EE207 Electrical PowerLecture 7 Power Electronics INTRODUCTION • Electronic system are used extensively in power systems and controls. • Basic operations of diodes and thyristors that are used to convert DC to AC and vice versa. • In power electronics , these devices acts as high speed switches. Rajparthiban Kumar EE207 Electrical Power 2 1 Diodes and diode circuits • A diode is an electronic device possessing two terminals anode (A) and cathode (K). it acts like a high speed switch whose contact open and close. Open circuit E =0 Rajparthiban Kumar EE207 Electrical Power 4 2 . it acts as an open circuit. It has no moving parts. A K E Rajparthiban Kumar EE207 Electrical Power 3 Rules of operation • 1)When no voltage is applied across the diode. the terminal short-circuits. the diodes remains an open circuit.7V or more is applied across the terminal. anode is negative compared to cathode. This is called forward bias. This is called reversed bias. anode more slightly positive compared to cathode. I E0 + Rajparthiban Kumar EE207 Electrical Power 5 • 3) If an inverse voltage is applied to the diode. E + Rajparthiban Kumar EE207 Electrical Power 6 3 . and a current I flows through it.• 2) If a forward voltage of 0. It acts like a closed switch. • Maximum temperature. PIV.Main characteristics of diode • Peak inverse voltage. Use heat sinks or cooling systems. • • • • • Battery charger with series resistor Battery charger with series inductor Single phase bridge rectifier Filters Others EE207 Electrical Power 8 Rajparthiban Kumar 4 . limit to the average current a diode can carry. • Maximum average current. Depends on construction. otherwise it will be destroyed. power dissipated as heat. can never exceed permissible limit. We will analyse the following circuits to illustrate the functions of diodes and also methods used in power technology. size and assembly. Rajparthiban Kumar EE207 Electrical Power 7 Application of diodes • Diodes have many application in electronic power circuits. Maximum inverse voltage before it breaks down. 60Hz 100V D 4 R=1 3 60V 1 Rajparthiban Kumar EE207 Electrical Power 9 Waveforms V 100 60 4 3 2 2 4 I 1 t 2 4 3 Rajparthiban Kumar EE207 Electrical Power 10 5 .Battery charger with series resistor 2 120V. Rajparthiban Kumar EE207 Electrical Power 11 • Resistor produces losses and are inefficient. 4 follows 3. • 4 is always more than 60V and 3 is always at 60V with respect to reference. when the switch is closed. Since no current flows through the circuit. 60Hz 100V 3 60V 1 Rajparthiban Kumar EE207 Electrical Power 12 4 D L= 3. • Positive current flows into the battery will charge it up. • Peak current is at 40A.3mH 6 .• Circuit analysis:• Diode is in forward bias only when potential at 2 is more than 4. 2 120V. Use inductor instead. • When the diode opens. Maximum current is 40A. • When the diode is open. current increases gradually. • The voltage accumulates. Rajparthiban Kumar EE207 Electrical Power 14 7 .V 2 4 3 2 4 I 1 t 2 4 3 Rajparthiban Kumar EE207 Electrical Power 13 • Circuit analysis:• Diode begins to conduct the same ways as the previous circuit. the inductor discharges. Single phase bridge rectifier • The circuit enables us to rectify both positive and negative half cycles of an ac source.9 Em. rms ia ib ia Rajparthiban Kumar EE207 Electrical Power 16 8 . to supply DC power to the load. G = Em ia ib I ib Rajparthiban Kumar R ia EE207 Electrical Power 15 Em Ed=0. Em Ed=0. so tends to maintain a constant power flow. • The purpose of this DC filters are to produce smooth power flow into a load. filters are used to smooth out the waveforms. and release the energy when voltage or current falls. G = Em ia ib I ib L R ia Rajparthiban Kumar EE207 Electrical Power 17 • The filter must be able to absord the energy whenever the dc voltage or current rises. rms ia ib Rajparthiban Kumar EE207 Electrical Power 18 9 .9 Em.Filters • To reduce the pulsation in voltage and current. Em R I C Rajparthiban Kumar EE207 Electrical Power 19 • The peak to peak ripple in percent is given by:- ripple = 5. and maintain a constant voltage.• Capacitor can be placed parallel to the load. If 60Hz source is used. The peak to peak current ripple is 10%. calculate: • The inductance of the inductor and peak to peak current ripple Rajparthiban Kumar EE207 Electrical Power 20 10 .5 P fWL • We wish to build a 135 V. 20A dc supply using a single phase bridge rectifier using an inductive filter. P = VI = 135(20) = 2700W 5.5 P 5.1× 20 = 2 A Fluctuates from 19A to 21A Rajparthiban Kumar EE207 Electrical Power 21 Three phase – 3pulse diode rectifier • Composed of 3 diodes connected in series with secondary windings of a three phase transformer • Em is the line-toneutral peak voltage 1 D1 I1 Em 2 D2 I2 D3 3 R Id Rajparthiban Kumar EE207 Electrical Power 22 I3 L 11 .75 J → L = 0.5 × 2700 WL = = = 24.124 H 2 2 I peak − peak = 0.75 J f × ripple 60(10) 1 1 W = LI 2 = L(20) 2 = 24. • For example if a 100KVA transformer is used. and the sudden jumps produce rapid fluctuations in the magnetic field. 6 pulse rectifier is developed.1 2 3 Ed=0. • Observe that the line current flow intermittently. it can deliver only 74kW without overheating.827Em i1 i2 i3 Rajparthiban Kumar EE207 Electrical Power 23 • Current flow one-third of the time in given winding. Rajparthiban Kumar EE207 Electrical Power 24 12 . which can cause noise. • To overcome the drawbacks. 654Em -0.6 pulse bridge rectifier Ia D1 Ib I1 I2 D2 I3 I5 D5 D3 L T Ic D4 I4 D6 I6 R Rajparthiban Kumar EE207 Electrical Power 25 1.35E 0 1.225 E Ed=1.5 Ed= 1.414 E 1 0.5 -1 0 60 120 180 240 300 360 420 480 540 600 660 i1 i6 Rajparthiban Kumar i2 i4 i3 i5 EE207 Electrical Power 26 13 . If a 600V.35 Rajparthiban Kumar EE207 Electrical Power 28 14 .35 Erms ripple = 0.17 P fWL 27 Rajparthiban Kumar EE207 Electrical Power Example • A 3 phase bridge rectifier has to supply power 360kW. and if a transformer of rating 100kVA is used.• Current flows two-third of the time.35 1. 60 Hz feeder is available. 240 V dc load. it can deliver 95kW. calculate • a) secondary line voltage E= Ed 240 = = 177V 1. Ed = 1. 3 phase. 225(177) = 217V Emax = 1.414 E Emin = 1.225E and 1.414(177) = 250V E peak −to − peak = 250 − 217 = 33V Rajparthiban Kumar EE207 Electrical Power 30 15 .• b) DC current per diode 360kW = 1500 A 240 dc current per diode = 1500/3 = 500A Id = Rajparthiban Kumar EE207 Electrical Power 29 • C) peak to peak output voltage – From graph the voltage fluctuates btw 1. • The effective current composed of fundamental component plus all the harmonics. • I=0. EE207 Electrical Power 31 Rajparthiban Kumar Harmonic content and THD • The rectangular current wave occurs very frequently in power electronics.816 Id and since IF=0.Fundamental line current IF • The line current have amplitude of Id. Total harmonic distortion = • THD = IH IF 32 Rajparthiban Kumar EE207 Electrical Power 16 . and are rectangular waves.955. 2 2 I 2 = IF + IH I F = rms of the fundamental line current I H = rms value of all the harmonics component combined I = rms value of the line current • The degree of distortion of an ac voltage or current is defined as the ratio of rms value of all the harmonic component divided by the rms value of the fundamental component.78 Id • • IF = 0.955I Power factor is 0. but no reactive power is absorbed because the current and voltage is still in phase. 6 pulse rectifier furnishes a dc current of 400A to the load.816 I d = 0. Estimate:• The rms value of line current I = 0. IF divided by the order of the harmonics.816(400) = 326 A Rajparthiban Kumar EE207 Electrical Power 34 17 . Rajparthiban Kumar EE207 Electrical Power 33 Example • The 3 phase.• Amplitudes of the harmonics components are equal to the amplitude of the fundamental. • The rms value of fundamental I F = 0.955I = 0.955(326) = 311A • The rms value of the 7th and 11th harmonic combined 311 IF = = 44 A 7 7 I 311 I H 11 = F = = 28 A 11 11 2 2 2 I H 7 + H 11 = I H 7 + I H 11 IH7 = H7 Rajparthiban Kumar+ H 11 I = 52 A Electrical Power EE207 35 18 .