be a great help from your side. – Ashutosh Yadav Sep 28 '16 at 19:15