Non-controlled Singlephase Rectifier Circuits

Transcription

1 Non-controlled Singlephase Rectifier Circuits Omar X. Avelar, Omar de la Mora & Diego I. Romero POWER ELECTRONICS (ESI 012A) Instituto Tecnológico y de Estudios Superiores de Occidente (ITESO) Departamento de Electrónica, Sistemas e Informática (DESI) SCHEMATICS Fig. 1: Half-wave rectifier with a resistive load. Fig. 4: Full-wave rectifier with a resistive load. Fig. 2: Half-wave rectifier with a resistive+inductive load. Fig. 5: Full-wave rectifier with a resistive+inductive load. Fig. 3: Half-wave rectifier with a motor equivalent as the load and a free-wheeling diode. Subject: Power Electronics - Page 1 of 8

2 CALCULATIONS For the direct current voltage, we start with the definition With this done, we can begin by calculating performance parameters such as T V CD = 1 T 0 And then substitute for both types a) Half-wave rectifier V p sin t d t (1) V DC = 1 V 2 p sin t d t = V p 0 Average output power AC output power Efficiency Form Factor P DC =V DC I DC P AC =V RMS I RMS = P DC P AC FF= V RMS V DC b) Full-wave rectifier V DC = 1 0 V p sin t d t = 2V p Ripple Factor Transf. Utility Factor 2 V RF= RMS V DC TUF= P CD V S I S 2 V DC We proceed to calculate the root mean square voltage, and by definition: Crest Factor CF = I p I s = V 1 RMS T T V p sin 2 t d t (2) 0 and then again for, a) Half-wave rectifier V RMS = b) Full-wave rectifier V p sin 2 t d t = V p 2 V RMS = 1 0 V p sin 2 t d t = V p 2 Fig. 6: Waveforms. Subject: Power Electronics - Page 2 of 8

3 We began by measuring the RMS voltage of our transformer to obtain the peak value of it. V ptx = [V ] = 18.17[V ] By putting a 100 resistor as a load and taking into consideration the voltage drop of the diodes, we have that: I DC = V DC V D R L and I RMS = V RMS V D R L And by applying our formulas from page 2, we can build a table of expected results which we will compare later on in the overview section. Half-wave rectifier Full-wave Rectifier V DC 5.82 V V I DC 50.2 ma ma V RMS 9.14 V V I RMS 83 ma 113 ma P DC 292 mw 1.16 W P AC 762 mw 1.46 W 38.31% 79.76% FF RF TUF CF Subject: Power Electronics - Page 3 of 8

4 SIMULATIONS Fig. 7 displays voltage and current curves of our half-wave rectifier without a free-wheeling diode and a resistive load (As in Fig. 1 in the schematics section) V(D5:2) V(D5:1)-V(D5:2) - I(R2) Fig. 7: (2) Voltage on the rectifier diode. (3) Current in the load. Next figure (Fig. 8) shows the same curves but with a resistive and inductive load (As in Fig. 2 in the schematics section). And finally the curves (Fig. 9) of the same resistive and inductive load but with a free-wheeling diode (As in Fig. 3 in the schematics section) V(R2:2) V(D5:1)-V(D5:2) - I(R2) Fig. 9: (2) Voltage on the rectifier diode. (3) Current in the load V(R2:1) -2 V(D5:1)-V(D5:2) - I(R2) Fig. 8: (2) Voltage on the rectifier diode. (3) Current in the load. We can notice the extended conduction state caused by the inductor; keeping the current flowing until it runs out of energy and causing voltage spikes at the load (this is known as flyback). Subject: Power Electronics - Page 4 of 8

5 Next to check the same parameters but for a full-wave rectifier with a resistive load. 1 DATA AND MEASURMENTS We built the circuits shown in the schematic section used the oscilloscope to gather some data. First, some captures of the half-wave rectifier with resistive load (Fig. 12). 5V -2V V(R1:2,D4:1) - I(R1) Fig. 10: (2) Current in the load. And for an inductive+resistive load, then, Fig. 12: Channel 2: Voltage on the load. 1 5V Then with the inductor+resistive load (Fig. 13). -2V V(R1:2,D4:1) - I(R1) Fig. 11: (2) Current in the load. The non rectifying diodes on each cycle from the full-wave rectifier act as a free-wheeling diodes and thus eliminating voltage spikes seen at the load. Fig. 13: Channel 2: Voltage on the load. Subject: Power Electronics - Page 5 of 8

6 And by adding the free-wheeling diode, then we remove the flyback effect. And by stopping our motor, we can see how it behaves as a merely resistive+inductive load (Fig. 16). Fig. 14: Channel 2: Voltage on the load. Fig. 16: Channel 2: Voltage on the stopped motor. And with a motor (Fig. 15), Fig. 15: Channel 2: Voltage on the motor. The voltage waveform on the motor has a DC component caused by Faraday's law of induction due to relative movement of a circuit and a magnetic field. Subject: Power Electronics - Page 6 of 8

7 And repeating the process for the full-rectifier bridge, we get the following figures. Now, to obtain the inductor value of the circuit, we will approximate the following equation (3) by using the Newton-Rhapson method. V p R i t = 2 [ sin t cos t e R L[ L L 1] R L t ] Choosing the value of our current, and measuring the elapsed time between it, for example the next figure displays the time ( t ) between i t =0. Fig. 17: Voltage on a resistive load.. Fig. 19: Elapsed time between current values of 0. And thus obtaining L 26.7[mH ], and the value according to the LCR bridge is L=12.28[mH ]. Summarizing the measurements values on the following table. Half-wave rectifier Full-wave Rectifier V DC 5.42 V 9.65 V I DC 51 ma 84.5 ma Fig. 18: Voltage on an inductive+resistive load. As we can see, there is no flyback effect. V RMS 12.9 V 10.8 V I RMS 121 ma 96 ma P DC mw mw P AC 1.56 W W 16.46% 78.64% FF RF TUF CF Subject: Power Electronics - Page 7 of 8

8 OVERVIEW The next table shows the difference between the performance parameters between the theoretical and implemented circuits. Half-wave rectifier Full-wave Rectifier Theoretical Implemented Theoretical Implemented V DC 5.82 V 5.42 V V 9.65 V I DC 50.2 ma 51 ma ma 84.5 ma V RMS 9.14 V 12.9 V V 10.8 V I RMS 83 ma 121 ma 113 ma 96 ma P DC 292 mw mw 1.16 W mw P AC 762 mw 1.56 W 1.46 W W 38.31% 16.46% 79.76% 78.64% FF RF TUF CF CONCLUSIONS We were able to apply the concepts of seen in class to compare two different rectifier topologies by checking the behavior of the voltage and current curves and analyzing the performance parameters. By using a merely resistive load and an inductive and resistive load we observed phenomenons such as flyback, this is caused when the coil or inductor at the load oppose a sudden change in current and keeping a small current flowing depending on the stored energy of the inductor forcing the rectifier diode to be conducting even if the forward voltage from the signal feeding the diode has not been reached. Flyback can be reduced by implementing a free-wheeling diode in the half-wave rectifier. We also noticed that for the full-wave rectifier, the same diodes acted as free-wheeling. We also noticed that the efficiency of the rectifier circuits were reduced (compared to the ideal values obtained in class) as we took into consideration the voltage drops from the diodes. Fig. 20: Voltage of a motor slowing down as the load at the output of a full-wave rectifier. This last figure was a very neat effect of the induced voltage in the motor by the rotation, it caused some diodes to block out and stopped acting as the free-wheeling diodes and forcing conduction as the coil from the motor keeping the current flowing even on the negative sinusoidal cycle. Subject: Power Electronics - Page 8 of 8