--- title: "Two semi converters for field and armature" subject: "Utilization of Electrical Energy" chapter: 3 key: "semester_7/utilization_of_electrical_energy:semi-converters-field-armature-firing-angle" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/semi-converters-field-armature-firing-angle site: BCT Engineering --- # Two semi converters for field and armature Utilization of Electrical Energy · Chapter 3: Control of Electric Drive - Asked in: 2081 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/semi-converters-field-armature-firing-angle - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3 ## Question A single phase 220V, 50Hz supply feeds a separately excited dc motor through two single phase semi converters, one for the field and another for armature. The firing angle of the field semi converters is zero, the field resistance is 250 Ω and the armature resistance is 0.2 Ω. The load torque is 150 Nm at 1500 rpm. The voltage constant is 0.8 V/A rad/s and torque constant is 0.8 Nm/A². Assuming armature and field current to be continuous and neglecting losses, determine: i) Field current. (ii) Firing of the converter in the armature circuit. iii) Power factor of the converter of the armature circuit. ## Short answer Field current with zero firing angle is 0.792 A. The required armature-circuit firing angle is 61.1°. The armature converter power factor is 0.82 lagging. ## Answer Data: $V_s = 220$ V, $V_m = \sqrt2 \times 220 = 311.13$ V, $R_f = 250\ \Omega$, $R_a = 0.2\ \Omega$, $T_L = 150$ N·m at 1500 rpm, $K_v = K_t = 0.8$. For a single-phase semi-converter (continuous current): $V_o = \dfrac{V_m}{\pi}(1+\cos\alpha)$. $$ \omega = \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s} $$ #### i) Field current With $\alpha_f = 0$: $$ \begin{aligned} V_f &= \frac{V_m}{\pi}(1+1) = \frac{2\times311.13}{\pi} = 198.07\ \text{V} \\ I_f &= \frac{V_f}{R_f} = \frac{198.07}{250} = 0.792\ \text{A} \end{aligned} $$ #### ii) Firing angle of armature converter $$ \begin{aligned} T &= K_t I_f I_a \Rightarrow I_a = \frac{150}{0.8 \times 0.792} = 236.66\ \text{A} \\ E_b &= K_v I_f \omega = 0.8 \times 0.792 \times 157.08 = 99.56\ \text{V} \\ V_a &= E_b + I_a R_a = 99.56 + 236.66\times0.2 = 146.89\ \text{V} \\ 1 + \cos\alpha_a &= \frac{\pi V_a}{V_m} = \frac{\pi\times146.89}{311.13} = 1.4832 \\ \cos\alpha_a &= 0.4832 \Rightarrow \alpha_a = 61.10^\circ \end{aligned} $$ #### iii) Input power factor of armature converter For a semi-converter with ripple-free current $I_a$, the supply current is a quasi-square wave of height $I_a$ flowing for $(\pi - \alpha)$ in each half cycle: $$ I_s = I_a\sqrt{\frac{\pi-\alpha}{\pi}} $$ Neglecting losses, input power = output power $= V_a I_a$: $$ \begin{aligned} PF &= \frac{V_a I_a}{V_s I_s} = \frac{\sqrt2(1+\cos\alpha)}{\sqrt{\pi(\pi-\alpha)}} \\ &= \frac{\sqrt2 \times 1.4832}{\sqrt{\pi \times (\pi - 1.0664)}} = 0.822\ \text{(lagging)} \end{aligned} $$ ($\alpha = 61.10^\circ = 1.0664$ rad.) | Quantity | Value | |---|---| | Field current $I_f$ | 0.792 A | | Armature current $I_a$ | 236.7 A | | Back emf $E_b$ | 99.56 V | | Armature voltage $V_a$ | 146.9 V | | Firing angle $\alpha_a$ | 61.1° | | Power factor | 0.822 lagging | **Answer: $I_f$ = 0.792 A, $\alpha_a$ = 61.1°, PF = 0.82 lagging.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the field current found with alpha_f = 0? A. If = Vs/Rf = 220/250 B. If = Vm/Rf C. If = Vf/Rf with Vf = 2Vm/pi D. If = Vf/Rf with Vf = Vm/(2pi) Answer: C. Semi-converter: Vf = (Vm/pi)(1+cos 0) = 198.07 V, so If = 0.792 A. (Key point: Field current) 2. How is the armature current found from the load torque? A. Ia = T/(Kt) with no field current B. Ia = T/(Kt If) = 150/(0.8 x 0.792) C. Ia = T x If D. Ia = T/(Kv w) Answer: B. T = Kt If Ia gives Ia = 236.7 A. (Key point: Armature current) 3. Which relation gives the armature converter firing angle? A. Va = (2Vm/pi) cos alpha B. Va = Vm cos alpha C. Va = (Vm/pi)(1 + cos alpha), with Va = Eb + Ia Ra D. Va = Vs/cos alpha Answer: C. Va = 99.56 + 236.66 x 0.2 = 146.9 V, so 1 + cos alpha = pi Va/Vm = 1.483 and alpha = 61.1°. (Key point: Firing angle) 4. In the semi-converter the supply current rms is Ia sqrt((pi - alpha)/pi). What does this imply? A. PF is 1 for any alpha B. PF falls as alpha is increased, giving a lagging PF (0.822 here) C. PF is leading D. PF is independent of Va Answer: B. PF = Va Ia/(Vs Is) = 0.822 lagging; a larger alpha shortens conduction and lowers PF. 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