Latching current for an SCR is 100 mA, a dc source of 200 V is also connected to the SCR which is supplying an R-L load. Compute the minimum width of the gate pulse required to turn on the device. Take L = 0.2 H & R = 20 ohm both in series.
Latching current for an SCR is 100 mA, a dc source of 200 V is also connected to the SCR which is supplying an R-L load. Compute the minimum width of the gate pulse required to turn on the device. Take L = 0.2 H & R = 20 ohm both in series.
Right Answer is:
100.5 μsec
SOLUTION
Expression of current in RL circuit as shown in the figure is
$\begin{array}{l}i(t) = \dfrac{V}{R}\left( {1 – {e^{ – t\frac{R}{L}}}} \right)\\\\100 \times {10^{ – 3}} = \dfrac{{200}}{{20}}\left( {1 – {e^{ – t\frac{{20}}{{0.5}}}}} \right)\end{array}$
Solving the value of t
t = 100.5 μs