BJT saturation explanation

In summary: Yes, this is correct. The electric field of the depletion region (between the two junctions) can accelerate the charge carriers in the same direction as the diffusion current, and this current will be the collector-emitter current.
  • #1
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TL;DR Summary
This thread is about my questions on the mechanism of how current passes from emitter to collector in a saturated BJT.
If I have an NPN transistor and let's say we set the base voltage higher than the collector voltage.(Emitter is connected to GND).There are 2 currents flowing in the base because we have two forward biased junctions inside the diode , 1 is the current flowing from emitter to base and 1 is the current flowing from collector to base.

What I don't understand is why the transistor acts like a short circuit. I know that if we connect an emitter resistor the total current flowing through that resistor will be equal to Vc-Ve / Re (saturation voltage is neglected here)however how do the charge carriers pass from the emitter to the collector at first place?

I mean in saturation mode both junctions(base-emitter and collector-base) are forward biased so therefore the electric field of the depletion region cannot accelerate the charge carriers in opposite direction of the diffusion current because there is a net voltage which opposes that so the collector emitter current cannot be current due to that but it can't also be diffusion current because electrons don't naturaly flow from P type to N type region (base to collector).What am I missing here?

I am sorry I ask this question again but the link provided didnt answer my questions I don't know why I submitted I did idk.
 
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  • #2
Simple answer: In Saturation both the E-B and the B-C diodes are forward biased, allowing high currents to flow.

Slightly more advanced, but not terribly satisfying:
https://learn.sparkfun.com/tutorials/transistors/operation-modes

Rather deep into the physics with many external links:
https://en.wikipedia.org/wiki/Bipolar_junction_transistor
Note especially the figure about 1/4 down the page, under the "Structure" heading.

(above found with:
https://www.google.com/search?&q=bjt+saturation+mode+condition)

Cheers,
Tom
 
  • #3
Helena Wells said:
What I don't understand is why the transistor acts like a short circuit.
No - it is not. The collector-emiiter voltage will still be app. Vce=0.2 V. (npn case).

Helena Wells said:
I mean in saturation mode both junctions(base-emitter and collector-base) are forward biased so therefore the electric field of the depletion region cannot accelerate the charge carriers in opposite direction of the diffusion current ...
The base region is extremely small and considerably less doped than the emitter (and collector) region. Therefore, the collector voltage (still above the emitter voltage) allows the majority of the electrons from the emitter (npn case) move to the collector - in addition to the carriers coming from the forward biased B-C junction.
 
  • #4
LvW said:
No - it is not. The collector-emiiter voltage will still be app. Vce=0.2 V. (npn case).The base region is extremely small and considerably less doped than the emitter (and collector) region. Therefore, the collector voltage (still above the emitter voltage) allows the majority of the electrons from the emitter (npn case) move to the collector - in addition to the carriers coming from the forward biased B-C junction.
Yes ok but I agree the base region is very small and electrons flowing in the base from the collector are very likely to be found at the collector. However why are electrons inside the BC junction move to the collector since there is nothing that will pull them to the collector the electric field inside the depletion region is countered by VBC voltage if they were to be pulled as minority charge carriers and they can't be majority charge carriers because electrons don't move from P to N type region.
 
  • #5
Helena Wells said:
However why are electrons inside the BC junction move to the collector since there is nothing that will pull them to the collector the electric field inside the depletion region is countered by VBC voltage if they were to be pulled as minority charge carriers and they can't be majority charge carriers because electrons don't move from P to N type region.
As you have said at the beginning - the B-C junction is forward biased. Is this not enough to accept a flow of electrons from B to C ?
 
  • #6
I ask for flow from E to C not from B to C.
 

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