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Effect of inductive anode circuit on SCR turn-on time: if intrinsic turn-on time is 6 μs, what happens when the anode circuit is inductive?

Difficulty: Easy

Correct Answer: More than 6 μs

Explanation:


Introduction / Context:
Turn-on of an SCR involves charge spreading and current build-up. External circuit inductance limits di/dt, stretching the apparent electrical turn-on interval. Recognizing this helps in snubber and gate design.


Given Data / Assumptions:

  • Intrinsic (device) turn-on time t_on,int ≈ 6 μs.
  • Anode circuit includes inductance L that limits current rise.
  • Gate drive is sufficient to trigger.


Concept / Approach:
Inductance resists rapid current change: v_L = L * di/dt. For a given applied voltage, higher L forces smaller di/dt, delaying attainment of the conduction level at which the device is considered fully on. Hence the effective turn-on time appears longer than the intrinsic value measured under low-inductance test conditions.


Step-by-Step Solution:

Assume same gate pulse; compare di/dt with and without L.With L present, di/dt decreases → longer time to reach rated conduction → apparent t_on increases.Therefore the practical turn-on time is > 6 μs.


Verification / Alternative check:

Manufacturers specify di/dt limits; series inductance or saturable reactors are sometimes added purposely to limit di/dt, which inherently increases turn-on interval.


Why Other Options Are Wrong:

Equal or less than 6 μs ignores the effect of L on current build-up; zero is impossible.


Common Pitfalls:

Confusing intrinsic charge-spread time with circuit-limited current rise; neglecting v = L di/dt.


Final Answer:

More than 6 μs
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