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An SCR having a turn ON time of 5 μsec, latching current of 50 mA and holding current of 40 mA is triggered by a short duration pulse and is used in the circuit shown in the figure given below. The minimum pulse width required to turn the SCR ON will be
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- 251 μsec
- 150 μsec
- 100 μsec
- 5 μsec
Correct Option: A
iR = | = | = 0.02 A | ||
R2 | 5 × 103 |
= | 1 - e{-20t / 0.5} | = 5(1 - e-40t) | |||
20 |
Then, iT = iL + iR = 0.02 + 5(1 – e – 40t)
It minimum pulse width is T for ia ≥ latching current.
Then, 0.02 + 5(1 – e – 40t) = 0.05
⇒ 5(1 – e – 0.40t) = 0.03
⇒ t = 150 μ sec.