Assertion–Reason on PIN Diodes: Applications and Structure Assertion (A): A PIN diode can be used as an RF/microwave attenuator and as a power limiter in receiver front-ends. Reason (R): A PIN diode has a thin intrinsic (i) layer between the p and n regions. Choose the most appropriate option about A and R.

Difficulty: Medium

Correct Answer: A is correct but R is wrong

Explanation:


Introduction / Context:
PIN diodes are widely used in RF and microwave systems as current-controlled resistors. Their unique structure allows applications such as variable attenuators, T/R switches, and power limiters that protect low-noise amplifiers from large incident signals.


Given Data / Assumptions:

  • We are evaluating an Assertion–Reason pair about a PIN diode's applications and the nature of its intrinsic layer.
  • No numerical computation is required; conceptual understanding is key.


Concept / Approach:

In a PIN diode, a deliberately thick intrinsic (i) region separates the p and n regions. Under forward bias at RF, the diode exhibits a resistance that depends on carrier lifetime and stored charge. Under reverse bias, it behaves like a small capacitor with low conductance. These behaviors enable attenuation control and RF limiting.


Step-by-Step Solution:

1) Check Assertion: PIN diodes are used as attenuators and limiters. This is true because biasing controls their RF resistance and they can shunt or series-attenuate signals, while carrier storage enables power limiting action.2) Check Reason: It claims the intrinsic layer is thin. In reality, the i-region is intentionally thick to reduce junction capacitance and increase carrier lifetime, which is essential to the RF behavior.3) Relate R to A: Since the i-layer is not thin, the Reason is false and cannot explain the Assertion.


Verification / Alternative check:

Standard PIN diode datasheets and RF textbooks describe the i-layer thickness (often several micrometers or more) and its role in RF resistance control and limiting behavior.


Why Other Options Are Wrong:

Option A: Wrong because R is false. Option B: Wrong since R is not even correct. Option D: Wrong because A is true. Option E: Wrong since A is true.


Common Pitfalls:

Confusing PIN with a regular pn diode; assuming the i-layer is always thin like in small-signal diodes; ignoring carrier lifetime effects at RF.


Final Answer:

A is correct but R is wrong.

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