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.
(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:
| Layer | Doping | Purpose |
|---|---|---|
| p-type substrate | moderate | Mechanical support and junction isolation |
| n+ buried layer | heavy | Low collector series resistance |
| n-type epitaxial layer | light | The 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.
Similar Questions
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The desired properties of metallization for integrated circuits are
A. low resistivity.
B. low conductivity.
C. easy to etch for pattern generation.
D. tough to etch for pattern generation.
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Statement I : Concentration of acceptor atoms in the region between isolation islands in a monolythic integrated circuit will be much higher then in a p-type substrate.
Statement II : The higher density is provided to prevent the depletion region of the reverse biased isolation to substrate junction from extending into the p+ type material.
In the light of the above statements, choose the most appropriate answer from the options given below :
Arrange the following fabrication steps in order needed to fabricate the IC :
(a) Oxidation (b) Etching (c) Metallization (d) Epitaxial growth
Codes :
Assertion (A) : In a bipolar IC fabrication, the diode isolation is used to isolate one device from the other.
Reason (R) : P+n diode used as isolation diode in IC fabrication.
Select your answer using the codes given below.
Assertion (A) : The value of thin film resistor can be changed even after fabrication.
Reason (R) : This is done by means of trimming.
Select your answer using the codes given below.
In an npn diffused junction transistor, the p-type base region is formed on the n-type collector region through the process of :
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