How does reverse bias affect leakage current in a diode?

AI Thread Summary
Reverse bias in a diode leads to leakage current primarily due to thermally generated electron-hole pairs. In reverse bias, electrons are removed from their parent atoms, creating holes, but their mobility is affected by the type of doping in the semiconductor. P-type doping results in low electron mobility, causing most electrons to recombine with holes before reaching the junction barrier. Conversely, n-type doping creates an excess of electrons, but holes also exhibit low mobility. Overall, the leakage current is influenced by the mobility of charge carriers, which varies with doping levels and semiconductor type.
Danish_Khatri
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My question is related to the leakage current occurring in a diode when connected in reverse bias mode. When the diode is forward biased, the source voltage first neutralizes the barrier potential and then the flow of current occurs. However, I am not able to understand how the current due minority charges flow through a diode when it is reerse bias.
I would be very grateful if there is someone who can help me with this confusion.
 
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In reverse, electron-hole pairs are being thermally generated just as in the forward case. For every electron removed from its parent atom, a hole is generated as well. The electrons move when reverse biased, and they encounter holes due to p-type doping. The thermally generated electrons & holes are equal in number, but doping a semiconductor with p-type impurites, i.e. acceptors, results in low electron mobility. The electrons recombine with the abundant holes and few make it to the junction barrier. Likewise, on the n-side, an excess of electrons occur due to doping, and holes have low mobility.

It's all about mobility of e- & h+ which varies with doping level and whether the semiconductor material is n-type, p-type, or intrinsic.

Claude
 
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