Consider the NPN BJT. Normally, if you want to use it as a diode, you short base to collector and treat the emitter as the cathode. But in principle, you could also use the base emitter junction by itself (collector floating), or even the base collector junction (emitter floating). I guess you could even tie the emitter and base together and treat the collector as the cathode. What would the properties be of these diodes? Anything that might be useful in special circumstances?


1 Answer 1


If you follow BJT modelling, then you can conclude these facts

  • Emitter is doped highest but has considerable bigger area than base and lesser than collector
  • Base is doped considerable more than collector but less than emitter and is having least area
  • Collector is doped least and has maximum area among all three.

Now this will lead to Band diagram as shown below.

Band diagram

Thus for considerable amount of drift & diffusion currents as per study from Energy levels , Base and collector must be shorted to make a P-type node of Diode and Emitter is heavily doped already hence it will make N-type node of Diode

NOTE : I might be not able to explain well , but I hope this facts come useful to understand

added later on

Why to short Base-Collector , Instead of Base-Emitter

Lets say following abbrevations for sake of explanation
BE: Base Emitter Junction
CB: Base Collector Junction

Consider the BE as single entity,Now since emitter is doped more heavy compared to base BE junction will act as N-Type element, since u are left with collector which is already N-Type element , Thus it will result in bad diode.
Now on other hand for CB as single entity, base is more heavy doped compared to collector, Hence CB will act as P-Type element, you are left with emitter which is already comparative doped as N-Type element. Thus this can result in well functional Diode.

More ....

$$ I_E = I_B +\ I_C \\ I_C = \beta I_B \\ I_C = \alpha I_E \\ I_E = (\beta +\ 1)I_B \\ I_C=I_s e^{\left(\frac{V_{BE}}{V_T}\right)} \\ \text{Normal values of}~\beta = 99 ~ \& ~ \alpha = 0.01\\ I_E \approx I_C \approx \text{few hundreds}~mA \\ I_B \approx \text{few}~\mu A $$ On study of above equations, we can have few conclusions

  • emitter & collector contribute major amount of currents. compared to base
  • base has to be in picture as its connecting link for collector & emitter

PS: this lab document has some good insights

  • \$\begingroup\$ I like the answer but I would welcome a better explanation why base and collector actually "have to" be shorted. Why not just use base and emitter? Maybe someone else can contribute some more details. \$\endgroup\$
    – Rev
    Commented Dec 18, 2014 at 8:00
  • \$\begingroup\$ @Rev1.0 I think the later added explanation can help.. \$\endgroup\$ Commented Dec 18, 2014 at 9:21
  • \$\begingroup\$ Maybe my question wasn't clear. That connecting BC still results in a p-region is understandable, there are simply more positive carriers in the base than negative in the collector region. My actual question was more: Why not ignore the collector? E-B should make a better diode than E-BC. That was essentially also part of the original question. \$\endgroup\$
    – Rev
    Commented Dec 18, 2014 at 10:01
  • \$\begingroup\$ @JigarGandhi I am very interesting about this subject since this technick used very often inside of ICs. So please try again and edit. \$\endgroup\$
    – GR Tech
    Commented Dec 18, 2014 at 11:04
  • \$\begingroup\$ i am not able to get exact reason i feel.. hope someone gets exact reason. \$\endgroup\$ Commented Dec 18, 2014 at 11:57

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