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TMS320F28027: calculation formula

Part Number: TMS320F28027
Other Parts Discussed in Thread: C2000WARE

Hi team,

Here's an issue from the customer may need your help:

A single phase inverting UPS that customers are doing now.

The peak of the output at no load corresponds to the value of the H-bridge Vbus (which was developed with 48V) multiplied by the modulation ratio, but with a pure resistance of 25R, it was found that the peak of the inverting AC wave was about 5 V lower than the product of the Vbus and modulation ratio.

a. What does this drop relate to?

b. Is power control calculated by apparent power or average power?

c. Where can they find the calculation of power in my code?

d. Is the CBC function enabled when there is also overcurrent? How is overcurrent controlled?

e. How is the power calculated when the load is an inductive load? Does the code for this project have a calculation of the phase?

Could you help check this case? Thanks.

Best Regards,

Cherry

  • Hi Cherry, Thanks for reaching out.  Please expect a delay in reply due the long weekend in observance of Dr. Martin Luther King day.  Regards, Krishna

  • Please check you the input waveform to your LC filter. I assume your inverter has an LC filter stage at its output. Then check the output of the LC. We have a reference design on single phase inverter. Pls check the details on this (link below). The power calculation uses the power measurement library inside C2000Ware.

    https://www.ti.com/tool/TIDM-HV-1PH-DCAC

    Shamim

  • Hi Krishna, thanks for your remind!

    Hi Shamim,

    Thanks for your support! Hers's the feedback from the customer:

    The input waveform of the filter is a SPWM wave with equal amplitude, which is equal to the supply voltage of the H-bridge. How is the voltage drop of the filter inductor calculated? Is the voltage drop across the filtered inductor calculated as 50Hz or 20KHZ?

    20KHZ is the switching frequency of the H-bridge and takes the form of a unipolar SPWM wave, regardless of the filtering mode of the band load, there must be a voltage drop. However, in the Voltage Source Inverter Reference Design document 3.1.2.3.1.5 Running the Code Item 14, The valid values of the waveforms of the DC380V modulation ratio of 0.5 output should be somewhat different from the calculations, and should not be very close to each other. Why is the loss not calculated with a 100 ohm load?

    "For the DC bus of 380 V, and a 100-Ω output with a 0.5 inverter modulation index, the output voltage
    will be close to guiVrms = 133"

    The customer personally think that 380 * 0.5 / 1.414 = 134 is just a theoretical value, and actually it should be around 134 – the loss voltage (10 ~ 30) = 124 ~ 104 ,which is correct.

    Thanks and regards,

    Cherry

  • Why is the loss not calculated with a 100 ohm load?

    You can calculate the loss with the highest load current. That means the lowest load resistor.

    Also there will be reduced voltage at the filter input because of the dead-time and the switching delay in the FET

    You have to subtract all of these voltage drop plus the drop across filter inductor to calculate the output.

  • Hi,

    Thanks for response.

    Are there formulas or technical articles for these voltage drops? And here's some additional questions:

    What does the H-bridge-to-inductor current waveform look like?

    The H-bridge is used as a cross-variant power supply, is the output current waveform of the H-bridge a sinusoidal or sinusoidal wave plus a variety of harmonics current forms?  What is the calculation of the output current versus the load current in parallel with the load as a capacitor? What is the voltage waveform on both ends of the inductor? Are there any articles for reference?

    Thanks and Regards,

    Cherry

  • Hi,

    May I know is there any updates? Thanks.

    Best Regards,

    Cherry