Why Is One Side of a Diode Heavily Doped?

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In diodes, one side is heavily doped to ensure effective electrical contact and improve conductivity, particularly at the junction where metal contacts connect to the semiconductor. The heavy doping on the p-side in standard diodes and the n-side in LEDs facilitates Ohmic contacts, which are essential for optimal device performance. This doping reduces the depletion region's impact, allowing for better current flow and efficiency. Additionally, in bipolar junction transistors, heavy doping enhances minority carrier injection, improving current gain. Overall, the primary reason for heavy doping is to achieve reliable electrical connections and enhance device functionality.
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Can someone explain why in diodes one side of the junction is heavily doped, namely the p-side in a standard diode, resulting in a p+n junction? Or in an led where the n-side is heavily doped? I can't seem to find the reasoning behind this anywhere.
 
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Doping is what gives a material (silicon, for example) it's "semiconducting" properties... If you didn't dope it, it'd be more or less an insulator.
 
Then you don't have to worry about depletion on the heavily doped side, i.e. it's negligible
 
elimenohpee said:
Can someone explain why in diodes one side of the junction is heavily doped, namely the p-side in a standard diode, resulting in a p+n junction? Or in an led where the n-side is heavily doped? I can't seem to find the reasoning behind this anywhere.

Sigh. How about a real answer.

The heavier doping, you will notice, is typically in drain and source areas where metal contacts are connecting them to other parts of the circuit. Or for other technologies like bipolar, you'll have an extra implant of heavier doping at all the contact points.

http://wpcontent.answcdn.com/wikipedia/en/thumb/6/62/Cmos_impurity_profile.PNG/500px-Cmos_impurity_profile.PNG

http://sub.allaboutcircuits.com/images/03302.pngSo, primarily the heavy doping is to assure an Ohmic contact, rather than a rectifying contact. Metal-Semiconductor junctions are diodes with depletions layers, etc., after all. Find the section in your semiconductor physics text about Ohmic contacts and it will become clear.

The other reason can be seen in the bipolar example: the buried N+ layer for improving collector resistance: highly doped == better conductivity. Sometimes you'll see "sinker" implants which are an extra N+ implant from the collector's contact N+ down to the buried layer. In this HBT patent, #106 is a sinker to contact the buried layer #102 for the collector implant #104. The actual collector contact would go on top of #106 overlapping a bit of the oxide #110.

http://www.freepatentsonline.com/7183627.htmlOne place where this heavy doping actually hurts is in the drain because you'll have a tiny depletion layer which means extremely high fields. This causes Hot Carrier Injection to occur which damages the MOSFET gate oxide over time (and is the primary failure mechanism above ~200 nm design rules). For this reason, Vdd values were dropped from 5 V to 3.3 V and lower years ago and things like LDD and DDD implants became standard to spread out the drain depletion layer and reduce the HCI creating fields.

You'll see heavier doping in bipolar emitters. The reason for this is that it improves the asymmetry of minority carrier injection which directly affects the beta or current gain of the device. The high doping assures that reverse injected minority carriers are quickly recombined. The lower base doping gives the minority carriers more chance to reach the base-collector depletion layer and become collector current. In the EM or GP models, this means the reverse characteristics are attenuated giving better overall forward device performance.

These are secondary to the ohmic contact reason though obviously important.

In the HBT example above, using a heterojunction Base-Emitter diode kicks butt on a simply PN junction Base-Emitter so the N+ emitter doping becomes moot.
 
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