Question

A power electronic configuration has 30 SCRs connected in parallel combination for a circuit to bear a load current of 960 A. If the rating of each SCR is 400 V, 40 A, then how many SCRs should be excluded from parallel path such that derating factor decreases by 50%.

Answer 1

We know,
$Stringefficiency,\eta =\frac{Currentratingofconfiguration}{Currentratingof1SCR\times NumberofSCRsinstring}...\left(i\right)$

$\eta =\frac{960}{40\times 30}=\frac{24}{30}=0.80$

Derating factor = 1 - string efficiency
= 1 - 0.8 = 0.2
For 50% decrease in derating factor

$\text{New derating factor}=\frac{0.2\times 50}{100}=0.1$
$\therefore$ New efficiency = 0.9

Using equation (i), we get

$0.9=\frac{960}{40\times n}$

$\Rightarrow$ $n=\frac{960}{40\times 0.9}=26.66\approx 27$
Hence, decrease is no. of parallel SCRs= 30 - 27 = 3.