A string of three series connected thyristors is used to withstand an off-stage voltage of 9 kV- Each thyristor is provided with static and dynamic equalizing circuit- The static equalizing resistance is 8- and dynamic equalizing circuit has Rc- 50 - and capacitance- C- 0-04 - F- If the thyristors T1- T2 and T3 have leakage current of 25 mA- 22 mA and 20 mA respectively- then ratio of discharge current of capacitor connected to thyristor T1 to discharge current in capacitor connected to thyristor T3 during turn-on will be

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Series and Parallel Connection of SCR - II

A string of t...

Question

A string of three series connected thyristors is used to withstand an off-stage voltage of 9 kV. Each thyristor is provided with static and dynamic equalizing circuit. The static equalizing resistance is $8 Ω$ and dynamic equalizing circuit has $R_{c} = 50 Ω a n d c a p a c i t a n c e; C = 0.04 μ F .$ If the thyristors $T_{1}, T_{2} a n d T_{3}$ have leakage current of 25 mA, 22 mA and 20 mA respectively, then ratio of discharge current of capacitor connected to thyristor $T_{1}$ to discharge current in capacitor connected to thyristor $T_{3}$ during turn-on will be __________

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Solution

Assuming I be the current flowing in string of thyristor in off-state
$∴ V o l t a g e a c r o s s S C R - 1 = (I - 0.025) \times 8000 = V_{1}$
$V o l t a g e a c r o s s S C R - 2 = (I - 0.022) \times 8000 = V_{2}$
$V o l t a g e a c r o s s S C R - 3 = (I - 0.020) \times 8000 = V_{3}$

The voltage across string,
$V_{s} = 9 k V$
$V_{s} = V_{1} + V_{2} + V_{3}$
$9000 = 8000 (I - 0.0025 + I - 0.022 + I - 0.020]$
$1.125 = 3 I - 0.067$
3I = 1.192
I = 0.3973 A

$V o l t a g e a c r o s s S C R - 1 = (0.3973 - 0.025) \times 8000$
= 2.9784 kV
$V o l t a g e a c r o s s S C R - 2 = (0.3973 - 0.022) \times 8000$
= 3.0024 kV
$V o l t a g e a c r o s s S C R - 3 = (0.3973 - 0.020) \times 8000$
= 3.0184 kV
$D i s c h a r g e c u r r e n t t h r o u g h S C R - 1 d u r i n g t u r n - o n$
$I_{D 1} = \frac{V_{1}}{R C}$
Discharge current through SCR-3 during turn -on
$I_{D 3} = \frac{V_{3}}{R C}$
$∴ R a t i o o f d i s c h a r g e c u r r e n t o f T_{1} a n d T_{3}$
$\frac{I_{D 1}}{I_{D 3}} = \frac{V_{1}}{V_{3}} . . . [a s R a n d C i s s a m e f o r e a c h t h y r i s t o r]$
$= \frac{2.9784}{3.0184} = 0.9867 ≃ 0.99$

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