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Starting induction motor with variable frequency

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.

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.

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≈E1=4.44fN1kwϕ⇒ϕ∝VfV \approx E_1 = 4.44 f N_1 k_w \phi \Rightarrow \phi \propto \frac{V}{f}
  • Constant V, low f: ϕ\phi becomes many times rated; the core saturates, the magnetising current and iron loss rise sharply, the motor overheats. Also reactances X=2πfLX = 2\pi f L fall, so the current rises further.
  • High slip at start on full frequency: at 50 Hz direct start, s=1s = 1, rotor current is 5–7 times rated while torque is small (rotor pf low).
  • Stator resistance effect: at low frequency R1R_1 becomes comparable to the reactance; with pure V/f, the I1R1I_1R_1 drop is a large fraction of VV, so flux and maximum torque fall:
Tmax=32ωs⋅V2R1+R12+(X1+X2′)2T_{max} = \frac{3}{2\omega_s}\cdot\frac{V^2}{R_1 + \sqrt{R_1^2 + (X_1 + X_2')^2}}

At rated frequency R1≪XR_1 \ll X and Tmax∝(V/f)2T_{max} \propto (V/f)^2; at low ff, the R1R_1 terms do not reduce with ff, so TmaxT_{max} drops.

Technique to overcome these problems

  1. Constant V/f control: change voltage in proportion to frequency, V/fV/f = constant, so flux stays at rated value and TmaxT_{max} is (nearly) the same at every frequency.
  2. Low-frequency voltage boost (IR compensation): add a fixed voltage at low frequency,
V=V0+kf,V0≈I1R1V = V_0 + k f, \quad V_0 \approx I_1 R_1

so that E1/fE_1/f, not V/fV/f, is held constant. 3. Start at a frequency giving maximum torque at standstill. Slip for maximum torque (approx.) sm=R2′/X2′s_m = R_2'/X_2'. With X2′∝fX_2' \propto f, choose the start frequency fstf_{st} such that sm=1s_m = 1:

fst=fr R2′X2r′f_{st} = f_r\,\frac{R_2'}{X_{2r}'}

The motor then starts with TmaxT_{max} at modest current. The frequency is ramped up (soft start) so that slip speed sωss\omega_s stays small and current stays near rated. 4. Closed-loop slip/current limit: the controller limits slip frequency ωsl\omega_{sl}; since T∝ϕ2ωslT \propto \phi^2\omega_{sl} and II depends on ωsl\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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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