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Series connection of two full converters

Short answer

Two single-phase full converters with dc outputs in series give high dc voltage and better power factor. Total output is (2Vm/pi)(cos alpha1 + cos alpha2); with one converter at zero and the other controlled, Vo = (2Vm/pi)(1+cos alpha2), maximum 4Vm/pi. Input primary current is a stepped wave with displacement factor cos(alpha/2).

Explain the series connection of two single phase full converter with necessary circuit diagram and waveforms.

Two single-phase full converters are fed from two separate transformer secondaries and their d.c. outputs are connected in series. This is used for high-voltage d.c. output (for example traction) and, with sequence control, to improve the input power factor and reduce harmonics.

Circuit

              +------------------+
  ~ ---+  S1  |  Full converter 1|--+ +
       |  ))  |  T1 T3 / T4 T2   |  |  Vo1
  pri  |      +------------------+  |
 (one  |                            +----+
 trans-|      +------------------+  |    | load
 former|  S2  |  Full converter 2|--+    | (RLE)
  two  |  ))  |  T1'T3'/T4'T2'   |  |    |
  sec.)|      +------------------+  |    |
  ~ ---+                            +----+
                                       -

Each secondary has rms voltage VsV_s and peak VmV_m. The total output is

Vo=Vo1+Vo2=2Vmπ(cos⁡α1+cos⁡α2)V_o = V_{o1} + V_{o2} = \frac{2V_m}{\pi}(\cos\alpha_1 + \cos\alpha_2)

The maximum output is 4Vmπ\frac{4V_m}{\pi} (both at α=0\alpha = 0).

Rectification mode (power flow a.c. to d.c.)

Converter 1 is kept at α1=0∘\alpha_1 = 0^\circ (output +2Vmπ+\frac{2V_m}{\pi}) and converter 2 is phase-controlled, α2\alpha_2 from 180∘180^\circ down to 0∘0^\circ:

Vo=2Vmπ(1+cos⁡α2)V_o = \frac{2V_m}{\pi}(1 + \cos\alpha_2)

At α2=180∘\alpha_2 = 180^\circ the two outputs cancel (Vo=0V_o = 0); at α2=0∘\alpha_2 = 0^\circ, Vo=4VmπV_o = \frac{4V_m}{\pi}. So the output is varied smoothly from 0 to 4Vmπ\frac{4V_m}{\pi} with power flowing from the a.c. supply to the load.

Inversion mode (power flow d.c. to a.c.)

The load must contain a d.c. source EE of reversed polarity (for example a motor in regenerative braking). One converter is held at α1=180∘\alpha_1 = 180^\circ (full negative voltage, −2Vmπ-\frac{2V_m}{\pi}) and the other is controlled between 0∘0^\circ and 180∘180^\circ:

Vo=2Vmπ(cos⁡α2−1)V_o = \frac{2V_m}{\pi}(\cos\alpha_2 - 1)

so VoV_o ranges from 0 to −4Vmπ-\frac{4V_m}{\pi}; current direction stays the same, so power flows back to the a.c. supply. (In practice the angle is kept below about 165∘165^\circ to allow a margin for commutation.)

Waveforms (rectification, α1=0\alpha_1 = 0, α2=α\alpha_2 = \alpha, constant load current IoI_o)

 vo1  /\  /\  /\     full-wave, alpha1 = 0
 ----/--\/--\/--\--
 vo2   _    _    _   full converter, alpha2,
 -----|-\-.|-\-.|-\  negative from pi to pi+a
 is1  ____      ____
 ----|    |____|      square, in phase
 is2    ____      __
 ------|    |____|    square, lags by alpha
 is   (sum, on primary):
          ____      level 2Io from a to pi
 ----.___|    |___.   level 0 from 0 to a
  • vo=vo1+vo2v_o = v_{o1} + v_{o2}: during 00 to α\alpha of each half cycle vo2v_{o2} is negative and reduces vov_o; from α\alpha to π\pi both add.
  • The primary current is=is1+is2i_s = i_{s1} + i_{s2} (equal turns) is a stepped wave: zero from 00 to α\alpha and 2Io2I_o from α\alpha to π\pi. Its fundamental lags by only α/2\alpha/2, so the displacement factor cos⁡(α/2)\cos(\alpha/2) is better than cos⁡α\cos\alpha of a single converter.

Advantages

  • High output voltage with devices of half the voltage rating.
  • Better input power factor and less harmonic content than a single converter of the same rating, because only one converter is phase-controlled at a time (the other runs at 0∘0^\circ or 180∘180^\circ).

✓ Answer checked by Claude Opus 5.5 (high effort)

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