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Simple half-wave rectifier D1
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Same arrangement with free-wheeling diode
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Figure 1117 Recti er connected to an inductive load
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vL (t) A VL 0 vL vAC A VL = t
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(and assuming that both D1 and D2 are ideal), we conclude that the DC component of the load voltage, VL , must appear across the load resistor, R (no steady-state DC voltage can appear across the inductor, since vL = LdiL /dt) Thus, the approximate DC current owing through the load is given by A IL = R (116)
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iL (t) IL IL = 0 VL R t
since the average output voltage of a half-wave recti er is A/ V for an AC source of peak amplitude A (see 8) The AC component of the load current (or ripple current) is not as simple to compute, since it is due to the AC component of vL , which is not a pure sinusoid The exact analysis would require the use of a Fourier series expansion For the purposes of this discussion, it is not unreasonable to assume that most of the energy is at a frequency equal to that of the AC source: iL (t) IL + IAC cos ( t + ) (117)
Figure 1118 Operation of a free-wheeling diode
where IL is the average load current, IAC is the peak value of the ripple current, and is its phase An acceptable approximation from which the amplitude of IAC
11
Power Electronics
may be computed is vL (t) A A + sin t 2 2 (118)
Figure 1119 graphically illustrates the extent of this approximation
Approximate voltage
1 08 06 04 02 0 02 04 06 08 1
Load voltage (V)
Actual rectified voltage
15 20 t (ms)
Figure 1119 Approximation of ripple voltage for a half-wave recti er
A common alternative to the half-wave recti er is the full-wave recti er, which was discussed in 8
EXAMPLE 114 Recti ers and Inductive Loads
Problem
Analyze the circuit depicted in the lower half of Figure 1117 to determine the rms amplitude of the ripple in the load voltage
Solution
Known Quantities: Source voltage; load resistance and inductance
Find: Vripple
Schematics, Diagrams, Circuits, and Given Data: vAC = 100 V rms, 60 Hz The load is a DC motor with Ra = 11 and La = 0001 H The motor speci cations may be found in the DC motor template in Electronics WorkbenchTM Assumptions: Ignore the DC motor mechanical load Focus on Computer-Aided Solutions: The analysis of this design has been conducted in simulation, using Electronics WorkbenchTM The simulation of this circuit may be found in the accompanying CD-ROM
Multisim
Comments: What happens if the freewheeling diode is not in the circuit You may try to
run a simulation without diode D2 in the circuit
Part II
Electronics
Three-Phase Recti ers It is important to realize that the same type of circuit that can be used for singlephase recti ers can also be employed to design multiphase recti ers Recall the analysis of three-phase AC power systems in Section 74 In many high-power applications, three-phase voltages need to be recti ed to give rise to a single DC supply; such recti cation can be achieved by means of an extension of the bridge recti er Consider the balanced three-phase circuit shown in Figure 1120 The three-phase wye-connected source is connected to a resistive load by means of a three-phase transformer, with a delta-connected primary and a wye-connected secondary The circuit could also operate without the transformer The three secondary currents, ia , ib , and ic , ow through pairs of diodes D1 to D6 in a manner very similar to the single-phase bridge recti er described in Figure 845 The diodes will conduct in pairs depending on the relative line voltages, according to the following sequence: D1 -D2 , D2 -D3 , D3 -D4 , D4 -D5 , D5 -D6 , and D6 -D1 Recall from the analysis of Section 74, equation 754, that the line-to-line voltage is 3 times the phase voltage in a three-phase wye-connected source The instantaneous source voltages and the related diode conduction periods, as well as the load voltage, are shown in Figure 1121 It can be shown that the average output voltage is given by the expression: /6 2 3 3 VL = Vm = 1654Vm 3Vm cos t d( t) = (119) 2 /6 0 where Vm is the peak phase voltage The rms output voltage can be calculated to be /6 2 3 9 3 2 + 3Vm cos2 t d( t) = Vrms = (1110) 2 /6 0 2 4 Vm = 16554Vm
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