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In reverse bias situation electrons attracted towards the positive terminal and holes attracted towards the negative terminal widen the depletion region.

My question is why can't these hole combined and make a bond with the metal terminal because the negative terminal will give electrons to the p side [as battery always provides electrons]?

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  • \$\begingroup\$ "Bond" has a specific technical meaning, which holes can never partake in (since they don't actually exist). Please refine your question further. \$\endgroup\$ – Ignacio Vazquez-Abrams Mar 24 '14 at 7:18
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What you have to understand is holes don't actually exist, electrons are the only charge carriers in a semi-conductor but there are 2 types of electrons 1) free electrons and 2) bounded electrons.The "Electron flow" that is usually referred to in literature refers to the flow of free electrons and the "Hole flow" is the flow due to bounded electrons.

If a bounded electron moves in a particular direction it leaves a hole at the position it was previously at and it removes the hole at its new position.So it is equivalent to think of bounded electron flow as hole flow.

To understand the process better lets let b - bounded electron , h - hole and take the following illustration as a row of bounded electrons and a hole in a semi-conductor.

When there is a potential difference across semiconductor the bounded electrons will be attracted towards the positive terminal and so the bounded electron to the right of the hole will move left, this causes a hole to "move" right, the same process will repeat when the hole has moved to its new position, it will then seem as though the hole is the one moving.

$$ \text{+ b h b b b b -} \\ \text{+ b b h b b b -} \\ \text{+ b b b h b b -} \\ \text{+ b b b b h b -} \\ $$

Now with this knowledge in mind we can proceed to analyze a reverse biased semiconductor.Consider the following a negativily biased p-region (excess holes) next to a depletion region (all electrons bounded).

When a negative bias is applied on the p-type material holes will move left making the depletion region get bigger as it can be seen in this illustration.

originally: $$ \text{ b h b b b h} \hspace{0.3cm} \text{b b b b} \\ \text{ b b h b h b} \hspace{0.3cm} \text{b b b b} \\ \text{ b b b h b h} \hspace{0.3cm} \text{b b b b} \\ \text{ b b b b h b} \hspace{0.3cm} \text{b b b b} \\ $$

then $$ \text{- h b b b h b} \hspace{0.3cm} \text{b b b b} \\ \text{- b h b h b b} \hspace{0.3cm} \text{b b b b} \\ \text{- b b h b b b} \hspace{0.3cm} \text{b b b b} \\ \text{- b b b h b b} \hspace{0.3cm} \text{b b b b} \\ $$

then $$ \text{- b b b h b b} \hspace{0.3cm} \text{b b b b} \\ \text{- h b b h b b} \hspace{0.3cm} \text{b b b b} \\ \text{- b h b b b b} \hspace{0.3cm} \text{b b b b } \\ \text{- b b h b b b} \hspace{0.3cm} \text{b b b b } \\ $$

note how the depletion region has now increased by 2 columns, this is a simplified illustration of what happens.Hope the diagrams were clear enough, I know they are very crude.

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"why can't these hole combined and make a bond with the metal terminal"

Holes are just absence of electrons. Of course absence of electrons cannot take part in bonding.

because the negative terminal will give electrons to the p side

What you said is correct negative terminal of the battery gives electrons to P side. This electrons fill the holes (absence of electron) in the P region. Number of free charge carriers are thus reduced in the device. Region where there is no free charge carriers are called depletion region through which conduction is not possible. If electron flow is continued from the negative terminal, width of the depletion region is increased.

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