Electrical Drives (EX-801) - Unit 4 Short Notes

UNIT 4: ELECTRICAL DRIVES

1. FUNDAMENTALS OF DRIVE SYSTEMS & STABILITY

1.1 Block Diagram of Electrical Drives

A typical drive system consists of:

• Power Supply: AC or DC source.

• Power Converter: Modifies power (AC/DC, voltage/frequency) for motor control (e.g., rectifier, chopper, inverter).

• Motor: Converts electrical to mechanical energy.

• Load: The mechanical system being driven (constant torque, fan/pump, etc.).

• Controller: Generates commands for the converter based on reference (speed/torque) and feedback.

• Sensor: Measures motor variables (current, speed, position).

[!TIP] Exam Focus: Be prepared to draw and explain each block's function (Jun 2025).

1.2 AC vs. DC Drives & Advantages of Electrical Drives

Advantages of Electrical Drives:

• Precise speed/torque control over wide range.

• Fast starting, braking, and reversal.

• High efficiency, automatic protection.

• Can operate in all four quadrants.

• Clean, quiet, and suitable for automated environments.

1.3 Steady-State Stability

• Concept: A drive is stable if, after a small disturbance, it returns to its original operating point. The operating point is where Motor Torque (T_m) = Load Torque (T_L).

• Stability Criterion: For stability, as speed (ω_m) increases, the net torque (T_m - T_L) must become negative (decelerating). Mathematically:

$$ \frac{d(T_m - T_L)}{d\omega_m} < 0 \quad \text{at equilibrium point} $$

• Dependence: Stability depends on the relative slopes of T_m(ω_m) and T_L(ω_m) curves, not on individual characteristics (May 2024).

[!EXAMPLE] Typical Problem (May 2024):

Given: $$\displaystyle T_m = (1 + 2a_m) $$ and $$\displaystyle T_L = 3\sqrt{\omega_m} $$. Find equilibrium points and stability.

Solution: Set $$\displaystyle T_m = T_L $$, solve for ω_m. Check slope condition for each point.

1.4 Load Equalization & Flywheel Analysis (★ VERY HIGH FREQUENCY)

• Need: For intermittent loads (e.g., punch press), motor would be oversized for peak torque. A flywheel stores kinetic energy during light-load periods and releases it during heavy-load periods, allowing a smaller motor.

• Derivation of Flywheel Inertia (J_f):

  1. Consider motor torque-speed characteristic as linear: $$\displaystyle T_m = a - b\omega_m $$.

  2. During heavy load (duration $$\displaystyle t_1 $$), motor torque is max ($$\displaystyle T_{max} $$), speed drops from $$\displaystyle \omega_{n1} $$ to $$\displaystyle \omega_{n2} $$.

  3. During light load (duration $$\displaystyle t_2 $$), motor torque is min ($$\displaystyle T_{min} $$), speed rises back to $$\displaystyle \omega_{n1} $$.

  4. Energy balance: Energy deficit during heavy load = Energy surplus during light load.

$$ \left[ T_{max} - T_L(\text{heavy}) \right] \cdot \frac{\theta_1}{\omega_{avg1}} = \left[ T_L(\text{light}) - T_{min} \right] \cdot \frac{\theta_2}{\omega_{avg2}} + \frac{1}{2} J_{eq} (\omega_{n1}^2 - \omega_{n2}^2) $$

where $$\displaystyle J_{eq} = J_m + J_f $$ (referred to motor shaft).

5.  Solving for $$\displaystyle J_f $$:

$$ \boxed{J_f = \frac{2 \cdot \text{Area under } (T_L - T_m) \text{ vs. } \theta \text{ curve}}{(\omega_{n1}^2 - \omega_{n2}^2)} - J_m} $$

[!TIP] Common Problem (Nov 2022, May 2024): Given load torque pattern, motor torque limits, motor inertia, no-load speed, and slip at a known torque (to find slope b). Calculate required $$\displaystyle J_f $$. Assume linear T-ω.

1.5 Quadrant Operation & Multi-Quadrant Drives

• Four Quadrants:

  • Quadrant I: Motoring (T>0, ω>0) - Forward motion.

  • Quadrant II: Braking (T<0, ω>0) - Forward braking (regenerative or dynamic).

  • Quadrant III: Motoring (T>0, ω<0) - Reverse motion.

  • Quadrant IV: Braking (T<0, ω<0) - Reverse braking.

• Two-Quadrant Drive: Operates in Quadrants I & II (forward drive with braking) or I & IV (reversible drive with braking in one direction). Achieved using single converter (with reverse current capability) or dual converters.

• Four-Quadrant Drive: Operates in all quadrants. Requires two full converters (dual converter) or reversible chopper circuits for DC drives; for AC drives, requires VSI with regenerative capability.

1.6 Drive Classification & Selection

• Group Drive: One motor drives multiple machines via line shafts. Disadvantages: Low efficiency, high maintenance, no individual control. Use: Historical, small workshops.

• Individual Drive: One motor per machine. Advantages: Flexible, independent control, high efficiency. Use: Modern industries.

• Constant Torque Drive: Torque capability is constant over speed range (e.g., DC motor armature control, AC motor with constant V/f below base speed). $P \propto \omega$.

• Constant Power Drive: Power capability is constant over speed range (e.g., DC motor field control above base speed, AC motor with constant voltage above base speed). $T \propto 1/\omega$.

• Industrial Applications:

  • Cement: Raw mill (constant torque), kiln (constant torque with high starting torque), fan/pump (variable torque).

  • Steel: Rolling mills (constant torque, high power, four-quadrant), cranes (four-quadrant), fans/pumps (variable torque).

2. POWER ELECTRONIC CONVERTERS FOR DRIVES

2.1 DC Drives: Power Converters

A. Phase-Controlled Converters (AC-DC)

• Single-phase semi-controlled (half-controlled) converter: Uses 2 SCRs + 2 diodes. Provides one-quadrant operation (motoring). Output voltage $$\displaystyle V_{dc} = \frac{V_m}{\pi}(1 + \cos\alpha) $$ for α ≤ 90°.

• Single-phase fully-controlled converter: 4 SCRs. Two-quadrant operation (motoring + regenerative braking in one direction). $$\displaystyle V_{dc} = \frac{2V_m}{\pi}\cos\alpha $$. Can reverse polarity by α > 90°.

• Three-phase converters: Higher power, lower ripple. Half-wave (3 SCRs) or full-wave (6 SCRs). $$\displaystyle V_{dc} = \frac{3\sqrt{6}V_{LL}}{\pi}\cos\alpha $$ for full-wave.

• Key Equations (Separately Excited Motor):

$$ V_a = V_{dc} - I_a R_a $$

$$ T_e = K_t \phi I_a \quad (\text{DC motor: } \phi \text{ constant for shunt}) $$

$$ \omega_m = \frac{V_a}{K_e \phi} - \frac{R_a}{K_e K_t \phi^2} T_e $$

[!TIP] Numerical Pattern (Nov 2022): Given converter type, α, motor parameters (V, I, R, L), find T and n for continuous conduction. Use $$\displaystyle V_{dc} $$ formula, then $$\displaystyle I_a = (V_{dc} - E_b)/R_a $$, where $$\displaystyle E_b = K_e \phi n $$.

B. DC Choppers (DC-DC)

• Principle: High-frequency switching (MOSFET/IGBT) to control average output voltage.

• Types:

  • Step-down (Buck): $$\displaystyle V_o = D V_{in} $$ (0<D<1). Motoring.

  • Step-up (Boost): $$\displaystyle V_o = \frac{V_{in}}{1-D} $$. Regenerative braking.

  • Class A (One-quadrant): Single switch + diode. Motoring only.

  • Class B (Two-quadrant): Two switches. Motoring + regenerative braking (current reversible).

  • Class C (Two-quadrant): Two switches (common positive). Motoring + regenerative (voltage reversible).

  • Class D (Four-quadrant): Four switches (H-bridge). Full four-quadrant operation.

  • Class E (Four-quadrant, current source): Four switches + inductors.

• Motoring Control (Class A/B/C): Switch ON (t_on): $$\displaystyle i_a $$ rises. Switch OFF (t_off): $$\displaystyle i_a $$ freewheels through diode. $$\displaystyle V_{avg} = D V_s $$.

• Regenerative Braking (Class B/C/D): Motor acts as generator. Energy fed back to supply. Requires path for current reversal.

2.2 AC Drives: Inverters

A. Voltage Source Inverter (VSI)

• Principle: DC link with capacitor. Outputs variable voltage/frequency AC.

• Six-step (180° or 120° conduction): Each switch conducts 180°. Output voltage is quasi-square wave. Fundamental $$\displaystyle V_{o} = \frac{\sqrt{6} V_{dc}}{\pi} $$.

• PWM (Sinusoidal PWM): Switches at high frequency (e.g., 2-20 kHz). Controls fundamental voltage and harmonic content. $$\displaystyle V_{o1} = \frac{\sqrt{6} V_{dc}}{\pi} m_a $$ (modulation index).

• VSI-fed IM Drive: Standard for variable speed. Requires V/f control to maintain flux.

• VSI-fed SM Drive: Needs rotor position sensor for proper phasing. Can operate at leading PF (load-commutated).

B. Current Source Inverter (CSI)

• Principle: DC link with large inductor. Outputs nearly constant current.

• Power Circuit: Typically 6 SCRs with commutation capacitors (for load commutation). Requires overlap angle μ.

• CSI-fed IM Drive: Simple, robust, but poor dynamic response. Speed controlled by frequency. Torque ∝ slip frequency.

• Comparison with VSI:

  | VSI | CSI | |----------------------------------|----------------------------------| | Voltage source (capacitor filter)| Current source (inductor filter) | | Fast dynamic response | Slow dynamic response | | MOSFET/IGBT switches | SCRs (need commutation) | | Regenerative capability easy | Regenerative requires extra circuit | | More common | Less common, used for high-power |

2.3 Converter/Inverter Fed Drive Characteristics

• Continuous vs. Discontinuous Conduction (DC drives): Discontinuous conduction occurs at low current/high speed. Torque-speed curve nonlinear, poor speed regulation.

• Harmonics & Ripple: Cause motor losses, torque pulsations, acoustic noise. Use PWM to reduce harmonics.

• Effect on Motor: Voltage/frequency ratio must be controlled (for AC) to avoid saturation (overfluxing) or underfluxing.

3. DC MOTOR DRIVES (★ VERY HIGH FREQUENCY)

3.1 Starting & Starting Methods

• Need: To limit starting current ($$\displaystyle I_a = V/R_a $$, very high) which causes:

  • Heavy sparking at commutator.

  • Excessive voltage drop in supply.

  • High electromagnetic stress.

• Methods:

  • Series Resistor Starter: Add external resistance in armature circuit. Gradually cut out as motor speeds up.

  • Soft Starter (Solid-state): Use thyristor/chopper to gradually apply voltage.

  • Direct-on-Line (DOL): Only for small motors (low $$\displaystyle R_a $$).

3.2 Speed Control Methods

• Armature Voltage Control (Below Base Speed):

  • Why? Flux φ is constant (shunt motor). Speed $$\displaystyle n \propto V_a $$. Reducing $$\displaystyle V_a $$ reduces speed while maintaining constant torque capability ($$\displaystyle T \propto I_a $$). Gives constant torque region.

  • Implementation: Phase-controlled converter or chopper.

• Field Flux Control (Above Base Speed):

  • Why? Armature voltage already at max. Reducing field current φ increases speed ($n \propto 1/\phi$) while keeping $$\displaystyle I_a $$ (and thus torque) limited by armature current rating. Gives constant power region ($$\displaystyle P \approx V I_a \approx \text{const} $$).

  • Implementation: Field rheostat or field chopper.

• Combined Control: For wide speed range (e.g., 0.1 to 2 p.u.).

3.3 Braking Methods (★ VERY HIGH FREQUENCY)

Torque Expressions (Separately Excited):

• Plugging: $$\displaystyle T = -K_t \phi I_a $$ (negative). $$\displaystyle I_a = (E_b + V)/R_a $$ (large).

• Dynamic: $$\displaystyle T = -K_t \phi I_a $$. $$\displaystyle I_a = E_b / (R_a + R_{ext}) $$.

• Regenerative: $$\displaystyle T = -K_t \phi I_a $$. $$\displaystyle I_a = (E_b - V)/R_a $$ (flows back to source).

[!TIP] Plugging Numerical (Nov 2022, May 2022): Given initial speed, find:

1. Resistance to limit braking current: $$\displaystyle R_{tot} = (E_{b0} + V) / I_{a(limit)} $$, $$\displaystyle R_{add} = R_{tot} - R_a $$.

2. Initial braking torque: $$\displaystyle T = K_t \phi I_a $$ (use $$\displaystyle I_a $$ from step 1).

3. Torque at zero speed: $$\displaystyle E_b=0 $$, so $$\displaystyle I_a = V/R_{tot} $$, $$\displaystyle T = K_t \phi I_a $$.

3.4 Converter/Chopper Fed DC Drive Analysis

• Steady-State (Continuous Conduction):

  • Average armature voltage: $$\displaystyle V_a = V_{dc} - I_a R_a $$.

  • For chopper: $$\displaystyle V_{dc} = D V_s $$.

  • Back-EMF: $$\displaystyle E_b = K_e \phi n $$.

  • Torque: $$\displaystyle T = K_t \phi I_a $$.

  • Speed: $$\displaystyle n = \frac{V_{dc} - I_a R_a}{K_e \phi} $$.

• Discontinuous Conduction: Occurs when current falls to zero before end of switching period. Requires piecewise analysis. Speed regulation becomes poor.

3.5 Closed-Loop Control

• Current Limit Control (CLC): Inner current loop limits $$\displaystyle I_a $$ to safe value during transients (starting, braking). Outer speed loop may be open or closed.

• Closed-Loop Speed Control:

  • Block Diagram: Speed reference $$\displaystyle \omega^* $$ → Speed controller (PI) → Current limiter → Converter/chopper → Motor. Feedback: Tachometer (speed) and current sensor.

  • Transient Analysis: Use small-signal linear models. Time constants: electrical ($$\displaystyle T_e = L_a/R_a $$) and mechanical ($$\displaystyle T_m = J/B $$).

  • Steady-State: Speed error zero with integral controller. Current follows current limit during transients.

4. INDUCTION MOTOR DRIVES (★ VERY HIGH FREQUENCY)

4.1 Speed Control Methods Overview

1. Stator Voltage Control: For fan/pump loads ($$\displaystyle T_L \propto \omega^2 $$). Reduces voltage → reduces torque → speed drops.

2. Frequency Control (V/f): Main method for wide speed range.

3. Pole Changing: Multiple stator windings for discrete speeds (constant torque).

4. Slip Power Recovery: For high-power wound-rotor IM, recover slip power.

4.2 Variable Frequency Control (V/f Control) (★)

• Principle: To maintain constant air-gap flux $$\displaystyle \phi_m $$, keep $$\displaystyle V/f = \text{constant} $$.

  • Below base speed ($$\displaystyle f \leq f_b $$): $$\displaystyle V/f = \text{const} $$. Flux constant → max torque constant.

  • Above base speed ($$\displaystyle f > f_b $$): Voltage limited to $$\displaystyle V_b $$. Flux $\phi \propto V/f$ decreases → max torque decreases. Constant power region ($$\displaystyle T_{max} \propto 1/f $$).

• Range of Speed Control: Typically 1:10 (e.g., 5 Hz to 50 Hz).

• Torque-Speed Characteristics: For each frequency, a family of curves. Max torque point moves to lower slip as frequency increases above base speed.

• Mathematical Expression for Max Torque:

$$ T_{max} = \frac{3}{\omega_s} \frac{(V^2 / f^2)}{(R_s / s)^2 + (X_s + X_r')^2} \bigg|_{s=s_{max}} $$

With $$\displaystyle V/f = \text{const} $$ below base speed, $$\displaystyle T_{max} \propto (V/f)^2 = \text{const} $$. Above base speed, $$\displaystyle V = V_b = \text{const} $$, so $$\displaystyle T_{max} \propto 1/f^2 $$.

[!TIP] Numerical (Jun 2025): Fan load ($$\displaystyle T \propto \omega^2 $$). Given rated speed at rated V/f. Find V, I, T at reduced speed. Use similarity: $$\displaystyle T \propto (V/f)^2 $$, $I \propto V/f$ (approx).

4.3 Slip Power Recovery Drives (★)

A. Static Kramer Drive

• Principle: Slip power ($$\displaystyle s P_{gap} $$) from rotor is rectified (diode bridge) and fed back to supply via a line-commutated inverter. Only sub-synchronous speeds possible.

• Power Circuit: Wound rotor → diode rectifier → DC link (inductor) → line-commutated inverter (SCRs) → supply.

• Closed-Loop Operation: Speed signal controls inverter firing angle α, thus controlling DC link voltage and rotor current → speed control.

• Speed-Torque: Similar to wound-rotor with external resistance, but power is recovered.

B. Static Scherbius Drive

• Principle: Slip power is fed back to the rotor from the supply via a force-commutated inverter (VSI). Allows super-synchronous speeds as well as sub-synchronous.

• Power Circuit: Supply → cycloconverter or PWM inverter → rotor slip rings → rotor winding. (Modern: VSI directly connected to rotor).

• Conventional vs. Scherbius Drive:

  • Conventional: Rotor resistance control (rheostat). Losses in resistor.

  • Scherbius Drive: Slip energy recovered. Efficient, smooth control.

• Comparison with Kramer:

  | Kramer | Scherbius | |-------------------------------------|-------------------------------------| | Sub-synchronous only | Sub & Super-synchronous | | Diode rectifier (uncontrolled) | Force-commutated inverter (controlled) | | Simpler, cheaper | More complex, expensive |

4.4 Stator Voltage Control

• Applied to fan/pump loads where $$\displaystyle T_L \propto \omega^2 $$ or $$\displaystyle \omega^3 $$.

• Torque: $$\displaystyle T \propto V^2 $$ (for constant slip).

• Speed: Reduces significantly with voltage.

• Efficiency: Poor at low voltage/speed due to high slip and rotor losses.

4.5 Braking of Induction Motors

• Regenerative Braking: Occurs when motor speed > synchronous speed for given frequency. Slip becomes negative. Power flows from rotor to stator to supply. Requires VSI or full converter for AC supply.

• AC Dynamic Braking: Disconnect stator from supply, connect to single-phase supply (or two leads of three-phase). Creates stationary magnetic field. Motor acts as generator, energy dissipated in rotor. Torque zero at synchronous speed (which is zero for single-phase).

  Two-lead connection: Connect two stator terminals to single-phase supply. Third terminal left open.

• DC Dynamic Braking: Disconnect AC, connect DC to stator. Creates stationary field. High braking torque.

• Plugging: Reverse any two stator phases. Torque opposes rotation. High braking torque, high energy loss.

4.6 Starting of Induction Motors

• DOL: Full voltage. High starting current (5-8× rated), high starting torque (1.5-2× rated).

• Reduced Voltage:

  • Star-Delta: Starting current $\downarrow$ to 1/3, torque $\downarrow$ to 1/3.

  • Auto-transformer: Starting current/torque $\propto$ (tap voltage)².

• Starting Current & Torque Calculation (May 2024):

  Given: Starting current = 4× I_fl, starting torque = 1.5× T_fl.

  To get full-load torque at start: Required voltage $$\displaystyle V_{start} = V_{rated} \times \sqrt{\frac{T_{fl}}{T_{start}}} = V_{rated} \times \sqrt{\frac{1}{1.5}} $$.

  Starting current at this voltage: $$\displaystyle I_{start} = 4 I_{fl} \times \frac{V_{start}}{V_{rated}} = 4 I_{fl} \times \sqrt{\frac{1}{1.5}} $$.

4.7 Steady-State Analysis

• Equivalent Circuit: Per-phase, referred to stator.

$$ V_1 = I_1 (R_s + jX_s) + I_2' (R_r'/s + jX_r') $$

• Torque-Slip:

$$ T = \frac{3}{\omega_s} \frac{(V_1^2 R_r'/s)}{(R_s + R_r'/s)^2 + (X_s + X_r')^2} $$

• Effect of Frequency & Voltage Change (May 2023):

  • If $f$ changes but $V/f$ kept constant: $$\displaystyle \omega_s \propto f $$, $$\displaystyle R_r'/s $$ unchanged? Actually $$\displaystyle R_r' $$ is referred to stator, so slip for max torque $$\displaystyle s_{max} \propto R_r' / X_{total} \propto 1/f $$? Careful: $X \propto f$. So $$\displaystyle s_{max} $$ remains same if $V/f$ const. Speed at rated torque: $$\displaystyle n = n_s(1-s) = (120f/P)(1-s) $$.

  • Example (May 2023): 50 Hz, 1000 rpm no-load → $$\displaystyle n_s=1000 $$? Actually $$\displaystyle n_s=120f/P $$. If 1000 rpm no-load, likely 4-pole ($$\displaystyle n_s=1500 $$ at 50 Hz). At 80 Hz, $$\displaystyle n_s=120*80/4=2400 $$ rpm. At same torque (same slip), speed = 2400*(1-s).

5. SYNCHRONOUS MOTOR DRIVES

5.1 Self-Controlled Synchronous Motor Drives (Load-Commutated Inverter - LCI) (★)

• Principle: Synchronous motor fed from a Current Source Inverter (CSI). Motor's back-EMF (from rotor field) commutates the inverter SCRs. No external commutation needed.

• Power Circuit: DC source (or rectifier) → inductor (current source) → 6-pulse inverter → SM. Firing signals derived from rotor position (through shaft-mounted encoder).

• Firing Scheme: Firing angle α advanced as speed increases to maintain overlap angle μ.

• Torque Production: $$\displaystyle T_e = \frac{3}{\omega_s} \frac{V_t E_f}{X_s} \sin\delta $$ (for negligible R). δ is load angle between $$\displaystyle V_t $$ and $$\displaystyle E_f $$.

• Operation: Can operate at leading power factor (over-excited) or lagging (under-excited). Typically operated at unity or leading PF for inverter commutation.

• Comparison with VSI-fed SM:

  | LCI (Self-Controlled) | VSI-fed | |---------------------------------|---------------------------------| | Current source, low ripple | Voltage source, high ripple | | Natural commutation (load) | Forced commutation (device) | | Simple, robust for high power | More flexible, dynamic control | | Power factor controlled by field| Power factor controlled by modulation |

5.2 VSI-Fed Synchronous Motor Drives (★)

• Working Principle: VSI provides variable voltage/frequency. Motor speed controlled by frequency. Voltage controlled to maintain flux (V/f constant below base speed).

• Block Diagram (Closed-Loop):

  
  Speed Ref → Speed Controller → Current Controller → VSI (PWM) → SM
  
                            ↑              ↑
  
                        Speed Feedback   Current Feedback (d,q axes)
  
  

  Position sensor (resolver/encoder) provides rotor angle for coordinate transformation (d-q).

• Control Strategies:

  • V/f Control: Similar to IM, but field current also controlled to optimize PF.

  • Power Factor Control: Adjust field current to achieve desired stator PF (often unity).

5.3 Braking of Synchronous Motors

• Regenerative Braking with VSI: Reduce frequency below actual speed. Motor becomes generator, power fed back to DC link and supply. Speed-torque curve shifts to left.

• DC Braking: Inject DC into stator. Creates stationary field. Rotor damper winding induces current → braking torque.

• AC Dynamic Braking: Disconnect from VSI, connect stator to single-phase supply. Similar to IM.

5.4 Starting of Synchronous Motors

• Need for Damper Winding: Provides starting torque (like induction motor) and damps oscillations during disturbances.

• Pull-in Phenomenon: After starting as induction motor (via damper), DC field is applied. If speed is close to synchronous ($$\displaystyle \omega \approx \omega_s $$), rotor pulls into synchronism. If speed is too low, the torque pulsations cannot accelerate rotor to synchronous speed → pull-out.

[!TIP] Why no pull-in at low speed? At low speed, slip is large, the pulsating torque has low average component. The average torque must overcome load torque and inertia to accelerate to synchronism. If average torque < load torque, pull-in fails.

5.5 Numerical Problems

A. Reluctance Motor (May 2024):

Given: $$\displaystyle X_d $$, $$\displaystyle X_q $$, load torque $$\displaystyle T_L $$, negligible $$\displaystyle R_a $$.

• Power input $$\displaystyle P_{in} = 3 V_t I_t \cos\phi $$.

• Torque: $$\displaystyle T_e = \frac{3}{\omega_s} \frac{V_t^2}{2} \left( \frac{1}{X_q} - \frac{1}{X_d} \right) \sin 2\delta $$.

• Power factor: $$\displaystyle \cos\phi = \frac{V_t / X_d \cos\delta + V_t / X_q \sin\delta}{I_t} $$? Actually, for salient pole SM:

$$ I_d = \frac{V_t}{X_d} \sin\delta, \quad I_q = \frac{V_t}{X_q} \cos\delta - \frac{E_f}{X_q} $$

But for reluctance motor ($$\displaystyle E_f=0 $$): $$\displaystyle I_q = \frac{V_t}{X_q} \cos\delta $$.

Then $$\displaystyle I_t = \sqrt{I_d^2 + I_q^2} $$, $$\displaystyle \cos\phi = I_q / I_t $$.

• Steps: From $$\displaystyle T_e = T_L $$, solve for δ. Then compute $$\displaystyle I_d, I_q, I_t, \cos\phi $$.

B. Self-Controlled SM (May 2023):

Given: 500 kW, 6.6 kV, 60 Hz, 6-pole, $$\displaystyle X_{dm} $$, $$\displaystyle X_{sr} $$, n (turns ratio), PF=1, n=5 (assume field turns?).

• Operation: Constant V/f up to base speed, constant V above.

• At a given speed (e.g., 0.5 p.u.), find:

  • Internal angle ψ (between $$\displaystyle E_f $$ and $$\displaystyle I_a $$).

  • $$\displaystyle I_m $$ (air-gap power component), $$\displaystyle I_f $$ (field current).

  • $$\displaystyle T_e $$, $$\displaystyle P_m $$.

  • Constant power range: from base speed to speed where $$\displaystyle I_a $$ reaches rated.

6. SPECIAL MOTORS & THEIR DRIVES (★ INCREASINGLY FREQUENT)

6.1 Switched Reluctance Motor (SRM) Drives (★)

• Construction: Double salient (both stator and rotor have salient poles). No PMs, no windings on rotor. Stator has concentrated windings.

• Principle of Operation: Variable Reluctance. Torque produced by tendency of rotor to align with stator pole that is excited. $$\displaystyle \rightarrow $$ "Turn-on" when rotor pole approaches stator pole. $$\displaystyle \rightarrow $$ "Turn-off" before rotor pole passes.

• Torque Production: $$\displaystyle T = \frac{1}{2} i^2 \frac{dL(\theta)}{d\theta} $$. Positive $dL/d\theta$ (increasing inductance) produces positive torque.

• Power Converter: Asymmetric bridge per phase. Allows independent control of each phase.

  
  Phase A: T1, T2 + D1, D2
  
  Phase B: T3, T4 + D3, D4 ...
  
  

• Control: Turn-on angle (θ_on), turn-off angle (θ_off), current limit. Chopper used for current control.

• Advantages over other AC drives:

  • Simple, robust, low-cost construction.

  • High starting torque, wide speed range.

  • Fault-tolerant (phase failure doesn't stop motor).

  • No PMs (no demagnetization, low cost).

  • Disadvantages: High torque ripple, acoustic noise, needs position sensor.

6.2 Permanent Magnet Synchronous Motor (PMSM) Drives

• Construction: Rotor has permanent magnets (NdFeB). Stator has sinusoidally distributed windings.

• Types:

  • Surface-mounted (SPMSM): Magnets on surface. $$\displaystyle L_d = L_q $$ (non-salient).

  • Interior (IPMSM): Magnets buried. $$\displaystyle L_d < L_q $$ (salient, provides reluctance torque).

• Back-EMF: Sinusoidal. $$\displaystyle E = K_e \omega \sin(\theta) $$.

• Comparison with BLDC:

  | PMSM | BLDC | |----------------------------------|----------------------------------| | Sinusoidal back-EMF | Trapezoidal back-EMF | | Sinusoidal current | Rectangular (DC) current | | Field-oriented control (FOC) | Six-step commutation | | Smooth torque, low ripple | Higher torque ripple | | More complex control | Simpler control |

6.3 Brushless DC (BLDC) Motor Drives

• Construction: Stator with three-phase windings. Rotor with surface magnets (trapezoidal EMF). Hall effect sensors or back-EMF sensing for rotor position.

• Operation: 120° electrical commutation. At any time, two phases conduct (one high, one low). Third phase floating.

• Switching Scheme: Six-step operation. Each phase conducts for 120°. Sequence: A(+)-B(-), A(+)-C(-), B(+)-C(-), B(+)-A(-), C(+)-A(-), C(+)-B(-).

• Torque: $$\displaystyle T = K_t I_a $$ (constant for given current). Torque ripple due to commutation.

6.4 Stepper Motors

• Variable Reluctance (VR) Stepper:

  • Construction: Stator with multiple phases (e.g., 2-phase with 4 poles each). Rotor made of soft iron with teeth.

  • Operation: Energize stator phases sequentially. Rotor teeth align with energized stator teeth. Step angle = $$\displaystyle \frac{360}{m \cdot N_r} $$ where m=phases, $$\displaystyle N_r $$=rotor teeth.

  • Micro-stepping: Divide full step into smaller steps by controlling phase currents sinusoidally (e.g., 1/16 step). Reduces vibration, noise, improves resolution.

• Permanent Magnet (PM) Stepper: Rotor has PMs. Can be unipolar or bipolar. Higher holding torque.

• Hybrid Stepper: Combines VR and PM principles. High resolution, high torque.

• Load Angle Control: In closed-loop stepper systems (servo), load angle (θ) between stator field and rotor position is controlled to generate torque.

7. ADVANCED CONTROL & APPLICATIONS

7.1 Digital Control of Drives (★)

• Block Diagram:

  
  Speed Ref → Digital Controller (DSP/MCU) → PWM Generator → Power Converter → Motor
  
                            ↑               ↑
  
                        Speed/Position Sensor   Current Sensor
  
  

• Advantages over Analog:

  • Flexibility (reprogrammable control algorithms).

  • High accuracy, noise immunity.

  • Complex control (FOC, DTC) possible.

  • Self-diagnosis, protection, communication.

• Implementation:

  • Microcontrollers: General-purpose, lower cost.

  • DSPs: Optimized for math-intensive control (FOC).

  • FPGAs/PLDs: High-speed parallel processing, custom PWM.

• Functions: Speed/current control loops, PWM generation, flux estimation, protection (overcurrent, overvoltage), communication (CAN, Ethernet).

7.2 Industrial Drive Applications

• Cement Industry:

  • Raw mill: Constant torque, high inertia. Individual drive with VSI-fed IM or synchronous motor.

  • Kiln: Constant torque, high starting torque. May use DC drive or slip-power recovery.

  • Fan/Pump (preheater, cooler): Variable torque. V/f control IM drives.

  • Crusher/Conveyor: Constant torque, four-quadrant (reversal). DC or AC with regenerative capability.

• Steel Industry:

  • Rolling Mills: Very high power (MW), constant torque, four-quadrant (reversal, fast response). DC drives historically, now AC drives with vector control (PMSM or high-power IM).

  • Cranes/hoists: Four-quadrant, high starting torque. Regenerative braking essential.

  • Fans/Pumps: Variable torque, V/f control.

  • Electric Arc Furnace: Medium voltage drives, cycloconverter or LCI-fed synchronous motor.

7.3 Energy Recovery Systems (★)

• Concept: Capture energy during braking (regenerative) and feed back to supply or store for reuse.

• Regenerative Braking: Motor acts as generator. Energy flows from motor → converter → AC supply (if converter allows).

• Utilization:

  • Direct to Grid: Most efficient. Requires regenerative converter (full converter, VSI).

  • Resistive Bank: Energy dissipated in resistor (dynamic braking). Less efficient.

  • Energy Storage: Use battery, supercapacitor, or flywheel to store braking energy for later use (e.g., in elevators, cranes).

• Importance: Reduces energy consumption, especially in frequent start-stop applications (trams, elevators, cranes).

7.4 Computer Numerical Control (CNC) (★)

• Definition: Computer-controlled machine tool system. Uses stored program (G-codes) to control motion of tools and workpieces.

• Role in Drives:

  • Provides position/speed references to individual axis drives (usually servo drives: PMSM or AC servo motors).

  • Closed-loop control: Uses high-resolution encoders/resolvers for precise positioning.

  • Coordination: Multi-axis interpolation (linear, circular).

  • Integration: CNC acts as supervisory controller; drive systems are slave controllers executing motion commands.

• Drive Requirements for CNC: High precision, fast response, smooth motion, ability to hold position (zero speed holding torque).

8. NUMERICAL PROBLEM PATTERNS (KEY FOR EXAMS)

DC Drives:

1. Converter/Chopper Fed:

   • Given: $$\displaystyle V_s $$, $f$, α (or D), $$\displaystyle R_a $$, $$\displaystyle L_a $$, motor $$\displaystyle V_{rated} $$, $$\displaystyle n_{rated} $$, $$\displaystyle I_{rated} $$.

   • Find: $$\displaystyle V_{dc} $$, $$\displaystyle I_a $$, $$\displaystyle E_b $$, $n$, $T$ for given condition.

   • Key: $$\displaystyle V_{dc} = f(\alpha, V_s) $$. $$\displaystyle E_b = V_{dc} - I_a R_a $$. $$\displaystyle n = E_b / K_e \phi $$. $$\displaystyle T = K_t \phi I_a $$.

2. Plugging Braking:

   • Given: $$\displaystyle V_{rated} $$, $$\displaystyle R_a $$, $$\displaystyle n_0 $$ (initial), $$\displaystyle I_{a(limit)} $$.

   • Find: $$\displaystyle R_{add} $$ to limit current, $$\displaystyle T_{initial} $$, $$\displaystyle T_{at\ n=0} $$.

   • Key: $$\displaystyle E_{b0} = K_e \phi n_0 $$. $$\displaystyle I_a = (E_{b0} + V) / (R_a + R_{add}) $$. Set $$\displaystyle I_a = I_{limit} $$ → $$\displaystyle R_{add} $$. $$\displaystyle T = K_t \phi I_a $$.

3. Flywheel Inertia:

   • Given: $$\displaystyle T_L $$ pattern (heavy $$\displaystyle T_{L1} $$ for $$\displaystyle t_1 $$, light $$\displaystyle T_{L2} $$ for $$\displaystyle t_2 $$), $$\displaystyle T_{m,max} $$, $$\displaystyle T_{m,min} $$, $$\displaystyle J_m $$, $$\displaystyle n_{no-load} $$, slip at known torque → find slope $b$.

   • Key: Assume linear $$\displaystyle T_m = a - b \omega $$. Find $$\displaystyle \omega_{n1}, \omega_{n2} $$ from energy balance or area method. Use $$\displaystyle J_f = \frac{2 \cdot \text{Area}}{\omega_{n1}^2 - \omega_{n2}^2} - J_m $$.

IM Drives:

1. V/f Control:

   • Given: Rated $$\displaystyle V_b, f_b, n_{rated} $$. For speed $n$ (< base), find $f$, then $$\displaystyle V = (f/f_b) V_b $$.

2. Stator Voltage Control (Fan Load):

   • Given: Rated $V, I, n, T$ (fan: $$\displaystyle T \propto \omega^2 $$). At new speed $n'$, find $V', I', T'$.

   • Key: $$\displaystyle \frac{T'}{T} = \left( \frac{n'}{n} \right)^2 $$. Since $$\displaystyle T \propto V^2 $$ (approx), $$\displaystyle V' = V \sqrt{T'/T} $$. $I' \propto V'$ (approx).

3. Wound Rotor Calculations:

   • Given: $$\displaystyle n_{fl} $$, $f$, $P$. Find: poles ($$\displaystyle P = 120f/n_s $$), slip $$\displaystyle s = (n_s - n)/n_s $$.

   • If rotor resistance doubled: $$\displaystyle s_{new} \propto R_r' $$ → $$\displaystyle s_{new} = 2s $$ (if torque same). Rotor copper loss $$\displaystyle P_{rcl} = s P_{gap} $$. If $$\displaystyle P_{rcl,initial} $$ given, find new at same torque? Actually if $$\displaystyle R_r $$ doubled for same torque, slip doubles, $$\displaystyle P_{rcl} = s P_{gap} $$ but $$\displaystyle P_{gap} $$ changes? For same torque, $$\displaystyle P_{gap} \propto s / (R_r'/s)^2 $$? Better: At same torque, slip inversely proportional to $$\displaystyle R_r' $$? Actually from torque equation, for same T, $$\displaystyle s \propto R_r' $$ if $X$ constant. So $$\displaystyle s_{new}=2s $$. $$\displaystyle P_{gap} = T \omega_s / (1-s) $$. Approximately same if s small. So $$\displaystyle P_{rcl,new} = s_{new} P_{gap} \approx 2 P_{rcl,initial} $$.

4. Frequency Change:

   • Given: $$\displaystyle f_1 $$, $$\displaystyle n_{no-load1} $$, $$\displaystyle n_{fl1} $$. At $$\displaystyle f_2 $$, find $n$ at rated torque.

   • Key: $$\displaystyle n_s \propto f $$. Slip $s$ at rated torque depends on $$\displaystyle R_r'/X $$. If $V/f$ constant, $X \propto f$, so $$\displaystyle s_{rated} $$ remains same. Thus $$\displaystyle n_2 = n_{s2} (1 - s_{rated}) $$.

5. Starting:

   • Given: $$\displaystyle I_{start}/I_{fl} = k $$, $$\displaystyle T_{start}/T_{fl} = m $$. Find $$\displaystyle V_{start} $$ for $$\displaystyle T_{start} = T_{fl} $$, and $$\displaystyle I_{start} $$ at that voltage.

   • Key: $$\displaystyle T \propto V^2 $$, so $$\displaystyle V_{start} = V_{rated} \sqrt{1/m} $$. $$\displaystyle I_{start} = k I_{fl} \cdot (V_{start}/V_{rated}) = k I_{fl} \sqrt{1/m} $$.

Synchronous Motors:

1. Reluctance Motor: As in 5.5A.

2. Self-Controlled SM (May 2023):

   • Given: $$\displaystyle P_{rated} $$, $$\displaystyle V_{rated} $$, $$\displaystyle f_b $$, $P$, $$\displaystyle X_{dm} $$, $$\displaystyle X_{sr} $$, $n$ (turns ratio), PF=1.

   • At a given speed (e.g., 0.5 p.u. below base), find $$\displaystyle T_e $$, $$\displaystyle P_m $$, $$\displaystyle I_m $$, $$\displaystyle I_f $$, constant power range.

   • Key: Below base: $$\displaystyle V/f = \text{const} $$. $$\displaystyle E_f \propto f $$. $$\displaystyle I_a $$ angle δ from power equation. $$\displaystyle I_m = I_a \sin\delta $$? Actually for SM: $$\displaystyle P_m = 3 V_t I_a \cos(\theta - \delta) $$? Better use phasor diagram. With PF=1, $$\displaystyle \theta=0 $$. $$\displaystyle V_t = E_f - j I_a X_s $$ (approx). $$\displaystyle |V_t| \propto f $$. $$\displaystyle E_f \propto f I_f $$. So at constant $$\displaystyle I_a $$, $$\displaystyle E_f $$ and $$\displaystyle V_t $$ both $\propto f$. Solve for $$\displaystyle I_f $$ from $$\displaystyle V_t $$ magnitude. $$\displaystyle T_e = \frac{P_m}{\omega_s} $$.

General Stability:

• Given $$\displaystyle T_m(\omega) $$ and $$\displaystyle T_L(\omega) $$, find equilibrium points ($$\displaystyle T_m=T_L $$) and stability ($$\displaystyle d(T_m-T_L)/d\omega < 0 $$ for stable).

END OF UNIT 4 NOTES
Focus on derivations (flywheel, V/f max torque), numerical patterns, and diagram explanations.