Grid connected transistor bridge inverter

  • Thread starter Samtheguy
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Homework Statement



Hi there. I was just wondering if anyone can conform that my method for this problem is correct so far?

The transistor-bridge converter shown below is being used to connect a PV array to the grid.
For the purpose of this coursework, you may assume that:
• The PV array behaves as an ideal DC voltage source.
• The grid is an ideal sinusoidal voltage source at 50 Hz.
• The components are ideal and lossless.
• Bipolar switching is used.

The DC voltage (Vdc) is 458 V.
The inverter switches at 1800 Hz and its output inductor is 4 mH.
The modulation index is 0.59 at angle of 15 degrees relative to the grid voltage.
The AC grid voltage is 217 V.

Question - Calculate Vi (Magnitude and angle)

The Attempt at a Solution



The modulation index is defined as: Mi = Vi peak (of the fundamental component) / Vdc
Therefore; Vi = Mi x Vdc
Vi = 0.59 x 458
Vi = 270.22V

However this is a peak voltage. Hence the rms voltage is; 270.22/sqrt(2)
So Vi = 191.074 V (rms)

Since it is at an angle of 15 degrees to the grid voltage, use the sine rule to obtain the total length of the Vi phasor.

Vi / sin(90) = 191.074 / sin(75)
Vi = 197.814V

So we'll say Vi = 198V at an angle of 15 degrees to the grid voltage.
Is that correct??
 

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Answers and Replies

  • #2
rude man
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Looks right to me. I assume Vi is the square-wave voltage on the left-hand side of the inductor.
 
  • #3
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Hi and thanks for your reply.

My professor has since confirmed that this method is correct. Vi is the fundamental voltage of the waveform generated by the transistor bridge. The Vi waveform is a square wave where the pulse width changes in relation to the reference sine wave but frequency remains pretty constant. Fourier analysis shows that the fundamental component of the voltage is 50Hz with an amplitude related to the modulation index and the input dc voltage.
 

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