I'm working on an induction motor drive with a scalar controller. As inverter modulation, I utilized both SVM and SPWM. I discovered that if the frequency and output fundamental voltage from SVM and SPWM are the same, the THD is about the same. The result can be seen in the image attached herewith. Can anybody explain why the THD is the same? Is THD only determined by the amount of fundamental voltage generated? Any literary references would be much appreciated.
The only difference between what you call SPWM and SVM will be the distribution of the zero switching states, therefore the THD is very similar (but not identical).
A good book to recommend in my eyes would be "Pulse Width Modulation for Power Converters: Principles and Practice" from Holmes and Lipo, 2003.
To cite one part that seems very relevant to your question:
"The main benefit of SVM compared to simple sinusoidal or third-harmonic injected PWM can be seen by contrasting Figure 5.5(b) for a simple sinusoidal reference, and Figure 5.8(a) for one-sixth third-harmonic injected modulation, against Figure 6.10(b). Clearly, the effect of SVM is to transfer harmonic energy into the outer sideband harmonics and to some extent to the sidebands of the second carrier group, and reduce the inner sideband harmonics as a consequence. This reduces the overall WTHD of the switched output waveform."
Klaus Schuhmacher Sir,
Please correct me if I am wrong: So, during the production process of the same line voltage, active voltage vector time in SVM (in a sample time) is the same as active voltage production time in naturally sampled SPWM (in the equivalent same sample)? Furthermore, in naturally sampled SPWM, there is no control over the zero vector location, while SVM allows for this flexibility. Is it the cause for the almost same THD?
Also, could you maybe explain why the author chose WTHD as a performance metric in his book? Is it because it aids in the detection of harmonic loss in an inductive load? Also, what is the relationship between WTHD and THD?
If WTHD is a performance parameter then low value of WTHD means good performance. I have seen that with increment of modulation index THD performance improves, THD decrease with increment of modulation index.
But in the following figure it shows WTHD increases with increment of modulation index? Why is that?
Because of its mathematics, your suggested book is difficult to understand, yet it is a true treasure.
Hi , Zarzisur Rahman Rony
Yes, I have already worked on improving the power quality of three-phase inverters by comparing 3 methods (SPWM, SVPWM and SHE PWM).
the best method was the SHE PWM which eliminates harmonics and gives a very low THD
Concerning your question, in another work, I made a comparative study between the vector and sinusoidal PWM, I found almost the same value of distortion (THD) for different switching frequency, but in the SVPWM, the switching losses decreases which increases the value of the fondalmental (I made a simulation under MATLAB and I validated the results experimentally by the FPGA tool (RT-Lab).
Kind regards
Zarzisur Rahman Rony
Well, SVM is regularly sampled so the durations for the active voltage vectors will be the same as a REGULARLY sampled sinusoidal PWM. I would suggest taking a look at section 6.1.2 of the mentioned book. By the way, "conventional" SVM simply distributes the zero vectors equally so that t0=t7.
Yes, the WTHD is used because an inductive load is assumed (see p.69). Please check the book for the definitions of the factor, it's simply a division of each harmonic by its harmonic number.
You have to be careful with the figure you included. The WTHD0 is shown there, not the WTHD. According to the definitions in the book, the WTHD would be WTHD=WTHD0/M.
WTHD definition: page 63,
WTHD0 definition: page 92
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