Single phase AC to DC conversion

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 $V_s$ and peak $V_m$. The total output is

$$V_o = V_{o1} + V_{o2} = \frac{2V_m}{\pi}(\cos\alpha_1 + \cos\alpha_2)$$

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

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

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

$$V_o = \frac{2V_m}{\pi}(1 + \cos\alpha_2)$$

At $\alpha_2 = 180^\circ$ the two outputs cancel ($V_o = 0$); at $\alpha_2 = 0^\circ$, $V_o = \frac{4V_m}{\pi}$. So the output is varied smoothly from 0 to $\frac{4V_m}{\pi}$ 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 $E$ of reversed polarity (for example a motor in regenerative braking). One converter is held at $\alpha_1 = 180^\circ$ (full negative voltage, $-\frac{2V_m}{\pi}$) and the other is controlled between $0^\circ$ and $180^\circ$:

$$V_o = \frac{2V_m}{\pi}(\cos\alpha_2 - 1)$$

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

Waveforms (rectification, $\alpha_1 = 0$, $\alpha_2 = \alpha$, constant load current $I_o$)

 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

• $v_o = v_{o1} + v_{o2}$: during $0$ to $\alpha$ of each half cycle $v_{o2}$ is negative and reduces $v_o$; from $\alpha$ to $\pi$ both add.
• The primary current $i_s = i_{s1} + i_{s2}$ (equal turns) is a stepped wave: zero from $0$ to $\alpha$ and $2I_o$ from $\alpha$ to $\pi$. Its fundamental lags by only $\alpha/2$, so the displacement factor $\cos(\alpha/2)$ is better than $\cos\alpha$ 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 $0^\circ$ or $180^\circ$).

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 $V_s$ and peak $V_m$. The total output is

$$V_o = V_{o1} + V_{o2} = \frac{2V_m}{\pi}(\cos\alpha_1 + \cos\alpha_2)$$

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

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

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

$$V_o = \frac{2V_m}{\pi}(1 + \cos\alpha_2)$$

At $\alpha_2 = 180^\circ$ the two outputs cancel ($V_o = 0$); at $\alpha_2 = 0^\circ$, $V_o = \frac{4V_m}{\pi}$. So the output is varied smoothly from 0 to $\frac{4V_m}{\pi}$ 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 $E$ of reversed polarity (for example a motor in regenerative braking). One converter is held at $\alpha_1 = 180^\circ$ (full negative voltage, $-\frac{2V_m}{\pi}$) and the other is controlled between $0^\circ$ and $180^\circ$:

$$V_o = \frac{2V_m}{\pi}(\cos\alpha_2 - 1)$$

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

Waveforms (rectification, $\alpha_1 = 0$, $\alpha_2 = \alpha$, constant load current $I_o$)

 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

• $v_o = v_{o1} + v_{o2}$: during $0$ to $\alpha$ of each half cycle $v_{o2}$ is negative and reduces $v_o$; from $\alpha$ to $\pi$ both add.
• The primary current $i_s = i_{s1} + i_{s2}$ (equal turns) is a stepped wave: zero from $0$ to $\alpha$ and $2I_o$ from $\alpha$ to $\pi$. Its fundamental lags by only $\alpha/2$, so the displacement factor $\cos(\alpha/2)$ is better than $\cos\alpha$ 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 $0^\circ$ or $180^\circ$).

Given: single-phase full-wave diode rectifier, $V_s = 230$ V, 50 Hz; battery $E = 200$ V, internal resistance $R = 0.5\ \Omega$.

In a diode rectifier charging a battery of emf $E$ through resistance $R$, a diode pair conducts only when the instantaneous supply voltage exceeds $E$.

 v  ^      .--.          .--.
    |    .'    '.      .'    '.  vs (rectified)
  E |---+--------+----+--------+--- battery emf
    |  /|        |\  /|        |
    | / |  io    | \/ |        |
 ---+---+--------+----+--------+--> wt
       t1     pi-t1  pi+t1
 io  ^   .--.          .--.
     |  /    \        /    \
 ----+-+------+------+------+--> wt

Conduction starts at $\theta_1 = \sin^{-1}(E/V_m)$ and ends at $\pi - \theta_1$ in every half cycle. During conduction $i_o = \frac{V_m\sin\omega t - E}{R}$.

$$\begin{aligned} I_{avg} &= \frac{1}{\pi R}\left[2V_m\cos\theta_1 - E(\pi - 2\theta_1)\right] \\ I_{rms}^2 &= \frac{1}{\pi R^2}\left[\left(\frac{V_m^2}{2} + E^2\right)(\pi - 2\theta_1) + \frac{V_m^2}{2}\sin 2\theta_1 - 4V_mE\cos\theta_1\right] \end{aligned}$$

Power to the battery $= E I_{avg}$; loss in the resistor $= I_{rms}^2 R$; rectifier output $= E I_{avg} + I_{rms}^2 R$.

Conduction angle

$$\begin{aligned} V_m &= \sqrt{2} \times 230 = 325.27\ \text{V} \\ \theta_1 &= \sin^{-1}\frac{200}{325.27} = 37.94^\circ = 0.6622\ \text{rad}, \qquad \pi - 2\theta_1 = 1.8171\ \text{rad} \end{aligned}$$

(i) Average charging current

$$\begin{aligned} I_{avg} &= \frac{1}{\pi \times 0.5}\left[2 \times 325.27 \times \cos 37.94^\circ - 200 \times 1.8171\right] \\ &= \frac{513.03 - 363.43}{1.5708} = 95.24\ \text{A} \end{aligned}$$

(ii) Power supplied to the battery

$$P_b = E I_{avg} = 200 \times 95.24 = 19048\ \text{W} \approx 19.05\ \text{kW}$$

(iii) Gross power output of the rectifier

Using the rms formula with $\sin 2\theta_1 = 0.9698$, $\cos\theta_1 = 0.7886$:

$$I_{rms} = 137.75\ \text{A}, \qquad I_{rms}^2 R = 137.75^2 \times 0.5 = 9487.5\ \text{W}$$ $$P_{gross} = P_b + I_{rms}^2 R = 19048 + 9487.5 = 28535.5\ \text{W} \approx 28.54\ \text{kW}$$

(iv) Extra series resistance for 30% less current

$\theta_1$ depends only on $E$ and $V_m$, so $I_{avg} \propto 1/R_{total}$. For $I_{avg}' = 0.7 I_{avg}$:

$$\begin{aligned} R_{total} &= \frac{0.5}{0.7} = 0.7143\ \Omega \\ R_{add} &= 0.7143 - 0.5 = 0.2143\ \Omega \end{aligned}$$

(New $I_{avg} = 0.7 \times 95.24 = 66.67$ A.)

Answer: $I_{avg} = 95.24$ A; $P_b = 19.05$ kW; $P_{gross} = 28.54$ kW; additional resistance $= 0.214\ \Omega$.

Given: single-phase diode bridge, $V_s = 15$ V rms, 50 Hz; $R = 2\ \Omega$; battery $E = 12$ V.

(i) Waveforms

In a diode rectifier charging a battery of emf $E$ through resistance $R$, a diode pair conducts only when the instantaneous supply voltage exceeds $E$.

 v  ^      .--.          .--.
    |    .'    '.      .'    '.  vs (rectified)
  E |---+--------+----+--------+--- battery emf
    |  /|        |\  /|        |
    | / |  io    | \/ |        |
 ---+---+--------+----+--------+--> wt
       t1     pi-t1  pi+t1
 io  ^   .--.          .--.
     |  /    \        /    \
 ----+-+------+------+------+--> wt

Conduction starts at $\theta_1 = \sin^{-1}(E/V_m)$ and ends at $\pi - \theta_1$ in every half cycle. During conduction $i_o = \frac{V_m\sin\omega t - E}{R}$.

$$\begin{aligned} I_{avg} &= \frac{1}{\pi R}\left[2V_m\cos\theta_1 - E(\pi - 2\theta_1)\right] \\ I_{rms}^2 &= \frac{1}{\pi R^2}\left[\left(\frac{V_m^2}{2} + E^2\right)(\pi - 2\theta_1) + \frac{V_m^2}{2}\sin 2\theta_1 - 4V_mE\cos\theta_1\right] \end{aligned}$$

Power to the battery $= E I_{avg}$; loss in the resistor $= I_{rms}^2 R$; rectifier output $= E I_{avg} + I_{rms}^2 R$.

(ii) Average and rms charging current

$$\begin{aligned} V_m &= \sqrt{2} \times 15 = 21.21\ \text{V} \\ \theta_1 &= \sin^{-1}\frac{12}{21.21} = 34.45^\circ = 0.6013\ \text{rad}, \qquad \pi - 2\theta_1 = 1.9391\ \text{rad} \\ I_{avg} &= \frac{1}{2\pi}\left[2 \times 21.21 \times 0.8246 - 12 \times 1.9391\right] = \frac{34.99 - 23.27}{6.283} = 1.865\ \text{A} \end{aligned}$$

With $\sin 2\theta_1 = 0.9330$:

$$I_{rms}^2 = \frac{1}{4\pi}\left[(225 + 144)(1.9391) + 225(0.9330) - 4(21.21)(12)(0.8246)\right] = 6.826$$ $$I_{rms} = 2.613\ \text{A}$$

(iii) Power supplied to the battery

$$P_b = E I_{avg} = 12 \times 1.865 = 22.38\ \text{W}$$

(iv) Power output of the rectifier

$$P_o = P_b + I_{rms}^2 R = 22.38 + 2.613^2 \times 2 = 22.38 + 13.65 = 36.03\ \text{W}$$

(v) Efficiency of charging

$$\eta = \frac{P_b}{P_o} = \frac{22.38}{36.03} = 62.1\%$$

Answer: $I_{avg} = 1.865$ A, $I_{rms} = 2.613$ A, $P_b = 22.38$ W, $P_o = 36.03$ W, $\eta = 62.1\%$.

A single-phase full converter is a fully controlled bridge of four thyristors. With a highly inductive load, the load current is continuous and constant ($I_o$), and the output voltage can be positive or negative (two-quadrant converter).

        is      +------+------+
   +---->-------| T1   |   T3 |
   |            |  K   |    K |
  ~ Vs          +--+---+---+--+---- + 
   |               |       |      load
   |            | T4   |   T2 |   (highly
   +------------|      |      |   inductive)
                +------+------+---- -   Io

Operation

1. $\alpha$ to $\pi + \alpha$: in the positive half cycle, $T_1$ and $T_2$ are fired at $\omega t = \alpha$. Load current flows $T_1 \to$ load $\to T_2$, so $v_o = v_s$ and $i_s = +I_o$. After $\omega t = \pi$, $v_s$ is negative, but $T_1, T_2$ keep conducting because the inductance maintains the current; $v_o$ becomes negative.
2. At $\pi + \alpha$: $T_3$ and $T_4$ are fired. The supply voltage reverse biases $T_1, T_2$, which turn OFF (line commutation). Now $v_o = -v_s$ and $i_s = -I_o$.
3. The cycle repeats. Average output voltage:

$$V_o = \frac{1}{\pi}\int_{\alpha}^{\pi+\alpha} V_m \sin\omega t \, d(\omega t) = \frac{2V_m}{\pi}\cos\alpha$$

For $\alpha < 90^\circ$ the converter rectifies; for $\alpha > 90^\circ$ (with a suitable d.c. source in the load) it inverts.

Waveforms for $\alpha = 30^\circ$

 vs    .-.         .-.
      /   \       /   \
 ----/-----\-----/-----\---> wt
             '-'         '-'
 vo    _          _
      | '.       | '.
 -----+---'._.---+---'._.--> wt
      a    pi    pi+a  (negative
     30         210     part)
 is   +Io_________
      |           |
 -----+           |      +---> wt
                  |______|
                  -Io
      T1,T2 ON    T3,T4 ON

The input current $i_s$ is a square wave of amplitude $I_o$, lagging the supply voltage by $\alpha$.

Average output voltage at $\alpha = 30^\circ$

$$\begin{aligned} V_m &= \sqrt{2} \times 230 = 325.27\ \text{V} \\ V_o &= \frac{2V_m}{\pi}\cos\alpha = \frac{2 \times 325.27}{\pi}\cos 30^\circ = 207.07 \times 0.866 = 179.33\ \text{V} \end{aligned}$$

Other values: $I_o = 10$ A, so the load power is $179.33 \times 10 = 1793.3$ W; the rms input current is $I_s = I_o = 10$ A.

Answer: $V_o = 179.33$ V.

Given: single-phase diode bridge, $V_s = 30$ V rms; $R = 5\ \Omega$; battery $E = 24$ V.

In a diode rectifier charging a battery of emf $E$ through resistance $R$, a diode pair conducts only when the instantaneous supply voltage exceeds $E$.

 v  ^      .--.          .--.
    |    .'    '.      .'    '.  vs (rectified)
  E |---+--------+----+--------+--- battery emf
    |  /|        |\  /|        |
    | / |  io    | \/ |        |
 ---+---+--------+----+--------+--> wt
       t1     pi-t1  pi+t1
 io  ^   .--.          .--.
     |  /    \        /    \
 ----+-+------+------+------+--> wt

Conduction starts at $\theta_1 = \sin^{-1}(E/V_m)$ and ends at $\pi - \theta_1$ in every half cycle. During conduction $i_o = \frac{V_m\sin\omega t - E}{R}$.

$$\begin{aligned} I_{avg} &= \frac{1}{\pi R}\left[2V_m\cos\theta_1 - E(\pi - 2\theta_1)\right] \\ I_{rms}^2 &= \frac{1}{\pi R^2}\left[\left(\frac{V_m^2}{2} + E^2\right)(\pi - 2\theta_1) + \frac{V_m^2}{2}\sin 2\theta_1 - 4V_mE\cos\theta_1\right] \end{aligned}$$

Power to the battery $= E I_{avg}$; loss in the resistor $= I_{rms}^2 R$; rectifier output $= E I_{avg} + I_{rms}^2 R$.

(i) Average charging current

$$\begin{aligned} V_m &= \sqrt{2} \times 30 = 42.43\ \text{V} \\ \theta_1 &= \sin^{-1}\frac{24}{42.43} = 34.45^\circ = 0.6013\ \text{rad}, \qquad \pi - 2\theta_1 = 1.9391\ \text{rad} \\ I_{avg} &= \frac{1}{5\pi}\left[2 \times 42.43 \times 0.8246 - 24 \times 1.9391\right] \\ &= \frac{69.97 - 46.54}{15.708} = 1.492\ \text{A} \end{aligned}$$

(ii) Efficiency of the charging system

$$\begin{aligned} I_{rms}^2 &= \frac{1}{25\pi}\left[(900 + 576)(1.9391) + 900(0.9330) - 4(42.43)(24)(0.8246)\right] = 4.368 \\ I_{rms} &= 2.090\ \text{A} \\ P_b &= E I_{avg} = 24 \times 1.492 = 35.80\ \text{W} \\ P_R &= I_{rms}^2 R = 4.368 \times 5 = 21.84\ \text{W} \\ \eta &= \frac{P_b}{P_b + P_R} = \frac{35.80}{57.64} = 62.1\% \end{aligned}$$

Answer: $I_{avg} = 1.49$ A and charging efficiency $= 62.1\%$.

A single-phase full converter is a fully controlled bridge of four thyristors. With a highly inductive load, the load current is continuous and constant ($I_o$), and the output voltage can be positive or negative (two-quadrant converter).

        is      +------+------+
   +---->-------| T1   |   T3 |
   |            |  K   |    K |
  ~ Vs          +--+---+---+--+---- + 
   |               |       |      load
   |            | T4   |   T2 |   (highly
   +------------|      |      |   inductive)
                +------+------+---- -   Io

Operation

1. $\alpha$ to $\pi + \alpha$: in the positive half cycle, $T_1$ and $T_2$ are fired at $\omega t = \alpha$. Load current flows $T_1 \to$ load $\to T_2$, so $v_o = v_s$ and $i_s = +I_o$. After $\omega t = \pi$, $v_s$ is negative, but $T_1, T_2$ keep conducting because the inductance maintains the current; $v_o$ becomes negative.
2. At $\pi + \alpha$: $T_3$ and $T_4$ are fired. The supply voltage reverse biases $T_1, T_2$, which turn OFF (line commutation). Now $v_o = -v_s$ and $i_s = -I_o$.
3. The cycle repeats. Average output voltage:

$$V_o = \frac{1}{\pi}\int_{\alpha}^{\pi+\alpha} V_m \sin\omega t \, d(\omega t) = \frac{2V_m}{\pi}\cos\alpha$$

For $\alpha < 90^\circ$ the converter rectifies; for $\alpha > 90^\circ$ (with a suitable d.c. source in the load) it inverts.

Waveforms for $\alpha = 30^\circ$

 vs    .-.         .-.
      /   \       /   \
 ----/-----\-----/-----\---> wt
             '-'         '-'
 vo    _          _
      | '.       | '.
 -----+---'._.---+---'._.--> wt
      a    pi    pi+a  (negative
     30         210     part)
 is   +Io_________
      |           |
 -----+           |      +---> wt
                  |______|
                  -Io
      T1,T2 ON    T3,T4 ON

The input current $i_s$ is a square wave of amplitude $I_o$, lagging the supply voltage by $\alpha$.

Fundamental component of $i_s$ for $I_o = 20$ A, $\alpha = 30^\circ$

The Fourier series of a square wave of amplitude $I_o$ gives the fundamental peak value

$$I_{s1(peak)} = \frac{4 I_o}{\pi} = \frac{4 \times 20}{\pi} = 25.46\ \text{A}$$

so its rms value is

$$I_{s1} = \frac{4 I_o}{\pi\sqrt{2}} = \frac{2\sqrt{2}}{\pi} I_o = 0.9003 \times 20 = 18.01\ \text{A}$$

The fundamental lags the supply voltage by $\phi_1 = \alpha = 30^\circ$:

$$i_{s1}(t) = 25.46\sin(\omega t - 30^\circ)\ \text{A}$$

(For reference: $I_s = 20$ A rms, displacement factor $\cos 30^\circ = 0.866$, input power factor $= 0.9003 \times 0.866 = 0.78$ lagging.)

Answer: fundamental of $i_s$ = 18.01 A rms (25.46 A peak), lagging $v_s$ by $30^\circ$.

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 $V_s$ and peak $V_m$. The total output is

$$V_o = V_{o1} + V_{o2} = \frac{2V_m}{\pi}(\cos\alpha_1 + \cos\alpha_2)$$

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

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

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

$$V_o = \frac{2V_m}{\pi}(1 + \cos\alpha_2)$$

At $\alpha_2 = 180^\circ$ the two outputs cancel ($V_o = 0$); at $\alpha_2 = 0^\circ$, $V_o = \frac{4V_m}{\pi}$. So the output is varied smoothly from 0 to $\frac{4V_m}{\pi}$ 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 $E$ of reversed polarity (for example a motor in regenerative braking). One converter is held at $\alpha_1 = 180^\circ$ (full negative voltage, $-\frac{2V_m}{\pi}$) and the other is controlled between $0^\circ$ and $180^\circ$:

$$V_o = \frac{2V_m}{\pi}(\cos\alpha_2 - 1)$$

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

Example

With $V_s = 230$ V on each secondary, $V_m = 325.27$ V and the maximum d.c. voltage is $\frac{4 \times 325.27}{\pi} = 414.1$ V, twice that of a single full converter (207.1 V), while each thyristor still blocks only $V_m = 325.3$ V.

Advantages

• High output voltage with devices of half the voltage rating.
• Better input power factor and lower harmonics, because only one converter is phase-controlled at a time.

Given: single-phase full-bridge diode rectifier, $V_s = 230$ V, 50 Hz; $R = 10\ \Omega$ with large inductance, so the load current is constant ($I_o$).

       is  +----D1----+----D3----+
   +--->---+          |          |
   |       |          +-- (+) ---+
  ~ Vs     |         load R + L  |  Io
   |       |          +-- (-) ---+
   +-------+----D4----+----D2----+

$D_1, D_2$ conduct in the positive half cycle and $D_3, D_4$ in the negative half cycle, each for $180^\circ$. The input current is a square wave of $\pm I_o$, in phase with $v_s$.

$$V_m = \sqrt{2} \times 230 = 325.27\ \text{V}$$

(i) Average output voltage and current

$$\begin{aligned} V_o &= \frac{2V_m}{\pi} = \frac{2 \times 325.27}{\pi} = 207.07\ \text{V} \\ I_o &= \frac{V_o}{R} = \frac{207.07}{10} = 20.71\ \text{A} \end{aligned}$$

(ii) Diode currents

Each diode carries $I_o$ for half the period:

$$\begin{aligned} I_{D(avg)} &= \frac{I_o}{2} = 10.35\ \text{A} \\ I_{D(rms)} &= \frac{I_o}{\sqrt{2}} = \frac{20.71}{\sqrt{2}} = 14.64\ \text{A} \end{aligned}$$

(iii) RMS output and input currents

• Output current is constant: $I_{o(rms)} = I_o = 20.71$ A.
• Input current is a square wave of amplitude $I_o$: $I_s = I_o = 20.71$ A.

(iv) Supply power factor

$$\begin{aligned} I_{s1} &= \frac{2\sqrt{2}}{\pi}I_o = 0.9003 \times 20.71 = 18.64\ \text{A} \\ \text{DF} &= \cos 0^\circ = 1 \\ \text{PF} &= \frac{I_{s1}}{I_s}\cos\phi_1 = 0.9003 \times 1 = 0.9003 \end{aligned}$$

Check: $P = V_o I_o = 207.07 \times 20.71 = 4288$ W and $V_s I_s \times \text{PF} = 230 \times 20.71 \times 0.9003 = 4288$ W.

Answer: $V_o = 207.07$ V, $I_o = 20.71$ A; diode $I_{avg} = 10.35$ A, $I_{rms} = 14.64$ A; output and input rms currents $= 20.71$ A; supply PF $= 0.90$.

In a series (dual-transformer) full converter, two single-phase full converters are fed from two separate secondary windings of the same transformer, and their dc outputs are connected in series so that the load voltage is the sum of the two outputs: $v_o = v_{o1} + v_{o2}$.

 Sec 1 --> [Converter 1, a1] (+)--------+
                 (-)                    |
                  |    vo1              |
                 (+)                  Load
 Sec 2 --> [Converter 2, a2]          (Io)
                 (-)   vo2              |
                  +---------------------+
 Sec 1, Sec 2: two secondaries of one transformer
 vo = vo1 + vo2

Output voltage

With a continuous (highly inductive) load current, each converter gives $\frac{2V_m}{\pi}\cos\alpha$, so

$$V_{dc} = \frac{2V_m}{\pi}\left(\cos\alpha_1 + \cos\alpha_2\right)$$

The maximum output is $\frac{4V_m}{\pi}$ (both $\alpha = 0$), i.e. twice that of one converter. Hence the scheme is used for high-voltage loads.

Operation in rectification mode

• Converter 1 is kept fully ON at $\alpha_1 = 0$ (it acts like a diode bridge).
• Converter 2's firing angle $\alpha_2$ is varied from $0$ to $\pi$.
• The output then varies from $\frac{4V_m}{\pi}$ (at $\alpha_2 = 0$) down to $0$ (at $\alpha_2 = \pi$): $V_{dc} = \frac{2V_m}{\pi}(1 + \cos\alpha_2)$.
• During the interval $\pi$ to $\pi + \alpha_2$, $v_{o1} = +|v_s|$ and $v_{o2} = -|v_s|$, so $v_o = 0$: the load current effectively freewheels and no reactive power is drawn in that interval.

(For inversion, converter 1 is held at $\alpha_1 = \pi$ and $\alpha_2$ is varied.)

Advantages

• Higher output voltage with lower device voltage rating.
• Better input power factor and lower harmonics than a single converter at the same output, because only one converter is phase-controlled at a time.

A single-phase full controlled rectifier (full converter) uses four thyristors in a bridge. With a highly inductive load the load current is continuous and constant ($I_o = 25$ A), so each thyristor pair conducts for $180^\circ$.

              P (+)
        +-------+-------+
        |               |
       T1              T3
        |               |
 a o----+               +----o b
        |               |
       T4              T2
        |               |
        +-------+-------+
              N (-)
 ac supply vs between a and b (is into a)
 Load (R-L, Io) between P and N

i) Operation for α = 30°

Data: $V_s = 230$ V, $V_m = \sqrt2 \times 230 = 325.27$ V, $\alpha = 30^\circ$.

• At $\omega t = 30^\circ$, T1 and T2 are fired; the load is connected to $v_s$.
• After $180^\circ$, $v_s$ becomes negative but T1, T2 keep conducting because the inductance keeps the current flowing; $v_o$ goes negative for $30^\circ$.
• At $210^\circ$, T3 and T4 are fired. They apply reverse voltage to T1, T2, which turn off (natural/line commutation). The cycle repeats.
• Input current $i_s$ is a square wave of $\pm 25$ A, lagging $v_s$ by $\alpha = 30^\circ$.

 vs       ___               ___
         /   \             /
 -------/-----\-----------/----
               \___/
 vo: follows vs from 30 to 210 deg, then
     -vs from 210 to 390 deg (goes below
     zero for 30 deg in each half cycle)
 io: constant 25 A
 is
 +Io      +---------+         +---
          |         |         |
  0 ------+         |         |
                    +---------+
 -Io
         30        210       390   (deg)

ii) Average and RMS output voltage

Output repeats every $\pi$; $v_o = V_m\sin\omega t$ from $\alpha$ to $\pi + \alpha$:

$$\begin{aligned} V_{dc} &= \frac{1}{\pi}\int_{\alpha}^{\pi+\alpha} V_m \sin\omega t\, d(\omega t) = \frac{V_m}{\pi}\big[-\cos\omega t\big]_{\alpha}^{\pi+\alpha} \\ &= \frac{2V_m}{\pi}\cos\alpha \end{aligned}$$ $$\begin{aligned} V_{rms} &= \left[\frac{1}{\pi}\int_{\alpha}^{\pi+\alpha} V_m^2 \sin^2\omega t\, d(\omega t)\right]^{1/2} = \left[\frac{V_m^2}{2\pi}\int_{\alpha}^{\pi+\alpha}(1-\cos 2\omega t)\, d(\omega t)\right]^{1/2} \\ &= \frac{V_m}{\sqrt2} = V_s \end{aligned}$$

So the rms output equals the supply rms (230 V) for any $\alpha$ in continuous conduction.

iii) Average output voltage

$$\begin{aligned} V_{dc} &= \frac{2 \times 325.27}{\pi}\cos 30^\circ = 207.07 \times 0.866 \\ &= 179.33\ \text{V} \end{aligned}$$

Answer: $V_{dc} = 179.33$ V, $V_{rms} = 230$ V.

Symmetrical angle control is a power factor improvement method in which a forced-commutated switch is turned ON at $\alpha = (\pi-\beta)/2$ and turned OFF at $(\pi+\beta)/2$, so that the conduction interval $\beta$ is placed symmetrically about the peak of the supply voltage ($\omega t = \pi/2$). As a result, the fundamental input current is in phase with the supply voltage and the displacement factor is unity.

Why it is needed

In a phase-controlled (thyristor) converter the input current always lags the voltage by $\alpha$ (or $\alpha/2$ in a semiconverter). At large $\alpha$ the input power factor becomes very poor. Thyristors cannot be turned off at will, but GTOs, IGBTs or power transistors can, so the position of the current pulse can be chosen freely.

Circuit

              P (+)
        +-------+-------+-------+
        |       |       |       |
       S1      S2       |       |
        |       |      Dm     Load
 a o----+       +--o b  |    (Ia)
        |       |       |       |
       D1      D2       |       |
        |       |       |       |
        +-------+-------+-------+
              N (-)
 S1, S2: forced-commutated switches (GTO/IGBT)
 Dm: freewheeling diode

Operation (highly inductive load, current $I_a$ constant)

1. Positive half cycle: S1 is turned ON at $\omega t = (\pi-\beta)/2$. Current flows $a \to$ S1 $\to$ load $\to$ D2 $\to b$; $v_o = v_s$, $i_s = +I_a$.
2. S1 is turned OFF at $(\pi+\beta)/2$ by a gate signal. The load current transfers to the freewheeling diode Dm; $v_o = 0$, $i_s = 0$.
3. Negative half cycle: S2 is turned ON at $(3\pi-\beta)/2$ and OFF at $(3\pi+\beta)/2$; current flows $b \to$ S2 $\to$ load $\to$ D1 $\to a$; $v_o = -v_s$ (positive), $i_s = -I_a$.
4. Rest of the time the load freewheels through Dm.

 is
 +Ia       +-----+
           |     |
  0  ------+     +-------+     +------
                         |     |
 -Ia                     +-----+
     (90-b/2)(90+b/2) (270-b/2)(270+b/2)
 b = beta; vs is positive 0-180, negative 180-360

Output voltage

$$V_{dc} = \frac{2}{2\pi}\int_{(\pi-\beta)/2}^{(\pi+\beta)/2} V_m\sin\omega t\, d(\omega t) = \frac{2V_m}{\pi}\sin\frac{\beta}{2}$$ $$V_{rms} = \frac{V_m}{\sqrt2}\left[\frac{1}{\pi}(\beta + \sin\beta)\right]^{1/2}$$

Input current and power factor

• Input current: quasi-square wave of $\pm I_a$, width $\beta$, centred on the voltage peaks.
• RMS input current: $I_s = I_a\sqrt{\beta/\pi}$
• Fundamental (rms): $I_{s1} = \frac{2\sqrt2 I_a}{\pi}\sin\frac{\beta}{2}$, with displacement angle $\phi_1 = 0$.
• Displacement factor $= \cos 0 = 1$.
• Power factor:

$$PF = \frac{I_{s1}}{I_s}\cos\phi_1 = \frac{2\sqrt2\,\sin(\beta/2)}{\sqrt{\pi\beta}}$$

Merits

• Unity displacement factor at all output levels; only harmonic (distortion) factor reduces PF.
• Less reactive power drawn from the supply.
• Requires self-commutated devices and gate turn-off circuitry, so cost is higher.

In extinction angle control, the switches are turned ON at the voltage zero crossing ($\omega t = 0$) and turned OFF at $\omega t = \pi - \beta$, where $\beta$ is the extinction angle. GTOs are needed because the turn-off occurs while the supply voltage is still positive (forced turn-off). The input current then leads the voltage, which improves the power factor.

Data: $V_s = 230$ V, $V_m = \sqrt2 \times 230 = 325.27$ V, $I_o = 20$ A, $\beta = 30^\circ = \pi/6$.

              P (+)
        +-------+-------+
        |               |
       S1              S3
        |               |
 a o----+               +----o b
        |               |
       S4              S2
        |               |
        +-------+-------+
              N (-)
 S1..S4: GTOs (or IGBTs) with anti-parallel diodes
 vs = 230 V, 50 Hz between a and b
 Highly inductive load (Io) between P and N

i) Waveforms and operation

During the freewheeling intervals the load is shorted by two switches of one leg, so $v_o = 0$ and no current is drawn from the supply.

 vo  (load voltage, period 180 deg)
     _____          _____
    /     |        /     |
   /      |       /      |
 -/-------+------/-------+-----
  0    180-b   180    360-b  360
 io  ============================  Io (constant)
 is
 +Io +-------+
     |       |
  0 -+       +---+         +---
                 |         |
 -Io             +---------+
  0    180-b 180      360-b 360     (b = beta)

ii) Average and RMS output voltage

$$\begin{aligned} V_{dc} &= \frac{1}{\pi}\int_0^{\pi-\beta} V_m\sin\omega t\, d(\omega t) = \frac{V_m}{\pi}(1+\cos\beta) \\ &= \frac{325.27}{\pi}(1 + 0.866) = 103.54 \times 1.866 = 193.20\ \text{V} \end{aligned}$$ $$\begin{aligned} V_{rms} &= \left[\frac{1}{\pi}\int_0^{\pi-\beta} V_m^2\sin^2\omega t\, d(\omega t)\right]^{1/2} = \frac{V_m}{\sqrt2}\left[\frac{1}{\pi}\left(\pi-\beta+\frac{\sin2\beta}{2}\right)\right]^{1/2} \\ &= 230\left[\frac{1}{\pi}\left(2.618 + 0.433\right)\right]^{1/2} = 230 \times 0.9855 = 226.66\ \text{V} \end{aligned}$$

iii) Fundamental component of input current

Fourier analysis of $i_s$ (amplitude $I_o$, width $\pi-\beta$, starting at $0$):

$$\begin{aligned} a_1 &= \frac{2I_o}{\pi}\sin\beta = \frac{40}{\pi}(0.5) = 6.366\ \text{A} \\ b_1 &= \frac{2I_o}{\pi}(1+\cos\beta) = \frac{40}{\pi}(1.866) = 23.759\ \text{A} \\ I_{s1,peak} &= \sqrt{a_1^2+b_1^2} = \frac{4I_o}{\pi}\cos\frac{\beta}{2} = 24.597\ \text{A} \\ I_{s1} &= \frac{24.597}{\sqrt2} = 17.39\ \text{A (rms)} \\ \phi_1 &= \tan^{-1}\frac{a_1}{b_1} = \frac{\beta}{2} = 15^\circ\ \text{(leading)} \end{aligned}$$

So $i_{s1} = 24.60\sin(\omega t + 15^\circ)$ A.

Answer: $V_{dc} = 193.20$ V, $V_{rms} = 226.66$ V, $I_{s1} = 17.39$ A (rms) leading $v_s$ by $15^\circ$. (Displacement factor $= \cos 15^\circ = 0.966$ leading.)

A single-phase full converter uses four thyristors in a bridge. With a highly inductive load the current is continuous and constant ($I_o = 25$ A), so it can work as a rectifier ($\alpha < 90^\circ$) or inverter ($\alpha > 90^\circ$, with a dc source). The figure's bottom pair is taken as T4 (left) and T2 (right).

              P (+)
        +-------+-------+
        |               |
       T1              T3
        |               |
 a o----+               +----o b
        |               |
       T4              T2
        |               |
        +-------+-------+
              N (-)
 ac supply vs between a and b (is into a)
 Load (R-L, Io) between P and N

Operation for α = 30°

Data: $V_s = 230$ V, $V_m = 325.27$ V, $I_o = 25$ A.

• Positive half cycle: T1 and T2 are forward biased and are fired at $\omega t = \alpha$. Load is connected to the supply, $v_o = v_s$, $i_s = +I_o$.
• At $\omega t = \pi$, $v_s$ reverses, but the large inductance keeps $I_o$ flowing, so T1, T2 continue to conduct and $v_o$ becomes negative from $\pi$ to $\pi+\alpha$ (energy returns from the inductance to the supply).
• At $\omega t = \pi+\alpha$, T3 and T4 are fired. Since $v_s < 0$, they apply reverse voltage across T1, T2, which turn off by natural (line) commutation. Now $v_o = -v_s$, $i_s = -I_o$.
• At $2\pi + \alpha$ T1, T2 are fired again and the cycle repeats.
• Each pair conducts for $180^\circ$; the input current is a square wave of amplitude $I_o$ lagging the voltage by $\alpha$.

Waveforms

 vs       ___               ___
         /   \             /
 -------/-----\-----------/----
               \___/
 vo: follows vs from 30 to 210 deg, then
     -vs from 210 to 390 deg (goes below
     zero for 30 deg in each half cycle)
 io: constant 25 A
 is
 +Io      +---------+         +---
          |         |         |
  0 ------+         |         |
                    +---------+
 -Io
         30        210       390   (deg)

Fundamental component of input current

The input current is a square wave of amplitude $I_o$, displaced by $\alpha$ from $v_s$. Its Fourier series is

$$i_s(t) = \sum_{n=1,3,5,\ldots}\frac{4I_o}{n\pi}\sin(n\omega t - n\alpha)$$

so the fundamental has peak $\frac{4I_o}{\pi}$ and rms value

$$I_{s1} = \frac{4I_o}{\sqrt2\,\pi} = \frac{2\sqrt2}{\pi}I_o = 0.9003\,I_o, \quad \text{lagging } v_s \text{ by } \phi_1 = \alpha$$ $$\begin{aligned} I_{s1} &= 0.9003 \times 25 = 22.51\ \text{A (rms)} \\ I_{s1,peak} &= \frac{4 \times 25}{\pi} = 31.83\ \text{A} \end{aligned}$$

So $i_{s1} = 31.83\sin(\omega t - 30^\circ)$ A.

Also, $V_{dc} = \frac{2V_m}{\pi}\cos\alpha = 207.07 \times 0.866 = 179.33$ V.

Answer: Fundamental input current $I_{s1} = 22.51$ A (rms), lagging $v_s$ by $30^\circ$.