Control of Electric Drive

In an induction motor, the air-gap flux is $\phi \propto \dfrac{V}{f}$ (neglecting stator drop). If frequency is reduced to lower the speed while voltage is kept constant, the flux rises and the core saturates. So below base speed, voltage is varied in proportion to frequency (constant V/f), keeping the flux and hence the maximum torque nearly constant.

Torque-speed characteristics

Synchronous speed $N_s = 120f/P$ falls with $f$; the slip speed at maximum torque, $s_m\omega_s \approx (R_2'/X_2')\,\omega_s$, is the same at every frequency because $X_2' \propto f$. Therefore the curves shift left parallel to each other with almost the same maximum torque.

 T
  |     f4    f3    f2   f1 = base
  |   _.-.  _.-.  _.-.  _.-.
  |  /   | /   | /   | /   |
  | /    |/    |/    |/    |  Tmax ~ const
  |/     /     /     /     |
  +------+-----+-----+-----+--> speed
        Ns4   Ns3   Ns2   Ns1
  (at very low f, Tmax falls - dotted -
   because stator R drop is significant;
   add voltage boost)

• From base speed down to about 10 to 15 % of base frequency, $T_{max}$ is nearly constant: constant torque region.
• At very low frequencies the stator resistance drop becomes comparable to $V$, so flux and $T_{max}$ fall; a voltage boost (offset $V_0$) is added: $V = V_0 + kf$.
• Above base speed, $V$ is held at rated and $f$ increased: field weakening (constant power).

Open-loop constant V/f control scheme

 3-ph AC ->[Diode rect]--+--[L]--+--[PWM inverter]--> IM
                         |       C       ^
                     DC link             | gate pulses
 w* (speed ref) ->[Ramp]->[f*]->[V/f  ]-->[PWM generator]
                     |          [block]  V* and f*
                     +--------->(with low-speed boost)

1. Rectifier and DC link: three-phase diode rectifier with LC filter gives a constant DC voltage.
2. Speed reference $\omega^*$ sets the frequency command $f^*$ (since $N \approx N_s$ at small slip).
3. Ramp (soft start) block limits the rate of change of $f^*$ so that slip and current stay within limits during acceleration and deceleration.
4. V/f function generator gives $V^* = V_0 + kf^*$ (boost at low frequency), saturating at rated voltage above base frequency.
5. PWM generator and voltage source inverter produce three-phase output with amplitude $V^*$ and frequency $f^*$.
6. The motor runs near synchronous speed for that frequency; speed drops slightly with load (equal to slip speed), as no speed feedback is used. Slip compensation can be added using measured current.

Features

• Simple, cheap, no speed sensor; several motors can be run from one inverter (textile mills).
• Speed accuracy limited by slip; poor dynamic response; current limit needed for protection.
• Widely used for fans, pumps, conveyors and other general-purpose drives.

Data: $V_s = 220$ V, $V_m = \sqrt2\times220 = 311.13$ V; $R_f = 250\ \Omega$, $R_a = 0.2\ \Omega$; $T = 50$ N-m at $N = 1000$ rpm; $K_v = 0.8$ V/A-rad/s, $K_t = 0.8$ N-m/A². For a single-phase semiconverter with continuous current:

$$V_{dc} = \frac{V_m}{\pi}(1 + \cos\alpha)$$

i) Field current

Field firing angle $\alpha_f = 0$:

$$\begin{aligned} V_f &= \frac{311.13}{\pi}(1 + 1) = 198.07\ \text{V} \\ I_f &= \frac{V_f}{R_f} = \frac{198.07}{250} = 0.792\ \text{A} \end{aligned}$$

ii) Armature firing angle

Armature current from $T = K_tI_fI_a$:

$$I_a = \frac{50}{0.8\times0.7923} = 78.89\ \text{A}$$

Speed $\omega = \dfrac{2\pi\times1000}{60} = 104.72$ rad/s. Back emf:

$$E_g = K_v\,\omega\,I_f = 0.8\times104.72\times0.7923 = 66.37\ \text{V}$$

Armature voltage:

$$V_a = E_g + I_aR_a = 66.37 + 78.89\times0.2 = 82.15\ \text{V}$$

Firing angle:

$$\begin{aligned} 82.15 &= \frac{311.13}{\pi}(1 + \cos\alpha_a) \\ \cos\alpha_a &= \frac{82.15\pi}{311.13} - 1 = -0.1705 \\ \alpha_a &= 99.82^\circ \end{aligned}$$

iii) Power factor of armature converter

With ripple-free armature current $I_a$, the supply current is a quasi-square wave of height $I_a$ conducting for $(\pi - \alpha)$ in each half cycle.

$$\begin{aligned} I_s &= I_a\sqrt{\frac{\pi-\alpha}{\pi}} = 78.89\sqrt{\frac{180-99.82}{180}} = 52.65\ \text{A} \\ I_{s1} &= \frac{2\sqrt2}{\pi}I_a\cos\frac{\alpha}{2} = 0.9003\times78.89\times0.6440 = 45.74\ \text{A} \end{aligned}$$

Displacement factor $= \cos(\alpha/2) = \cos 49.91^\circ = 0.644$.

$$PF = \frac{I_{s1}}{I_s}\cos\frac{\alpha}{2} = \frac{45.74}{52.65}\times0.644 = 0.559\ \text{(lagging)}$$

Check by power balance (losses neglected): $PF = \dfrac{V_aI_a}{V_sI_s} = \dfrac{82.15\times78.89}{220\times52.65} = 0.559$.

Answer: (i) $I_f = 0.792$ A; (ii) $\alpha_a = 99.8^\circ$; (iii) PF $= 0.559$ lagging.

Data: $V_s = 400$ V, $R_a = 0.1\ \Omega$, $K_v = 4$ V-s/rad, $I_a = 150$ A (continuous, ripple-free), duty cycle $k = 0.6$. For a step-down chopper, $V_a = kV_s$ and the average source current is $kI_a$.

a) Input power

$$\begin{aligned} V_a &= kV_s = 0.6\times400 = 240\ \text{V} \\ P_i &= V_s(kI_a) = 400\times0.6\times150 = 36{,}000\ \text{W} = 36\ \text{kW} \end{aligned}$$

b) Motor speed

$$\begin{aligned} E_g &= V_a - I_aR_a = 240 - 150\times0.1 = 225\ \text{V} \\ \omega &= \frac{E_g}{K_v} = \frac{225}{4} = 56.25\ \text{rad/s} \\ N &= \frac{56.25\times60}{2\pi} = 537.1\ \text{rpm} \end{aligned}$$

c) Developed torque

$$T = K_vI_a = 4\times150 = 600\ \text{N-m}$$

Speed range for $k$ = 0.1 to 0.9 (same $I_a = 150$ A)

Answer: (a) 36 kW; (b) 56.25 rad/s = 537.1 rpm; (c) 600 N-m. Speed range: minimum 59.7 rpm (6.25 rad/s) at 10 %, maximum 823.6 rpm (86.25 rad/s) at 90 % duty cycle.

For a three-phase half-controlled (semi-converter) bridge with continuous current, line voltage $V_L$:

$$V_a = \frac{3\sqrt2\,V_L}{2\pi}(1 + \cos\alpha) = \frac{3V_{ml}}{2\pi}(1+\cos\alpha)$$ $$\frac{3\sqrt2\times400}{2\pi} = 270.09\ \text{V}$$

Data: $R_a = 0.2\ \Omega$, $K = 0.25$ V/rpm ($E_b = 0.25N$).

No-load speed ($I_a = 5$ A, $\alpha = 45^\circ$)

$$\begin{aligned} V_a &= 270.09\,(1 + \cos45^\circ) = 270.09\times1.7071 = 461.08\ \text{V} \\ E_b &= V_a - I_aR_a = 461.08 - 5\times0.2 = 460.08\ \text{V} \\ N_0 &= \frac{460.08}{0.25} = 1840.3\ \text{rpm} \end{aligned}$$

Firing angle for 1500 rpm at rated current (100 A)

$$\begin{aligned} E_b &= 0.25\times1500 = 375\ \text{V} \\ V_a &= 375 + 100\times0.2 = 395\ \text{V} \\ 1 + \cos\alpha &= \frac{395}{270.09} = 1.4624 \\ \cos\alpha &= 0.4624 \;\Rightarrow\; \alpha = 62.45^\circ \end{aligned}$$

Answer: no-load speed $\approx 1840$ rpm; firing angle for 1500 rpm at 100 A $\approx 62.5^\circ$.

A PID controller produces a control signal from the error $e(t)$ between reference and measured value using proportional, integral and derivative actions:

$$u(t) = K_pe(t) + K_i\int_0^te(\tau)\,d\tau + K_d\frac{de(t)}{dt}$$ $$G_c(s) = \frac{U(s)}{E(s)} = K_p + \frac{K_i}{s} + K_ds$$

PID speed and torque control of a DC motor (cascade control)

Motor equations (separately excited DC):

$$V_a = R_aI_a + L_a\frac{dI_a}{dt} + K\omega, \qquad T = KI_a = J\frac{d\omega}{dt} + B\omega + T_L$$

Torque is proportional to armature current, so torque control = current control.

 w*->(+)->[Speed PID]-Ia*->(+)->[Current PI]
      ^-   (limit)          ^-       |
      |                     |   firing/duty
      |                     |        v
      |                     |   [Converter]
      |                     |        | Va
      |                     |   [DC motor]--> w
      |                     |     |  |
      |  current sensor <---+-----+  |
      +---- tacho/encoder <----------+

1. Outer speed loop: speed error $\omega^* - \omega$ goes to the speed PID (usually PI). Its output is the current (torque) reference $I_a^*$, limited to the maximum allowed current, so the motor accelerates at maximum safe torque.
2. Inner current (torque) loop: current error $I_a^* - I_a$ goes to the current PI controller, which sets the converter firing angle (rectifier) or duty cycle (chopper). This loop is fast; it quickly corrects supply-voltage changes and limits current during starting, overload and stall.
3. Converter supplies $V_a$ to the armature.
4. Feedback: tachogenerator/encoder for speed, CT/Hall sensor/shunt for current.

For an induction motor the same structure is used in vector (field-oriented) control: an outer speed PI produces a torque-current command $i_{qs}^*$, and inner current PI loops control the inverter.

Behaviour

• On a sudden load increase, speed drops; the speed PID raises $I_a^*$, torque increases and speed returns to the reference with zero steady-state error (integral action).
• During starting, the speed error is large, $I_a^*$ saturates at the limit, so the motor accelerates at constant maximum torque.
• Derivative action is used sparingly (filtered) because of noise in speed measurement.

Advantages: accurate speed, fast torque response, inherent current protection, simple tuning of each loop separately.

In a slip-ring induction motor, part of the air-gap power, the slip power $sP_g$, appears in the rotor circuit. With rotor-resistance speed control this power is wasted as heat in external resistors, making low-speed operation inefficient. In a slip power recovery system, the slip power is taken out of the rotor through power converters and fed back to the supply (or to the shaft), so speed can be controlled efficiently.

Static Kramer drive (sub-synchronous)

 3-ph supply ---------+-----------------------+
                      |                       |
                 [Stator]              [Transformer]
                 Slip-ring IM                 |
                 [Rotor]                      |
                    | slip rings              |
               [Diode bridge]--[L]--[Line-commutated
                 (rectifier)  DC    thyristor inverter]
                              link

Working

1. Rotor voltage $sE_2$ at slip frequency is rectified by an uncontrolled diode bridge:

$$V_{d} = 1.35\,sE_2 \quad (E_2 = \text{standstill rotor line voltage})$$

2. A line-commutated thyristor inverter, operating with firing angle $90^\circ < \alpha < 180^\circ$, converts the DC power back to AC at supply frequency and returns it to the mains through a transformer (turns ratio adapts the voltage). Its DC voltage is

$$V_{I} = -1.35\,V_L'\cos\alpha$$

where $V_L'$ is the inverter-side AC line voltage.

3. Neglecting the drop in the DC-link inductor, $V_d = V_I$:

$$1.35\,sE_2 = -1.35\,V_L'\cos\alpha \;\Rightarrow\; s = -\frac{V_L'}{E_2}\cos\alpha$$

4. Changing $\alpha$ changes the counter-voltage seen by the rotor, so the slip, and hence speed, changes: at $\alpha = 90^\circ$, $s \approx 0$ (near synchronous speed); as $\alpha \to 180^\circ$, slip increases (lower speed).

5. Torque is proportional to DC-link current: $T \propto I_d$; the drive gives a shunt-like speed characteristic.

Features

• Speed control only below synchronous speed (power flows only out of the rotor through the diode bridge).
• High efficiency, since slip power is recovered; converter rating is proportional to the speed range (for pumps and fans with limited range, a small converter suffices).
• Drawbacks: poor power factor (inverter draws reactive power), harmonics; starting needs rotor resistors because the converter is rated for limited slip.

Static Scherbius drive

• The diode bridge and inverter are replaced by a bidirectional converter (cycloconverter, or back-to-back PWM converters, as in doubly-fed induction generators).
• Slip power can flow both ways, so speed control is possible below and above synchronous speed, with regenerative braking and power-factor control.

Applications

Large pumps, fans, compressors, cement-mill drives, and wind turbines (doubly-fed induction generator), where a limited speed range is needed with high efficiency.

Data: $V_1 = 200$ V, $I_L = 10$ A, $N_1 = 1800$ rpm, $R_a = 0.6\ \Omega$, $R_{sh} = 360\ \Omega$, constant load torque. Assumption: flux is proportional to field current (no saturation).

Initial condition

$$\begin{aligned} I_{f1} &= \frac{200}{360} = 0.5556\ \text{A}, \qquad I_{a1} = 10 - 0.5556 = 9.444\ \text{A} \\ E_{b1} &= 200 - 9.444\times0.6 = 194.33\ \text{V} \end{aligned}$$

(i) Speed at 180 V

The field is across the supply, so the field current (flux) also falls:

$$I_{f2} = \frac{180}{360} = 0.5\ \text{A}, \qquad \frac{\phi_1}{\phi_2} = \frac{0.5556}{0.5} = 1.111$$

Constant torque, $T \propto \phi I_a$:

$$I_{a2} = I_{a1}\frac{\phi_1}{\phi_2} = 9.444\times1.111 = 10.494\ \text{A}$$ $$E_{b2} = 180 - 10.494\times0.6 = 173.70\ \text{V}$$

Since $E_b \propto \phi N$:

$$\begin{aligned} N_2 &= N_1\frac{E_{b2}}{E_{b1}}\cdot\frac{\phi_1}{\phi_2} = 1800\times\frac{173.70}{194.33}\times1.111 \\ &= 1787.7\ \text{rpm} \end{aligned}$$

(ii) Starting resistance for 20 A line current at 180 V

At start $E_b = 0$. Field current $= 0.5$ A, so armature current

$$I_{a,st} = 20 - 0.5 = 19.5\ \text{A}$$ $$\begin{aligned} R_a + R_{ext} &= \frac{180}{19.5} = 9.231\ \Omega \\ R_{ext} &= 9.231 - 0.6 = 8.631\ \Omega \end{aligned}$$

Answer: (i) speed $\approx 1788$ rpm (it falls only slightly because the flux falls with the voltage); (ii) additional series resistance $\approx 8.63\ \Omega$.

A soft starter is a solid-state AC voltage controller placed between the supply and an induction motor. It starts the motor at a reduced voltage and raises the voltage smoothly to full value, so the starting current and torque rise gradually instead of in one jump.

Principle of operation

• Each phase has a pair of anti-parallel thyristors (or a triac) in series with the motor winding.
• By changing the firing angle $\alpha$ of the thyristors, only part of each half cycle of the supply is applied to the motor. A large $\alpha$ gives a low rms voltage; $\alpha$ near zero (or the conduction angle near 180°) gives full voltage.
• The frequency stays at 50 Hz; only the rms voltage changes.

 R ---[SCR pair]----+
 Y ---[SCR pair]----+---  3-phase
 B ---[SCR pair]----+     induction
        |                 motor
   firing control  <--- ramp / current
   (alpha: 150->0 deg)    feedback
        |
   bypass contactor closes at full V

How the ramp-up is produced

1. At start the controller fires at a large angle (say 120–150°), giving about 20–40 % of rated voltage.
2. A ramp generator reduces $\alpha$ step by step over a set time (typically 2–30 s), so the rms voltage rises linearly from the initial value to full voltage.
3. Many starters use current-limit control: the current is measured and $\alpha$ is adjusted so that it never exceeds a set value (e.g. 2–4 times full load).
4. When full voltage is reached, a bypass contactor shorts the thyristors to remove their losses.
5. A soft stop works the other way: voltage is ramped down so that pumps and conveyors stop gently.

Impact on motor performance

Motor torque is proportional to the square of voltage, $T \propto V^2$, and starting current is roughly proportional to $V$.

Limitations: since torque falls with $V^2$, a soft starter is not suited to loads needing high starting torque (e.g. loaded crushers). It does not control speed in steady state; it only controls starting and stopping. It is ideal for fans, pumps, compressors and conveyors.

Assumption: field current is constant (rated), so $E = K N$ and $T = K I_a$.

Key relations

• Chopper output: $V_a = \alpha V_s$
• Load torque proportional to speed: $T_L \propto N$. Since $T = K I_a$, the armature current is also proportional to speed: $I_a = c\,N$.
• Armature equation:

$$\begin{aligned} V_a &= E + I_a R_a = K N + R_a c N = (K + R_a c)\,N \end{aligned}$$

So the armature voltage is directly proportional to speed for this load, whatever the values of $K$ and $c$. Hence

$$\frac{\alpha_2}{\alpha_1} = \frac{V_{a2}}{V_{a1}} = \frac{N_2}{N_1}$$

Duty ratio at 800 rpm

$$\begin{aligned} \alpha_2 &= \alpha_1 \frac{N_2}{N_1} = 0.9 \times \frac{800}{1200} = 0.6 \end{aligned}$$

ON time

Chopping period:

$$T = \frac{1}{f} = \frac{1}{250} = 4\ \text{ms}$$ $$\begin{aligned} T_{on} &= \alpha_2 T = 0.6 \times 4\ \text{ms} = 2.4\ \text{ms} \\ T_{off} &= 4 - 2.4 = 1.6\ \text{ms} \end{aligned}$$

Check: $V_{a1} = 0.9 \times 220 = 198$ V at 1200 rpm and $V_{a2} = 0.6 \times 220 = 132$ V at 800 rpm; $132/198 = 800/1200$.

Note: the rated data (2.2 kW, 1000 rpm) are not needed, because both the back emf and the $I_a R_a$ drop scale with speed for a load with $T_L \propto N$.

Answer: Duty ratio = 0.6, ON time $T_{on}$ = 2.4 ms (OFF time 1.6 ms).

In a single-phase fully controlled bridge (four thyristors) feeding a separately excited dc motor, the average armature voltage is set by the firing angle $\alpha$, and the speed follows from the dc motor equations.

Circuit

        +----T1-----+----T3----+
  ac    |           |          |  +
 v_s ~  |           +---[ Ra La E ]  motor
 supply |           |          |  -
        +----T4-----+----T2----+
   T1,T2 fired at alpha; T3,T4 at pi+alpha

Derivation

Supply: $v_s = V_m \sin\omega t$. With continuous conduction, T1–T2 conduct from $\alpha$ to $\pi+\alpha$, so the armature voltage follows $v_s$ in this interval:

$$\begin{aligned} V_a &= \frac{1}{\pi}\int_{\alpha}^{\pi+\alpha} V_m \sin\omega t\, d(\omega t) \\ &= \frac{V_m}{\pi}\left[-\cos\omega t\right]_{\alpha}^{\pi+\alpha} = \frac{2V_m}{\pi}\cos\alpha \end{aligned}$$

Motor equations (constant field, $K = K_b$):

$$\begin{aligned} V_a &= E_b + I_a R_a, \quad E_b = K\omega_m, \quad T = K I_a \end{aligned}$$

(the inductor has zero average voltage in steady state). Therefore

$$\begin{aligned} \omega_m &= \frac{V_a - I_a R_a}{K} = \frac{2V_m\cos\alpha}{\pi K} - \frac{R_a}{K}I_a \\ \omega_m &= \frac{2V_m\cos\alpha}{\pi K} - \frac{R_a}{K^2}T \end{aligned}$$

In rpm, $N = \dfrac{60}{2\pi}\omega_m$. With $V_m = \sqrt2 V_s$:

$$N = \frac{60}{2\pi K}\left(\frac{2\sqrt2 V_s}{\pi}\cos\alpha - I_a R_a\right)$$

• $0 \le \alpha < 90^\circ$: $V_a > 0$, motoring (quadrant I).
• $90^\circ < \alpha < 180^\circ$: $V_a < 0$; with reversed back emf (load driving) the converter inverts, giving regenerative braking (quadrant IV).
• Speed falls linearly with torque; the slope $R_a/K^2$ is the same for every $\alpha$, so the speed–torque lines are parallel.

Waveforms (continuous conduction, fixed speed)

 v_s   /\        /\
      /  \      /  \
 ----/----\----/----\----> wt
          \  /      \  /
           \/        \/
 v_a  |alpha|
       __    __    __      (pieces of +|v_s|
 ---- /  \  /  \  /  \     from alpha to
      .   \/    \/    \    pi+alpha; dips
 E ----------------------  below 0 after pi)
 i_a  ~~~~~~~~~~~~~~~~~~   ripple about Ia
      ----------------->   continuous, > 0

• $v_a$ follows $+v_s$ from $\alpha$ to $\pi+\alpha$ and $-v_s$ from $\pi+\alpha$ to $2\pi+\alpha$; it goes negative for part of each half cycle.
• The back emf $E_b$ is a constant line (fixed speed).
• $i_a$ rises when $v_a > E_b$ and falls when $v_a < E_b$, but never reaches zero (continuous conduction), ripple frequency $2f$.

Data: $V_{ph} = 400/\sqrt3 = 230.94$ V, $r_1 = r_2' = 1\ \Omega$, $x_1 = x_2' = 2\ \Omega$, 6 poles, 50 Hz. Magnetising branch neglected (approximate circuit).

$$\begin{aligned} N_s &= \frac{120 f}{P} = \frac{120\times50}{6} = 1000\ \text{rpm} \\ s_{fl} &= \frac{1000-950}{1000} = 0.05 \\ s_b &= \frac{N_s + N}{N_s} = \frac{1000+950}{1000} = 1.95 \ \ (\text{plugging}) \end{aligned}$$

i) Initial braking current and torque

$$I = \frac{V_{ph}}{\sqrt{(r_1 + r_2'/s)^2 + (x_1+x_2')^2}}$$ $$\begin{aligned} I_{fl} &= \frac{230.94}{\sqrt{(1+20)^2 + 4^2}} = \frac{230.94}{21.378} = 10.80\ \text{A} \\ I_b &= \frac{230.94}{\sqrt{(1+0.5128)^2 + 4^2}} = \frac{230.94}{4.276} = 54.00\ \text{A} \\ \frac{I_b}{I_{fl}} &= 5.00 \end{aligned}$$

Torque $T = \dfrac{3 I^2 r_2'}{\omega_s s}$, so

$$\begin{aligned} \frac{T_b}{T_{fl}} &= \left(\frac{I_b}{I_{fl}}\right)^2 \frac{s_{fl}}{s_b} = 5.00^2 \times \frac{0.05}{1.95} = 0.641 \end{aligned}$$

Answer (i): $I_b = 5.0\, I_{fl}$ (54.0 A), $T_b = 0.641\, T_{fl}$.

ii) Rotor resistance for $I_b = 1.5\, I_{fl}$

The braking current is largest at the start ($s = 1.95$). Required impedance:

$$\begin{aligned} Z &= \frac{230.94}{1.5\times10.80} = 14.25\ \Omega \\ 1 + \frac{R_2'}{1.95} &= \sqrt{14.25^2 - 4^2} = 13.68 \\ R_2' &= 24.72\ \Omega \ \ (\text{total, referred to stator}) \end{aligned}$$

External resistance referred to stator $= 24.72 - 1 = 23.72\ \Omega$.

Referring to the rotor (stator : rotor turns = 2 : 3, so $R_{rotor} = R' \times (3/2)^2$):

$$R_{ext} = 23.72 \times 2.25 = 53.38\ \Omega \ \text{per phase}$$

New initial braking torque:

$$\begin{aligned} \frac{T_b'}{T_{fl}} &= \frac{(1.5 I_{fl})^2 (24.72/1.95)}{I_{fl}^2 (1/0.05)} = \frac{2.25 \times 12.68}{20} = 1.43 \end{aligned}$$

Answer (ii): insert 53.4 Ω per phase in the rotor (23.7 Ω referred to stator); initial braking torque = 1.43 × full-load torque.

Adding rotor resistance reduces the current but raises the torque, because the rotor power factor improves at high slip.

Data: $V_s = 220$ V, $V_m = \sqrt2 \times 220 = 311.13$ V, $R_f = 250\ \Omega$, $R_a = 0.2\ \Omega$, $T_L = 150$ N·m at 1500 rpm, $K_v = K_t = 0.8$.

For a single-phase semi-converter (continuous current): $V_o = \dfrac{V_m}{\pi}(1+\cos\alpha)$.

$$\omega = \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s}$$

i) Field current

With $\alpha_f = 0$:

$$\begin{aligned} V_f &= \frac{V_m}{\pi}(1+1) = \frac{2\times311.13}{\pi} = 198.07\ \text{V} \\ I_f &= \frac{V_f}{R_f} = \frac{198.07}{250} = 0.792\ \text{A} \end{aligned}$$

ii) Firing angle of armature converter

$$\begin{aligned} T &= K_t I_f I_a \Rightarrow I_a = \frac{150}{0.8 \times 0.792} = 236.66\ \text{A} \\ E_b &= K_v I_f \omega = 0.8 \times 0.792 \times 157.08 = 99.56\ \text{V} \\ V_a &= E_b + I_a R_a = 99.56 + 236.66\times0.2 = 146.89\ \text{V} \\ 1 + \cos\alpha_a &= \frac{\pi V_a}{V_m} = \frac{\pi\times146.89}{311.13} = 1.4832 \\ \cos\alpha_a &= 0.4832 \Rightarrow \alpha_a = 61.10^\circ \end{aligned}$$

iii) Input power factor of armature converter

For a semi-converter with ripple-free current $I_a$, the supply current is a quasi-square wave of height $I_a$ flowing for $(\pi - \alpha)$ in each half cycle:

$$I_s = I_a\sqrt{\frac{\pi-\alpha}{\pi}}$$

Neglecting losses, input power = output power $= V_a I_a$:

$$\begin{aligned} PF &= \frac{V_a I_a}{V_s I_s} = \frac{\sqrt2(1+\cos\alpha)}{\sqrt{\pi(\pi-\alpha)}} \\ &= \frac{\sqrt2 \times 1.4832}{\sqrt{\pi \times (\pi - 1.0664)}} = 0.822\ \text{(lagging)} \end{aligned}$$

($\alpha = 61.10^\circ = 1.0664$ rad.)

Answer: $I_f$ = 0.792 A, $\alpha_a$ = 61.1°, PF = 0.82 lagging.

Speed of an induction motor is $N = N_s(1-s) = \dfrac{120f}{P}(1-s)$. So speed can be changed by changing the slip $s$ (at fixed $N_s$), the supply frequency $f$, or the number of poles $P$. Slip is changed by stator voltage, rotor resistance or slip-power injection.

1. Stator voltage control

• AC voltage controller (anti-parallel thyristors) reduces the rms stator voltage at fixed frequency. Since $T \propto V^2$, the torque curve shrinks and the operating point moves to higher slip.
• Simple and cheap; used for fans and pumps ($T \propto N^2$).
• Narrow range, poor efficiency at low speed (rotor loss $= sP_g$), harmonics.

2. Pole changing

• Stator winding reconnected (e.g. Dahlander) or two separate windings, giving 2–4 fixed speeds. Only for squirrel cage motors; speed changes in steps.

3. Variable frequency (V/f) control

• An inverter supplies variable voltage and frequency, keeping $V/f$ constant so that air-gap flux stays constant.
• Wide, smooth speed range above and below base speed, high efficiency, good starting torque; most common modern method (VSI/PWM or CSI drives).

 3-ph ac -> [Rectifier] -> [DC link L/C] -> [Inverter] -> IM
                                 V/f controller ^

4. Rotor resistance control (slip ring motor)

• External resistance in rotor circuit raises slip at a given torque; static version uses a diode bridge, chopper and one resistor.
• Simple, high starting torque, but slip power is wasted as heat.

 rotor -> [diode bridge] -> L -> [chopper || R]

5. Slip power recovery (slip ring motor)

• Static Kramer drive: rotor power rectified, inverted and fed back to the supply; sub-synchronous speeds only.
• Static Scherbius drive: cycloconverter or back-to-back converter in rotor; power flows both ways, giving sub- and super-synchronous speeds.
• Efficient; used for large pumps, fans, compressors.

6. Vector (field oriented) control

• Stator current split into flux and torque components and controlled separately, so the induction motor behaves like a separately excited dc motor; fast response for high-performance drives.

Four-quadrant operation means the dc motor can work with either direction of speed and either direction of torque. The speed–torque plane has four quadrants, and a four-quadrant drive can run the motor forward or reverse and can both drive (motor) and brake in each direction.

            +speed
   II: Fwd braking  |  I: Fwd motoring
   T<0, w>0         |  T>0, w>0
   ------------------+------------------ torque
   III: Rev motoring |  IV: Rev braking
   T<0, w<0         |  T>0, w<0
            -speed

How it gives motoring and braking

For a separately excited motor, speed sets the sign of back emf $E$ and torque sets the sign of armature current $I_a$ ($T = K I_a$).

• Motoring: $E$ and $I_a$ have the same sign, so $E I_a > 0$ and power flows from supply to load.
• Braking (regenerative): the converter makes the terminal voltage less than $E$ (or reverses current), so $I_a$ reverses while speed is the same direction. Now $E I_a < 0$; the motor acts as a generator and kinetic or potential energy is returned to the supply (or burnt in a resistor in dynamic braking).

A four-quadrant converter (dual converter, or four-quadrant H-bridge chopper) can reverse both the voltage and the current, so a hoist, for example, raises the load in quadrant I and lowers it under control in quadrant IV, and a reversing mill can brake quickly and reverse.

Raising the load

Power needed to lift the load:

$$P_{load} = mgv = 400 \times 9.81 \times 2.5 = 9810\ \text{W}$$

Motor output (hoist efficiency 80 %) and input (motor efficiency 85 %):

$$\begin{aligned} P_{out} &= \frac{9810}{0.80} = 12262.5\ \text{W} \\ P_{in} &= \frac{12262.5}{0.85} = 14426.5\ \text{W} \\ I &= \frac{P_{in}}{V} = \frac{14426.5}{220} = 65.57\ \text{A} \end{aligned}$$

Lowering at 2.5 m/s by rheostatic braking

When lowering, the falling load drives the motor, which runs as a generator disconnected from the supply with a resistor across its armature (field stays excited). Power now flows the other way, so each efficiency reduces it:

$$\begin{aligned} \text{Shaft power to machine} &= 9810 \times 0.80 = 7848\ \text{W} \\ \text{Electrical power generated} &= 7848 \times 0.85 = 6670.8\ \text{W} \end{aligned}$$

The speed is the same as when raising and the field is unchanged, so the generated emf is taken as about the supply voltage, $E \approx 220$ V (armature resistance not given; its drop is neglected). All this power is dissipated in the braking resistance:

$$\begin{aligned} I_b &= \frac{6670.8}{220} = 30.32\ \text{A} \\ R &= \frac{E^2}{P} = \frac{220^2}{6670.8} = 7.26\ \Omega \end{aligned}$$

Answer: Supply current while raising = 65.6 A; braking resistance for lowering at 2.5 m/s ≈ 7.26 Ω.

(If the armature resistance $R_a$ were known, the external resistance to add would be $7.26 - R_a$.)

Data: $V = 200$ V, $N = 875$ rpm, $I_a = 150$ A, $R_a = 0.06\ \Omega$; supply 220 V, 50 Hz, $V_m = 311.13$ V.

For a single-phase full converter (continuous conduction):

$$V_a = \frac{2V_m}{\pi}\cos\alpha = 198.07\cos\alpha$$

Motor constant

At rated condition:

$$\begin{aligned} E &= V - I_a R_a = 200 - 150\times0.06 = 191\ \text{V} \\ K &= \frac{E}{N} = \frac{191}{875} = 0.21829\ \text{V/rpm} \end{aligned}$$

Rated torque means rated current, so $I_a R_a = 9$ V in every part.

i) Rated torque, 750 rpm

$$\begin{aligned} E &= 0.21829 \times 750 = 163.71\ \text{V} \\ V_a &= 163.71 + 9 = 172.71\ \text{V} \\ \cos\alpha &= \frac{172.71}{198.07} = 0.8720 \Rightarrow \alpha = 29.31^\circ \end{aligned}$$

ii) Rated torque, −500 rpm

Speed reversed, torque positive: regenerative braking in quadrant IV (e.g. an overhauling load).

$$\begin{aligned} E &= 0.21829 \times (-500) = -109.14\ \text{V} \\ V_a &= -109.14 + 9 = -100.14\ \text{V} \\ \cos\alpha &= \frac{-100.14}{198.07} = -0.5056 \Rightarrow \alpha = 120.37^\circ \end{aligned}$$

iii) Speed at $\alpha = 160^\circ$ and rated torque

$$\begin{aligned} V_a &= 198.07\cos160^\circ = -186.12\ \text{V} \\ E &= V_a - I_a R_a = -186.12 - 9 = -195.12\ \text{V} \\ N &= \frac{-195.12}{0.21829} = -893.9\ \text{rpm} \end{aligned}$$

Answer: (i) 29.3°, (ii) 120.4°, (iii) −894 rpm (reverse direction, regenerative braking).

The Ward-Leonard system controls the speed of a separately excited dc motor by varying its armature voltage, which is obtained from a separately excited dc generator driven at constant speed by an induction (or synchronous) motor. The motor field is kept constant.

Diagram

 3-ph   +------+  shaft  +-----+      +-----+
 ac ----|  IM  |=========|  G  |------|  M  |===> load
        +------+         +-----+      +-----+
                           |            |
                        gen. field   motor field
                        (variable,   (constant,
                         reversible)  from exciter)
           exciter --[rheostat + reversing switch]

• M–G set: induction motor (IM) drives the dc generator G at constant speed.
• The generator armature is connected directly to the motor M armature.
• Generator field is fed from an exciter through a potential divider and reversing switch.

Working

1. Generator voltage $V_g \propto \phi_g$ (speed constant). Changing the generator field current changes $V_g$ smoothly from zero to rated value.
2. Motor speed $N \propto \dfrac{V_g - I_a R}{\phi_m}$, so speed rises from zero to base speed as $V_g$ is raised (constant torque region).
3. Reversing the generator field reverses $V_g$, so the motor reverses without switching the armature circuit.
4. Above base speed, the motor field can be weakened (constant power region).
5. Regenerative braking: when $V_g$ is reduced below the motor emf, current reverses; the motor acts as a generator, drives G as a motor, and G drives the IM above synchronous speed, returning energy to the supply. Hence four-quadrant operation is possible.

Advantages

• Very wide and smooth speed control (about 1 : 10 or more) in both directions.
• Speed from zero upward; good starting, no starting resistance.
• Inherent regenerative braking and four-quadrant operation.
• Speed regulation good; small control currents (field circuits only).
• With a synchronous motor or flywheel (Ilgner set), power factor and load peaks on supply are improved.

Disadvantages

• Three machines of nearly the same rating: high cost, large floor space and heavy foundation.
• Low overall efficiency (about 60–70 %), especially at light load.
• Noise, maintenance of commutators and brushes.
• Slower response than static drives.

It is used for rolling mills, elevators, mine winders, paper machines and cranes; today it is largely replaced by the static Ward-Leonard (thyristor converter) drive.

Data: $V = 220$ V, $P = 20$ kW, $N = 1200$ rpm, $R_a = 0.1\ \Omega$, $\eta = 88\ \%$. Shunt field current is not given, so it is neglected ($I_a \approx I_L$).

Rated current and back emf

$$\begin{aligned} I_a &= \frac{P}{\eta V} = \frac{20000}{0.88\times220} = 103.31\ \text{A} \\ E_b &= V - I_a R_a = 220 - 10.33 = 209.67\ \text{V} \\ \omega &= \frac{2\pi\times1200}{60} = 125.66\ \text{rad/s} \\ K\phi &= \frac{E_b}{\omega} = \frac{209.67}{125.66} = 1.6685\ \text{V s/rad} \end{aligned}$$

i) Series resistance for $I_b = 2I_a$

In reverse current braking (plugging) the armature is reversed, so the supply voltage and back emf add:

$$\begin{aligned} R_a + R &= \frac{V + E_b}{2I_a} = \frac{220 + 209.67}{206.61} = 2.0796\ \Omega \\ R &= 2.0796 - 0.1 = 1.98\ \Omega \end{aligned}$$

ii) Initial braking torque

$$T_b = K\phi\, I_b = 1.6685 \times 206.61 = 344.7\ \text{N·m}$$

iii) Torque at 400 rpm

$$\begin{aligned} E &= 209.67\times\frac{400}{1200} = 69.89\ \text{V} \\ I &= \frac{220 + 69.89}{2.0796} = 139.40\ \text{A} \\ T &= 1.6685 \times 139.40 = 232.6\ \text{N·m} \end{aligned}$$

Answer: (i) 1.98 Ω, (ii) 344.7 N·m, (iii) 232.6 N·m.

The braking torque falls as the motor slows, but does not become zero at standstill ($I = 220/2.0796 = 105.8$ A), so the supply must be switched off at zero speed or the motor will start in reverse.

Data: $V_L = 400$ V, $R_a = 0.2\ \Omega$, $I_{rated} = 100$ A, $K = 0.25$ V/rpm.

For a three-phase fully controlled bridge (continuous conduction):

$$V_a = \frac{3\sqrt2 V_L}{\pi}\cos\alpha = \frac{3\sqrt2\times400}{\pi}\cos\alpha = 540.19\cos\alpha$$

No-load speed at $\alpha = 45^\circ$, $I_a = 5$ A

$$\begin{aligned} V_a &= 540.19 \cos45^\circ = 381.97\ \text{V} \\ E &= V_a - I_a R_a = 381.97 - 5\times0.2 = 380.97\ \text{V} \\ N_0 &= \frac{E}{K} = \frac{380.97}{0.25} = 1523.9\ \text{rpm} \end{aligned}$$

Firing angle for 1500 rpm at rated current

$$\begin{aligned} E &= 0.25 \times 1500 = 375\ \text{V} \\ V_a &= 375 + 100\times0.2 = 395\ \text{V} \\ \cos\alpha &= \frac{395}{540.19} = 0.7312 \Rightarrow \alpha = 43.01^\circ \end{aligned}$$

Answer: No-load speed = 1523.9 rpm; firing angle for 1500 rpm at 100 A = 43.0°.

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

 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 V 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:

$$V_i = -1.35\, V_L' \cos\alpha$$

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: converters are rated only for the slip power (e.g. 30 % range needs about 30 % rating), so it is economical for limited speed ranges on large pumps, fans and compressors. Drawbacks: poor power factor (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 power can flow both ways, i.e. the static Scherbius drive (cycloconverter or two back-to-back thyristor/PWM converters, as in a doubly fed induction machine).

Data: $V = 220$ V, $N = 535$ rpm, $R_a = 0.086\ \Omega$, $I_a = 150$ A (full load), limit $I_b = 200$ A.

Back emf at full-load speed

$$\begin{aligned} E_b &= V - I_a R_a = 220 - 150\times0.086 = 207.1\ \text{V} \\ \omega &= \frac{2\pi\times535}{60} = 56.03\ \text{rad/s} \\ K\phi &= \frac{207.1}{56.03} = 3.6966\ \text{V s/rad} \end{aligned}$$

Series resistance for plugging

At the instant of plugging, the reversed supply voltage and the back emf act in the same direction round the armature circuit:

$$\begin{aligned} R_a + R &= \frac{V + E_b}{I_b} = \frac{220 + 207.1}{200} = 2.1355\ \Omega \\ R &= 2.1355 - 0.086 = 2.05\ \Omega \end{aligned}$$

Initial braking torque

$$T_{b0} = K\phi\, I_b = 3.6966 \times 200 = 739.3\ \text{N·m}$$

Torque at half speed (267.5 rpm)

$$\begin{aligned} E &= \frac{207.1}{2} = 103.55\ \text{V} \\ I &= \frac{220 + 103.55}{2.1355} = 151.51\ \text{A} \\ T &= 3.6966 \times 151.51 = 560.1\ \text{N·m} \end{aligned}$$

Answer: series resistance = 2.05 Ω; initial braking torque = 739.3 N·m; torque at half speed = 560.1 N·m.

(Full-load torque for comparison $= 3.6966 \times 150 = 554.5$ N·m, so braking starts at 1.33 × full-load torque.)

In variable frequency control, the stator supply frequency is changed by an inverter so that the synchronous speed $N_s = 120f/P$ changes, and the motor speed changes with it at small slip. To keep the flux constant, the voltage is changed in proportion to frequency (V/f control).

Why voltage must change with frequency

Stator induced emf: $E_1 = 4.44\, f\, N_1 \phi\, k_w$, so

$$\phi \propto \frac{E_1}{f} \approx \frac{V}{f}$$

• If $f$ is reduced at constant $V$, flux rises, the core saturates and magnetising current becomes very large.
• If $f$ is raised at constant $V$, flux falls and torque capability falls. Hence below base frequency $V/f$ is kept constant.

Torque relations

Neglecting stator impedance, slip speed $\omega_{sl} = s\omega_s$:

$$T = \frac{3}{\omega_s}\frac{V^2 (R_2'/s)}{(R_2'/s)^2 + X_2'^2}$$

With $X_2' \propto f$, the maximum torque is

$$T_{max} = \frac{3}{2\omega_s}\frac{V^2}{X_2'} \propto \left(\frac{V}{f}\right)^2$$

So with constant $V/f$, $T_{max}$ is constant at all frequencies, and the slip speed at $T_{max}$, $s_m\omega_s = R_2'\omega_s/X_2'$, is constant. The speed–torque curves simply shift parallel to each other.

Operating regions

 V, T, P
  |        ____________ V (rated)
  |      /|
  |    /  |  T falls ~1/f
  |  /    |____________ P constant
  |/______|____________ T constant
  +-------+----------> f
       base f (50 Hz)
 constant torque | constant power
 (V/f const)     | (V fixed, field weak)

• Below base speed: $V/f$ constant, constant torque region.
• Above base speed: voltage cannot exceed rated, so $V$ is fixed and $f$ increased; flux falls ($\phi \propto 1/f$), giving the constant power (field weakening) region with $T_{max} \propto 1/f^2$.
• At very low frequency the stator $I_1R_1$ drop is significant, so a voltage boost is added to keep the flux.

Converter

 ac -> [rectifier] -> [dc link] -> [PWM inverter] -> IM
                         V/f reference and firing logic

A PWM voltage source inverter (or a cycloconverter for very low speeds) supplies variable voltage and frequency.

Advantages: wide speed range above and below synchronous speed, high efficiency (low slip), smooth starting with full torque at low current, regenerative braking possible; suitable for cage motors. Drawbacks: costly power electronics and harmonics.