Three phase AC to DC conversion

A three-phase full-wave bridge rectifier uses six diodes: D1, D3, D5 form the positive (common-cathode) group and D4, D6, D2 the negative (common-anode) group. At any instant the diode of the most positive phase in the top group and the diode of the most negative phase in the bottom group conduct, so the load gets the highest line voltage.

        P (+) o-----+------+------+-----+
                    |      |      |     |
                   D1     D3     D5     |
                    |      |      |   Load
 a o----------------+      |      |     |
 b o-----------------------+      |     |
 c o------------------------------+     |
                    |      |      |     |
                   D4     D6     D2     |
                    |      |      |     |
        N (-) o-----+------+------+-----+
 (phase a joins D1-D4, b joins D3-D6, c joins D5-D2)

Operation

• Each diode conducts for $120^\circ$; a new pair takes over every $60^\circ$. Conduction sequence: D1D2, D2D3, D3D4, D4D5, D5D6, D6D1 (numbering = order of conduction).
• Output voltage is the upper envelope of the six line voltages $v_{ab}, v_{ac}, v_{bc}, v_{ba}, v_{ca}, v_{cb}$: a six-pulse waveform with ripple frequency $6f = 300$ Hz.
• Each segment lasts $60^\circ$, varying from $\frac{\sqrt3}{2}V_{mL}$ to $V_{mL}$ and back.

 vo  vab  vac  vbc  vba  vca  vcb
    _/\__/\__/\__/\__/\__/\_   <- peak VmL
                                  min 0.866 VmL
    30   90   150  210  270  330  390 (deg)
    six 60-degree caps per cycle

Average output voltage

Let $V_{mL} = \sqrt3 V_m$ be the peak line voltage. Taking one $60^\circ$ segment of a line voltage $V_{mL}\sin\theta$ from $\theta = \pi/3$ to $2\pi/3$:

$$\begin{aligned} V_{dc} &= \frac{6}{2\pi}\int_{\pi/3}^{2\pi/3}V_{mL}\sin\theta\,d\theta = \frac{3V_{mL}}{\pi}\left[\cos\frac{\pi}{3} - \cos\frac{2\pi}{3}\right] \\ &= \frac{3V_{mL}}{\pi} = \frac{3\sqrt3\,V_m}{\pi} = 1.654\,V_m = 1.35\,V_L \end{aligned}$$

RMS output voltage

$$\begin{aligned} V_{rms}^2 &= \frac{3}{\pi}\int_{\pi/3}^{2\pi/3}V_{mL}^2\sin^2\theta\,d\theta = \frac{3V_{mL}^2}{2\pi}\left[\theta - \frac{\sin2\theta}{2}\right]_{\pi/3}^{2\pi/3} \\ &= \frac{3V_{mL}^2}{2\pi}\left[\frac{\pi}{3} + \frac{\sqrt3}{2}\right] = 3V_m^2\left[\frac{1}{2} + \frac{3\sqrt3}{4\pi}\right] \end{aligned}$$ $$V_{rms} = V_m\left[\frac{3}{2} + \frac{9\sqrt3}{4\pi}\right]^{1/2} = 1.6554\,V_m$$

The rms value is almost equal to the average (form factor $= 1.6554/1.6540 = 1.0009$, ripple factor about 4%), so the output is very smooth.

Example: for a 400 V line supply, $V_{dc} = 1.35 \times 400 = 540$ V.

A three-phase single-way (half-wave, three-pulse) controlled rectifier uses three thyristors, one per phase, connected to a star-connected supply (or transformer secondary) with the load returned to the neutral. The output voltage is controlled by delaying the firing of each thyristor by $\alpha$ from its natural commutation point.

          T1
 a o------>|-----+
          T2     |
 b o------>|-----+------o P (+)
          T3     |        |
 c o------>|-----+       Load
                          |
 n o----------------------o N (-)
 a, b, c, n: star-connected secondary
 va = Vm sin wt, vb lags 120, vc lags 240

Operation

• Each thyristor's anode goes to one phase and the cathodes are joined to the positive output. A thyristor can conduct only when its phase is the most positive and it has been fired.
• The firing angle $\alpha$ is measured from the natural commutation point (the instant a diode would start conducting), i.e. from $30^\circ$ after the zero crossing of its phase voltage. So T1 is fired at $\omega t = 30^\circ + \alpha$, T2 at $150^\circ + \alpha$, T3 at $270^\circ + \alpha$.
• When T2 is fired, $v_b > v_a$, so T1 becomes reverse biased and turns off (natural/line commutation).
• With a highly inductive load, the load current is continuous and constant, so each thyristor conducts for exactly $120^\circ$, even if its phase voltage becomes negative. The output follows the phase of the conducting thyristor.

Waveform for α = 60°

 phase:      va           vb           vc
 fired at:   90           210          330   (deg)
 ends at:    210          330          450
 each phase is used from 90 to 210 deg of its
 own sine; vo goes negative for the last
 30 deg of every 120 deg pulse

 vo
      |`-.           |`-.           |`-.
      |   `-.        |   `-.        |   `-.
 -----+------`-.-----+------`-.-----+------`-.
                `-.  |         `-.  |         `-.
      90       180 210         330           450
 each pulse: Vm (at 90) -> 0 (at 180)
 -> -0.5 Vm (at 210), then jumps to next phase

Average output voltage

Taking the conduction of T1 (the waveform repeats every $2\pi/3$):

$$\begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\left(\frac{\pi}{6}+\alpha\right) - \cos\left(\frac{5\pi}{6}+\alpha\right)\right] \\ &= \frac{3V_m}{2\pi}\cdot 2\sin\left(\frac{\pi}{2}+\alpha\right)\sin\frac{\pi}{3} = \frac{3V_m}{2\pi}\cdot\sqrt3\cos\alpha \end{aligned}$$ $$V_{dc} = \frac{3\sqrt3\,V_m}{2\pi}\cos\alpha = 0.827\,V_m\cos\alpha$$

where $V_m$ is the peak phase voltage. Maximum value (at $\alpha = 0$) is $V_{dm} = 0.827V_m$, so $V_{dc} = V_{dm}\cos\alpha$.

• $0 \le \alpha < 90^\circ$: $V_{dc} > 0$, rectifier mode.
• $\alpha = 90^\circ$: $V_{dc} = 0$.
• $90^\circ < \alpha < 180^\circ$: $V_{dc} < 0$, inverter mode (needs a dc source in the load).

For $\alpha = 60^\circ$: $V_{dc} = 0.827V_m\cos60^\circ = 0.4135\,V_m$. (E.g. with 230 V per phase, $V_m = 325.27$ V and $V_{dc} = 134.50$ V.)

A three-phase single-way (half-wave, three-pulse) uncontrolled rectifier uses three diodes, one in each phase of a star-connected supply (or transformer secondary), with the load connected between the common cathode point and the neutral.

          D1
 a o------>|-----+
          D2     |
 b o------>|-----+------o P (+)
          D3     |        |
 c o------>|-----+       Load
                          |
 n o----------------------o N (-)
 a, b, c, n: star-connected secondary
 va = Vm sin wt, vb lags 120, vc lags 240

Operation and waveforms

• Each diode's anode is connected to one phase; the cathodes are joined to form the positive output. The diode connected to the most positive phase conducts and the other two are reverse biased.
• Measured from the zero crossing of $v_a$, $v_a$ is the highest from $30^\circ$ to $150^\circ$, $v_b$ from $150^\circ$ to $270^\circ$ and $v_c$ from $270^\circ$ to $390^\circ$. These crossover points ($30^\circ$, $150^\circ$, $270^\circ$) are the points of natural commutation.
• Each diode conducts for $120^\circ$; the output is the upper envelope of the three phase voltages, with three pulses per cycle (ripple frequency $3f = 150$ Hz), never falling below $V_m/2$.
• Load current returns through the neutral; each diode must block a peak inverse voltage of $\sqrt3 V_m$ (peak line voltage).

 vo     va        vb        vc
      _----_    _----_    _----_
     /      \__/      \__/      \
    /  D1     ^   D2    ^   D3    \
   30        150       270       390  (deg)
 vo = upper envelope; minimum Vm/2 at
 crossover points, peak Vm

Average output voltage

The output consists of three identical pulses per cycle; taking D1's pulse ($v_a$ from $30^\circ$ to $150^\circ$):

$$\begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6}^{5\pi/6}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\frac{\pi}{6} - \cos\frac{5\pi}{6}\right] \\ &= \frac{3V_m}{2\pi}\left[\frac{\sqrt3}{2} + \frac{\sqrt3}{2}\right] = \frac{3\sqrt3\,V_m}{2\pi} = 0.827\,V_m \end{aligned}$$

RMS output voltage

$$\begin{aligned} V_{rms}^2 &= \frac{3}{2\pi}\int_{\pi/6}^{5\pi/6}V_m^2\sin^2\omega t\,d(\omega t) = \frac{3V_m^2}{4\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\pi/6}^{5\pi/6} \\ &= \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} - \frac{1}{2}\left(-\frac{\sqrt3}{2} - \frac{\sqrt3}{2}\right)\right] = \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} + \frac{\sqrt3}{2}\right] \\ &= V_m^2\left[\frac{1}{2} + \frac{3\sqrt3}{8\pi}\right] \end{aligned}$$ $$V_{rms} = V_m\left[\frac{1}{2} + \frac{3\sqrt3}{8\pi}\right]^{1/2} = 0.8407\,V_m$$

Here $V_m$ is the peak phase voltage. Form factor $= 0.8407/0.8270 = 1.017$ and ripple factor $= \sqrt{FF^2-1} = 18.3\%$, much better than single-phase rectifiers (48% for full wave).

Example: with 230 V per phase, $V_m = 325.27$ V, $V_{dc} = 268.99$ V and $V_{rms} = 273.45$ V.

A three-phase fully controlled bridge rectifier (six-pulse full converter) uses six thyristors. With a resistive load the output current follows the output voltage, so conduction is continuous for $\alpha \le 60^\circ$ and discontinuous for $\alpha > 60^\circ$ (the line voltage reaches zero before the next thyristor is fired).

        P (+) o-----+------+------+-----+
                    |      |      |     |
                   T1     T3     T5     |
                    |      |      |     R
 a o----------------+      |      |     |
 b o-----------------------+      |     |
 c o------------------------------+     |
                    |      |      |     |
                   T4     T6     T2     |
                    |      |      |     |
        N (-) o-----+------+------+-----+
 (a joins T1-T4, b joins T3-T6, c joins T5-T2)

Operation (3 marks)

• T1, T3, T5 form the positive group and T4, T6, T2 the negative group. Thyristors are fired in the sequence T1, T2, T3, T4, T5, T6 at $60^\circ$ intervals; two thyristors (one from each group) conduct at a time, so the load sees a line voltage.
• $\alpha$ is measured from the natural commutation point, $30^\circ$ after the zero crossing of the phase voltage: T1 is fired at $\omega t = 30^\circ + \alpha$ (with respect to $v_a$).
• Each thyristor needs a double pulse (or a $60^\circ$+ wide pulse) so that the incoming thyristor and its partner are both on at start-up.

• With R load, for $\alpha \le 60^\circ$ each segment of $v_o$ stays positive and each thyristor conducts $120^\circ$.
• For $\alpha > 60^\circ$, the line voltage becomes zero inside the segment, the current drops to zero and the thyristors turn off; $v_o = 0$ until the next pair is fired.

Average and RMS output voltage (3 marks)

For $0 \le \alpha \le 60^\circ$ (continuous):

Using a line voltage $v_{ab} = V_{mL}\sin\theta$ ($V_{mL} = \sqrt3V_m$); the $60^\circ$ segment runs from $\theta = \pi/3 + \alpha$ to $2\pi/3 + \alpha$:

$$\begin{aligned} V_{dc} &= \frac{3}{\pi}\int_{\pi/3+\alpha}^{2\pi/3+\alpha}V_{mL}\sin\theta\,d\theta = \frac{3V_{mL}}{\pi}\left[\cos\left(\frac{\pi}{3}+\alpha\right) - \cos\left(\frac{2\pi}{3}+\alpha\right)\right] \\ &= \frac{3V_{mL}}{\pi}\cdot 2\sin\left(\frac{\pi}{2}+\alpha\right)\sin\frac{\pi}{6} = \frac{3V_{mL}}{\pi}\cos\alpha = \frac{3\sqrt3V_m}{\pi}\cos\alpha \end{aligned}$$ $$\begin{aligned} V_{rms}^2 &= \frac{3}{\pi}\int_{\pi/3+\alpha}^{2\pi/3+\alpha}V_{mL}^2\sin^2\theta\,d\theta = \frac{3V_{mL}^2}{2\pi}\left[\frac{\pi}{3} - \frac{\sin\left(\frac{4\pi}{3}+2\alpha\right) - \sin\left(\frac{2\pi}{3}+2\alpha\right)}{2}\right] \\ &= \frac{3V_{mL}^2}{2\pi}\left[\frac{\pi}{3} + \frac{\sqrt3}{2}\cos2\alpha\right] \end{aligned}$$ $$V_{rms} = V_{mL}\left[\frac{1}{2} + \frac{3\sqrt3}{4\pi}\cos2\alpha\right]^{1/2} = \sqrt3V_m\left[\frac{1}{2} + \frac{3\sqrt3}{4\pi}\cos2\alpha\right]^{1/2}$$

For $60^\circ < \alpha \le 120^\circ$ (discontinuous): each segment runs from $\pi/3 + \alpha$ to $\pi$:

$$V_{dc} = \frac{3V_{mL}}{\pi}\left[1 + \cos\left(\alpha + \frac{\pi}{3}\right)\right], \qquad V_{rms} = V_{mL}\left[\frac{3}{2\pi}\left(\frac{2\pi}{3} - \alpha + \frac{1}{2}\sin\left(2\alpha + \frac{2\pi}{3}\right)\right)\right]^{1/2}$$

Output becomes zero at $\alpha = 120^\circ$.

Waveforms for α = 0° and α = 30° (2 marks)

 alpha = 0 (same as diode bridge)
 vo  vab  vac  vbc  vba  vca  vcb
    _/\__/\__/\__/\__/\__/\_    peak VmL,
   30   90   150  210  270  330  min 0.866 VmL

 alpha = 30
 vo  each 60-deg segment starts at VmL (peak of
     the line voltage) and falls to 0.5 VmL:
    |`-.|`-.|`-.|`-.|`-.|`-.
   60   120  180  240  300  360  (deg)

• $\alpha = 0^\circ$: $V_{dc} = \frac{3V_{mL}}{\pi} = 1.35V_L$ (same as diode bridge); six pulses per cycle, ripple at 300 Hz.
• $\alpha = 30^\circ$: each pulse is the falling part of a line voltage from its peak $V_{mL}$ to $0.5V_{mL}$; $V_{dc} = 1.35V_L\cos30^\circ = 1.169V_L$.

A three-phase single-way (half-wave, three-pulse) controlled rectifier uses three thyristors, one per phase, connected to a star-connected supply (or transformer secondary) with the load returned to the neutral. The output voltage is controlled by delaying the firing of each thyristor by $\alpha$ from its natural commutation point.

          T1
 a o------>|-----+
          T2     |
 b o------>|-----+------o P (+)
          T3     |        |
 c o------>|-----+       Load
                          |
 n o----------------------o N (-)
 a, b, c, n: star-connected secondary
 va = Vm sin wt, vb lags 120, vc lags 240

Operation

• Each thyristor's anode goes to one phase and the cathodes are joined to the positive output. A thyristor can conduct only when its phase is the most positive and it has been fired.
• The firing angle $\alpha$ is measured from the natural commutation point (the instant a diode would start conducting), i.e. from $30^\circ$ after the zero crossing of its phase voltage. So T1 is fired at $\omega t = 30^\circ + \alpha$, T2 at $150^\circ + \alpha$, T3 at $270^\circ + \alpha$.
• When T2 is fired, $v_b > v_a$, so T1 becomes reverse biased and turns off (natural/line commutation).
• With a highly inductive load, the load current is continuous and constant, so each thyristor conducts for exactly $120^\circ$, even if its phase voltage becomes negative. The output follows the phase of the conducting thyristor.

Average output voltage

Taking the conduction of T1 (the waveform repeats every $2\pi/3$):

$$\begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\left(\frac{\pi}{6}+\alpha\right) - \cos\left(\frac{5\pi}{6}+\alpha\right)\right] \\ &= \frac{3V_m}{2\pi}\cdot 2\sin\left(\frac{\pi}{2}+\alpha\right)\sin\frac{\pi}{3} = \frac{3V_m}{2\pi}\cdot\sqrt3\cos\alpha \end{aligned}$$ $$V_{dc} = \frac{3\sqrt3\,V_m}{2\pi}\cos\alpha = 0.827\,V_m\cos\alpha$$

where $V_m$ is the peak phase voltage. Maximum value (at $\alpha = 0$) is $V_{dm} = 0.827V_m$, so $V_{dc} = V_{dm}\cos\alpha$.

• $0 \le \alpha < 90^\circ$: $V_{dc} > 0$, rectifier mode.
• $\alpha = 90^\circ$: $V_{dc} = 0$.
• $90^\circ < \alpha < 180^\circ$: $V_{dc} < 0$, inverter mode (needs a dc source in the load).

Waveform for α = 90°

 phase:      va           vb           vc
 fired at:   120          240          360   (deg)
 ends at:    240          360          480
 each phase is used from 120 to 240 deg of its
 own sine; positive and negative areas are
 equal, so Vdc = 0

 vo
      |`-.            |`-.            |`-.
 -----+---`-.-----+---+---`-.-----+---+---`-.
              `-. |           `-. |
      120  180  240           360
 each pulse: +0.866 Vm -> 0 (at 180)
 -> -0.866 Vm (at 240); equal areas

At $\alpha = 90^\circ$: $V_{dc} = 0.827V_m\cos90^\circ = 0$. The output is a train of $120^\circ$ pulses, each going from $+0.866V_m$ to $0$ to $-0.866V_m$; the positive and negative areas cancel, so the average output voltage and the average power are zero, although the load current still flows (it is sustained by the inductance and, in practice, requires a dc source in the load).

A three-phase single-way (half-wave, three-pulse) controlled rectifier uses three thyristors, one per phase, connected to a star-connected supply (or transformer secondary) with the load returned to the neutral. The output voltage is controlled by delaying the firing of each thyristor by $\alpha$ from its natural commutation point. Because it uses three thyristors it is also called a three-pulse midpoint converter. The load is assumed highly inductive so that the current is continuous.

          T1
 a o------>|-----+
          T2     |
 b o------>|-----+------o P (+)
          T3     |        |
 c o------>|-----+       Load
                          |
 n o----------------------o N (-)
 a, b, c, n: star-connected secondary
 va = Vm sin wt, vb lags 120, vc lags 240

Operation

• Each thyristor's anode goes to one phase and the cathodes are joined to the positive output. A thyristor can conduct only when its phase is the most positive and it has been fired.
• The firing angle $\alpha$ is measured from the natural commutation point (the instant a diode would start conducting), i.e. from $30^\circ$ after the zero crossing of its phase voltage. So T1 is fired at $\omega t = 30^\circ + \alpha$, T2 at $150^\circ + \alpha$, T3 at $270^\circ + \alpha$.
• When T2 is fired, $v_b > v_a$, so T1 becomes reverse biased and turns off (natural/line commutation).
• With a highly inductive load, the load current is continuous and constant, so each thyristor conducts for exactly $120^\circ$, even if its phase voltage becomes negative. The output follows the phase of the conducting thyristor.

 vo (alpha = 30 deg shown)
        .-.          .-.          .-.
      /    \       /    \       /    \
      |     \      |     \      |     \
 -----+------+-----+------+-----+------+--
      60    180   180    300   300    420
 iT1  +------+  (120 deg block of Io)
 For alpha > 30 the end of each pulse dips
 below zero; at alpha = 90 the average is 0

Average output voltage

Taking the conduction of T1 (the waveform repeats every $2\pi/3$):

$$\begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\left(\frac{\pi}{6}+\alpha\right) - \cos\left(\frac{5\pi}{6}+\alpha\right)\right] \\ &= \frac{3V_m}{2\pi}\cdot 2\sin\left(\frac{\pi}{2}+\alpha\right)\sin\frac{\pi}{3} = \frac{3V_m}{2\pi}\cdot\sqrt3\cos\alpha \end{aligned}$$ $$V_{dc} = \frac{3\sqrt3\,V_m}{2\pi}\cos\alpha = 0.827\,V_m\cos\alpha$$

where $V_m$ is the peak phase voltage. Maximum value (at $\alpha = 0$) is $V_{dm} = 0.827V_m$, so $V_{dc} = V_{dm}\cos\alpha$.

• $0 \le \alpha < 90^\circ$: $V_{dc} > 0$, rectifier mode.
• $\alpha = 90^\circ$: $V_{dc} = 0$.
• $90^\circ < \alpha < 180^\circ$: $V_{dc} < 0$, inverter mode (needs a dc source in the load).

RMS output voltage

$$\begin{aligned} V_{rms}^2 &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m^2\sin^2\omega t\,d(\omega t) = \frac{3V_m^2}{4\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\pi/6+\alpha}^{5\pi/6+\alpha} \\ &= \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} - \frac{\sin\left(\frac{5\pi}{3}+2\alpha\right) - \sin\left(\frac{\pi}{3}+2\alpha\right)}{2}\right] = \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} + \frac{\sqrt3}{2}\cos2\alpha\right] \end{aligned}$$

(using $\sin\left(\frac{5\pi}{3}+2\alpha\right) - \sin\left(\frac{\pi}{3}+2\alpha\right) = 2\cos(\pi+2\alpha)\sin\frac{2\pi}{3} = -\sqrt3\cos2\alpha$)

$$V_{rms} = V_m\left[\frac{1}{2} + \frac{3\sqrt3}{8\pi}\cos2\alpha\right]^{1/2}$$

At $\alpha = 0$ these reduce to the diode rectifier values $0.827V_m$ and $0.8407V_m$.

Thyristor ratings

Average thyristor current $= I_o/3$, rms $= I_o/\sqrt3$, peak reverse (and forward) voltage $= \sqrt3V_m$.

A three-phase single-way (half-wave, three-pulse) uncontrolled rectifier uses three diodes, one in each phase of a star-connected supply (or transformer secondary), with the load connected between the common cathode point and the neutral. It is the simplest three-phase rectifier and is also called the three-pulse midpoint (M3) rectifier.

          D1
 a o------>|-----+
          D2     |
 b o------>|-----+------o P (+)
          D3     |        |
 c o------>|-----+       Load
                          |
 n o----------------------o N (-)
 a, b, c, n: star-connected secondary
 va = Vm sin wt, vb lags 120, vc lags 240

Operation

• Each diode's anode is connected to one phase; the cathodes are joined to form the positive output. The diode connected to the most positive phase conducts and the other two are reverse biased.
• Measured from the zero crossing of $v_a$, $v_a$ is the highest from $30^\circ$ to $150^\circ$, $v_b$ from $150^\circ$ to $270^\circ$ and $v_c$ from $270^\circ$ to $390^\circ$. These crossover points ($30^\circ$, $150^\circ$, $270^\circ$) are the points of natural commutation.
• Each diode conducts for $120^\circ$; the output is the upper envelope of the three phase voltages, with three pulses per cycle (ripple frequency $3f = 150$ Hz), never falling below $V_m/2$.
• Load current returns through the neutral; each diode must block a peak inverse voltage of $\sqrt3 V_m$ (peak line voltage).

 vo     va        vb        vc
      _----_    _----_    _----_
     /      \__/      \__/      \
    /  D1     ^   D2    ^   D3    \
   30        150       270       390  (deg)
 vo = upper envelope; minimum Vm/2 at
 crossover points, peak Vm

Expression for average output voltage

Since the three pulses are identical, the average over one pulse ($2\pi/3$) equals the average over a cycle:

$$\begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6}^{5\pi/6}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\frac{\pi}{6} - \cos\frac{5\pi}{6}\right] \\ &= \frac{3V_m}{2\pi}\left[\frac{\sqrt3}{2} + \frac{\sqrt3}{2}\right] = \frac{3\sqrt3\,V_m}{2\pi} = 0.827\,V_m \end{aligned}$$

In terms of rms values: $V_{dc} = \frac{3\sqrt3}{2\pi}\sqrt2\,V_{ph} = 1.17\,V_{ph} = 0.675\,V_L$.

Remarks

• Each diode carries the full load current for $120^\circ$: $I_{D,avg} = I_o/3$, $I_{D,rms} = I_o/\sqrt3$ (inductive load).
• Peak inverse voltage per diode $= \sqrt3V_m$.
• The transformer secondary carries dc current (unidirectional), which can saturate the core; a zig-zag secondary or a bridge circuit avoids this.
• Example: $V_{ph} = 230$ V gives $V_{dc} = 0.827 \times 325.27 = 268.99$ V.

A three-phase single-way controlled rectifier has three thyristors, one per phase of a star-connected supply, with the load between the common cathode and the neutral. With a purely resistive load the current has the same shape as the voltage, so a thyristor turns off as soon as its phase voltage reaches zero. Two cases arise: continuous conduction for $\alpha \le 30^\circ$ and discontinuous conduction for $\alpha > 30^\circ$.

          T1
 a o------>|-----+
          T2     |
 b o------>|-----+------o P (+)
          T3     |        |
 c o------>|-----+       Load
                          |
 n o----------------------o N (-)
 a, b, c, n: star-connected secondary
 va = Vm sin wt, vb lags 120, vc lags 240

Operation

• $\alpha$ is measured from the natural commutation point ($30^\circ$ after the phase zero crossing): T1 is fired at $30^\circ+\alpha$, T2 at $150^\circ+\alpha$, T3 at $270^\circ+\alpha$.
• $\alpha \le 30^\circ$: when T2 is fired, $v_a$ is still positive but $v_b > v_a$, so T1 is commutated. Each thyristor conducts for $120^\circ$ and $v_o$ stays positive (continuous).
• $\alpha > 30^\circ$: $v_a$ reaches zero at $180^\circ$ before T2 is fired; the current falls to zero and T1 turns off. $v_o = 0$ until the next firing (discontinuous). Each thyristor conducts from $30^\circ + \alpha$ to $180^\circ$, i.e. $(150^\circ - \alpha)$.

Waveforms

 vo
        .-.          .-.          .-.
      /    \       /    \       /    \
      |     \      |     \      |     \
 -----+------+-----+------+-----+------+--
      60    180   180    300   300    420
 each pulse: 0.866 Vm -> Vm (at 90) -> 0

 vo (alpha = 60, R load)
      |`-.           |`-.           |`-.
      |   `-.        |   `-.        |   `-.
 -----+------`+------+------`+------+------`+
      90     180    210     300    330     420
 each pulse: Vm (at 90) -> 0 (at 180), then
 vo = 0 until the next thyristor is fired

Average and RMS output voltage

Case 1: $0 \le \alpha \le 30^\circ$ (T1 conducts $\pi/6+\alpha$ to $5\pi/6+\alpha$):

$$\begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\left(\frac{\pi}{6}+\alpha\right) - \cos\left(\frac{5\pi}{6}+\alpha\right)\right] \\ &= \frac{3V_m}{2\pi}\cdot 2\sin\left(\frac{\pi}{2}+\alpha\right)\sin\frac{\pi}{3} = \frac{3V_m}{2\pi}\cdot\sqrt3\cos\alpha \end{aligned}$$ $$V_{dc} = \frac{3\sqrt3\,V_m}{2\pi}\cos\alpha = 0.827\,V_m\cos\alpha$$ $$V_{rms} = V_m\left[\frac{1}{2} + \frac{3\sqrt3}{8\pi}\cos2\alpha\right]^{1/2}$$

Case 2: $30^\circ < \alpha \le 150^\circ$ (T1 conducts $\pi/6+\alpha$ to $\pi$):

$$\begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{\pi}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[1 + \cos\left(\alpha + \frac{\pi}{6}\right)\right] \end{aligned}$$ $$\begin{aligned} V_{rms}^2 &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{\pi}V_m^2\sin^2\omega t\,d(\omega t) = \frac{3V_m^2}{4\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\pi/6+\alpha}^{\pi} \\ &= \frac{3V_m^2}{4\pi}\left[\frac{5\pi}{6} - \alpha + \frac{1}{2}\sin\left(\frac{\pi}{3} + 2\alpha\right)\right] \end{aligned}$$ $$V_{rms} = V_m\left[\frac{3}{4\pi}\left(\frac{5\pi}{6} - \alpha + \frac{1}{2}\sin\left(\frac{\pi}{3}+2\alpha\right)\right)\right]^{1/2}$$

Numerical values

Assumption: supply 400 V line, i.e. $V_{ph} = 230$ V, $V_m = 325.27$ V (no voltage is given).

$$\begin{aligned} \alpha = 30^\circ: \quad V_{dc} &= 0.827 \times 325.27 \times \cos30^\circ = 232.96\ \text{V} \\ V_{rms} &= 325.27\left[0.5 + 0.2067 \times 0.5\right]^{1/2} = 252.66\ \text{V} \\ \alpha = 60^\circ: \quad V_{dc} &= \frac{3 \times 325.27}{2\pi}\left[1 + \cos90^\circ\right] = 155.30\ \text{V} \\ V_{rms} &= 325.27\left[\frac{3}{4\pi}\left(\frac{5\pi}{6} - \frac{\pi}{3} + \frac{1}{2}\sin\pi\right)\right]^{1/2} = 325.27\sqrt{0.375} = 199.19\ \text{V} \end{aligned}$$

Answer: $V_{dc} = 0.716V_m$ (232.96 V) at $\alpha = 30^\circ$ and $0.477V_m$ (155.30 V) at $\alpha = 60^\circ$.

A three-phase full bridge rectifier with six diodes (or six thyristors at $\alpha = 0$) connects the load at every instant to the largest line voltage: the most positive phase through the upper group (D1, D3, D5) and the most negative phase through the lower group (D4, D6, D2).

        P (+) o-----+------+------+-----+
                    |      |      |     |
                   D1     D3     D5     |
                    |      |      |     R
 a o----------------+      |      |     |
 b o-----------------------+      |     |
 c o------------------------------+     |
                    |      |      |     |
                   D4     D6     D2     |
                    |      |      |     |
        N (-) o-----+------+------+-----+
 (a joins D1-D4, b joins D3-D6, c joins D5-D2)

Operation and output waveform (R load)

 line voltages (peak VmL), six per cycle:
 vo  vab  vac  vbc  vba  vca  vcb
    _/\__/\__/\__/\__/\__/\_    max VmL
                                 min 0.866 VmL
   30   90   150  210  270  330  390 (deg)
 io = vo / R (same shape)

• Each diode conducts for $120^\circ$; a commutation occurs every $60^\circ$.
• Output is a six-pulse wave (ripple frequency $6f = 300$ Hz for 50 Hz), varying between $0.866V_{mL}$ and $V_{mL}$; with R load the current has the same shape.
• Each phase current is two $120^\circ$ pulses (positive and negative) per cycle, so no dc flows in the supply.

Average output voltage

Taking one segment of $v_{ab} = V_{mL}\sin\theta$ from $\pi/3$ to $2\pi/3$ (period $\pi/3$):

$$\begin{aligned} V_{dc} &= \frac{3}{\pi}\int_{\pi/3}^{2\pi/3}V_{mL}\sin\theta\,d\theta = \frac{3V_{mL}}{\pi}\left[-\cos\theta\right]_{\pi/3}^{2\pi/3} = \frac{3V_{mL}}{\pi}\left[\frac{1}{2} + \frac{1}{2}\right] \\ &= \frac{3V_{mL}}{\pi} = \frac{3\sqrt2\,V_L}{\pi} = 1.35\,V_L = 1.654\,V_m \end{aligned}$$

If the bridge uses thyristors fired at $\alpha$ (R load), $V_{dc} = 1.35V_L\cos\alpha$ for $\alpha \le 60^\circ$ and $V_{dc} = 1.35V_L\left[1+\cos(\alpha+60^\circ)\right]$ for $60^\circ < \alpha \le 120^\circ$.

Example: 400 V, 50 Hz supply gives $V_{dc} = 1.35 \times 400 = 540$ V, and $I_{dc} = 540/R$.

A three-phase single-way (half-wave, three-pulse) controlled rectifier uses three thyristors, one per phase, connected to a star-connected supply (or transformer secondary) with the load returned to the neutral. The output voltage is controlled by delaying the firing of each thyristor by $\alpha$ from its natural commutation point. The load is highly inductive, so the load current is continuous and constant.

          T1
 a o------>|-----+
          T2     |
 b o------>|-----+------o P (+)
          T3     |        |
 c o------>|-----+       Load
                          |
 n o----------------------o N (-)
 a, b, c, n: star-connected secondary
 va = Vm sin wt, vb lags 120, vc lags 240

Operation

• Each thyristor's anode goes to one phase and the cathodes are joined to the positive output. A thyristor can conduct only when its phase is the most positive and it has been fired.
• The firing angle $\alpha$ is measured from the natural commutation point (the instant a diode would start conducting), i.e. from $30^\circ$ after the zero crossing of its phase voltage. So T1 is fired at $\omega t = 30^\circ + \alpha$, T2 at $150^\circ + \alpha$, T3 at $270^\circ + \alpha$.
• When T2 is fired, $v_b > v_a$, so T1 becomes reverse biased and turns off (natural/line commutation).
• With a highly inductive load, the load current is continuous and constant, so each thyristor conducts for exactly $120^\circ$, even if its phase voltage becomes negative. The output follows the phase of the conducting thyristor.

Waveform for α = 30°

 phase:      va           vb           vc
 fired at:   60           180          300   (deg)
 ends at:    180          300          420
 each phase is used from 60 to 180 deg of its
 own sine; vo just touches zero at the end
 of each pulse (boundary of continuity)

 vo
        .-.          .-.          .-.
      /    \       /    \       /    \
      |     \      |     \      |     \
 -----+------+-----+------+-----+------+--
      60    180   180    300   300    420
 each pulse: 0.866 Vm -> Vm (at 90) -> 0

The thyristor currents are $120^\circ$ blocks of the constant load current $I_o$.

Average output voltage

Taking the conduction of T1 (the waveform repeats every $2\pi/3$):

$$\begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\left(\frac{\pi}{6}+\alpha\right) - \cos\left(\frac{5\pi}{6}+\alpha\right)\right] \\ &= \frac{3V_m}{2\pi}\cdot 2\sin\left(\frac{\pi}{2}+\alpha\right)\sin\frac{\pi}{3} = \frac{3V_m}{2\pi}\cdot\sqrt3\cos\alpha \end{aligned}$$ $$V_{dc} = \frac{3\sqrt3\,V_m}{2\pi}\cos\alpha = 0.827\,V_m\cos\alpha$$

where $V_m$ is the peak phase voltage. Maximum value (at $\alpha = 0$) is $V_{dm} = 0.827V_m$, so $V_{dc} = V_{dm}\cos\alpha$.

RMS output voltage

$$\begin{aligned} V_{rms}^2 &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m^2\sin^2\omega t\,d(\omega t) = \frac{3V_m^2}{4\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\pi/6+\alpha}^{5\pi/6+\alpha} \\ &= \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} - \frac{\sin\left(\frac{5\pi}{3}+2\alpha\right) - \sin\left(\frac{\pi}{3}+2\alpha\right)}{2}\right] = \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} + \frac{\sqrt3}{2}\cos2\alpha\right] \end{aligned}$$

(using $\sin\left(\frac{5\pi}{3}+2\alpha\right) - \sin\left(\frac{\pi}{3}+2\alpha\right) = 2\cos(\pi+2\alpha)\sin\frac{2\pi}{3} = -\sqrt3\cos2\alpha$)

$$V_{rms} = V_m\left[\frac{1}{2} + \frac{3\sqrt3}{8\pi}\cos2\alpha\right]^{1/2}$$

Calculation for α = 30°

Assumption: phase voltage 230 V (400 V line), $V_m = \sqrt2 \times 230 = 325.27$ V.

$$\begin{aligned} V_{dc} &= \frac{3\sqrt3 \times 325.27}{2\pi}\cos30^\circ = 268.99 \times 0.866 = 232.96\ \text{V} \\ V_{rms} &= 325.27\left[0.5 + \frac{3\sqrt3}{8\pi}\cos60^\circ\right]^{1/2} = 325.27 \times 0.7768 = 252.66\ \text{V} \end{aligned}$$

Answer: $V_{dc} = 0.716V_m$ = 232.96 V (for 230 V per phase).