Dosen Pembina MK : I Nyoman Wahyu Satiawan, ST, M.Sc. Ph.D NIP. 197009081998021001 Jurusan Teknik Elektro Fakultas Teknik Universitas Mataram Normal rectifiers are considered as uncontrolled rectifiers. Once the source and load parameters are established, the dc level of the output and power transferred to the load are fixed quantities. A way to control the output is to use SCR instead of diode. Two condition must be met before SCR can conduct: ◦ The SCR must be forward biased (VSCR>0) ◦ Current must be applied to the gate of SCR A phase-controlled thyristor is turned on by applying a short pulse to its gate, and is turned off due to natural or line commutation. They are used extensively in industrial applications, especially in variable speed drives. 1. dc motor speed control systems, widely used in steel mills, paper mills, and such. 2. Electrochemical and electro-metallurgical processes. 3. Magnet power supplies. 4. Converters at the input end of dc transmission lines. 5. Portable hand tool drives 3 The phase-control converters are classified as: ◦ Single-phase converters ◦ Three-phase converters Each type is divided into: ◦ Semiconverter ◦ Full converter ◦ Dual converter 4 A semiconverter is a one-quadrant converter and it has one polarity of output voltage and current. A full converter is a two-quadrant converter and the polarity of its output voltage can be either positive or negative. A dual converter can operate in four quadrants, and both the output voltage and the output current can be either positive or negative. 5 Operation Principle Consider the following circuit, where α is the delay or firing (ignition) angle. 6 7 If the frequency of the supply is fs, the lowest frequency of the output ripple voltage will be fs. Now: Vm 1 Vdc Vm sin t d (t ) (1 cos ) 2 2 The dc voltage can be varied from Vm/π to 0 by varying from 0 to π. The average output voltage will be maximum when α is 0. 8 The rms output voltage is: 1 2 2 Vrms [ Vm sin t d (t )]1 / 2 2 Vrms Vm 1 sin 2 1 / 2 [ ( )] 2 2 The rms load current is : I rms Vrms R The average power absorbed by load (resistor) : 9 Design a circuit to produce an average voltage of 40V across 100 load resistor from a 120Vrms 60 Hz ac source. Determine the power absorbed by the resistor and the power factor. Briefly describe what happen if the circuit is replaced by diode to produce the same average output. Solution In such that to achieved 40V average voltage, the delay angle must be Vs [1 cos ] 2 120 2 40 [1 cos ] 2 61.2o 1.07 rad Vo V 2 rms 75.6 2 P 57.1W R 100 Vo , rms Vm sin( 2 ) 1 2 2 120 2 1.07 sin 2(1.07) 1 2 2 75.6V pf 57.1 0.63 75.6 (120) 100 Solution •If an uncontrolled diode is used, the average voltage would be Vo Vs 2 (120) 54V • That means, some reducing average resistor to the design must be made. • A series resistor or inductor could be added to an uncontrolled rectifier, while controlled rectifier has advantage of not altering the load or introducing the losses •The analysis of the circuit is very much similar to that of uncontrolled rectifier. t V m for t [sin( wt ) sin( )e i ( wt ) Z 0 otherwise and L L Z R 2 (L) 2 , tan 1 , R R rms current , I rms 1 2 2 1 2 i ( t ) d ( t ) i (t ) d (t ) 2 2 and , average current , 1 Io i (t ) d (t ) 2 The average output voltage, Vm [cos cos ] 1 Vm sin( t )dt Vo 2 2 The average power absorbed by load , 2 P I rms R ; The circuit for a single-phase full converter is shown below. The load is assumed to be highly inductive, so the load current is continuous and ripple free. The converter operates in the rectification and inversion modes. 16 17 The average output voltage is: 2 Vdc 2 Vm sin t d (t ) 2Vm cos The dc voltage can be varied from 2Vm/π to -2Vm/π by varying α from 0 to π. 18 The average output voltage will be maximum when α is 0. The rms output voltage is: Vrms 2 2 2 [ Vm sin t d (t )]1/ 2 2 Vrms Vm Vs 2 19 The operation of the converter can be divided into two identical modes. Mode 1 is when T1 and T2 conduct, and mode 2 is when T3 and T4 conduct. 20 Summary of some important points in analysis When analyzing a thyristor circuit, start from a diode circuit with the same topology. The behavior of the diode circuit is exactly the same as the thyristor circuit when firing angle is 0. Take different principle when dealing with different load – For resistive load: current waveform of a resistor is the same as the voltage waveform –For inductive load with a large inductor: the inductor current can be considered constant I Nyoman Wahyu Satiawan FT UNRAM Beroperasi dgn sumber AC 3 –fase. Dapat menghasilkan tegangan keluaran DC yang lebih besar. Menghasilkan daya DC yang lebih besar. Frekwensi riple yang lebih tinggi pada tegangan keluaran. Memerlukan filter yang lebih sederhana untuk menghaluskan tegangan dan arus beban. Digunakan secara luas pada high power variable speed industrial dc drives. # tegangan fase ke netral di definisikan sbb : vRN van Vm sin t ; Vm Max. Phase Voltage vYN 2 vbn Vm sin t 3 Vm sin t 1200 vBN 2 vcn Vm sin t 3 Vm sin t 1200 Vm sin t 2400 Tegangan keluaran vo merupakan penjumlahan dari tegangan keluaran dari masing-masing fase T1 di triger pada t (30 ) 6 5 T2 di triger pada t (150 ) 6 7 T3 di triger pada t (270 ) 6 2 Setiap Thyristor konduksi pada 120 atau radian 3 Jika tegangan fase ref. adalah vRN = van = Vm sin ωt, maka tegangan keluaran DC atau tegangan rata2 untuk arus beban kontinyu dihitung dengan : 56 3 Vdc V sin td t m 2 6 56 3Vm Vdc sin td t 2 6 5 3Vm 6 ( cos t ) Vdc 2 6 Vdc 3Vm 2 5 cos cos 6 6 Dengan menerapkan aturan trigonometri, Cos (A+B) = (Cos A. Cos B – Sin A. Sin B) , maka 3Vm Vdc 2 5 5 cos 6 cos( ) sin 6 sin( ) cos 6 cos( ) sin 6 sin( ) 3Vm cos(150 ) cos( ) sin( 150 ) sin( ) cos(30 ) cos( ) sin( 30 ) sin( ) 2 3Vm Vdc cos(180 30 ) cos( ) sin( 180 30 ) sin( ) cos(30 ) cos( ) sin( 30 ) sin( ) 2 Note : cos(180 30 ) cos(30 ) dan sin( ) cos(180 30 ) sin( 30 ) Vdc Vdc 3Vm cos(30 ) cos( ) sin( 30 ) sin( ) cos(30 ) cos( ) sin( 30 ) sin( ) 2 3Vm Vdc 2 cos 30 cos( ) 2 3Vm 3 Vdc cos( ) 2 x 2 2 3Vm 3 3 Vdc 3 cos( ) cos( ) 2 2 3V Vdc Lm 3 cos( ) 2 Dimana, VLm 3Vm tegangan masukan line to line #Tegangan keluaran DC atau tegangan rata-rata maks. pada sudut penyalaan α= 0 adalah Vdc(max) 3 3Vm Vdm 2 #Tegangan keluaran rata-rata yang dinormalisasi adalah Vdcn Vdc Vn cos Vdm # Nilai rms dari tegangan keluaran dicari dengan persamaan : Vo ( RMS ) 3 2 5 6 2 2 V sin t . d ( t ) m 6 1 2 Shg didapat, Vo ( RMS ) 1 3 3Vm cos 2 6 8 1 2 Van Vbn Vcn V0 =300 =30 0 0 30 0 60 90 0 0 120 0 150 0 180 0 210 Van 0 240 0 270 0 300 0 330 0 360 Vbn 0 390 =60 0 30 0 60 90 0 0 120 0 150 0 180 0 210 0 240 t 0 420 Vcn V0 0 0 0 270 0 300 0 330 0 360 0 390 0 420 0 =600 t Vbn Van Vcn V0 =900 =90 0 0 0 30 0 60 0 90 0 120 0 0 0 0 0 0 0 0 150 180 210 240 270 300 330 360 0 0 390 420 t Dikenal sebagai konverter / penyearah 6pulsa. Digunakan pada aplikasi dengan daya output tinggi s.d. 120kW. Dapat melakukan operasi dua kuadran. Thyristor2 di triger dgn interval /3. Frequency ripple tegangan keluaran 6fS. T1 ditriger pada t = (/6 + ), T6 sudah dlm keadaan konduksi ketika T1 ON. Pada interval (/6 + ) sampai (/2 + ), T1 dan T6 konduksi bersamaan & tegangan keluaran sama dengan vab = (van – vbn) T2 di trigger pada t = (/2 + ), dan T6 OFF secara alamiah ketika dlm kondisi as it is reverse biased ketika T2 di trigger. Selama interval (/2 + ) sampe (5/6 + ), T1 dan T2 konduksi bersama & tegangan pada beban vO = vac = (van – vcn) Thyristor2 di berikan nomor sesuai dengan urutan dimana mereka di triger. Urutan triger dari thyristor adalah 12, 23, 34, 45, 56, 61, 12, 23, 34, ……… Kita definisikan tegangan 3 fase (tegangan fase ke netral) sbb; vRN van Vm sin t ; 2 ) Vm sin( t 120) 3 2 vBN vcn Vm sin( t ) Vm sin( t 120) 3 .......................................... Vm sin( t 240) vYN vbn Vm sin( t Vm adalah teg angan fase puncak untuk beban yang terhubung bintang Tegangan antar fase dari sumber adalah; vRY vab (van vbn ) 3Vm sin( t ); 6 vYB vbc (vbn vcn ) 3Vm sin( t ); 2 vBR vca (vcn van ) 3Vm sin( t ); 2 Tegangan keluaran pada beban terdiri dari 6 pulsa tegangan pada setiap periode 2 radians, sehingga tegangan keluaran rata-rata dihitung dengan : VO dc 6 Vdc 2 2 6 vO .d t ; vO vab 3Vm sin t 6 2 3 Vdc 3Vm sin( t )t 6 6 Vdc 3 3Vm cos dimana VmL 3 Vm adalah teg angan antar fase masukan Tegangan keluaran DC rata-rata diperoleh pada sudut penyalaan α= 0 Vdc(max) Vdc 3 3Vm 3VmL Tegangan keluaran DC rata-rata yang dinormalisasi adalah Vdcn Vdc 3 3Vm 3VmL Nilai rms tegangan keluaran adalah Vo ( rms) 6 2 V d t o 6 2 1 2 VO rms VO rms VO rms 6 2 3 2 2 v . d t ab 6 2 1 2 2 2 3 V sin t . d t m 6 6 2 1 3 3 3Vm cos 2 2 4 1 2 1 2 ECE 442 47 1 Vdc T T f (t )dt 0 2 Vdc 2 6 Vdc 3 3 6 3Vm cos td (t ) 0 Vm 1.654Vm Vrms 2 2 6 6 2 2 0 3Vm cos td (t ) 1 2 Vrms Vrms 3 9 3 Vm 2 4 1.6554Vm 1 2 3Vm Im peak R 6 4 2 2 Ir I m cos td (t ) 2 0 I r 0.5518 I m ECE 442 1 2 51