In class B commutation (resonant pulse commutation), L = 5 $μ$ H and C = 20 $μ$ F are used. If the initial voltage across the capacitor is $V_{S}$ = 230 V, then the value of conduction time for auxiliary thyristor used and peak current, $I_{P}$ will be

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Solution

In class B commutation

Peak current through capacitor,
$I_{P} = V_{S} \sqrt{\frac{C}{L}} = 230 \sqrt{\frac{20}{5}} = 460 A$
The conduction time of auxiliary thyristor,

$T_{A} = \frac{t}{ω_{r}}$
The resonant frequency,

$ω_{r} = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{5 \times 10^{- 6} \times 20 \times 10^{- 6}}}$

$= \frac{1}{10 \times 10^{- 6}} = 0.1 \times 10^{6} r a d / s$

$T_{A} = \frac{π}{ω} = \frac{π}{0.1 \times 10^{6}} = 10 π μ sec$

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In class B commutation (resonant pulse commutation), L = 5 μH and C = 20 μF are used. If the initial voltage across the capacitor is VS = 230 V, then the value of conduction time for auxiliary thyristor used and peak current, IP will be

In class B commutation Peak current through capacitor, IP=VS√CL=230√205=460A The conduction time of auxiliary thyristor, TA=tωr The resonant frequency, ωr=1√LC=1√5×10−6×20×10−6 =110×10−6=0.1×106rad/s TA=πω=π0.1×106=10πμsec

In class B commutation (resonant pulse commutation), L = 5 $μ$ H and C = 20 $μ$ F are used. If the initial voltage across the capacitor is $V_{S}$ = 230 V, then the value of conduction time for auxiliary thyristor used and peak current, $I_{P}$ will be