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:
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:
Common Pitfalls:
Final Answer:
A is correct but R is wrong
Discussion & Comments