The three-phase inverter shown in figure operates such that each switch gets a gating signal for 120°

Question

The three-phase inverter shown in figure operates such that each switch gets a gating signal for 120°. The gating sequence is: S₁: 0120, S2: 60-180, S3: 120 240 … and so on. The source voltage is 300V and the load is a resistance of 5092/phase. 1) Complete the table below. 2) Sketch the following (showing all values, angles and conducting devices on the curves): – – – Solution Interval 0 →60 60 120 120 →180 180→240 V 240 →300 300-360 a S₁ D, Conducting devices Va DA The gating signals for the six switches The three phase voltages: Va, Vb, Vc The three line-line voltages: Vab, Vbe, Vea 3) Calculate the rms values of the phase and line-line voltages. 4) Find the power delivered to the load. + 1 ia Va b O N D S So Do S₂ D₂ Vb + 1 Ve Vp ib – C n DS Z ic Ve Z.

Answer

1. Completing the table:

Solution IntervalPhase Voltage (Va)Conducting DevicesLine Voltage (VL)
0° → 60°0 VD1, D6Vab
60° → 120°300 VD1, D3Vab + Vc
120° → 180°0 VD3, D6Vc
180° → 240°300 VD3, D5Vc + Vab
240° → 300°0 VD5, D6Vab
300° → 360°300 VD5, D1Vab + Vc

2. Sketching the curves:

a. Phase Voltage (Va):

  • A triangular wave with amplitude 300 V.
  • Conducting devices:
    • During 0° – 60° and 240° – 300°, D1 conducts (Va = 0).
    • During 60° – 120° and 180° – 240°, D3 conducts (Va = 300 V).

b. Gating Signals for Switches:

  • Three square waves with 120° phase shift.
  • S1: High for 0° – 60° and 240° – 300°.
  • S2: High for 60° – 180°.
  • S3: High for 120° – 240°.

c. Three-Phase Voltages (Va, Vb, Vc):

  • Three shifted copies of Va, with phase differences of 120°.
  • Vb lags Va by 120°.
  • Vc lags Vb by another 120°.

d. Three Line-Line Voltages (Vab, Vbc, Vca):

  • Obtained by subtracting phase voltages.
  • Vab = Va – Vb.
  • Vbc = Vb – Vc.
  • Vca = Vc – Va.

3. Calculating RMS Values:

  • Phase voltage RMS: Vrms = Vpeak / √2 = 300 V / √2 ≈ 212.13 V.
  • Line voltage RMS: Vrms = √3 * Vphase RMS ≈ 367.39 V.

4. Calculating Power Delivered to Load:

  • Per-phase power: P = Vrms^2 / R = (212.13 V)^2 / 5092 Ω ≈ 8.88 W.
  • Total power: Ptotal = 3 * P = 3 * 8.88 W ≈ 26.64 W.

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