Assertion–Reason on SCR turn-on behavior Assertion (A): As an SCR turns on and anode current rises, the anode–cathode voltage drop decreases. Reason (R): During turn-on, the reverse-biased junction inside the SCR breaks down in avalanche.

Difficulty: Medium

Correct Answer: A is correct but R is wrong

Explanation:


Introduction / Context:
Understanding the internal physics of SCR turn-on helps explain observed waveforms: a rapid drop in anode–cathode voltage accompanies a rise in current after triggering. It is important to distinguish regenerative carrier injection from avalanche breakdown.



Given Data / Assumptions:

  • Silicon-controlled rectifier (four-layer p-n-p-n structure).
  • Gate-triggered turn-on under normal operating conditions.



Concept / Approach:
When properly triggered, the SCR turns on via regenerative action between its two internal transistor equivalents. Carrier injection forward-biases the central junction; the device transitions from blocking to conducting with a steep decrease in VAK. This is not due to destructive avalanche breakdown under normal operation.



Step-by-Step Solution:
Apply gate current, initiating conduction in the p-n junction near the gate.Mutual transistor action causes rapid increase in carriers; the middle junction switches from reverse to effectively forward-biased by injection.The device enters the low on-state resistance region; VAK collapses while IA rises.Therefore A is true (VAK falls as IA rises), but R is false because normal turn-on is not avalanche breakdown.



Verification / Alternative check:
Typical SCR V–I characteristics and turn-on waveforms show a transition to a low on-state voltage with moderate slope resistance; avalanche is associated with overvoltage, not standard gate turn-on.



Why Other Options Are Wrong:

  • (a), (b) would require R to be a correct explanation, which it is not.
  • (d) A is well-known to be true.
  • (e) Both being wrong contradicts basic SCR behavior.



Common Pitfalls:

  • Confusing “breakover” with “avalanche”; controlled turn-on uses gate injection below breakover voltage.



Final Answer:
A is correct but R is wrong


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