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1 the Single-Phase Full-Bridge Diode Rectifier

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1 the Single-Phase Full-Bridge Diode Rectifier ENEL 585 Experiment 1 The Single-Phase Full-Bridge Diode Rectifier Purpose: To become familiar with the single-phase diode bridge rectifier. Operation with three types of passive filtering (L, C, and LC) is investigated. Introduction: In this experiment you will investigate an old method of obtaining a low ripple dc voltage source from an ac supply. A pure resistor is employed as the load. The voltage and current waveforms in this experiment are sampled using National Instruments 1 MSps data acquisition hardware and the Lab Windows CVI software. The waveforms are multiplex sampled, hence the sampling frequency decreases as more waveforms are displayed. The lab 1 hardware and software, employed in this lab, were kindly developed by Sean Victor Hum, Ed Evanik, Jason Pannell, Richard Galambos and Arif Al-Judi. The hardware for Labs 2 onward was kindly developed at the U of Minnesota. Tutorial: Shown in Fig. 1 (a) is a schematic of a full-bridge diode rectifier with single-phase ac source, and shown in Fig. 1 (b) is a schematic of a full-bridge diode rectifier with single-phase ac source. Yup, they are identical. In ENEL 585 we will use the Fig. 1 (b) representation. The use of a four diode full-bridge implies full-wave ie, 120Hz ripple, rectification. The reverse relationship is not necessarily true: Full-wave rectification may also be accomplished using a half-bridge diode rectifier fed by a 2-phase ac source (a.k.a. split ac), as may be obtained from a single-phase source using a centre- tapped transformer. There is another type of half-bridge diode rectifier, consisting of a rectifying diode and free-wheeling diode, fed directly by a single- phase source. This converter provides half-wave rectification, ie, with 60Hz ripple. Of course a single diode may be used for half-wave rectification, generally employed for low-inductance or pure resistive loads. Interestingly the single diode rectifier is not referred to as a quarter-bridge Rectifier (Topologically, we could refer to the single diode rectifier as a single-phase two- wire half-bridge rectifier. What a mouthful huh?). – The Single-Phase Full-Bridge Diode Rectifier Rectifier Diode Full-Bridge –Single-Phase The (a) (b) Figure 1: Bridge diode rectifier with possible L and/or C filter component 1 EXPERIMENT: 1 For completeness of discussion, we can note that there exists an analogy between single-phase and 3-phase diode bridges. A single-phase full-bridge (2-wire 4 diode) rectifier produces 120Hz ripple. A 3-phase full- bridge (3-wire 6 diode) rectifier produces 360Hz ripple. A single-phase half-bridge (2-wire 1 or 2 diode) rectifier produces 60Hz ripple. A 3-phase half-bridge (4-wire 3 diode) rectifier produces 180Hz ripple. To understand the single-phase full-bridge operation, refer to Fig. 1 and consider the case where the line voltage, aka the source voltage, vs , goes positive. For the moment, assume no filter components are present and the load is purely resistive. It makes sense that diode, D1 , will tend to turn on (since its anode is going positive). At the same time, diode, D4, will tend to turn on (since its cathode is going negative). Thus, with both D1 and D4 on, the output voltage is positive. During the next half cycle, ie, the ac source goes negative, diodes, D2 and D3 , now turn on simultaneously, and again the output voltage is positive. In this way, the current that flows into the load is always positive. Referring to Fig. 1 (a), can you see the symmetry of circuit operation? If no filter components (L f or Cf) are present in the converter, then the load voltage, vo , will simply be a rectified sine wave, having an amplitude equal to the ac source (less two diode drops). Assuming a resistive load, R, then having the filter inductor alone present will produce first-order filtering (with time constant Lf / R) of the rectified sine wave, giving a lower ripple (ie, peak-to-peak) output voltage. One might think that having the filter capacitor alone present would also produce first-order filtering (with time constant RC f). This is not the case. To better understand this circuit consider its operation in the time-domain, that is, as the filter capacitor is charged up, the diode pairs, D1 D4 and D2 D3, need only turn on near the peaks of the ac source to just "top off" the voltage on the capacitor (remember the load is continually "bleeding" the capacitor). As a result, the output voltage (ie, the capacitor voltage) is approximately equal to the peak line voltage and has a more or less saw-tooth ripple component (can you see what the actual shape of the ripple component is, assuming a resistive load?). It is also possible to build the rectifier with both filter inductor and filter capacitor present, in which case you roughly obtain 2nd-order filtering of the rectified sine wave (especially if Lf is big enough). When do you use each type of filtering? This is a function of the load requirement. You use as little filtering as possible, but generally, at lower power levels (less than 1 kW) a single filter capacitor is effective, filter- wise and cost-wise. Whereas at higher power levels, a single filter inductor may payoff, since the peak diode current will be reduced in this case, permitting the use of less expensive diodes. Both Lf and Cf are used only if a lower ripple voltage is needed, and the peak input current is a concern. In the lab, we will be calculating input power factor (c.f. Lecture Notes). You may also wish to calculate some average and rms quantities (c.f. Lecture Notes) and compare to the software calculation. Pre-Lab (one per lab group collected at the start of the lab): 1. Consider a single-phase diode bridge rectifier having an ac source of 10V amplitude, a resistive load of 22Ω, and no filtering components. Assuming ideal diodes, calculate (approximately if necessary) the following quantities. Include well-labeled sketches of waveforms for the calculations. Please calculate: average load voltage, peak-to-peak (ie, ripple) load voltage, rms load voltage, Rectifier Diode Full-Bridge –Single-Phase The peak diode current, rms diode current, rms line (or source) current, input power factor. 1 EXPERIMENT: 2 2. Repeat the analysis of problem 1, if a filter capacitor of 4, 700 µF is added to the converter circuit. Hint: If you have trouble finding the conduction angle of the diodes, ie, the Δω t that each diode is on for, you may use the fact that it is about 25 ° (this value is not exact, but allows you to make calculations). 3. Repeat the analysis of problem 1, if a filter inductor of 0.6H is added to the converter circuit (ie, no filter capacitor). Hints: To find the output current ripple (and hence the ripple voltage), first realize that the output of the bridge itself is a rectified sine wave having dc and ac components. Note that the dc component appears entirely across the resistance in the circuit (this doesn't preclude an ac component across the resistance as well). Note that the ac component has a fundamental at 120Hz. Assume the peak-to-peak voltage of the 120Hz fundamental is roughly equal to the peak-to-peak voltage of the ac component itself. Now - think of the inductor as an impedance: Hint: ZLf ≈ 2πf Lf = 2ωLf , where f=120Hz and ω = 377rad/s (this is not a typo, the power industry often sets ω = 2π60Hz = 377rad/s as a constant) . What is the impedance of the series LR combination? You should now be able to approximately determine the peak-to- peak output ripple current and then use Ohm’s Law to find peak-to-peak ripple load voltage. 4. Repeat the analysis of problem 1, if a filter inductor of 0.6H and a filter capacitor of 4,700µF are used in the converter circuit. To simplify this analysis, assume the capacitor is ideal (in practice its equivalent series resistance would be considered), assume the inductor is ideal, and let the load resistance be 22Ω. There are two possible ways to analyze the problem. One way is to proceed along the lines of problem 3, noting that the load voltage may be approximated as pure dc. Once the inductor current ripple is obtained, it may be integrated to obtain the capacitor (ie, load) voltage ripple. Another analysis approach is to consider the LCR elements as a passive low-pass filter having a 3dB corner frequency of f3dB =1/2π LC and an attenuation slope of 40dB per decade. If you perform both analyses, you can check your results. Procedure (a lab book is not required, but neat notes must be taken): 1. A single-phase diode bridge module will be provided to your lab group. This module is supplied by the 120V (rms of course) mains, stepped down by a 120:6.3 transformer (In days of old, 6.3Vrms was a very common AC Voltage used to heat the filament of a vacuum tube.). The step down transformer provides isolation, and the desired ac voltage, vs(t), indicated in Fig. 1. For this lab, we can assume that the line inductance is negligible. There is a power switch on the transformer box. Turn power off whenever making any circuit changes. 2. Connect a resistive load of 22Ω across the load output terminals of the rectifier module. If unsure about the wiring have it checked.
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