Question

A $3-\varphi$ 3 pulse converter, fed from $3-\varphi$, 400 V, 50 Hz supply, has a load $R=2\Omega$, E = 200 V and large inductance so that load current is constant at 20 A. If source has an inductance of 2 mH, then the value of overlap angle for inverter operation is

• A. ${3.15}^{\circ }$
• B. ${1.7}^{\circ }$
• C. ${8.54}^{\circ }$
• D. ${2.1}^{\circ }$

Answer 1

The correct option is A ${3.15}^{\circ }$

When converter circuit is working in an inverter mode,

$V_0=-E_0+I_{0}R$

$\frac{3V_{ml}}{2\pi }cos\alpha -\frac{3\omega L_s}{2\pi }{I}_0=-E_0+I_{0}R$

$270.094cos\alpha =-154$

$\alpha =co{s}^{-1}\left(\frac{-154}{270.094}\right)$

$\alpha ={124.76}^{\circ }$

For a $3-\varphi$ converter,

$I_0=\frac{V_{mL}}{2\omega L_s}[cos\alpha -cos(\alpha +\mu \left)\right]$

$20=\frac{\sqrt{2}\times 400}{2\times 2\pi \times 50\times 2\times {10}^{-3}}[-0.5701-cos(\alpha +\mu \left)\right]$

$=450.1581(-0.5701-cos(\alpha +\mu \left)\right)$

$cos(\alpha +\mu )=-0.6145$

$\mu ={127.91}^{\circ }-{124.76}^{\circ }$

$\mu ={3.15}^{\circ }$