Behaviour of intrinsic semiconductor to an electric field

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In a p-i semiconductor junction under forward bias, current can flow due to the presence of thermally generated electron-hole pairs, even in intrinsic silicon (Si). The depletion region forms at the junction between the intrinsic and doped regions, where free carriers are absent, leaving behind charged ions. In intrinsic Si, while it is undoped, thermal generation of carriers occurs, meaning there are free carriers except in the depletion zone. The electric field distribution across the intrinsic region is influenced by the junction with the doped area. Overall, the discussion clarifies that intrinsic Si does have free carriers, contradicting the notion that it lacks them entirely.
arunkumarcea
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If I have a semiconductor junction of p-i and an electric field applied to it in forward bias, will the current flow?. How is the depletion region formed in intrinsic region since there are no free carriers there.What will be the electric field distribution through the intrinsic region?
 
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What makes you believe there are no free carriers in intrinsic Si? Intrinsic Si has not been doped with any acceptor or donor elements, ideally it is pure Si. Electron-hole pairs are being thermally generated constantly just like with n and p type Si. The i layer has free carriers, except of course in a depletion zone. Did this help?

Claude
 
How can a depletion region be formed in intrinsic Si?Will they leave behind negative ions or positive ions?
 
arunkumarcea said:
How can a depletion region be formed in intrinsic Si?Will they leave behind negative ions or positive ions?

When an intrinsic region adjoins a doped region, there can be depletion. If a slab of intrinsic is energized w/o any p-n junctions then there is no depletion.

Claude
 
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