Six-pulse converter: supply voltage and firing range

Interpretation: the motor's rated armature (dc) voltage is 400 V, i.e. at rated speed and torque with $\alpha = 0$ the converter gives 400 V dc. Ideal supply and thyristors; continuous conduction.

For a 3-phase six-pulse fully controlled bridge:

$$V_a = \frac{3\sqrt2}{\pi}V_L\cos\alpha = 1.3505\,V_L\cos\alpha$$

RMS supply voltage

With $\alpha = 0$, $V_a = 400$ V:

$$V_L = \frac{400\,\pi}{3\sqrt2} = 296.19\ \text{V (line, rms)}$$

(phase rms $= 296.19/\sqrt3 = 171.0$ V.)

Back emf

$$\begin{aligned} E_{1800} &= V_a - I_a R_a = 400 - 60\times0.5 = 370\ \text{V} \\ E_{900} &= 370 \times \frac{900}{1800} = 185\ \text{V} \end{aligned}$$

Full-load torque is held throughout, so $I_a = 60$ A and $I_aR_a = 30$ V.

Firing angle at 900 rpm

$$\begin{aligned} V_a &= 185 + 30 = 215\ \text{V} \\ \cos\alpha &= \frac{215}{400} = 0.5375 \Rightarrow \alpha = 57.49^\circ \end{aligned}$$

At 1800 rpm, $V_a = 400$ V and $\alpha = 0^\circ$.

Answer: supply voltage = 296.2 V (line rms); firing angle range = 0° (1800 rpm) to 57.5° (900 rpm).