In an \$p-n-p\$ transistor, \$10^{10}\$ electrons enter the emitter in \$10^{-6}s\$ Calculate the emitter current.

My problem:

The lame way to attempt this question is to say that:

$$\text{current}=\frac{\text{charge flown}}{\text{time taken}}=\frac{\text{electrons flown}\cdot\text{charge on one electron}}{\text{time taken}}$$

But! We were taught in junction diodes, that holes and electrons both contribute to the total current. There will be as many equal and opposite holes flowing from emitter to base as many electrons enter the emitter. Hence, the correct answer should be:

$$\text{current}=\frac{2\cdot\text{electrons flown}\cdot\text{charge on one electron}}{\text{time taken}}$$

with the extra factor of \$2\$. But, my book has taken the first answer as correct. What is the fault in my logical reasoning?

  • \$\begingroup\$ I am new here, I checked this page, can anyone tell why we need to escape the $ sign with a backslash while writing inline mathjax here. It isn't the same as on Math.SE or Chem.Se. Thank you! \$\endgroup\$ – Gaurang Tandon Feb 17 '18 at 13:58
  • \$\begingroup\$ I agree that both electrons and holes contribute to the total current flow. Question is, is the ratio of holes and electrons in a PNP's BE junction the same as in a diode? In a PNP the base is smaller (shorter) and it has a lower doping level than the emitter. So what does that mean for the amount of holes and electrons in Base and Emitter? \$\endgroup\$ – Bimpelrekkie Feb 17 '18 at 14:35
  • \$\begingroup\$ @Bimpelrekkie Acc to me, it should mean that the the current in the transistor is majorly due to the majority carries, and that minority current can be neglected. Also, in a simple diode, both drift current and diffusion current are equal in magnitude at thermal equilibrium. So, while these two situations of a diode and of a transistor look identical, they actually aren't and my trying to relate them is fruitless. Is my interpretation correct? Thanks! \$\endgroup\$ – Gaurang Tandon Feb 17 '18 at 14:43

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