Lecture 21. Single-phase SPWM inverter switching schemes 21.1 Single-phase SPWM Inverter with Unipolar Switching Scheme In this scheme, switches T1 and T2 or T3 and T4 are not switched on together. Instead, the load current is allowed to circulate through a diode and the remaining switch whenever the control voltage is lower than the saw-tooth carrier. When this happens, zero voltage is applied to the load, resulting in three-level output voltage. The local circulation of the load current, without going through the DC source, means that load current ripple is smaller for the same switching frequency. The switching signals for the unipolar single-phase inverter can be obtained from the following: when e c > v tri, when e c < v tri, T1 is on and T4 is OFF T1 is OFF and T4 is ON when e c > v tri, T3 in ON and T2 is OFF when e c < v tri, T3 is OFF and T2 is ON Lecture 21 Single-phase SPWM Inverters 21-1 F. Rahman
i d V d T 1 D 1 A R i o T 3 L B + v o D 3 T 4 D 4 T 2 D 2 Figure 21.1 Note that the output of each pulse-width modulator drives one leg of the inverter only in a complementary manner. Whenever two upper or lower switches in the two legs of the inverter are simultaneously ON, the output voltage across the load is zero. Unlike the bipolar scheme, the output voltage now has three states, namely, +Vd, Vd and zero. The output voltage contains harmonics of order, n = j 2mf ± k. Note that harmonic sidebands around the carrier frequency do not exist, the lowest order harmonics appear around 2m f. The output voltage waveform also shows that even though the inverter switching frequency is m f f o, the effective switching frequency of the output is 2m f f o. Lecture 21 Single-phase SPWM Inverters 21-2 F. Rahman
Figure 21.2 f s = 450 Hz, f o = 50 Hz, m = 0.8 Figure 21.2 f s = 900 Hz, f o = 50 Hz, m = 0.8 Lecture 21 Single-phase SPWM Inverters 21-3 F. Rahman
Figure 21.3 f s = 2500 Hz, f o = 50 Hz, m = 0.8 0.8 V on max V d m = 0.8 1 m f 2m f 2m f - 1 2m f +1 3m f 4m f Harmonic order of v o Figure 21.4 Lecture 21 Single-phase SPWM Inverters 21-4 F. Rahman
Approximate analysis of single-phase SPWM inverter voltage waveforms Consider the output voltage waveform of a single-phase unipolar SPWM inverter as shown in figure 21.5. The number of output pulses per half cycle, k = 5. e c v tri +V d - V d t 0 T/2 T Figure 21.5 The centers of the output voltage pulses are at T t i (i0.5) i0,1,2,3,..., k 1 2k (21.1) The duration of an output voltage pulse, t pi, is given by T tpi m(sinot i ) 2k (21.2) Lecture 21 Single-phase SPWM Inverters 21-5 F. Rahman
The delta (or impulse) functions representing the voltage pulses located at t i and of duration t pi can be analysed by Fourier series, the coefficients of which are given by T/2 4 bni Vd tpi ti sinno tdt T (21.3) o B n 4mV 2k d k1 sinoi t sin noi t (21.4) i0 For k = 5, n 3 3n sin sin sin sin 10 10 10 10 5 5n 7 7n Bn 0.4mVd sin sin sin sin 10 10 10 10 9 9n sin sin 10 10 (21.5) Table 2 shows the output voltage harmonic content up to the order 21. Lecture 21 Single-phase SPWM Inverters 21-6 F. Rahman
Table 2 Coefficient Value s B 1 mv d B 3 0 B 5 0 B 7 0 B 9 mv d B 11 -mv d B 13 0 B 15 0 B 17 0 B 19 -mv d B 21 -mv d From an exact Fourier series analysis we would however obtain for m = 0.2 and k = 5, the harmonics as in Table 3: Table 3 Coefficient Value s B 1 0.1999V d B 3 0.0003V d B 5 0.0000V d B 7 0.0016V d B 9 0.1921V d B 11-0.1883V d B 13-0.0053V d B 15-0.0002V d B 17 B 19 Lecture 21 Single-phase SPWM Inverters 21-7 F. Rahman
Note that, from Fourier analysis, the 3 rd, 5 th and the 7 th harmonics are small but not zero. The dominant output harmonics are the 9 th and the 11 th which are more easily filtered out than the lower order terms. Lecture 21 Single-phase SPWM Inverters 21-8 F. Rahman
The DC link current (i d ) of a single-phase inverter Bipolar switched full-bridge SPWM inverter When T1 and T2 are OFF and the load current is positive (from A to B or left to right in the load), it is forced by the load inductance to flow back into the DC source. The DC link current i d is then negative. Similarly, when T3 and T4 are OFF and the load current is negative, it is forced to flow back into the DC source. The DC link current waveform for an inverter with f s = 900 Hz (i.e., k = 9) is indicated in Figure 21.6. Note that the i d waveform (bottom figure) has DC and AC components. If we assume that the switching frequency of the inverter is very (infinitely) high and LC filters exist at the DC input and AC output terminals, a simple expression for the DC link can be found. Note that when f s approaches infinity, the LC filter components are so small that they can be ignored in the analysis. v o & i o i d Figure 19.30 (19.35) v v V sin t o o1 o1max o Figure 21.6 I d Lecture 21 Single-phase SPWM Inverters 21-9 F. Rahman
io io1 Io1max sin ot (21.6) Analysis of the DC link current v o1 i o1 2 Figure 21.7 Equating input DC power to the inverter and the output AC power to the load, p d p (21.6) o Assuming the fundamental output current lags the fundamental output voltage with a phase angle, and that the power associated with output harmonic components are negligible (for a inverter with infinitely high switching frequency) Vi ddvo1max sin ot Io1max sin ot (21.7) V I o1 o1 o1 o1 i d cos cos2ot V d V I V d (21.8) Lecture 21 Single-phase SPWM Inverters 21-10 F. Rahman
Note the second term on the RHS of equation (21.8) has zero average value. The average value of the DC link current is I d V I V o1 o1 d cos (21.9) The above analysis has neglected the switching frequency related ripples in the DC link current. Lecture 21 Single-phase SPWM Inverters 21-11 F. Rahman
DC-link current in the unipolar switched SPWM inverter The DC link current for this inverter does not have the reverse current flow back into the DC mains when two top or bottom switches are ON simultaneously. During such times, the load current circulates through a diode and a switch, removing the negative current pulses from i d. I d Figure 21.8 Note that the analysis of the previous section (equations 21.8 and 21.9) equally holds for the unipolar switched inverter. Lecture 21 Single-phase SPWM Inverters 21-12 F. Rahman
Device ratings and the DC Link current. When the device current waveforms are known, as shown earlier, the device (diode and transistor) current ratings can be found accurately. In order to arrive at the ratings of the switches and the diodes in quantitative terms and to determine the dc link current we will now assume that the load draws a sinusoidal current. We will consider only one arm of an inverter comprising of two switches and two diodes as shown in Figure 21.9. i d V d T 1 D 1 A R i o T 3 L B + v o D 3 T 4 D 4 T 2 D 2 Figure 21.9 v o1 i o1 2 Figure 21.10 Lecture 21 Single-phase SPWM Inverters 21-13 F. Rahman
Assuming that the switching frequency is high, the average current carried by the switch over any switching period is given by the duty cycle of the ON and OFF times of the switch. The average switch current is thus given by, 1 ITDC Io1 max sin t d t 2 I o1max The switch RMS current, 1 cos 2 (21.10) 1 2 2 ITRMS Io1 max sin t d t 2 I TRMS Io1max 1 sin2 2 2 (21.11) Similarly, the diode average current, 1 IDDC Io1max sin t d t 2 1 cos 2 Io1max (21.12) The diode RMS current, Lecture 21 Single-phase SPWM Inverters 21-14 F. Rahman
1 IDRMS Io1max sin t d t 2 2 2 Io1max 1 sin2 2 2 (21.13) Note that the in-phase component (power component) of the load current is I o1 cos. Thus, for sinusoidal load currents the supply DC current is related to the in-phase component of the load current. The mean supply current is Id ITDC IDDC I o1max cos 2I o1 cos (21.14) Note that the I d in equation 21.14 is contributed by one switch and diode for one leg of an inverter. For a fullbridge inverter, the above must be multiplied by a factor of two for the two legs. For a three-phase inverter, which is covered in a latter section, the contribution to I d from each inverter leg must be multiplied by a factor of three. Lecture 21 Single-phase SPWM Inverters 21-15 F. Rahman
Single-phase inverter with high-frequency transformer. The square-wave or SPWM inverters which produce output at 50 Hz or so cannot take advantage of the reduction in the size associated with high switching frequency even if the PWM carrier frequency is high. Figure 21.11 Power converter circuits which utilize a high frequency AC link solve this problem. In this scheme, a squarewave single-phase inverter with high-frequency output (in excess of 10s or 100s of khz) supplies a highfrequency transformer, followed by a rectifier and the low frequency inverter. Phase angle control can be exercised in the square-wave high-frequency inverter for transformer input voltage control. Additionally, PWM control can also be exercised in the low frequency inverter. Such a scheme is indicated in figure 21.12. Lecture 21 Single-phase SPWM Inverters 21-16 F. Rahman
Rectifier without transfromer HF Squarewave Inverter Rectifier Inverter Load Phase angle control PWM control Figure 21.12 Lecture 21 Single-phase SPWM Inverters 21-17 F. Rahman