Power electronics, notes (11 to 12)_ part_3, Lecture notes of Power Electronics

Download Power electronics, notes (11 to 12)_ part_3 and more Lecture notes Power Electronics in PDF only on Docsity! These Slides are Prepared By: Professor Dr Md Abdur Razzak Head of the Department, EEE Independent University of Bangladesh Lecture – 11 Controlled Rectifier Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 1/33 EEE 332 / 413 (POWER ELECTRONICS AND DRIVES) Lesson Outcomes After completing this lecture, students will be able to  list the types and explain the operation & characteristics of controlled rectifiers.  calculate the performance parameters of controlled rectifiers.  analyze the design of controlled rectifier circuits.  evaluate the effects of load inductance on the load current. Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 4/33 Semiconverter If Vm is the peak input voltage, the average output voltage is The rms output voltage is given by        cos1 2 )(sin 2 1   m mdc V tdtVV           2 2sin1 2 sin 2 1 22       m mrms V ttdVV Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 5/33 Example: 11-1 Problem: If the converter shown in figure has a purely resistive load of R and the delay angle /2, determine (a) the rectification efficiency, (b) the form factor (FF), (c) the ripple factor (RF), (d) the peak inverse voltage (PIV) of thyristor T1. Solution:          2 cos1 2 )(sin 2 1 mm mdc VV tdtVV  R V I mdc 2  222 1 22 2sin1 2 mmm rms VVV V                    2 2 1 22         mmm dcdcdc V RR VV IVP R V I mrms 22  R V R VV IVP mmmrmsrmsac 82222 2  Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 6/33 Example: 11-1 (contd..) (a) (b) (c) (d) The peak inverse voltage (PIV) of thyristor T1, PIV = Vm 22.2 2222 FF factor, Form    mm dc rms VV V V %26.202026.0 4 8 ,efficiencyion Rectificat 2    ac dc P P   %19898.112.221FFRF factor, Ripple 22  Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 9/33 Example: 11-2 (contd..) Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 10/33 Example: 11-2 (contd..) Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 11/33 Example: 11-2 (contd..) Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 14/33 Dual converter (contd..) Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 15/33 Dual converter (contd..) For α1 = 0, only the converter 1 operates; for α1 = π, only the converter 2 operates. For 0  α1 < π/2, the converter 1 supplies a positive load current +io and thus the circulating current can only be positive. For π/2 < α1  π, the converter 2 supplies a negative load current -io and thus only a negative circulating current can flow. At α1 = π/2, the converter 1 supplies positive circulating during the first half-cycle, and the converter 2 supplies negative circulating during the second half-cycle. The instantaneous circulating current depends on the delay angle. For α1 = 0, its magnitude becomes minimum when ωt = nπ, n = 0, 2, 4…. and maximum when ωt = nπ, n = 1, 3, 5…. If the peak load current is Ip, one of the converters that controls the power flow may carry a peak current of (Ip + 4Vm/ωLr). Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 16/33 Dual converter (contd..) The dual converters can be operated with or without a circulating current. In case of operation without circulating current, only one converter operates at a time and carries the load current, and the other converter is completely blocked by inhibiting gate pulses. However, the operation with circulating current has the following advantages: 1. The circulating current maintains continuous conduction of both converters over the whole control range, independent of the load. 2. Because one converter always operates as a rectifier and the other converter operates as an inverter, the power flow in either direction at any time is possible. 3. Because both converters are in continuous conduction, the time response for changing from one quadrant operation to another is faster. Gating sequence. The gating sequence is as follows: Gate the positive converter with a delay angle of α1 = α. Gate the negative converter with a delay angle of α2 = π - α through gate isolating circuits. Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 19/33 Three-phase half-wave converter  Three-phase converters are extensively used in industrial applications up to the 120-kW level, where a two-quadrant operation is required. The thyristors are turned on at an interval of π/3. Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 20/33 Three-phase full converter  Three-phase converters are extensively used in industrial applications up to the 120-kW level, where a two-quadrant operation is required. The thyristors are turned on at an interval of π/3. Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 21/33 Three-phase full converter (contd..)  The frequency of output ripple voltage is 6fs and the filtering requirement is less than that of half-wave converters.  If the line-to-neutral voltages are defined as,  The corresponding line-to-line voltages are: The gating sequence is as follows: Generate a pulse signal at positive zero crossing of the phase voltage van. Delay pulse by the desired angle α + π/6 and apply it to the gate and cathode terminals of T1 through a gate-isolating circuit. Generate five more pulses each delayed by π/6 from each other for gating T2, T3,...T6, respectively, through gate-isolating circuits. Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 24/33 Example: 11-4 (contd..) Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 25/33 Example: 11-5 Problem: The input current of a three-phase full converter as shown in Figure is continuous with a negligible ripple content. (a) Express the input current in Fourier series, and determine the HF of input current, the DF, and the input PF. (b) If the delay angle α = π/3, calculate Vn, HF, DF, and PF. Solution: Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 26/33 Example: 11-5 (contd..) Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 29/33 Assignment: 11-1 Problem: A three-phase semiconverter shown in Figure is operated from a three- phase Y-connected 208V, 60Hz supply and the load resistance is R = 10Ω. If it is required to obtain an average output voltage of 50% of the maximum possible output voltage, calculate (a) the delay angle α, (b) the rms and average output currents, (c) the average and rms thyristor currents, (d) the rectification efficiency, (e) the TUF, and (f) the input PF. Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 30/33 Three-phase dual converter Prof. Abdur Razzak EEE 332/413 Lecture 11 Page 31/33 Three-phase dual converter (contd..)