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Figure 1141 Half-bridge voltage source inverter
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0 VS 2 0 VS D1 2 vo io S1
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Figure 1142 Half-bridge voltage source inverter waveforms
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vo = VS /2 The switching sequence of S1 and S2 is shown in Figure 1142 It is important that each switch be turned off before the other is turned on; otherwise, the DC supply would be short-circuited Since the load is always going to be inductive in the case of a motor drive, it is important to observe that the load current, io , will lag the voltage waveform, as shown in Figure 1142 As shown in this gure, there will be some portions of the cycle in which the voltage is positive but the current is negative The function of diodes D1 and D2 is precisely to conduct the load current whenever it is of direction opposite to the polarity of the voltage Without these diodes, there would be no load current in this case Figure 1142 also shows which element is conducting in each portion of the cycle A full-bridge version of the VSI can also be designed as shown in Figure 1143; the associated output voltage waveform is shown in Figure 1144 The operation of this circuit is analogous to that of the half-bridge VSI; switches S1 and S2 are red during the rst half-cycle, and switches S3 and S4 during the second half Note that the full-bridge con guration allows the output voltage to swing from VS to VS The diodes provide a path for the load current whenever the load voltage and current are of opposite polarity
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+ VS S4 D4
+ vo Load io S2 D2
Figure 1143 Full-bridge voltage source inverter
vo VS 0 VS
Figure 1144 Half-bridge voltage source inverter output voltage waveform
A three-phase version of the VSI is shown in Figure 1145 Once again, the operation is analogous to that of the VSI circuits just presented The related waveforms are shown in Figure 1146 The top three waveforms depict the pole voltages, which are referenced to the DC supply neutral point, o The pole voltages are obtained by ring the switches S1 through S6 at appropriate times For example, if S1 is red at t = 0, then pole a is connected to the positive side of the DC supply, and vao = VS /2; if S4 is subsequently turned on at t = , then pole a is connected to the negative side of the DC supply, and vao = VS /2 The other pairs of switches are then red in an analogous sequence, shifted by 120 electrical degrees with respect to each other, to obtain the waveforms shown in the top three graphs of Figure 1146 The line voltages are obtained from the pole voltages using the following relations: vab = vao vbo vbc = vco vco vca = vco vao and are shown in the second set of three diagrams in Figure 1146 These are also phase-shifted by 120 Now, we can also express the pole voltages in terms of the (1123)
Part II
Electronics
+ VS 2 o + VS 2
a S4 D4
b S6 D6
c S2 D2
a Load
VS 2 0 V S 2
Figure 1145 Three-phase voltage source inverter
load phase voltages, van , vbn , and vcn : vao = van vno vbo = vbn vno vco = vcn vno and since we must have van + vbn + vcn = 0 for balanced operation (see 7), we can derive the following relationship for the DC supply neutral (o) to load neutral (n) voltage: vno = vao + vbo + vco 3 (1125) (1124)
VS 2 0 VS 2 VS 2 0 V S 2
vab VS 0 VS vbc VS 0 VS vca VS 0 VS vno VS 6 V S 6 1 V 3 S t t t
This voltage is also shown to be a square wave switching three times as fast as the inverter output voltage Finally, to obtain the phase voltages, we make use of the relations van = vao vno = 2 vao 1 (vbo + vco ) 3 3 vbn = vbo vno =
2 v 3 bo
1 (v 3 ao
+ vco )
(1126)
vcn = vco vno = 2 vbo 1 (vao + vbo ) 3 3 Only one phase voltage, van , is shown in the picture; however, it is straightforward to construct the other two phase voltages using equation 1126 Note that the load phase voltage waveform shown in Figure 1146 is a coarse stepwise approximation of a sinusoidal waveform; the corresponding load current, ia , is a ltered version of the load voltage, since the load is inductive in nature, and is therefore somewhat smoothed with respect to the voltage waveform The discontinuous nature of these waveforms creates a signi cant higher harmonic spectrum (see the box Fourier Analysis in 6), at frequencies that are integer multiples of the inverter output frequency; this is an unavoidable property of all inverters that employ switching circuits, but the problem can be reduced by using more complex switching schemes Another major shortcoming of this AC supply is that if the DC supply is xed, the amplitude of the inverter output is xed
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