Skip to main content

3-phase half controlled bridge: no-load speed, firing angle

Short answer

For a three-phase semiconverter, Va = 270.09(1+cos alpha). At alpha=45 and Ia=5A, Eb=460.08V, giving no-load speed 1840 rpm. For 1500 rpm at 100A, Va=395V and alpha = 62.5 degrees.

A three phase half controlled thyristor bridge with 400V, 3-phase, 50Hz supply is feeding a separately excited dc motor. Armature resistance is 0.2Ω, 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 5A and firing angle is 45°. Also determine firing angle to obtain a speed of 1500 rpm at rated current.

For a three-phase half-controlled (semi-converter) bridge with line voltage VLV_L and continuous current:

Va=32 VL2π(1+cos⁡α)=3Vml2π(1+cos⁡α),Vml=2 VLV_a = \frac{3\sqrt2\,V_L}{2\pi}(1 + \cos\alpha) = \frac{3V_{ml}}{2\pi}(1+\cos\alpha), \qquad V_{ml} = \sqrt2\,V_L 32×4002π=270.09 V\frac{3\sqrt2\times400}{2\pi} = 270.09\ \text{V}

Data: Ra=0.2 ΩR_a = 0.2\ \Omega, K=0.25K = 0.25 V/rpm (Eb=0.25NE_b = 0.25N).

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

Va=270.09 (1+cos⁡45∘)=270.09×1.7071=461.08 VEb=Va−IaRa=461.08−5×0.2=460.08 VN0=460.080.25=1840.3 rpm\begin{aligned} V_a &= 270.09\,(1 + \cos45^\circ) = 270.09\times1.7071 = 461.08\ \text{V} \\ E_b &= V_a - I_aR_a = 461.08 - 5\times0.2 = 460.08\ \text{V} \\ N_0 &= \frac{460.08}{0.25} = 1840.3\ \text{rpm} \end{aligned}

Firing angle for 1500 rpm at rated current (100 A)

Eb=0.25×1500=375 VVa=375+100×0.2=395 V1+cos⁡α=395270.09=1.4624cos⁡α=0.4624  ⇒  α=62.45∘\begin{aligned} E_b &= 0.25\times1500 = 375\ \text{V} \\ V_a &= 375 + 100\times0.2 = 395\ \text{V} \\ 1 + \cos\alpha &= \frac{395}{270.09} = 1.4624 \\ \cos\alpha &= 0.4624 \;\Rightarrow\; \alpha = 62.45^\circ \end{aligned}

Answer: no-load speed ≈1840\approx 1840 rpm; firing angle for 1500 rpm at 100 A ≈62.5∘\approx 62.5^\circ.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Utilization of Electrical Energy chapter 3 questions →