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3-phase thyristor bridge: no-load speed and firing angle

Short answer

No-load speed at α = 45° and 5 A is 1523.9 rpm. Firing angle to obtain 1500 rpm at rated armature current (100 A) is 43.0°.

A 3 phase fully controlled Thyristor Bridge with 400 V, 3 phase, 50 Hz supply is feeding a separately excited DC motor. Armature resistance is 0.2 Ω, armature rated current is 100 A and back emf constant is 0.25 V/rpm. Determine no-load speed if no-load armature current is 5 A and firing angle is 45°. Also determine firing angle to obtain a speed of 1500 rpm at rated current.

Data: VL=400V_L = 400 V, Ra=0.2 ΩR_a = 0.2\ \Omega, Irated=100I_{rated} = 100 A, K=0.25K = 0.25 V/rpm.

For a three-phase fully controlled bridge (continuous conduction):

Va=32VLπcos⁡α=32×400πcos⁡α=540.19cos⁡αV_a = \frac{3\sqrt2 V_L}{\pi}\cos\alpha = \frac{3\sqrt2\times400}{\pi}\cos\alpha = 540.19\cos\alpha

No-load speed at α=45∘\alpha = 45^\circ, Ia=5I_a = 5 A

Va=540.19cos⁡45∘=381.97 VE=Va−IaRa=381.97−5×0.2=380.97 VN0=EK=380.970.25=1523.9 rpm\begin{aligned} V_a &= 540.19 \cos45^\circ = 381.97\ \text{V} \\ E &= V_a - I_a R_a = 381.97 - 5\times0.2 = 380.97\ \text{V} \\ N_0 &= \frac{E}{K} = \frac{380.97}{0.25} = 1523.9\ \text{rpm} \end{aligned}

Firing angle for 1500 rpm at rated current

E=0.25×1500=375 VVa=375+100×0.2=395 Vcos⁡α=395540.19=0.7312⇒α=43.01∘\begin{aligned} E &= 0.25 \times 1500 = 375\ \text{V} \\ V_a &= 375 + 100\times0.2 = 395\ \text{V} \\ \cos\alpha &= \frac{395}{540.19} = 0.7312 \Rightarrow \alpha = 43.01^\circ \end{aligned}

Answer: No-load speed = 1523.9 rpm; firing angle for 1500 rpm at 100 A = 43.0°.

✓ Answer checked by Claude Opus 5.5 (high effort)

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