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Question

Assertion (A) : The epitaxial layer growth possesses the same crystal structure to that of the substrate on which it is grown.

Reason (R) : It can only be of similar conductivity type but possesses different value of resistivity to that of the substrate.

Select your answer using the codes given below.

This question was previously asked in
UGC NET 2015 Paper 1 Question Paper (27-Dec-2015)
The correct answer is

(A) is true, but (R) is false

Examine the assertion — true, and it is the definition of the word. "Epitaxy" comes from the Greek epi (upon) and taxis (ordered arrangement). In epitaxial growth the deposited atoms arrange themselves in registry with the underlying lattice, so the new layer continues the substrate's single-crystal structure with the same orientation. If it did not, the deposit would be polycrystalline or amorphous and would not be called epitaxial at all.

Examine the reason — false, and the error is in the words "can only be of similar conductivity type". An epitaxial layer may be doped independently of the substrate, in both magnitude and type. Growing an n-type layer on a p-type substrate is not an exception; it is the foundation of the standard bipolar process:

LayerDopingPurpose
p-type substratemoderateMechanical support and junction isolation
n+ buried layerheavyLow collector series resistance
n-type epitaxial layerlightThe collector region — devices are built here

The reverse-biased junction between the n-epi pockets and the p-substrate is exactly what isolates one device from the next. So opposite-type epitaxy is not merely possible, it is the norm.

The second half of the reason — that the layer has a different resistivity from the substrate — is perfectly true and is indeed the main point of using epitaxy. But a statement that is half right is still false, and the word "only" is what breaks it.

Assemble. (A) is true and (R) is false, which is code 3.

Why epitaxy is worth the trouble. It decouples two requirements that would otherwise conflict. A bipolar transistor wants a lightly doped collector for high breakdown voltage, but a heavily doped path underneath for low series resistance and to avoid saturation. Growing a light n layer over a heavy n+ buried layer delivers both. The same trick gives CMOS its latch-up immunity: a thin lightly doped epi over a heavily doped substrate provides a low-resistance path that starves the parasitic thyristor.

How it is done. Vapour-phase epitaxy at 1100–1200 °C reduces silicon tetrachloride or pyrolyses silane, with dopant gases such as phosphine or diborane added to set the layer's type and level — entirely independently of what lies beneath.

Hence, (A) is true, but (R) is false.

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