--- title: "Single-phase fully controlled converter DC motor speed" subject: "Utilization of Electrical Energy" chapter: 3 key: "semester_7/utilization_of_electrical_energy:full-converter-dc-motor-speed" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/full-converter-dc-motor-speed site: BCT Engineering --- # Single-phase fully controlled converter DC motor speed Utilization of Electrical Energy · Chapter 3: Control of Electric Drive - Asked in: 2081 Baisakh, 8 marks - Video: [Single phase Fully Controlled Rectifier fed DC Motor Drive](https://www.youtube.com/watch?v=qRNl1QjCNbA) — Kiwi Tuition Academy - 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/full-converter-dc-motor-speed - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3 ## Question Derive an expression for motor speed in terms of supply voltage, firing angle and motor constants of a single-phase fully controlled separately excited dc motor. Draw waveform of supply voltage, armature voltage and current at a fixed speed under continuous conduction. ## Short answer The average armature voltage is Va = (2√2 Vs/π) cosα. Motor speed in rpm is N = (60/(2πK))(Va – Ia Ra) = (60/(2πK))((2√2 Vs/π) cosα – Ia Ra). Continuous conduction yields parallel speed-torque lines for different firing angles. ## Answer In a single-phase fully controlled bridge (four thyristors) feeding a separately excited dc motor, the average armature voltage is set by the firing angle $\alpha$, and the speed follows from the dc motor equations. #### Circuit ```text P (+) +---+---+ | | T1 T3 | | A -----+ +----- B | | | | | T4 T2 | | | | | | +---+---+ | | N (-) | +------- ~ v_s -------+ P ---[ Ra ]---[ La ]---( E )--- N motor armature T1,T2 fired at alpha; T3,T4 at pi+alpha ``` #### Derivation Supply: $v_s = V_m \sin\omega t$. With continuous conduction, T1–T2 conduct from $\alpha$ to $\pi+\alpha$, so the armature voltage follows $v_s$ in this interval: $$ \begin{aligned} V_a &= \frac{1}{\pi}\int_{\alpha}^{\pi+\alpha} V_m \sin\omega t\, d(\omega t) \\ &= \frac{V_m}{\pi}\left[-\cos\omega t\right]_{\alpha}^{\pi+\alpha} = \frac{2V_m}{\pi}\cos\alpha \end{aligned} $$ Motor equations (constant field, $K = K_b$): $$ \begin{aligned} V_a &= E_b + I_a R_a, \quad E_b = K\omega_m, \quad T = K I_a \end{aligned} $$ (the inductor has zero average voltage in steady state). Therefore $$ \begin{aligned} \omega_m &= \frac{V_a - I_a R_a}{K} = \frac{2V_m\cos\alpha}{\pi K} - \frac{R_a}{K}I_a \\ \omega_m &= \frac{2V_m\cos\alpha}{\pi K} - \frac{R_a}{K^2}T \end{aligned} $$ In rpm, $N = \dfrac{60}{2\pi}\omega_m$. With $V_m = \sqrt2 V_s$: $$ N = \frac{60}{2\pi K}\left(\frac{2\sqrt2 V_s}{\pi}\cos\alpha - I_a R_a\right) $$ - $0 \le \alpha < 90^\circ$: $V_a > 0$, motoring (quadrant I). - $90^\circ < \alpha < 180^\circ$: $V_a < 0$; with reversed back emf (load driving) the converter inverts, giving regenerative braking (quadrant IV). - Speed falls linearly with torque; the slope $R_a/K^2$ is the same for every $\alpha$, so the speed–torque lines are parallel. #### Waveforms (continuous conduction, fixed speed) ```text v_s | .-. .-. | / \ / \ --+-/-----\---------/-----\--------> wt |/ \ / \ | '-._.-' '-._ 0 pi 2pi v_a (alpha about 45 deg) | |'-. |'-. | | \ | \ --+-----+----\--------+----\-------> wt | alpha \_| pi+alpha \_| E_b - - - - - - - - - - - - - - - - - (constant level) i_a | ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ripple about Ia, at 2f --+--------------------------------> wt (never reaches zero) ``` - $v_a$ follows $+v_s$ from $\alpha$ to $\pi+\alpha$ and $-v_s$ from $\pi+\alpha$ to $2\pi+\alpha$; it goes negative for part of each half cycle (from $\pi$ to $\pi+\alpha$, and so on). - The back emf $E_b$ is a constant line (fixed speed). - $i_a$ rises when $v_a > E_b$ and falls when $v_a < E_b$, but never reaches zero (continuous conduction); the ripple frequency is $2f$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a single-phase full converter with continuous conduction, the average armature voltage is: A. (Vm/pi)(1+cos alpha) B. (2Vm/pi) cos alpha C. (Vm/pi) cos alpha D. (2Vm/pi) sin alpha Answer: B. Integrating Vm sin(wt) from alpha to pi+alpha over pi gives (2Vm/pi) cos alpha. (Key point: Average armature voltage) 2. Which expression gives the motor speed? A. wm = (Va + Ia Ra)/K B. wm = Va K - Ia Ra C. wm = (Va - Ia Ra)/K = 2Vm cos(alpha)/(pi K) - (Ra/K) Ia D. wm = Va/(K Ia Ra) Answer: C. From Va = Eb + Ia Ra with Eb = K wm, wm = (Va - Ia Ra)/K. (Key point: Speed expression) 3. What is true of the speed-torque lines for different firing angles? A. They are parallel, with slope Ra/K^2 B. They all meet at one point C. They are horizontal D. Their slope increases with alpha Answer: A. wm = 2Vm cos(alpha)/(pi K) - (Ra/K^2)T, so alpha only shifts the line; the slope is the same. (Key point: Speed-torque lines) 4. For 90° < alpha < 180° with reversed back emf (load driving the motor), what happens? A. The motor stalls B. The converter inverts and gives regenerative braking C. The motor runs forward at higher speed D. Armature current becomes discontinuous Answer: B. Va is negative, so the converter works as an inverter, returning power to the supply (quadrant IV). 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