Question

Ten thyristors are used in a string that withstands a dc voltage of $V_s=15kV$. The maximum leakage current is 10 mA and equalizing resistance R = $56k\Omega$. Voltage sharing of each thyristor and the steady state voltage derating factor are respectively __________

Answer 1

For equal voltage sharing, resistance R is connected across thyristor.

$R=\frac{n{V}_{lm}-V_s}{(n-1)\Delta I_B}$

Maximum steady state voltage $\left(V_{lms}\right)$

$V_{lms}=\frac{R(n-1)\Delta I_b+V_s}{n}$

$V_{lms}=\frac{56\times {10}^3\times 9\times 10\times {10}^{-3}+15\times {10}^3}{10}=2.004kV$

The steady state voltage derating factor (DF) is given by,

$DF=1-\eta$

$=1-\frac{V_{dc}}{n\times V_{bm}}=1-\frac{15}{10\times 2.004}$

$DF=0.2515$