DC Drives Chapter 5 EET 428 Power Electronics

DC Drives Chapter 5 EET 428 – Power Electronics & Drive Semester 2 – 2017/2018

5. 1 Speed Control of Separately-excited & Shunt DC Motor �The general equation of separately-excited motor is: �For a given torque, the motor speed is a function of: 1) Resistance in armature circuit – when a resistance is inserted in the armature circuit, the speed drop, ∆ω increases & the motor speed decreases. 2) Terminal voltage (armature voltage) – reducing the armature voltage, Vt of the motor reduces the motor speed. 3) Field flux (or field voltage) – reducing the field voltage reduces the flux, φ & the motor speed increases. 2

5. 2 By Adding Resistance �An additional resistance can be added to the armature circuit as shown below. 3

5. 2 By Adding Resistance (cont’d) �Assuming the load torque is unidirectional & constant. The field & armature voltages are also constant. �Point 1 is when there is no additional resistance added into the armature. �If Radd 1 is added to the armature, the motor operates at point 2, where ω2 is: 4

5. 2 By Adding Resistance (cont’d) �By adding a resistance to an armature, the speed drop, ∆ω is increased. �If the added resistance keeps increasing, the motor will operates at point 4. �Point 4 is a holding operation where the motor speed is: 5

5. 2 By Adding Resistance (cont’d) �Example 1: A 150 V, DC shunt motor drives a constant torque load at a speed of 1200 rpm. The armature & field resistances are 1 Ω & 150 Ω respectively. The motor draws a line current of 10 A at the given load. The field current stays constant. a) b) c) Calculate the resistance that should be added to the armature circuit to reduce the speed by 50%. ( 7. 83 Ω) Assuming the rotational losses to be 100 W, calculate the efficiency of the motor without & with the added resistance. ( 77. 93% & 35. 65%) Calculate the resistance that must be added to the armature circuit to operate the motor at the holding condition. ( 15. 67 Ω) 6

5. 3 By Adjusting Armature Voltage �A basic configuration of controlling speed by adjusting the armature voltage is shown below. 7

5. 3 By Adjusting Armature Voltage (cont’d) �Assuming the field voltage & load torque are kept constant. �By adjusting the armature voltage, Vt it will not affect the speed drop, ∆ω but it will affect the no-load speed, ω0. �Hence, by adjusting Vt, the speed-torque characteristic is shifted in parallel. 8

5. 3 By Adjusting Armature Voltage (cont’d) �If the armature voltage keeps decreasing, the motor will operates at point 4. �Point 4 is a holding operation where the motor speed is: 9

5. 3 By Adjusting Armature Voltage (cont’d) �Example 2: For a 20% increase in the armature voltage, calculate the percentage change in the no-load speed, ω0 assuming the load torque & field voltage are kept constant for separately-excited DC motor. ( 20% increase) 10

5. 4 By Adjusting Field Voltage �A basic configurations of controlling speed by adjusting the field voltage are shown below. OR 11

5. 4 By Adjusting Field Voltage (cont’d) �Assuming the armature voltage & load torque are kept constant. �By adjusting the field voltage, Vf it will affect both the speed drop, ∆ω & no-load speed, ω0. �Considerations when adjusting the field voltage: 1) Vf must not exceed the absolute maximum rating. 2) Large reductions in field may result in excessive speed. 3) If the load is constant, large reductions in field may result in excessive Ia. 12

5. 4 By Adjusting Field Voltage (cont’d) �Example 3: For a 20% decrease in the field voltage, calculate the percentage change in the no-load speed, ω0 & speed drop, ∆ω assuming the load torque & armature voltage are kept constant for separately-excited DC motor. ( 25% increase) 13

5. 4 By Adjusting Field Voltage (cont’d) �Example 4: A 150 V, separately-excited DC motor drives a constant torque load at a speed of 1200 rpm. The armature & field resistances are 1 Ω & 150 Ω, respectively. The field voltage is the same as armature voltage (Vt). The armature current is 9 A. A resistance is added in the field circuit to reduce the field current by 20% while the torque is kept constant. Assume that the total current sourcing from the field voltage source is also kept constant. Calculate the new armature current, motor speed, value of the added shunt field resistance and its losses. ( Ia 2 = 11. 25 A; 23% increase) ( ω2 = 154. 57 rad/sec OR 1476. 1 rpm; 23% increase) ( Rf add = 600 Ω) ( Pf added losses = 37. 5 W) 14

5. 5 Single-phase, Half-wave Drives �A single-phase, half-wave drive is connected to the armature circuit of the motor as shown below. �The armature winding inductance, La needs to be included in the circuit due to variation of armature current (Ia is not constant). 15

5. 5 Single-phase, Half-wave Drives (cont’d) �Before the triggering of the SCR at α, the instantaneous voltage of Vt is equal to Ea assuming the motor is operating at steady-state. �During the SCR conduction, Vt is equal to the instantaneous source voltage, Vs. �The voltage across the resistive component of the armature winding is identical in shape to the instantaneous armature current. 16

5. 5 Single-phase, Half-wave Drives (cont’d) �The instantaneous terminal voltage is: where: TET 411 Semester 1 - 2012/2013 17

5. 5 Single-phase, Half-wave Drives (cont’d) �The instantaneous current is: where: 18

5. 5 Single-phase, Half-wave Drives (cont’d) �The speed is (taking the average value in one period): 19

5. 5 Single-phase, Half-wave Drives (cont’d) �Example 5: A 1 hp, shunt DC motor is loaded by a constant torque of 10 Nm. The armature resistance of the motor is 2 Ω & the field constant Kφ = 2. 5 Vsec. The motor is driven by a half-wave SCR converter. The power source is 240 Vrms, 50 Hz. The triggering angle of the converter is 30˚ & the conduction period is 90˚. Calculate the motor speed & the developed power delivered to the load. ( ω = 18. 833 rad/sec OR 179. 84 rpm) ( Pdev = 188. 33 W; 25% of motor rating) 20

5. 6 Single-phase, Full-wave Drives �A single-phase, full-wave drive is connected to the armature circuit of the motor as shown below. 21

5. 6 Single-phase, Full-wave Drives (cont’d) �The circuit consists of four (4) SCRs. The switching of SCRs is dependent on the polarity of the source voltage, VAC. �The current i 1 flows when the AC source voltage is in the positive half-cycle & SCRs S 1 & S 2 are triggered. �The current i 2 flows when the AC source voltage is in the negative half-cycle & SCRs S 3 & S 4 are triggered. 22

3. 6 Single-phase, Full-wave Drives (cont’d) �The waveforms of the circuit is shown below. �The speed is: 23

5. 6 Single-phase, Full-wave Drives (cont’d) �Example 6: A 1 hp, shunt DC motor is loaded by a constant torque of 10 Nm. The armature resistance of the motor is 2 Ω & the field constant Kφ = 2. 5 Vsec. The motor is driven by a full-wave SCR converter. The power source is 240 Vrms, 50 Hz. The triggering angle of the converter is 30˚ & the conduction period is 90˚. Calculate the motor speed & the developed power delivered to the load. ( ω = 25. 233 rad/sec OR 240. 96 rpm) ( Pdev = 252. 33 W; 33. 82% of motor rating) 24

5. 7 Continuous Armature Current �For heavily loaded motors, the conduction period may equal or exceed 180˚ which may result in an overlap of i 1 & i 2 & causing a continuous motor current. �The speed is: 25

5. 7 Continuous armature current (cont’d) �Example 7: A separately-exited DC motor has a constant torque of 10 Nm. The motor is driven by a full-wave converter connected to an AC supply of 240 Vrms. The field constant of the motor Kφ = 2. 5 Vsec & the armature resistance is 2 Ω. Calculate the triggering angle, α for the motor to operate at 200 rpm. The motor current is continuous. ( α = 73. 78°) 26