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I wanted to start this question by giving my understanding of the make up of a PN junction. I do this so that you can pick me up on anything that I may be getting wrong in the lead up to my question.

I have been doing some research into PN junction diodes and understand that you have one junction which is formed by a p-type region and an n-type region. These two regions have been doped with impurities to give an excess of electrons (free electrons) in the n-type region and a deficit of electrons (free holes) in the p-type region. However due to imperfect manufacturing techniques, we also get minority carriers in the p and n type regions which means that the n-type has some atoms which have a deficit of electrons (free holes) and in the p-type we have some atoms which have an excess of electrons (fee electrons).

Now my question is this - how is it possible that these minority carriers exist? Wouldn't they just combine with the majority carriers?

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    \$\begingroup\$ Not instantly, they need to run into each other. Like if there's a terrorist in town and 10,000 police, will he just vanish? \$\endgroup\$ – Fizz Nov 16 '15 at 23:55
  • \$\begingroup\$ I don't quite understand you analogy. A minority carrier is a free hole or a free electron (depending on what type of semiconductor material we are in). However if we have free electrons from the majority carriers and free holes from the minority carriers, wouldn't the free electrons combine with the free holes? I understand that nothing actually vanishes, but rather just combines. \$\endgroup\$ – BradSlattery Nov 17 '15 at 2:17
  • \$\begingroup\$ I tried to give you an intuition. If you want the math: iue.tuwien.ac.at/phd/park/node32.html \$\endgroup\$ – Fizz Nov 17 '15 at 3:51
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The minority carriers in a semiconductor are not from manufacturing defects, but from spontaneous generation. Basically, in a semiconductor, electron-hold pairs are continuously being generated (using thermal energy). At equilibrium, these are also recombining when they (later) run into each other. A balance is created with the number or carriers being a strongly increasing function of temperature.

Now, when the semiconductor is doped, the number of majority carriers will be very high, and so when minority carriers get spontaneously generated (which still occurs), they have a much higher chance of recombining with a majority carrier quickly. This means that the number of minority carriers will decrease, but will not become zero.

There is an equation that describes this (Law of Mass Action). Basically, the product of Majority*Minority carriers is a constant (at a given temperature). When doped, the Majority increases and the minority decrease.

Undoped, this gives ne = nh; ne x nh = constant. When doped to Ne, you get ne x nh = Ne x Nh.

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  • \$\begingroup\$ Indeed: en.wikipedia.org/wiki/… \$\endgroup\$ – Fizz Nov 17 '15 at 3:15
  • \$\begingroup\$ I think I am getting what you're saying. So let's say we have some semiconductor material (un-doped) at a constant temperature (let's just say room temperature). This provides the semiconductor material with a constant fixed supply of energy in the form of heat. This heat energy causes some electron-hole bonds to break into a free hole and a free electron (thereby creating both minority and majority carriers). However after a period of time, a free electron will combine with that hole. Does that sound about right? \$\endgroup\$ – BradSlattery Nov 17 '15 at 4:10
  • \$\begingroup\$ Yes, although it's a probabilistic process; the electron and hole that recombine are not necessarily from a pair that split. While it does take energy to create the pair, this is released when they recombine, so no net energy is required to sustain a given level of carriers. \$\endgroup\$ – jp314 Nov 17 '15 at 4:22
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The surge power & surge current capability os a TVS is proportional to junction area. TVS diodes are constructed with large cross sectional area junctions to absorb high transients.

VI characteristics is similar to that of Zener but TVS are designed & tested for transient suppression.

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  • \$\begingroup\$ It looks like you posted this answer on the wrong question! \$\endgroup\$ – Kevin Reid Nov 17 '15 at 14:49

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