--- title: "Slip power recovery: static Kramer drive" subject: "Utilization of Electrical Energy" chapter: 3 key: "semester_7/utilization_of_electrical_energy:slip-power-recovery-static-kramer" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/slip-power-recovery-static-kramer site: BCT Engineering --- # Slip power recovery: static Kramer drive Utilization of Electrical Energy · Chapter 3: Control of Electric Drive - Asked in: 2079 Bhadra, 8 marks - Video: [Speed Control of Induction Motor | Kramer System of Speed Control](https://www.youtube.com/watch?v=dCv9cCHuZic) — Poly Tutorials Engineering hours - 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/slip-power-recovery-static-kramer - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3 ## Question What is Slip Power Recovery Scheme? Describe in detail about Slip Power Recovery Scheme using Static Kramer drive. Can supersynchronous speed be achieved through it? Explain. ## Short answer Static Kramer drive rectifies rotor slip power with a diode bridge and returns it to the supply through a line-commutated inverter; firing angle sets slip and speed below synchronous speed. Supersynchronous speed is not possible because the diode bridge only allows power to flow out of the rotor; a bidirectional converter such as a Scherbius drive is required. ## Answer **Slip power recovery** is a speed control method for wound-rotor (slip ring) induction motors in which the slip power $sP_g$, which would be wasted as heat in rotor resistance control, is taken out of the rotor and returned to the supply (or to the shaft). Speed is reduced by extracting power at slip frequency, so efficiency stays high. #### Static Kramer drive ```text 3-ph supply ----+------------------+ | | stator [step-up/match [ WRIM ] transformer] rotor ^ | | [diode bridge] -- Ld -- [line-commutated (rectifier) smoothing thyristor inverter] Vd = 1.35 s E2 Vi = -1.35 VL' cos(alpha) ``` #### Working 1. Rotor voltage $sE_2$ at slip frequency ($E_2$ = standstill rotor line voltage) is rectified by an uncontrolled diode bridge: $$ V_d = \frac{3\sqrt2}{\pi}\, s E_2 = 1.35\, s E_2 $$ 2. A thyristor bridge working as an inverter ($90^\circ < \alpha < 180^\circ$) returns this dc power to the ac mains through a transformer ($V_L'$ = inverter-side ac line voltage): $$ V_i = -1.35\, V_L' \cos\alpha $$ 3. Neglecting the drop in the dc link, $V_d = V_i$: $$ s E_2 = V_L' |\cos\alpha| \quad\Rightarrow\quad s = \frac{V_L'}{E_2}|\cos\alpha| $$ So the inverter firing angle sets the slip and hence the speed. At $\alpha = 90^\circ$ the counter voltage is zero and the motor runs near synchronous speed (as if the rotor were shorted); increasing $\alpha$ towards 180° increases the slip and lowers the speed. 4. Torque is proportional to the dc link current, $T \propto I_d$, so the drive behaves like a separately excited dc motor with armature voltage control. 5. The recovered power $\approx sP_g$ goes back to the supply, so overall efficiency is high. **Features:** the converters are rated only for the slip power (e.g. a 30 % speed range needs about 30 % rating), so the drive is economical for limited speed ranges on large pumps, fans and compressors. Drawbacks: poor power factor (the inverter draws reactive power), harmonics, and a separate starting resistance is needed because the converter is not rated for full rotor voltage at standstill. #### Can super-synchronous speed be obtained? **No.** The diode bridge in the rotor allows power to flow only **out of** the rotor. Super-synchronous speed requires slip power to be fed **into** the rotor (negative slip). Therefore the static Kramer drive gives only sub-synchronous speed control in one direction (motoring, quadrant I). For super-synchronous operation the diode bridge must be replaced by a controlled converter so that power can flow both ways, i.e. the **static Scherbius drive** (a cycloconverter or two back-to-back thyristor/PWM converters, as in a doubly fed induction machine). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does slip power recovery do? A. Wastes slip power in rotor resistance B. Returns slip power from the rotor to the supply instead of wasting it as heat C. Increases the stator voltage D. Removes the slip rings Answer: B. Slip power s Pg is extracted from the rotor and fed back, keeping efficiency high. (Key point: Concept) 2. What sets the slip in a static Kramer drive? A. Diode bridge rating B. Inverter firing angle: s = (V'/E2)|cos alpha| C. Stator frequency D. Motor poles Answer: B. Balancing Vd = 1.35 s E2 with the inverter voltage gives s proportional to |cos alpha|. (Key point: Slip control) 3. Can the Kramer drive give super-synchronous speed? A. Yes, with alpha < 90° B. Yes, if the transformer ratio is changed C. No, the diode bridge passes power only out of the rotor D. Yes, at light load Answer: C. Super-synchronous needs power fed into the rotor; a diode bridge can't, so Scherbius is needed. (Key point: Super-synchronous) 4. In the Kramer drive torque is proportional to: A. Slip B. Stator voltage C. DC link current Id D. Firing angle Answer: C. T is proportional to Id, so the drive behaves like a separately excited dc motor. 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