What happens when the CB junction is in forward bias and theBE junction is in reverse bias in an NPN bipolar transistor?

I thought of two possibilities:

  1. The NPN transistor acts as it is but in reverse (I mean the I_c flows from collector to emitter.)
  2. Then I realized that the concertation of donor atoms in the emitter is greater than collector and came up with a different explanation with a different result.

I think that the concentration of donor atoms is large in the emitter compared to the collector because in active mode (CB junction is in reverse bias and BE junction is in forward bias) to increase the electron current in to base when this huge amount of electron current enters into the base region when the electrons enter from emitter to base then a lot of electrons go intothe CB depletion region because the doping of the collector and base is small so the depletion region contains a low concentration of electrons on the base side compared to the electrons coming from the emitter so due to the concentration gradient between electrons coming from the emitter region and the electrons in the CB depletion region, the electrons will diffuse into the depletion region (this is the what I understood from the Razavi's video. He gives the reason at 49:30 minutes. Tell me if I misunderstood the reason why electrons flow from the base to the collector.) The remaining small portion of electrons are attracted to the base terminal (because the base terminal is positive and its width is small so only small portion will go into the base region) but when the CB junction is in forward bias and the BE junction is in reverse bias then we can say that electrons will flow from the collector region to the base region but they are no large and from base to the emitter its not possible to go because now the EB depletion region has high concentration of electrons on the base side so now the diffusion current is not possible so the diode will not allow the current.

I don't know which of my assumptions is true. I ran a simulation on in this site and another one in LTspice. I got very little current (like -ve pico amperes) in this site's simulation and in LTspice I got unexpected values. Can you explain why I got these results and explain these questions?

  1. Explain how electrons flow from the base region (which came from emitter to base region) to the depletion region at CB junction?
  2. Why do we heavily dope the emitter? (I know it is to get a large amount of majority carriers but I want to know if there is any other reason and if there is any relation to the question 1.)
  3. Why do the widths of the collector, the emitter, and the base follow this order?
  4. (Optional) Recommend a book which contains diodes, MOSFETs and bioplar transistors and amplifiers and opamps and deals with conceptual contions and explains why and how things work instead of just solving some example problems.

LTspice simulation: lt spice sim

CircuitLAB simulation:


simulate this circuit – Schematic created using CircuitLab


1 Answer 1


It's the reverse active region. You still get transistor action just like in the forward active region but since in "normal" operation the collector is lightly doped to increase the collector-base breakdown voltage and the emitter is heavily doped to increase current gain, you flip it around and end up with a transistor that has low current gain and low C-B breakdown voltage.

1: Minority carriers diffuse across the base due to the concentration gradient set up by injection from the emitter and recombination + removal at the collector-base junction

2: The emitter is heavily doped to enhance carrier injection into the base.

3: I'm assuming you mean collector width greater than emitter width greater than base width. Base width must be small to minimize recombination, collector width is large because the depletion region extends far into the collector, and emitter width can be less because its depletion region is narrower.

  • \$\begingroup\$ can you answer my refer me a website or an article that explains about reverse active mode \$\endgroup\$
    – Qwe Boss
    Commented Apr 17 at 13:48
  • \$\begingroup\$ and can you answer my remaining questions \$\endgroup\$
    – Qwe Boss
    Commented Apr 17 at 13:49
  • 2
    \$\begingroup\$ You can achieve a low saturation voltage this way, occasionally useful. \$\endgroup\$
    – John Doty
    Commented Apr 17 at 14:10

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