Question

A three phase fully controlled bridge converter with 405 V supply, 0.04 $\Omega$ resistance per phase and 0.25 $\Omega$ reactance per phase is operating in the inverting mode at a firing advance angle of 35°. The mean generator voltage when the current is leveled at 80 A is (Thyristor voltage drop is 1.5 V)

Answer 1

Since, the firing advance angle is ${35}^{\circ }$.
Therefore, firing angle $\alpha ={180}^{\circ }-{35}^{\circ }={145}^{\circ }$ $V_{0x}=\frac{3V_{mL}}{\pi }\cos \alpha -2I_0{R}_s-2V_T-\frac{3\omega L_s{I}_0}{\pi }$

$=\frac{3\times \sqrt{2}\times 415}{\pi }\cos 145{}^{\circ }-(2\times 80\times 0.04)-(2\times 1.5)-\left(\frac{3\times 0.25\times 80}{\pi }\right)$

$=-487.5898=-E$

$\therefore E=487.5898\approx 487.6V$