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I was studying LED Heterostructures and I found out that my book is always giving reference to p-p-n type heterostructure. So I looked up into another book in order to find other type of heterostructures like npn, pnp, nnp etc. but that book also contained same example. Is there any specific reason behind using ppn heterostructure for LED. Can't I use nnp, npn or ppp?

thanks for giving your precious time.

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  • $\begingroup$ Double Heterostructures are more common. Here's an old paper discussing single vs. double and a variety of example structures. www3.alcatel-lucent.com/bstj/vol58-1979/articles/… $\endgroup$
    – user6972
    Commented Mar 22, 2014 at 8:26
  • $\begingroup$ That was really a good paper but it didn't answer my question. That paper explained the advantages of SH over DH LED. But I want to know why ppn DH LED, why not other types like nnp, nnn type DH? $\endgroup$
    – celos
    Commented Mar 22, 2014 at 14:45
  • $\begingroup$ Are you learning about band gap theory or quantum well theory? Can you edit your question with the example? Basically LED's are pn junctions with other things doped in between. As shown in the paper there are many considerations as to structure and doping. It has less to do with nnp, ppn, etc. Perhaps this will provide some more background ee.sc.edu/personal/faculty/simin/ELCT566/… $\endgroup$
    – user6972
    Commented Mar 22, 2014 at 18:21
  • $\begingroup$ For example here is a resonant cavity led structure. ecse.rpi.edu/~schubert/Light-Emitting-Diodes-dot-org/chap23/… or a current blocking layer ecse.rpi.edu/~schubert/Light-Emitting-Diodes-dot-org/chap23/… $\endgroup$
    – user6972
    Commented Mar 22, 2014 at 18:46
  • $\begingroup$ Can you post more information about the specific structure that is confusing you. As user6972 correctly suggests LEDs are basically a forward biased pn-junction. Any deviation from that basic structure is just an optimisation, nothing fundamental. $\endgroup$
    – boyfarrell
    Commented Mar 23, 2014 at 15:52

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A common structure for LEDs the ppin-junction,

[p-layer][p electron blocking layer][i layer][n layer]

The p-layer, i-layer and n-layer is just your standard pin-junction structure. The p and n layers provide an electric field, which under forward bias will drive electron and hole towards i-layer where they can recombine radiatively. Normally the p and n layers will be made of a higher band gap material (e.g. AlGaAs) and form a barrier around the lower band gap i-layer (e.g. GaAs). This helps the i-layer act as a radiative recombination centre to improve efficiency.

However, it is possible for electrons and holes to pass over the i-region without recombining radiative, in LEDs this is loss mechanism. Electrons are much more mobile than holes (by about an order of magnitude), so if we can prevent electrons from escaping we should be able to improve the efficiency of the LED (i.e. the electrons are the rate limiting carrier type). This is the role of the second p-layer or electron blocking layer.

If the electron blocking layer has a wider band gap (so AlGaAs with a higher Al fraction than the n and p layers), and has more p-dopant, then it is possible to align the valence band with the p-layer (obviously there is a space charge at the hetero-interface where the bands are not flat). This gives a large barrier only in the conduction band which prevents electrons from crossing the junction (and escaping) but does not restrict the hole current flow, thus improving the device efficiency. So this is why ppn or ppin structures are used in practice. It's an optimisation.

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  • $\begingroup$ So, the bottom line is that the asymmetry in selection of pp vs nn is because of the asymmetry in electron and hole mobility, right? $\endgroup$
    – Ruslan
    Commented Mar 29, 2014 at 13:19
  • $\begingroup$ There are probably other factor as well such as sheet and contact resistance, these devices operate with quite high currents so all resistances need to be very small to avoid losses, the doping will influence this. For these reason it would be best to grow on a p-substrate and have highly doping n-contacts at the surface. So this is probably why you see both ppn and nnp structures. But you're right, which ever route you take you want to target the electron escape. $\endgroup$
    – boyfarrell
    Commented Mar 29, 2014 at 13:58

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