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What would happen to a BJT if both junctions of the transistor are forward biased? Is there any literature on this? By literature I mean characteristics/equations. Any links would help. I have also drawn circuit diagrams to make my point clear.

schematic

simulate this circuit – Schematic created using CircuitLab

Where I was taught in college, we worked on

  1. active region (BE-forward biased, CB-reverse biased)
  2. cutoff region (BE-reverse biased, CB-reverse biased)
  3. saturation region (BE-forward biased, CB-on the verge of a reverse bias)
  4. What happens if CB and BE are both forward biased? Is there any literature or quantitative analysis method for that? What happens if BE is reverse biased and BC is forward biased? Is there any literature or operation zone defined for that.

The reason I want to inquire is because I was recently working on an Op-amp circuit attached to a BJT and there was a case where both the junctions were forward biased and I could not fathom how to analyze it because of basic complexity, in fact I did not even consider the case that both the junctions could be simultaneously forward biased and I only came to that conclusion when I simulated my circuit, hence I want to know how the regions operate so that in future I would not face problems with my circuits.

ubuntu_noob
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    It's called a diode. –  Jun 26 '14 at 17:26
  • It's saturation region. Nothing special. – Spehro Pefhany Jun 26 '14 at 17:28
  • but even in the saturation region, the current Ic flows into the collector in case of an n-p-n and out of the collector in case of an p-n-p, but what if the direction of Ic gets reversed? Is it still saturation? – ubuntu_noob Jun 26 '14 at 17:31
  • @ubuntu_noob It's saturated if the both junctions are forward biased, according to my definition. The situation you describe (Ic flowing out of an NPN, Ie flowing into an NPN, both junctions forward biased) is saturated in the reverse mode. IIRC, Vec(sat) in that mode can be lower than the normal Vce(sat) forward mode, but of course hfe is much lower than in the forward mode (less than 10 typically unless you have a (rare) transistor type designed to work in reverse). – Spehro Pefhany Jun 26 '14 at 17:43
  • See the [wiki](http://en.wikipedia.org/wiki/Bipolar_junction_transistor#Regions_of_operation). Figure on the right. – sherrellbc Jun 26 '14 at 19:42
  • Related: http://electronics.stackexchange.com/questions/12477/how-do-bjt-transistors-work-in-a-saturated-state http://electronics.stackexchange.com/questions/241194/why-does-the-collector-current-direction-remain-the-same-in-saturation-and-activ – Incnis Mrsi Aug 28 '16 at 13:51

5 Answers5

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That appears to be just an extension of the saturation region if CB and BE are both forward biased: http://en.wikipedia.org/wiki/Bipolar_junction_transistor#Regions_of_operation

It looks like your definition of saturation region is overly narrow.

If BE is reverse biased and BC is forward biased, then it still acts like a transistor, but because of the physical way bjt's are setup, you have much less gain than forward active.

horta
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What would happen to a BJT if both junctions of the transistor are forward biased?

Without reverse bias across base-collector region, the device no-longer behaves like a transistor and the two PN junctions act as forward biased diodes. A bit boring really.

Andy aka
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  • What exactly does your “bias” denote? The voltage (specifically, its sign) between terminals attached to **p** and **n** sides of the junction? – Incnis Mrsi Aug 28 '16 at 13:54
  • @IncnisMrsi I have no idea why you asked that question(s) or what your question means. – Andy aka Aug 28 '16 at 16:31
  • Note that @The Photon [pointed to your misconceptions](http://electronics.stackexchange.com/questions/241194/why-does-the-collector-current-direction-remain-the-same-in-saturation-and-activ) since. A forward-biased junction (namely, C–B) doesn’t yet imply that the junction is already *on* (where the BJT behaves like two diodes). – Incnis Mrsi Aug 28 '16 at 17:01
  • @IncnisMrsi please feel free to leave a more in-depth answer. That's what this site is about. – Andy aka Aug 28 '16 at 17:15
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schematic

simulate this circuit – Schematic created using CircuitLab

The transistors will behave as a pair of diodes with a common anode (NPN) or common cathode (PNP).

Transistor
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  • In which voltage range? We may not ignore p–n junction effects when B–C voltage is comparable to, or less, than the band gap. You can look for some insightful pictures at http://www.iitg.ac.in/apvajpeyi/ph218/Lec-5.pdf (although the region Ī̲ refer to isn’t pictured explicitly). – Incnis Mrsi Aug 29 '16 at 14:55
  • This circuit with resistors can’t bring a BJT to the region where B–C is forward biased (not yet on) but the collector current flows as in the forward active. To effect it, collector and emitter shouldn’t be powered from the same source. This answer covers only a part of the saturation region, and the least interesting part. – Incnis Mrsi Aug 29 '16 at 15:09
  • So, as the original poster proposed namely this “dull” configuration, Ī̲ don’t insist on answer. – Incnis Mrsi Aug 29 '16 at 16:08
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The main purpose of a transistor in electronics is to give output based on the base current. But, if both sides are forward biased, we will receive output and if both are reverse biased, we won't receive output, irrespective of base current.

BTitan
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As Transistor noticed, the BJT is in the saturation region. More specifically, in a hard saturation (both junctions are on); see the book below.

R. S. Ramshaw
Power Electronics Semiconductor Switches
Springer Science & Business Media, 2013
ISBN 1475762194

This article can also be helpful in explaining regions between such an abnormal BJT state (as in the circuit presented) and the forward-active mode. Although graphs don’t show the on–on state, only transition to it, the low-voltage forward region (where VBC has an abnormal, forward bias, but VCE is still right) is shown, at least partially.

Incnis Mrsi
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