Don't know if you mean P/N or PIN so I'll cover them both.
A P/N diode contains a p/n silicon junction. The p layer has mostly positive charge carriers and the n layer mostly negative carriers. This is the basic junction found in bipolar transistors. Ref. 1 provides an explanation of how this junction works.
A PIN diode has an intrinsic (I) silicon layer between the P- and N-regions. It is used as a variable resistor at rf frequencies. See ref. 2.
Silicon diodes have a reverse breakdown voltage, which remains relatively constant over a wide range of reverse current. A zener diode is essentially a silicon diode made with a controlled thickness to provide breakdown at a voltage that is a useful value, and therefore can be used as a voltage regulator and limiter. See ref. 3.
2006-11-16 23:31:15
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answer #1
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answered by kirchwey 7
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zener diode is a special type of pn junction diode .usually pn diode allows the current pass when that is forward biased but in case of zener diode it passes currents through it when reverse biased.this can be attributed to the fact that the zener diode works in the zener breakdown region . zener breakdown occurs in the diode when the heavily doped p&n type of semiconductors are joined . so diffusion of holes and electrons are high . and this causes so high electric field at junction to break the atom . atoms are broken into -ve an +ve ions . so the diode has sufficient charge carriers to let high electricity flow even in reverse biased.
moreover zener diode regulates the same voltage across it . no matter how much current is flowing through it. that is why this is used as voltage regulator in the circuit where the constant voltage is desired
2006-11-17 00:33:33
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answer #2
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answered by rahul 1
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Schottky diode has the lowest. Zener and ordinary PN junction diodes have the same voltage drop, which is higher than that of Schottky diode.
2016-05-21 22:22:15
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answer #3
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answered by Anonymous
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PN = positive-Negative.
Zener = nEUTRAL.
2006-11-16 23:11:04
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answer #4
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answered by Anonymous
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This pdf file might help..
2006-11-16 23:19:52
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answer #5
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answered by Debbie M 4
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you try by google search engine
2006-11-16 23:11:53
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answer #6
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answered by Anonymous
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