UNIT 5: ELECTRICAL DRIVES - EXAM-FOCUSED SHORT NOTES

A. FUNDAMENTALS & SYSTEM OVERVIEW

Electrical Drive System Block Diagram

A typical drive system consists of:

1. Power Supply: AC or DC source.

2. Power Converter: Modifies electrical power (AC/DC, voltage/frequency) for the motor.

3. Motor: Converts electrical power to mechanical.

4. Load: The mechanical system being driven.

5. Controller: Generates commands for the converter based on references and feedback.

6. Sensor: Measures variables (speed, current, position) for feedback.

[!TIP] Exam Focus: Be prepared to draw and label this block diagram. Know the function of each block, especially the power converter (key component distinguishing drive types).

AC vs. DC Drives & Advantages of Electrical Drives

Advantages of Electrical Drives: Precise speed/torque control, high efficiency, fast dynamic response, easy automation, regenerative braking capability, remote control.

Classification of Electrical Drives

• By Motor Type: DC drives, AC induction drives, AC synchronous drives (wound/PM), Special motors (SRM, stepper, BLDC).

• By Operating Quadrant:

  • Single Quadrant: Motoring only (1st quadrant: +ω, +T).

  • Two Quadrant: Motoring (1st) + Regenerative braking (2nd quadrant: +ω, -T).

  • Four Quadrant: Full reversible operation (1st & 2nd for forward, 3rd & 4th for reverse).

• By Control: Open-loop, Closed-loop, Digital (microprocessor/DSP based).

Load Torque & Speed Characteristics

[!TIP] Utility: Matching drive characteristics to load type is critical for efficient design. V/f control is ideal for fan/pump loads.

Steady-State Stability of Drives

• Concept: An operating point (ω₀, T₀) is stable if a small disturbance causes the system to return to ω₀.

• Criterion: Stability depends on the relative slopes of motor torque ($$\displaystyle T_m $$) and load torque ($$\displaystyle T_L $$) curves.

  Stable if: $$\displaystyle \frac{d(T_m - T_L)}{d\omega} < 0 $$ at equilibrium ($$\displaystyle T_m = T_L $$).

  Unstable if: $$\displaystyle \frac{d(T_m - T_L)}{d\omega} > 0 $$.

• Graphical Interpretation: At stable point, $$\displaystyle T_m $$ curve must fall faster than $$\displaystyle T_L $$ curve as ω increases.

• Example: Given $$\displaystyle T_m = 100 - 0.1\omega $$ and $$\displaystyle T_L = 50 + 0.05\omega $$. Equilibrium at $$\displaystyle 100 - 0.1\omega = 50 + 0.05\omega \Rightarrow \omega = 333.3 $$. Slope: $$\displaystyle d(T_m-T_L)/d\omega = -0.1 - 0.05 = -0.15 < 0 $$ → Stable.

Load Equalization (Flywheel Application)

• Need: For intermittent heavy loads (e.g., punching, shearing), a large motor would be needed to supply peak torque. A flywheel stores kinetic energy during light-load periods and releases it during heavy-load periods, allowing a smaller motor.

• Derivation (Key Formula):

  Consider a motor with inertia $$\displaystyle J_m $$ and torque-speed characteristic: $$\displaystyle T_m = T_{m0} - \beta \omega $$ (linear approx).

  Load torque $$\displaystyle T_L $$ is periodic: high $$\displaystyle T_h $$ for duration $$\displaystyle t_h $$, low $$\displaystyle T_l $$ for $$\displaystyle t_l $$.

  Let $$\displaystyle \omega_1 $$ = min speed (at end of high-load period), $$\displaystyle \omega_2 $$ = max speed (at end of low-load period). Speed swing $$\displaystyle \Delta\omega = \omega_2 - \omega_1 $$.

  From energy balance during high-load period:

$$ \frac{1}{2}(J_m + J_f)(\omega_2^2 - \omega_1^2) = \int_{\omega_1}^{\omega_2} (T_m - T_L) d\omega \approx \text{Area between curves} $$

For linear $$\displaystyle T_m $$ and constant $$\displaystyle T_h $$, solving gives:

$$ J_f = \frac{(T_h - T_{m0})t_h - (T_l - T_{m0})t_l}{\omega_2^2 - \omega_1^2} \cdot \frac{(J_m + J_f)}{2} \approx \text{(Solve iteratively)} $$

**Simplified for small swing**: $$\displaystyle J_f \approx \frac{(T_h - T_{m0})t_h - (T_l - T_{m0})t_l}{\beta (\omega_2 - \omega_1)} $$.

• Key Assumption: Motor torque-speed characteristic is linear in the operating range.

[!TIP] Exam Trap: The flywheel inertia $$\displaystyle J_f $$ is added to the total inertia. The motor torque limit $$\displaystyle T_{m0} $$ is the maximum allowable torque. Always check units (N-m, sec, rad/s).

B. DC MOTOR DRIVES

Starting & Starters

• Why Starter? To limit the huge starting current ($$\displaystyle I_a = V/R_a $$, very large as $$\displaystyle R_a $$ is small) which damages commutator and supply.

• Types:

  1. DOL (Direct-On-Line): Not used for DC due to high current.

  2. Series Resistor Starter: Resistors in series with armature, cut out step-by-step as motor picks up speed.

  3. Face Plate Starter: Includes no-volt and overload protection.

Speed Control Methods

[!TIP] Why? Armature voltage control maintains constant flux (Φ rated), so torque is proportional to $$\displaystyle I_a $$. Field weakening reduces Φ, so for same $$\displaystyle I_a $$, torque drops but speed rises, keeping power ~constant.

Converter-Fed DC Drives

1. Single-Phase Semi-Controlled Converter (Rectifier)

• Circuit: Two SCRs (T1, T2) + two diodes (D1, D2) in bridge.

• Operation (Continuous Conduction, Motoring):

  • For $$\displaystyle \alpha < 90° $$, SCRs conduct from $$\displaystyle \omega t = \alpha $$ to $\pi+\alpha$.

  • Average armature voltage: $$\displaystyle V_a = \frac{V_m}{\pi}(1 + \cos\alpha) $$.

  • Where $$\displaystyle V_m = \sqrt{2} V_{rms} $$ (peak AC voltage).

  • Speed control by varying firing angle $\alpha$ ($0 \le \alpha \le 90°$ for motoring).

• Waveforms: Show $$\displaystyle v_s $$, $$\displaystyle v_a $$, $$\displaystyle i_a $$ (continuous, ripple).

2. Single-Phase Fully Controlled Converter

• Circuit: Four SCRs (T1-T4).

• Operation:

  • For $$\displaystyle \alpha < 90° $$, operates as rectifier (motoring).

  • For $$\displaystyle \alpha > 90° $$, operates as inverter (regenerative braking). $$\displaystyle V_a $$ becomes negative, power flows from motor to AC supply.

  • $$\displaystyle V_a = \frac{2V_m}{\pi}\cos\alpha $$.

  • Inversion limit: $$\displaystyle \alpha_{max} \approx 180° - \delta $$, where $\delta$ is overlap angle.

Chopper Control of DC Motors

• Principle: High-frequency ON-OFF switching of a DC source. Average voltage $$\displaystyle V_{avg} = \alpha V_s $$, where $\alpha$ = duty cycle ($$\displaystyle T_{on}/T $$).

• Motoring (Step-Down Chopper):

  • Switch (MOSFET/IGBT) ON: $$\displaystyle V_s $$ applied to motor, current rises.

  • Switch OFF: Motor current freewheels through diode, $$\displaystyle v_a = 0 $$.

  • $$\displaystyle V_{avg} < V_s $$, speed control below base.

  • Circuit: DC source $$\displaystyle V_s $$, switch, diode parallel to motor, motor $$\displaystyle E, R_a, L_a $$.

• Regenerative Braking (Step-Up Chopper):

  • Motor acts as generator ($$\displaystyle E > V_{avg} $$).

  • Switch ON: Motor current charges inductor, motor isolated.

  • Switch OFF: Stored inductor energy + motor EMF fed back to supply via diode.

  • $$\displaystyle V_{avg} = \frac{V_s}{1-\alpha} > V_s $$.

  • Circuit: Motor, switch in series with supply, diode from switch-motor junction to +ve supply.

[!TIP] Key Difference: Motoring chopper is step-down ($$\displaystyle V_{avg} < V_s $$), Regenerative chopper is step-up ($$\displaystyle V_{avg} > V_s $$).

Braking Methods

[!TIP] Critical Points:

• Plugging: Torque at zero speed is zero (since $$\displaystyle I_a = (V - E)/R_a $$, at ω=0, E=0, I_a = V/R_a but direction? Actually at ω=0, E=0, if supply reversed, I_a = -V/R_a, torque is negative but magnitude is max? Wait: At ω=0, E=0, with reversed voltage, I_a = -V/R_a, so torque is negative (braking) but speed is zero, so no braking action—torque is zero at ω=0 because there's no relative motion? No, torque is produced by current, so T = KφI_a. At ω=0, I_a is finite (V/R_a), so T is finite but negative. However, once speed is zero, braking should stop. In plugging, the torque is opposite to rotation, so it decelerates to zero. At exactly zero speed, if voltage is still reversed, current would try to accelerate in reverse direction. In practice, plugging is stopped at zero speed by disconnecting supply. So torque at ω=0 is not zero but the net effect is zero because speed is zero. Many textbooks say torque becomes zero at final speed because the drive is disconnected. Be careful.)

• Regenerative Condition: $$\displaystyle \omega > 0 $$ and $$\displaystyle V_a > E $$ (for separately excited, $$\displaystyle E = K\phi\omega $$). Actually for regeneration, we need $$\displaystyle E > V_a $$ so current reverses. So condition is $$\displaystyle \omega > V_a/(K\phi) $$.

• Dynamic Braking: Requires separate resistor $$\displaystyle R_b $$. Torque proportional to speed ($$\displaystyle T_b \propto -\omega $$) because $$\displaystyle I_a \approx E/R_b = K\phi\omega/R_b $$.

Numerical Problems (DC Drives)

Typical Problem Types:

1. Converter-fed (Continuous Conduction):

   • Given: $$\displaystyle V_{rms} $$, $$\displaystyle R_a $$, $$\displaystyle E = K\phi\omega $$, $\alpha$.

   • Find: $$\displaystyle V_a = \frac{2V_m}{\pi}\cos\alpha $$ (full) or $$\displaystyle \frac{V_m}{\pi}(1+\cos\alpha) $$ (semi). Then $$\displaystyle I_a = (V_a - E)/R_a $$, $$\displaystyle T = K\phi I_a $$, $$\displaystyle \omega = E/(K\phi) $$.

   • Check: $$\displaystyle I_a > 0 $$ for motoring.

2. Plugging:

   • Initial speed $$\displaystyle \omega_i $$, $$\displaystyle E_i = K\phi\omega_i $$.

   • Plugging: Supply reversed → $$\displaystyle V_a $$ becomes $-V$ (if original +V).

   • Total resistance in armature circuit = $$\displaystyle R_a + R_{add} $$.

   • Initial braking current: $$\displaystyle I_{ab} = \frac{(-V) - (-E_i)}{R_{total}} = \frac{E_i - V}{R_{total}} $$ (magnitude).

   • Braking torque: $$\displaystyle T_b = K\phi I_{ab} $$ (negative).

   • At zero speed ($$\displaystyle \omega=0, E=0 $$): $$\displaystyle I_a = -V/R_{total} $$, $$\displaystyle T_b = -K\phi V/R_{total} $$ (finite, but plugging is usually disconnected at ω=0 to avoid reverse rotation).

3. Shunt Motor Voltage Drop:

   • Constant torque load → $$\displaystyle I_a $$ constant (since $$\displaystyle T = K\phi I_a $$, Φ constant for shunt).

   • $V$ drops → $$\displaystyle E = V - I_a R_a $$ drops → $$\displaystyle \omega = E/(K\phi) $$ drops proportionally.

   • $$\displaystyle \omega_2 = \omega_1 \cdot (V_2 - I_a R_a)/(V_1 - I_a R_a) $$.

C. INDUCTION MOTOR DRIVES

Speed Control Fundamentals

• Synchronous Speed: $$\displaystyle n_s = \frac{120f}{P} $$ (rpm), where $f$ = supply frequency, $P$ = number of poles.

• Slip: $$\displaystyle s = \frac{n_s - n}{n_s} $$.

• Torque-Speed Characteristic:

  • Stable region: $s \approx 0$ to $$\displaystyle s = s_{max} $$ (stable if $$\displaystyle dT/ds > 0 $$).

  • Unstable region: $$\displaystyle s > s_{max} $$.

  • Maximum torque (breakdown torque): $$\displaystyle T_{max} \propto \frac{V^2}{s_{max}} $$ (approx).

Stator Voltage Control

• Principle: Reduce $V$ → reduces torque ($$\displaystyle T \propto V^2 $$ for constant slip). Used for fan/pump loads where torque $$\displaystyle \propto \omega^2 $$.

• Application: Speed control below base speed for quadratic loads. Poor torque capability at low voltage.

• Numerical Problem:

  Given motor parameters ($$\displaystyle R_s, R_r', X_s, X_r', X_m $$), rated $V$, $\omega$, load torque $$\displaystyle T_L $$ at new speed $$\displaystyle \omega_{new} $$.

  • Find slip $$\displaystyle s = (n_s - n)/n_s $$.

  • Use Thevenin equivalent: $$\displaystyle V_{th} = V \cdot \frac{jX_m}{R_s + jX_s + jX_m} $$, $$\displaystyle Z_{th} = \frac{(jX_s)(jX_m)}{R_s + j(X_s+X_m)} $$.

  • Torque: $$\displaystyle T = \frac{3}{\omega_s} \cdot \frac{|V_{th}|^2 \cdot (R_r'/s)}{(R_{th} + R_r'/s)^2 + (X_{th} + X_r')^2} $$.

  • Solve for $s$ given $T$, then $$\displaystyle n = n_s(1-s) $$.

  • Current: $$\displaystyle I_2 = \frac{V_{th}}{(R_{th} + R_r'/s) + j(X_{th}+X_r')} $$, stator current $$\displaystyle I_s \approx I_2 $$ (approx).

Variable Frequency Control (V/f Control)

• Goal: Maintain constant air-gap flux $\Phi$ to avoid saturation (below base) or under-excitation (above base).

• Below Base Speed ($$\displaystyle f < f_{base} $$):

  • Maintain constant V/f ratio. $$\displaystyle \Phi \propto V/f = \text{constant} $$.

  • $$\displaystyle T_{max} \propto (V/f)^2 = \text{constant} $$. So max torque remains constant across frequencies.

  • Speed range: 0 to $$\displaystyle n_s $$ (synchronous speed at $$\displaystyle f_{base} $$).

• Above Base Speed ($$\displaystyle f > f_{base} $$):

  • Voltage limited to rated $$\displaystyle V_{base} $$. So V constant, f increases → V/f ratio decreases → flux weakens.

  • $$\displaystyle T_{max} \propto (V/f)^2 \propto 1/f^2 $$ → decreases.

  • Power approximately constant ($$\displaystyle P \propto T_{max} \cdot \omega \propto (1/f) \cdot f = \text{const} $$). So constant power region.

• Derivation of $$\displaystyle T_{max} $$:

  From torque equation: $$\displaystyle T = \frac{3}{\omega_s} \frac{|V_{th}|^2}{\frac{R_{th}}{s} + R_r'} \cdot \frac{R_r'/s}{(R_{th}+R_r'/s)^2 + (X_{th}+X_r')^2} $$.

  At $$\displaystyle s = s_{max} = R_r' / \sqrt{R_{th}^2 + (X_{th}+X_r')^2} $$,

$$ T_{max} = \frac{3}{2\omega_s} \frac{|V_{th}|^2}{R_{th} + \sqrt{R_{th}^2 + (X_{th}+X_r')^2}} $$

Since $$\displaystyle V_{th} \propto V $$, $$\displaystyle \omega_s \propto f $$, so $$\displaystyle T_{max} \propto V^2/f $$.

Hence, constant V/f → constant $$\displaystyle T_{max} $$.

Current Source Inverter (CSI) Fed IM Drives

• Why CSI? Suitable for high power (≥ 1 MW) applications. Provides constant current, robust against short circuits.

• Power Circuit:

  • DC Link: Large inductor $$\displaystyle L_d $$ to maintain constant current $$\displaystyle I_d $$.

  • Inverter: Thyristor bridge (6-pulse) with forced commutation (since load is inductive).

  • Motor: 3-phase IM.

• Operation:

  • Six-step (120° conduction) operation.

  • Output current is quasi-square wave, magnitude ≈ $$\displaystyle I_d $$.

  • Speed control by varying frequency $f$ (by changing firing sequence of inverter).

  • Voltage magnitude adjusts automatically with frequency due to motor impedance.

• Disadvantages: Requires complex forced commutation circuit, torque pulsations.

Slip Power Recovery Drives

Principle: Instead of wasting slip power in rotor resistance (as in wound-rotor motor with external resistors), recover it and feed back to supply or motor shaft → higher efficiency.

1. Static Kramer Drive

• Application: Subsynchronous speed control ($$\displaystyle n < n_s $$).

• Circuit:

  1. Rotor AC → Diode bridge (uncontrolled rectifier) → DC.

  2. DC → Inverter (controlled, usually line-commutated) → feeds power back to supply (or sometimes to motor shaft via another machine).

• Operation: By controlling inverter firing angle, we control DC link voltage → controls rotor current → controls slip → controls speed.

• Speed Range: $0$ to $$\displaystyle n_s $$ (but not supersynchronous).

• Closed-Loop Block Diagram:

  
  Speed Ref → Controller → Inverter Firing Angle → Inverter → DC Link → Rotor Circuit → IM → Speed
  
                            ↑
  
                          Speed Sensor (Feedback)
  
  

  Power flows: Rotor → Diode Bridge → DC Link → Inverter → AC Supply.

2. Static Scherbius Drive

• Application: Supersynchronous speed control ($$\displaystyle n > n_s $$) and also subsynchronous.

• Circuit: Same as Kramer but inverter is fully controlled (thyristor bridge) and firing angle can be > 90°.

• Operation:

  • For $$\displaystyle n > n_s $$ (negative slip), rotor EMF polarity reverses. Diode bridge still rectifies. Inverter firing angle $$\displaystyle \alpha > 90° $$ makes it operate as inverter (power from DC link to AC supply).

  • By varying $\alpha$, we control DC link voltage magnitude and polarity → controls slip magnitude and sign → controls speed above or below $$\displaystyle n_s $$.

• Comparison:

  | Feature | Static Kramer | Static Scherbius | | :--- | :--- | :--- | | Speed Range | Subsynchronous only | Subsynchronous & Supersynchronous | | Inverter | Line-commutated (α < 90°) | Forced-commutated (α > 90° possible) | | Flexibility | Less | More |

AC Dynamic Braking (Two-Leg Connection)

• Circuit: Disconnect motor from supply. Connect two stator terminals (say U, V) to a braking resistor $$\displaystyle R_b $$. W terminal left open.

• Operation:

  • Rotor still rotating, induces EMF in stator.

  • Two-phase circuit forms a single-phase circuit with resistor. Currents flow in two phases, creating a stationary magnetic field.

  • This field cuts rotating rotor → induces rotor currents → produces braking torque (opposes rotation).

• Torque-Speed Characteristic:

  • Braking torque exists in 2nd quadrant (ω > 0, T < 0).

  • At $$\displaystyle \omega = 0 $$, torque = 0 (no relative motion).

  • Maximum braking torque occurs at some speed > 0.

• Advantage: Simple, no extra equipment needed beyond resistor.

Starting & Plugging for IM

• DOL Starting: Direct connection to supply. Starting current $$\displaystyle I_{st} \approx 6-8 \times I_{fl} $$, starting torque $$\displaystyle T_{st} \approx 1.5-2.5 \times T_{fl} $$.

• Reduced Voltage Starting (to limit $$\displaystyle I_{st} $$):

  • $$\displaystyle T \propto V^2 $$, $I \propto V$.

  • To get rated torque at start: need $$\displaystyle V_{st} = V_{rated} $$ → defeats purpose.

  • To limit current to $$\displaystyle k \times I_{fl} $$: $$\displaystyle V_{st} = \frac{V_{rated}}{k} $$ → starting torque becomes $$\displaystyle T_{st} = \frac{T_{rated}}{k^2} $$.

  • Example: If $$\displaystyle k=2 $$ (current halved), $$\displaystyle T_{st} = T_{rated}/4 $$ → may not start heavy loads.

• Plugging (IM):

  • Swap any two stator phases while motor running.

  • Rotating magnetic field reverses direction → slip $$\displaystyle s > 1 $$ → torque reverses → braking.

  • High braking torque, but high energy loss in rotor (all slip power dissipated as heat).

  • Stopped at zero speed to avoid reverse rotation.

Numerical Problems (IM)

1. V/f Control at Different Frequency:

Given: $$\displaystyle P=4 $$, $$\displaystyle f_1=50Hz $$, $$\displaystyle n_{fl}=1370 $$ rpm (so $$\displaystyle n_{s1}=1500 $$, $$\displaystyle s_{fl}=0.02 $$). Parameters: $$\displaystyle R_s=2\Omega $$, $$\displaystyle R_r'=5\Omega $$, $$\displaystyle X_s=X_r'=5\Omega $$, $$\displaystyle X_m=80\Omega $$. Load: fan ($$\displaystyle T \propto \omega^2 $$). At rated $V$ (400V), motor runs at rated speed (1370 rpm). Find $V$, $I$, $T$ at $$\displaystyle n=1200 $$ rpm with $$\displaystyle f=80Hz $$.

• Solution:

  • $$\displaystyle n_{s2} = 120 \times 80 / 4 = 2400 $$ rpm.

  • At $$\displaystyle n=1200 $$, $$\displaystyle s_2 = (2400-1200)/2400 = 0.5 $$.

  • Since $$\displaystyle T \propto \omega^2 $$ (fan load), $$\displaystyle T_2 = T_{rated} \times (1200/1370)^2 $$.

  • Need $$\displaystyle V_2 $$ such that motor can develop $$\displaystyle T_2 $$ at $$\displaystyle s=0.5 $$.

  • Use Thevenin equivalent. $$\displaystyle V_{th} \propto V_2 $$, $$\displaystyle Z_{th} $$ independent of $V$.

  • Torque equation: $$\displaystyle T_2 = \frac{3}{\omega_{s2}} \frac{|V_{th2}|^2 (R_r'/s_2)}{(R_{th}+R_r'/s_2)^2 + (X_{th}+X_r')^2} $$.

  • Compute $$\displaystyle R_{th}, X_{th} $$ from $$\displaystyle R_s, X_s, X_m $$.

  • Since $$\displaystyle T_{rated} $$ known at $$\displaystyle s_{fl}=0.02 $$, $$\displaystyle f_1=50Hz $$, $$\displaystyle V_1=400V $$, can compute constant $$\displaystyle K = T \cdot \omega_s / |V_{th}|^2 $$.

  • Then $$\displaystyle V_{th2} = V_2 \cdot \frac{X_m}{\sqrt{R_s^2 + (X_s+X_m)^2}} $$.

  • Solve for $$\displaystyle V_2 $$ from $$\displaystyle T_2 $$ equation.

  • Then $$\displaystyle I_2 \approx |I_2| = \frac{|V_{th2}|}{\sqrt{(R_{th}+R_r'/s_2)^2 + (X_{th}+X_r')^2}} $$.

  • Stator current $$\displaystyle I_s \approx I_2 $$ (approx).

2. Slip for Maximum Torque with Changed Frequency: $$\displaystyle T_{max} \propto (V/f)^2 $$. If $V/f$ constant, $$\displaystyle T_{max} $$ constant. Slip at max torque: $$\displaystyle s_{max} = R_r' / \sqrt{R_{th}^2 + (X_{th}+X_r')^2} $$ → independent of $V$ and $f$ if $$\displaystyle R_r', X $$'s are constant. So $$\displaystyle s_{max} $$ same for all frequencies in constant V/f control.

3. Rotor Copper Loss Problem:

Given: Wound rotor IM, $$\displaystyle n_{fl}=288 $$ rpm, $$\displaystyle f=50Hz $$.

• Poles: $$\displaystyle n_s = 120f/P $$. $$\displaystyle n_s > 288 $$. Try $$\displaystyle P=4 \Rightarrow n_s=1500 $$, $$\displaystyle s=0.808 $$ (too high). $$\displaystyle P=6 \Rightarrow n_s=1000 $$, $$\displaystyle s=0.712 $$. $$\displaystyle P=8 \Rightarrow n_s=750 $$, $$\displaystyle s=0.616 $$. $$\displaystyle P=10 \Rightarrow n_s=600 $$, $$\displaystyle s=0.52 $$. $$\displaystyle P=12 \Rightarrow n_s=500 $$, $$\displaystyle s=0.424 $$. $$\displaystyle P=16 \Rightarrow n_s=375 $$, $$\displaystyle s=0.232 $$. $$\displaystyle P=20 \Rightarrow n_s=300 $$, $$\displaystyle s=0.04 $$. $$\displaystyle P=24 \Rightarrow n_s=250 $$, $$\displaystyle s=-0.152 $$ (impossible). So likely $$\displaystyle P=20 $$? But 288 rpm for 50Hz, 4-pole is 1500 rpm, slip 0.808 is possible but high. Usually full load slip is 2-5%. So 288 rpm is close to 300 rpm for 20-pole? 20-pole: $$\displaystyle n_s=120*50/20=300 $$ rpm, $$\displaystyle s=(300-288)/300=0.04 $$ (4%) → reasonable. So P=20 poles.

• Slip at full load: $$\displaystyle s_{fl} = 0.04 $$.

• If rotor resistance doubled: $$\displaystyle R_r' \rightarrow 2R_r' $$. $$\displaystyle s_{max} \propto R_r' $$ → doubles. But operating slip for same load torque? Load torque constant? Not specified. Usually for constant torque load, slip increases. From torque equation, for same $T$, if $$\displaystyle R_r' $$ doubles, $s$ must increase (roughly double if $s$ small and $$\displaystyle R_{th} $$ negligible). So new slip $$\displaystyle s_{new} \approx 2s_{fl} = 0.08 $$ (if $$\displaystyle R_{th} \ll R_r'/s $$).

• Rotor copper loss $$\displaystyle P_{rcl} = s \cdot P_{rinput} $$. $$\displaystyle P_{rinput} $$ (air-gap power) for same torque and speed? Actually if load torque constant, $$\displaystyle T \propto I_2^2 R_r'/s $$. With $$\displaystyle R_r' $$ doubled, to keep $T$ constant, $$\displaystyle I_2^2 $$ must halve approximately, so $$\displaystyle P_{rinput} = 3 I_2^2 R_r'/s $$ might stay similar? Let's derive: $$\displaystyle T = \frac{3}{\omega_s} \frac{R_r'/s}{(R_{th}+R_r'/s)^2} |V_{th}|^2 $$. For small $s$, $$\displaystyle R_{th}+R_r'/s \approx R_r'/s $$, so $$\displaystyle T \approx \frac{3}{\omega_s} \frac{s}{R_r'} |V_{th}|^2 $$. So for constant $T$, $$\displaystyle s/R_r' = \text{const} $$. If $$\displaystyle R_r' $$ doubles, $s$ doubles. Then $$\displaystyle P_{rinput} = T \omega_s / s $$? Actually $$\displaystyle P_{rinput} = T \omega_s / (1-s) \approx T \omega_s $$ for small $s$. So $$\displaystyle P_{rinput} $$ roughly constant. Then $$\displaystyle P_{rcl} = s \cdot P_{rinput} $$ → doubles. So rotor copper loss doubles.

D. SYNCHRONOUS MOTOR DRIVES

Self-Controlled Synchronous Motor (Brushless DC / LCI)

• Principle: Motor shaft position fed back to inverter firing circuit. Inverter output frequency locked to rotor speed → $$\displaystyle f = p \cdot n / 120 $$. Motor behaves like a DC motor with constant torque angle $\delta$ (load angle).

• Power Circuit:

  • CSI (Load-Commutated Inverter - LCI): Thyristor bridge (6-pulse) with DC link inductor. Commutation provided by motor's own back-EMF (synchronous machine acts as synchronous generator). Requires separate excitation (DC field).

  • VSI (PWM): For PMSM/BLDC, uses rotor position sensors (Hall effect/encoders) to commutate.

• Comparison with Separate Excitation:

  • Separate Excitation: Inverter frequency independent of rotor position → needs rotor position sensor for starting and to maintain synchronism. Can operate at any frequency up to limit.

  • Self-Controlled: Frequency always = $p n / 120$. No risk of loss of synchronism. Starting requires separate means (e.g., damper winding or auxiliary motor) to bring near synchronous speed before applying DC field.

VSI Fed Synchronous Motor Drives

• Circuit: Three-phase VSI (PWM or six-step) feeding synchronous motor (wound-rotor or PM).

• Operation:

  • PWM VSI: Provides variable voltage and frequency. For PMSM/BLDC, requires rotor position to commutate thyristors/MOSFETs at correct instant.

  • Six-step VSI: Voltage is quasi-square wave, fundamental voltage $$\displaystyle V_1 \propto V_{dc} $$.

• Regenerative Braking: Motor overspeeds ($$\displaystyle \omega > \omega_{ref} $$), back-EMF $$\displaystyle E > V_{dc} $$? Actually for VSI, if motor acts as generator, power flows from AC motor terminals to DC link, then from DC link back to supply via inverter operating in inversion mode (requires DC link capacitor and active front end or braking resistor).

• Closed-Loop Block Diagram:

  
  Speed Ref → Speed Controller → Current/Voltage Ref → PWM Modulator → VSI → Synchronous Motor → Speed
  
                              ↑
  
                            Current/Speed Sensor (Feedback)
  
  

Permanent Magnet Synchronous Motor (PMSM)

• Construction: Stator with 3-phase windings, Rotor with permanent magnets (surface-mounted or interior).

• Operation: Rotating magnetic field from stator interacts with constant rotor flux → torque $$\displaystyle T = \frac{3}{\omega_s} \frac{V E_f}{X_s} \sin\delta $$ (for salient or non-salient). Requires rotor position for commutation (like BLDC but with sinusoidal back-EMF).

• Comparison with Wound-Rotor SM:

  • PMSM: No field winding, no slip rings, higher efficiency, higher power density, but constant flux (no field weakening unless special design).

  • Wound-Rotor: Field current controllable → wider constant power speed range (field weakening possible).

Regenerative Braking in Synchronous Motors

• Condition: Motor speed exceeds synchronous speed corresponding to terminal voltage frequency ($$\displaystyle \omega > \omega_s $$). Then $\delta$ becomes negative → torque negative (braking).

• With VSI: If DC link has active front end (PWM rectifier), power can be fed back to AC supply. Otherwise, energy dissipated in braking resistor across DC link.

• Speed-Torque Characteristic: Shows stable operation in 2nd quadrant (ω > 0, T < 0). Pull-out torque decreases as speed increases above synchronous.

Numerical Problems (SM)

1. Reluctance Motor (Salient Pole) Load Angle:

Given: $$\displaystyle V=400V $$, $$\displaystyle f=50Hz $$, $$\displaystyle P=4 $$, star-connected, $$\displaystyle R_s \approx 0 $$, $$\displaystyle X_d=8\Omega $$, $$\displaystyle X_q=22\Omega $$, $$\displaystyle T_L=80 $$ Nm.

• Torque equation for reluctance motor (neglecting $$\displaystyle R_s $$):

$$ T = \frac{3V^2}{2\omega_s} \left( \frac{1}{X_q} - \frac{1}{X_d} \right) \sin 2\delta $$

Where $$\displaystyle \omega_s = 2\pi f = 314 $$ rad/s (electrical).

*   Compute constant: $$\displaystyle \frac{3V^2}{2\omega_s} \left( \frac{1}{X_q} - \frac{1}{X_d} \right) = \frac{3 \times (400/\sqrt{3})^2}{2 \times 314} \left( \frac{1}{22} - \frac{1}{8} \right) $$.

*   Solve for $\delta$ from $$\displaystyle T=80 $$.

• Line Current: $$\displaystyle I_a = \frac{V}{\sqrt{(X_d \sin\delta)^2 + (X_q \cos\delta)^2}} $$? Actually for salient pole, $$\displaystyle I_d = \frac{V}{X_d} \sin\delta $$, $$\displaystyle I_q = \frac{V}{X_q} \cos\delta $$, so $$\displaystyle I = \sqrt{I_d^2 + I_q^2} $$.

• Power Factor: $$\displaystyle \cos\phi = \frac{V}{I \cdot Z} $$? Actually $$\displaystyle \cos\phi = \frac{P}{\sqrt{3} V I} $$, and $$\displaystyle P = T \omega_m $$. $$\displaystyle \omega_m = \omega_s / (p/2) $$? For 4-pole, $$\displaystyle p=2 $$, $$\displaystyle \omega_m = \omega_s / 2 = 157 $$ rad/s (mechanical). So $$\displaystyle P = 80 \times 157 = 12560 $$ W. Then $$\displaystyle \cos\phi = P / (\sqrt{3} \times 400 \times I) $$.

2. Variable Frequency Control Above Base Speed:

Given: Synchronous motor, constant V/f up to base speed, then constant V above. At operation above base speed (constant V), flux weakens. Torque $$\displaystyle T \propto I_a \cdot \Phi \propto I_a / f $$ (since $\Phi \propto 1/f$). For given $T$, $$\displaystyle I_a \propto f $$. Power $$\displaystyle P = T \omega \propto T \cdot f $$ → constant if $T \propto 1/f$. So constant power region.

E. SPECIAL MOTORS & STEPPER MOTORS

Switched Reluctance Motor (SRM)

• Construction: Salient poles on both stator and rotor. No PMs, no windings on rotor. Stator has concentrated windings (each phase on a pair of poles).

• Principle: Variable Reluctance. Torque produced by tendency of rotor to align with excited stator pole to minimize reluctance. Each phase excited sequentially.

• Power Circuit: Typically asymmetric bridge per phase (two switches and two diodes per phase). Allows independent control and regenerative braking.

• Switching Scheme:

  • Rotor position sensor (optical/ Hall) detects rotor pole position.

  • Controller excites stator phase when rotor pole is approaching alignment position.

  • Turn-off before misalignment to avoid negative torque.

• Advantages:

  • Simple, robust, low-cost construction.

  • High starting torque (can start at zero speed).

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

  • Wide speed range.

  • No PMs → no demagnetization issue.

Stepper Motors

1. Variable Reluctance (VR) Stepper

• Construction: Stator with multiple phases (e.g., 3-phase), rotor made of ferromagnetic material with no PM, teeth on both stator and rotor.

• Operation: When stator phase excited, rotor teeth align to minimize reluctance. Step angle determined by number of rotor teeth and stator phases: $$\displaystyle \theta_s = \frac{360°}{N_r \cdot N_s} $$? Actually for VR: $$\displaystyle \theta_s = \frac{(N_s - N_r) \times 360°}{N_s \cdot N_r} $$ where $$\displaystyle N_s $$ = stator teeth, $$\displaystyle N_r $$ = rotor teeth.

• Characteristics: Detent torque when unexcited, no holding torque without excitation.

2. Permanent Magnet (PM) Stepper

• Construction: Rotor made of permanent magnet (cylindrical or disc with poles). Stator has multiple phases.

• Operation: Interaction between stator magnetic field and rotor PM. Step angle determined by number of stator phases and rotor poles: $$\displaystyle \theta_s = \frac{360°}{\text{number of rotor poles}} $$ for two-phase? Actually for PM: $$\displaystyle \theta_s = \frac{360°}{N_r} $$ for two-phase on two-pole rotor? Standard: For a 4-pole rotor (2 pole pairs) and two-phase, step angle = 45°? Better: Step angle = $360°/(\text{number of steps per revolution})$. Typically $$\displaystyle \theta_s = \frac{360°}{N_r \cdot N_s} $$? No, for PM stepper with $p$ pole pairs and $m$ phases, $$\displaystyle \theta_s = \frac{360°}{p \cdot m} $$? Actually common: Two-phase, 50-tooth rotor → step 1.8° (200 steps/rev). Formula: $$\displaystyle \theta_s = \frac{360°}{N_r} $$ for PM stepper? Not exactly. Standard: Step angle = $360° / (\text{number of rotor teeth})$ for VR? Let's clarify:

  • VR: $$\displaystyle \theta_s = \frac{360° \cdot (N_s - N_r)}{N_s \cdot N_r} $$.

  • PM: $$\displaystyle \theta_s = \frac{360°}{N_r} $$ (if $$\displaystyle N_r $$ = number of rotor teeth/poles? Actually for PM with toothed rotor, it's similar to VR. For smooth rotor PM, step angle determined by stator winding arrangement. For two-phase with two stator poles per phase, and rotor with 2 poles (1 pair), step = 90°. So generally $$\displaystyle \theta_s = \frac{360°}{\text{number of steps per rev}} $$.

  • Hybrid: Combines VR and PM principles → smaller step angles (e.g., 1.8°).

3. Hybrid Stepper

• Combines toothed rotor (like VR) and PM rotor (like PM). Provides small step angles (1.8°, 0.9°) and high torque.

4. Micro-Stepping

• Concept: Subdivide a full step into smaller increments by controlling phase currents sinusoidally (instead of full-on/full-off).

• How: Currents in phases are proportioned to produce a magnetic field vector at an angle between full-step positions.

• Advantages:

  • Smooth motion, reduced resonance.

  • Higher resolution.

  • Reduced acoustic noise.

  • Improved low-speed performance.

Brushless DC Motor (BLDC)

• Construction: PM rotor (usually surface-mounted), stator with three-phase windings (similar to PMSM). Trapezoidal back-EMF waveform.

• Operation: Electronic commutation based on rotor position sensors (Hall effect). Each phase conducts for 120° electrical. Torque is ripply due to trapezoidal current and back-EMF.

• Power Circuit: Three-phase VSI (six-step or PWM).

• Comparison with PMSM:

  | Feature | BLDC | PMSM | | :--- | :--- | :--- | | Back-EMF | Trapezoidal | Sinusoidal | | Current | Rectangular (120° conduction) | Sinusoidal | | Control | Simple (6-step) | Requires sinusoidal PWM (FOC) for high performance | | Torque Ripple | Higher | Lower (with FOC) | | Applications | Fans, pumps, appliances | High-performance servo, robotics |

F. CLOSED-LOOP & DIGITAL CONTROL

Closed-Loop Drive Systems

• Need: Improve dynamic response, reduce steady-state error, reject disturbances (load changes), achieve precise speed/position control.

• Typical Structure:

  1. Inner Current Loop (fast, ~1-2 ms):

     • Measures armature current $$\displaystyle I_a $$.

     • Compares with current reference $$\displaystyle I_{a,ref} $$ (from outer loop).

     • PI controller → PWM/converter firing → controls $$\displaystyle I_a $$ quickly. Provides current limiting and torque control (since $$\displaystyle T \propto I_a $$ for DC, or for AC with vector control).

  2. Outer Speed Loop (slower, ~10-100 ms):

     • Measures speed $\omega$.

     • Compares with speed reference $$\displaystyle \omega_{ref} $$.

     • PI controller → outputs $$\displaystyle I_{a,ref} $$ to current loop.

• Block Diagram (DC Drive):

  
  ω_ref → (+) → Speed Controller (PI) → I_a_ref → Current Controller (PI) → PWM/Firing → Converter → DC Motor → ω
  
            ↑                                                                 |
  
          ω (Tachometer)                                                     I_a (CT)
  
  

• Block Diagram (AC Drive with Vector Control):

  Similar, but current loop controls $$\displaystyle I_d $$ (flux) and $$\displaystyle I_q $$ (torque) components after coordinate transformation.

Digital Control of Drives

• Advantages over Analog:

  • Flexibility (easy algorithm changes, multi-function).

  • Accuracy (no component drift).

  • Complex control strategies (vector control, DTC, adaptive control).

  • Self-diagnostics, communication (fieldbus).

  • Cost reduction at scale.

• Typical Architecture:

  • Microprocessor/DSP/FPGA: Executes control algorithm (sampling, calculation, PWM generation).

  • Sensors: Encoders, resolvers, current/voltage transducers.

  • Power Interface: Gate drivers for IGBTs/MOSFETs.

  • ADC/DAC: Analog-to-digital conversion for sensor signals, digital-to-analog for references if needed.

• Implementation Aspects:

  • Sampling: Must sample faster than system dynamics (typically 10-20 kHz for current loop).

  • Quantization: ADC resolution affects accuracy.

  • Delays: Computation delay + PWM update delay must be accounted for in stability design (often modeled as transport delay).

G. INDUSTRIAL APPLICATIONS & SPECIAL TOPICS

Drive Schemes for Specific Industries

• Cement Industries:

  • Applications: Raw mills, kilns, cement mills, fans, conveyors.

  • Drives: Large slip power recovery drives (Static Kramer) for wound-rotor IM in mills (high torque, speed control). VFDs (VSI) for fans and pumps (energy saving). High-power CSI for mills.

• Steel Industries:

  • Applications: Rolling mills (high torque, high power), cranes, conveyors, fans.

  • Drives: Historically DC drives (series/compound) for rolling mills (excellent torque-speed control). Now large AC drives with vector control (DTC or FOC) for induction or synchronous motors. Regenerative braking essential for reversing mills and cranes.

Group & Individual Drives

• Group Drive: One motor drives multiple machines via line shaft/belts. Disadvantages: No individual speed control, efficiency low if not all machines running, fault affects all. Use: Small workshops, legacy systems.

• Individual Drive: One motor per machine. Advantages: Independent speed control, high efficiency, flexibility, fault isolation. Use: Modern industries, CNC machines.

• Selection: Individual drives preferred for flexibility and efficiency; group drives only for simple, always-on applications.

Constant Torque & Constant Power Drives

• Constant Torque Drive: Torque capability independent of speed. Achieved by armature voltage control in DC drives (below base speed) or constant V/f in AC drives (below base speed). Used for hoists, conveyors, extruders.

• Constant Power Drive: Power capability independent of speed → torque $\propto 1/\omega$. Achieved by field weakening in DC drives (above base speed) or constant voltage, increasing frequency in AC drives (above base speed). Used for machine tools (milling, turning).

• Range: DC: Wide constant torque (0 to base), constant power (base to max). AC V/f: Constant torque (0 to base), constant power (base to ~2-3×base, limited by voltage and current).

Energy Recovery Systems

• Regenerative Braking: Motor acts as generator, feeding power back to supply or grid. Used in:

  • Elevators/Cranes: Descending loads generate power.

  • Electric Trains: Braking deceleration.

  • Downhill Conveyors: Continuous generation.

• Systems: Requires reversible converter (fully controlled converter or VSI with active front end). Energy can be fed back to grid (preferred) or dissipated in resistor if grid not available.

Transient & Steady-State Analysis of DC Drives

• Steady-State:

  • Equivalent circuit: $$\displaystyle V_a = E + I_a R_a $$.

  • $$\displaystyle E = K\phi \omega $$.

  • $$\displaystyle T = K\phi I_a $$.

  • So $$\displaystyle \omega = \frac{V_a - I_a R_a}{K\phi} $$, $$\displaystyle T = K\phi \frac{V_a - K\phi \omega}{R_a} $$.

• Transient:

  • Mechanical: $$\displaystyle J \frac{d\omega}{dt} = T_m - T_L $$ (J = total inertia).

  • Electrical: $$\displaystyle L_a \frac{di_a}{dt} = V_a - E - I_a R_a $$ (if inductance considered).

  • For analysis, often assume electrical time constant $$\displaystyle L_a/R_a $$ << mechanical time constant $$\displaystyle J/T_{load} $$ → decoupled analysis.

  • Linearization: For small perturbations around operating point $$\displaystyle (I_{a0}, \omega_0) $$:

$$ \Delta T_m = K\phi \Delta I_a, \quad \frac{d(\Delta\omega)}{dt} = \frac{K\phi}{J} \Delta I_a - \frac{1}{J} \frac{dT_L}{d\omega} \Delta\omega $$

    Stability requires $$\displaystyle dT_L/d\omega > 0 $$ (typical for most loads) and positive damping.

CNC (Computer Numerical Control) in Drives

• Role: Drives are the actuators for CNC axes (X, Y, Z, spindle). They execute motion commands from CNC controller.

• Requirements:

  • Precise speed and position control (accuracy ±0.001 mm).

  • Fast response (high bandwidth).

  • Smooth motion (low ripple).

  • Feedback: High-resolution encoders/resolvers.

• Typical Drive Configurations:

  • Spindle Drive: AC drive (VSI-fed PMSM or IM) with speed control.

  • Axis Drives: Servo drives (DC or AC with vector control) with position control. Often PMSM or linear motors for high precision.

Electric Traction Drives

• Requirements:

  • High starting torque.

  • Wide speed range (0 to 200+ km/h).

  • Regenerative braking (energy saving).

  • Robustness, reliability.

  • High power-to-weight ratio.

• Typical Drive Systems:

  1. DC Series Motor: Historically used (simple, high starting torque). Now obsolete.

  2. AC Induction Motor with Vector Control: Most common today (robust, low maintenance). Uses IGBT-VSI with DTC/FOC.

  3. PMSM: Increasing use (higher efficiency, power density). Used in high-speed trains, metros.

• Braking Methods: Regenerative braking (primary), plugging (emergency), dynamic braking (resistor).

KEY FORMULAS & CONCEPTS BOX

[!TIP] Final Exam Strategy:

1. Draw Diagrams: For every "explain" question, draw the relevant circuit/block diagram first.

2. Derivations: Focus on load equalization and V/f control max torque derivation—they are 7-mark favorites.

3. Comparisons: AC vs DC, Kramer vs Scherbius, BLDC vs PMSM, braking methods.

4. Numericals: Practice converter/chopper-fed DC drive calculations (find $$\displaystyle V_a $$, $$\displaystyle I_a $$, $T$, $\omega$) and IM V/f control (find $V$, $I$, $T$ at new $f$ or $n$). Plugging problems are common.

5. Block Diagrams: Closed-loop drives (DC and AC), Static Kramer drive.

6. Special Motors: SRM principle/advantages, Stepper types/micro-stepping, PMSM construction.

7. Industrial Apps: Cement (slip power recovery), Steel (rolling mills, regenerative).