--- title: "Starting induction motor with variable frequency" subject: "Utilization of Electrical Energy" chapter: 3 key: "semester_7/utilization_of_electrical_energy:induction-motor-start-variable-frequency" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/induction-motor-start-variable-frequency site: BCT Engineering --- # Starting induction motor with variable frequency Utilization of Electrical Energy · Chapter 3: Control of Electric Drive - Asked in: 2070 Chaitra - 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/induction-motor-start-variable-frequency - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3 ## Question What happen if an induction motor is started with variable frequency? What is the technique to overcome the problems associated with variable frequency? Justify your answer with suitable mathematical expression. ## Short answer Starting at low frequency raises flux (φ ∝ V/f), causing saturation and high magnetizing current if voltage is not reduced. The problem is overcome by constant V/f control with low-frequency voltage boost (IR compensation) and starting at a frequency where slip for maximum torque equals 1, then ramping up frequency. ## Answer Starting an induction motor from a variable frequency supply (low frequency at first, then raised) gives a **high starting torque with low starting current**, because at low frequency the synchronous speed is low and the rotor slip frequency is small. But if only frequency is reduced while voltage is kept constant, the motor is **over-fluxed** and draws a very high magnetising current. #### What happens at low frequency Stator emf: $$ V \approx E_1 = 4.44 f N_1 k_w \phi \Rightarrow \phi \propto \frac{V}{f} $$ - **Constant V, low f**: $\phi$ becomes many times rated; the core saturates, the magnetising current and iron loss rise sharply, the motor overheats. Also reactances $X = 2\pi f L$ fall, so the current rises further. - **High slip at start on full frequency**: at 50 Hz direct start, $s = 1$, rotor current is 5–7 times rated while torque is small (rotor pf low). - **Stator resistance effect**: at low frequency $R_1$ becomes comparable to the reactance; with pure V/f, the $I_1R_1$ drop is a large fraction of $V$, so flux and maximum torque fall: $$ T_{max} = \frac{3}{2\omega_s}\cdot\frac{V^2}{R_1 + \sqrt{R_1^2 + (X_1 + X_2')^2}} $$ At rated frequency $R_1 \ll X$ and $T_{max} \propto (V/f)^2$; at low $f$, the $R_1$ terms do not reduce with $f$, so $T_{max}$ drops. #### Technique to overcome these problems 1. **Constant V/f control**: change voltage in proportion to frequency, $V/f$ = constant, so flux stays at rated value and $T_{max}$ is (nearly) the same at every frequency. 2. **Low-frequency voltage boost (IR compensation)**: add a fixed voltage at low frequency, $$ V = V_0 + k f, \quad V_0 \approx I_1 R_1 $$ so that $E_1/f$, not $V/f$, is held constant. 3. **Start at a frequency giving maximum torque at standstill.** Slip for maximum torque (approx.) $s_m = R_2'/X_2'$. With $X_2' \propto f$, choose the start frequency $f_{st}$ such that $s_m = 1$: $$ f_{st} = f_r\,\frac{R_2'}{X_{2r}'} $$ The motor then starts with $T_{max}$ at modest current. The frequency is ramped up (soft start) so that slip speed $s\omega_s$ stays small and current stays near rated. 4. **Closed-loop slip/current limit**: the controller limits slip frequency $\omega_{sl}$; since $T \propto \phi^2\omega_{sl}$ and $I$ depends on $\omega_{sl}$, the motor accelerates at controlled torque and current. Thus V/f control with boost and a frequency ramp gives a smooth start with high torque and low current, which a direct-on-line start cannot give. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What happens if frequency is reduced but voltage is kept constant? A. Flux falls and torque drops B. Flux rises well above rated (phi proportional to V/f), causing saturation, high magnetising current and overheating C. Nothing, as the speed adjusts D. Slip becomes zero Answer: B. Since phi is proportional to V/f, lower f with constant V over-fluxes the core. (Key point: Constant V at low f) 2. Why is a voltage boost V = V0 + kf used at low frequency? A. To compensate the stator IR drop (V0 about I1R1) so E1/f stays constant B. To increase slip C. To reduce the magnetising current to zero D. To raise the base frequency Answer: A. At low f the IR drop is a large share of V, reducing flux and torque; the boost restores the air-gap emf. (Key point: Low-frequency voltage boost) 3. Why does Tmax fall at low frequency even with constant V/f? A. Reactances increase at low f B. Supply voltage rises C. Rotor resistance becomes zero D. R1 does not reduce with f while reactances do, so the R1 term in the Tmax denominator becomes significant Answer: D. At low f, X = 2πfL falls but R1 stays fixed, so Tmax falls from the (V/f)^2 relation. (Key point: Why Tmax drops) 4. How is the start frequency chosen for maximum starting torque? A. So that slip for maximum torque sm = R2'/X2' equals 1, i.e. f_st = f_r R2'/X2r' B. So that the slip is zero C. At the full 50 Hz D. So that voltage is zero Answer: A. Since X2' is proportional to f, lowering f until sm = 1 gives maximum torque at standstill with modest current. 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