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While I was studying about the working of transistors I came across a term called the depletion region which stops the flow of electrons from one place to another. A doubt that came to me was why couldn't the majority carriers form a depletion region?

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  • \$\begingroup\$ Because it's the majority carriers that are being depleted. \$\endgroup\$ Nov 26, 2014 at 6:56
  • \$\begingroup\$ @IgnacioVazquez-Abrams:Thanks for your comment.But why couldn't the majority carriers form a barrier potential because they too possess an positive and negative potential isn't it? \$\endgroup\$
    – justin
    Nov 26, 2014 at 6:59
  • \$\begingroup\$ They possibly could. But then it wouldn't be a "depletion region". \$\endgroup\$ Nov 26, 2014 at 7:00
  • \$\begingroup\$ @IgnacioVazquez-Abrams:why didn't the majority carriers formed a potential difference or what stopped them from producing one? \$\endgroup\$
    – justin
    Nov 26, 2014 at 7:14
  • \$\begingroup\$ I would say that it's the fact that they'd rather be on the other side of the interface. \$\endgroup\$ Nov 26, 2014 at 7:16

2 Answers 2

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You don't mention a specific type of transistor so I am going to explain generally only the physical principles.

You can think of a transistor as two connected pn junctions.

You have holes flowing from the p-type region to the n-type region. Also you have electrons flow from the n-type to the p-type. When holes reach the n-type region they will disappear with electrons. So in the n-type region which was electrically neutral now there are going to be positive charges. The same thing happen in the p-type only with negative charges.

More precisely when you have the holes flow from p to n upon arrival in the n region they will be the minority carriers and the electrons the majority carriers. When there were no holes the n-type region was electrically neutral. Now the minority carriers (holes) disappear (connect) with the majority carriers (electrons) so now the n-type region is positive charged. As I mentioned the same goes for the p-type region but with negative charge at the end.

The formation of the depletion region in a pn junction:

enter image description here

A pn junction in forward bias:

enter image description here

And in reverse bias:

enter image description here

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  • \$\begingroup\$ when does this happen i.e is it when the transistor is in off or on state? \$\endgroup\$
    – justin
    Nov 5, 2014 at 10:50
  • \$\begingroup\$ This is when the transistor is in open circuit. In any case either on or off you will have a depletion region. But it will be of different width. \$\endgroup\$
    – Adam
    Nov 5, 2014 at 10:53
  • \$\begingroup\$ does that mean that they would be a depletion region even if you don't apply an bias voltage. \$\endgroup\$
    – justin
    Nov 5, 2014 at 10:56
  • \$\begingroup\$ Yes it would be. Also it would be even if you apply bias or forward voltage but with different width in each case. \$\endgroup\$
    – Adam
    Nov 5, 2014 at 10:59
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    \$\begingroup\$ As you have a lot of elementary questions and there is an extensive discussion in comments which isn't appropriate I would recommend to you to start reading the physical principles behind diodes and build up until transistors. There are a lot of excellent books concerning electronics like Sedra Smith or Jaeger. :) \$\endgroup\$
    – Adam
    Nov 5, 2014 at 11:32
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I think the answer for this is in this imageenter image description here

This shows that only when the majority carrier(s) transfer from one region to another,minory carriers in effect really produce the depletion region.

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