As far as I know, for an NPN-BJT, when both (base-emitter and base-collector) are forward biased; we consider the operation to be in saturation mode. Here, the transistor functions as a short circuit between emitter and collector. [SEE IMAGE BELOW]

enter image description here

However, I expect quite different.

enter image description here

With the electric field directions as shown above;

I expect that a large number of electrons present in the emitter would now easily cross the emitter-base junction due to a reduced depletion region and supporting electric field.

Similar should be the case for the collector-base junction. (Electrons from collector would now be arriving at base)

So now, we have electrons coming in from emitter to base, and collector to base junction.

We therefore expect the direction of currents to be base to collector, and base to emitter.

Overall, we have a large current passing to the base, which is supplying current to emitter and collector.

Now, this is nowhere similar to a short circuit situation.

Where have I gone wrong?

Also, please explain the actual physical mechanism of the functioning of this BJT in saturation mode.

  • \$\begingroup\$ Basically your first diagram cannot happen. \$\endgroup\$ – Andy aka Oct 19 at 21:34
  • \$\begingroup\$ I think it's closer to say that saturation leads to forward bias of the other (BC) junction. \$\endgroup\$ – Brian Drummond Oct 19 at 21:39
  • \$\begingroup\$ @Andy aka. That diagram is what I found most often in literature around. I have not used that while suggesting what I expect out of this situation. Where have I gone wrong in my analysis, that's the problem. \$\endgroup\$ – Yashkalp Sharma Oct 19 at 21:43
  • \$\begingroup\$ @Brian, can you please explain your comment in more detail. I rather thought that having forward bias if BC junction is one of the condition of saturation. \$\endgroup\$ – Yashkalp Sharma Oct 19 at 21:47
  • \$\begingroup\$ I cannot understand why you are speaking of a "short circuit". We have a rather large base current (larger than in normal operation) - but that`s all. This has nothing to do with a short circuit. The base voltage drives a current through two pn junctions - hence, the base current is split into two currents. Where is something like a "short"? \$\endgroup\$ – LvW Oct 20 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.