AC voltage controller

A single-phase AC voltage controller changes the rms voltage applied to a load by delaying the firing of two anti-parallel thyristors (or a TRIAC) by an angle $\alpha$ in each half cycle (phase control). The output frequency is the same as the input frequency.

Circuit

        T1 --->|---
   +---|          |----+
   |    ---|<---  |    |
   |       T2          |
 ( ~ ) vs = Vm sin wt   R   vo
   |                   |
   +-------------------+

T1 handles the positive half cycle and T2 the negative half cycle.

Operation

1. $0 < \omega t < \alpha$: T1 is forward biased but not fired; no current; $v_o = 0$.
2. $\omega t = \alpha$: T1 is fired; it conducts and $v_o = v_s$.
3. $\pi$: supply and load current fall to zero together (resistive load), so T1 turns off naturally (line commutation).
4. $\pi < \omega t < \pi + \alpha$: T2 is forward biased but not yet fired; $v_o = 0$.
5. $\omega t = \pi + \alpha$: T2 is fired; $v_o = v_s$ (negative) until $2\pi$.

The current has the same shape as the voltage: $i_o = v_o/R$.

Waveforms

 vs    _               _
     /   \           /
 ---/-----\---------/----
            \     /
              \_/
 vo      .
        | \
 -------|--\--------.------
 0      a   pi      |    / 2pi
                    |  /
                    |/
                   pi+a
 gates: T1 at a, T2 at pi+a

The output is an AC waveform with chopped beginnings of each half cycle.

RMS output voltage

With $v_s = V_m \sin\omega t$ and by symmetry of the two halves:

$$\begin{aligned} V_{o} &= \left[\frac{1}{\pi}\int_{\alpha}^{\pi} V_m^2 \sin^2\omega t\; d(\omega t)\right]^{1/2} \\ &= \left[\frac{V_m^2}{2\pi}\int_{\alpha}^{\pi} (1-\cos 2\omega t)\; d(\omega t)\right]^{1/2} \\ &= \left[\frac{V_m^2}{2\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)\right]^{1/2} \end{aligned}$$

Since $V_s = V_m/\sqrt{2}$:

$$V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}$$

• $\alpha = 0$: $V_o = V_s$ (full voltage).
• $\alpha = \pi$: $V_o = 0$.

So the rms output is controlled smoothly from $V_s$ to 0 as $\alpha$ goes from 0 to $\pi$. The input power factor is $V_o/V_s$, which falls as $\alpha$ increases.

Applications: lamp dimmers, heater control, speed control of fans and small induction motors, soft starters.

An AC voltage controller is a thyristor (or TRIAC) circuit that varies the rms value of AC voltage applied to a load at the same frequency, by phase control or integral cycle control. In micro hydro (MHP), it is the heart of the electronic load controller (ELC), which keeps the generator load constant.

Operation of the AC voltage controller

        T1 --->|---
   +---|          |----+
   |    ---|<---  |    |
   |       T2          |
 ( ~ ) vs              R   vo
   |                   |
   +-------------------+

• T1 is fired at $\alpha$ in the positive half cycle and conducts until $\pi$; T2 is fired at $\pi + \alpha$ and conducts until $2\pi$.
• Before firing, the output is zero; after firing, $v_o = v_s$.
• With a resistive load:

$$V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}, \qquad P = \frac{V_o^2}{R}$$

So by changing $\alpha$ from 0 to $\pi$, the power taken by the load changes smoothly from full to zero.

Why an ELC is needed in micro hydro

• Small MHP plants (up to about 100 kW in Nepal) usually have no mechanical governor, because it is costly and slow.
• The turbine gets a fixed water flow, so it always produces nearly constant power.
• When consumer load drops, the extra power would speed up the generator, raising frequency and voltage and damaging appliances.

ELC principle

The generator always supplies a constant total power:

$$P_{gen} = P_{consumer} + P_{ballast} = \text{constant}$$

The ELC sends the surplus power to a ballast (dump) load, usually resistive water or air heaters, through an AC voltage controller.

 Turbine-Gen ---+------> Consumer load
     |          |
     |          +--[T1/T2 ACVC]--> Ballast R
     |                  ^
     +--> f / V sense --+
          controller (alpha)

Working

1. The controller measures the generator frequency (or voltage).
2. If consumer load falls, frequency starts to rise. The controller reduces $\alpha$, so more power goes to the ballast.
3. If consumer load rises, frequency starts to fall. The controller increases $\alpha$, so less power goes to the ballast.
4. Thus the total electrical load and hence speed, frequency and voltage stay nearly constant.

Features

• Cheap, simple, no moving parts, fast response.
• Ballast heat can be used for water or space heating.
• Phase control causes harmonics and poor power factor at mid firing angles; for this reason some ELCs use several binary-weighted ballast steps or integral cycle control.

A single-phase bridge cycloconverter converts AC at frequency $f_i$ directly to AC at a lower frequency $f_o = f_i/n$, using two full-bridge thyristor converters connected back to back: the positive (P) converter supplies the positive half of the output and the negative (N) converter the negative half.

Circuit

        P-converter          N-converter
     +--P1---P3--+        +--N1---N3--+
 vs  |           |  load  |           |
 ~ --+           +--[R]---+           +-- ~
     |           |        |           |
     +--P4---P2--+        +--N4---N2--+

• P-converter: P1, P2 (conduct when supply positive) and P3, P4 (conduct when supply negative); output current flows A to B through the load (+ve).
• N-converter: N1, N2 and N3, N4 connected in reverse, so load current flows B to A (−ve).
• Only one converter is enabled at a time (circulating-current-free mode).

Operation for $f_o = f_i/2$ (example, $\alpha = 0$, R load)

For the first output half cycle (two input half cycles), the P-converter acts as a full-wave rectifier, so both input half cycles appear positive on the load. For the next two input half cycles the N-converter works, and both appear negative.

 vs   /\    /\    /\    /\
     /  \  /  \  /  \  /  \
 ---/----\/----\/----\/----\-
              (each hump is
               one half cycle)
 vo   /\  /\
     /  \/  \
 ---/--------\--------/-----
              \  /\  /
               \/  \/
     |<--P--->|<--N--->|
       To = 2 Ti

So the output period is twice the input period, i.e. $f_o = f_i/2$ (25 Hz from 50 Hz). With firing angle $\alpha$, each hump starts at $\alpha$, reducing the rms output.

Conduction sequence ($f_o = f_i/2$)

For $f_o = f_i/3$, each converter works for three input half cycles (P1P2, P3P4, P1P2, then N3N4, N1N2, N3N4 style alternation), giving 3 humps per output half cycle.

RMS output (R load)

$$V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}$$

Uses: low-speed, high-power AC drives (cement mills, ship propulsion), induction heating supplies.

A single-phase step-down cycloconverter is a direct AC-to-AC converter that gives an output frequency $f_o = f_i/n$ (n = 2, 3, 4 ...) from a fixed-frequency supply, without an intermediate DC link. It uses naturally (line) commutated thyristors.

Circuit (mid-point / centre-tapped type)

          a o----+---P1->|---+
            |    +---|<-N1---+
  ~ vs    Tr|                |
  (input) O o----[ Load ]----+  (common point K)
            |                |
          b o----+---P2->|---+
                 +---|<-N2---+

• Positive group: P1, P2 — load current from K to O (positive output).
• Negative group: N1, N2 — load current from O to K (negative output).
• Let $v_{aO}$ be positive in the positive half cycle of the supply.

Operation for $f_o = f_i/4$ (e.g. 50 Hz to 12.5 Hz), R load, $\alpha = 0$

Positive output half cycle (4 input half cycles):

• Supply +ve: P1 conducts, load gets $+v_{aO}$.
• Supply −ve: P2 conducts, load gets $+v_{bO}$ (b is now positive).
• This repeats, so 4 positive humps appear across the load.

Negative output half cycle (next 4 input half cycles):

• Supply +ve: N2 conducts (b is negative, current O to K): load voltage negative.
• Supply −ve: N1 conducts: load voltage negative.

Waveforms

 vs  /\  /\  /\  /\  /\  /\  /\  /\
    /  \/  \/  \/  \/  \/  \/  \/  \
 (alternate humps are negative)

 vo  /\/\/\/\
    /        \
 --/----------\----------/--
               \        /
                \/\/\/\/
    |<-- P grp ->|<-- N grp ->|
    P1 P2 P1 P2   N2 N1 N2 N1

All thyristors switch at natural current zeros (R load), so no forced commutation is needed. Output period = 4 × input period.

Conduction table ($f_o = f_i/4$)

Control of output voltage

Delaying each firing by $\alpha$ reduces the rms output (R load):

$$V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}$$

where $V_s$ is the rms voltage of each half-winding. For a closer approximation to a sine wave, $\alpha$ is made large at the ends and small in the middle of each output half cycle.

Features: only step-down frequency with natural commutation; output contains harmonics; used in low-speed large AC motor drives.

A three-phase to single-phase cycloconverter produces a single-phase, variable (lower) frequency output from a three-phase supply using two three-phase thyristor bridges connected in anti-parallel. One bridge (P) carries positive load current and the other (N) carries negative load current.

Circuit (bridge type, 12 thyristors)

  a b c                     a b c
  | | |                     | | |
 +-------+                 +-------+
 | P-bridge|               | N-bridge|
 | P1..P6 |               | N1..N6 |
 +-------+                 +-------+
   +|  |-                    -|  |+
    |  +-------[ Load ]-------+  |
    +---------------------------+
   P: io > 0          N: io < 0

• Each bridge is a 3-phase fully controlled converter, with average output $V_{dc} = \dfrac{3\sqrt{2}V_L}{\pi}\cos\alpha$.
• The two bridges are connected so that they drive current in opposite directions through the load.

Working

1. Positive half of output: the P-bridge is fired. Its firing angle $\alpha_P$ is varied gradually: large near the start (low voltage), small in the middle (high voltage), large again at the end. The mean output thus rises and falls like a positive half sine wave.
2. Negative half of output: the P-bridge is blocked and the N-bridge is fired in the same way, giving the negative half.
3. The time for which each bridge works sets the output frequency $f_o$; the depth of $\alpha$ variation sets the output amplitude.
4. For a sinusoidal output, cosine-wave crossing control is used:

$$\alpha_P = \cos^{-1}(r\sin\omega_o t), \qquad \alpha_N = \pi - \alpha_P$$

where $r$ is the voltage ratio (0 to 1). Then the mean output is $V_o(t) = \dfrac{3\sqrt{2}V_L}{\pi}\,r\sin\omega_o t$.

Waveform

 vo (made of 6-pulse line segments)
      _/\_/\_
    /\       /\
 --/-----------\-------------/--
                 \/\_   _/\/
                     \/\/
   |<- P-bridge ->|<- N-bridge ->|
   alpha: 90..0..90  90..180..90 (inverted)

The output is built from pieces of the six line voltages; its fundamental is at $f_o$.

Modes

• Circulating-current-free: only one bridge fired at a time; a short dead time at current zero avoids a short circuit between bridges.
• Circulating-current mode: both fired with $\alpha_P + \alpha_N = 180°$ and an inter-group reactor limits the circulating current; smoother output, no dead time.

Points

• Practical output frequency is limited to about one-third of the input frequency for acceptable harmonics.
• With inductive loads each bridge may operate in rectifying or inverting mode depending on the sign of $v_o$ and $i_o$ (four-quadrant).
• Applications: low-speed high-power drives (ball mills, ship propulsion, rolling mills).

A single-phase AC voltage controller controls the rms voltage across a load by phase-delayed firing of two anti-parallel thyristors, T1 for the positive half and T2 for the negative half.

Operation with resistive load

        T1 --->|---
   +---|          |----+
   |    ---|<---  |    |
   |       T2          |
 ( ~ ) 220 V, 50 Hz    R   vo
   |                   |
   +-------------------+

• $0$ to $\alpha$: no thyristor on, $v_o = 0$.
• At $\alpha$, T1 is fired; $v_o = v_s$ until $\pi$, where current becomes zero and T1 turns off naturally.
• $\pi$ to $\pi+\alpha$: $v_o = 0$.
• At $\pi + \alpha$, T2 is fired; $v_o = v_s$ until $2\pi$.

 vo (alpha = 90 deg)
        |\
        | \
 -------+--\-------+-------
 0     90  180     |   /360
                   |  /
                   | /
                  270

Each half cycle the load gets only the second quarter of the sine wave. The output frequency stays 50 Hz.

RMS output voltage formula

$$\begin{aligned} V_o &= \left[\frac{1}{\pi}\int_{\alpha}^{\pi} (\sqrt{2}V_s\sin\omega t)^2\, d(\omega t)\right]^{1/2} \\ &= V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} \end{aligned}$$

Calculation

Given $V_s = 220$ V, $\alpha = 90° = \pi/2$, so $\sin 2\alpha = \sin 180° = 0$:

$$\begin{aligned} V_o &= 220\sqrt{\frac{1}{\pi}\left(\pi - \frac{\pi}{2} + 0\right)} \\ &= 220\sqrt{0.5} = 220 \times 0.7071 \\ &= 155.56\ \text{V} \end{aligned}$$

Answer: $V_{o,rms} = 155.56$ V (half the power of full conduction, since $V_o^2/V_s^2 = 0.5$).

In integral cycle (on-off) control, the thyristors are fired at the zero crossings so that complete cycles are passed to the load for $n$ cycles and blocked for $m$ cycles. The duty ratio is

$$k = \frac{n}{n+m} = \frac{6}{6+4} = 0.6$$

Given: $V_s = 230$ V, $f = 50$ Hz, $R = 15\ \Omega$, $V_m = \sqrt{2}\times 230 = 325.27$ V.

(i) RMS output voltage

$$V_o = V_s\sqrt{k} = 230\sqrt{0.6} = 230 \times 0.7746 = 178.16\ \text{V}$$

Load current $I_o = V_o/R = 178.16/15 = 11.88$ A; load power $P = V_o^2/R = 2116$ W.

(ii) Input power factor

$$pf = \frac{P}{V_s I_o} = \frac{V_o^2/R}{V_s\,V_o/R} = \frac{V_o}{V_s} = \sqrt{k} = \sqrt{0.6} = 0.775$$

(iii) Thyristor currents

Peak current $I_m = V_m/R = 325.27/15 = 21.68$ A. Each thyristor conducts one half-cycle in each ON cycle.

$$\begin{aligned} I_{T,avg} &= \frac{k\,I_m}{\pi} = \frac{0.6 \times 21.68}{\pi} = 4.14\ \text{A} \\ I_{T,rms} &= \frac{I_m}{2}\sqrt{k} = \frac{21.68}{2}\times 0.7746 = 8.40\ \text{A} \end{aligned}$$

(iv) Waveforms

 vs  /\  /\  /\  /\  /\  /\  /\  /\  /\  /\
      \/  \/  \/  \/  \/  \/  \/  \/  \/  \/
     |<-------- 10 cycles = 200 ms -------->|

 vo  /\  /\  /\  /\  /\  /\
      \/  \/  \/  \/  \/  \/ ________________
     |<----- 6 ON (120 ms) ->|<- 4 OFF 80ms->|

• Input current has the same shape as $v_o$ (R load), flowing only in the 6 ON cycles.
• Thyristors switch at zero voltage, so little EMI; suited to heaters with large thermal time constants.

Answer: $V_o = 178.16$ V, pf = 0.775, $I_{T,avg} = 4.14$ A, $I_{T,rms} = 8.40$ A.

In a single-phase AC voltage controller, the two anti-parallel SCRs do not share a common cathode, so their gate circuits must be electrically isolated from each other (and from the power circuit).

        +---T1 ->|---+
  A o---+            +---o B --[Load]
        +---|<- T2---+
   K(T1) = B,   K(T2) = A
   G1 referred to B, G2 referred to A

 Gate pulse ->[Pulse Tr 1]-> G1,K1
 Gate pulse ->[Pulse Tr 2]-> G2,K2

Reasons:

• A gate signal is always applied between the gate and its own cathode. Cathode of T1 is at point B and cathode of T2 is at point A.
• Points A and B differ by the voltage across the controller, which can be the full peak supply voltage (325 V for 230 V supply) when both SCRs are off.
• If both firing circuits used one common reference, A and B would be connected through the trigger circuit, shorting the SCRs and supply through the gate circuit, damaging it, and causing false firing.

Method: use a separate pulse transformer (or opto-coupler) for each SCR, so each gate–cathode pair floats at its own potential while the control circuit stays at ground potential.

For an AC voltage controller with R-L load, current continues after the voltage zero until the extinction angle $\beta$, found from the load current equation.

Given: $V_s = 230$ V, $R = 2\ \Omega$, $X_L = 2\ \Omega$, $\alpha = 45°$.

$$\begin{aligned} Z &= \sqrt{2^2 + 2^2} = 2.828\ \Omega \\ \phi &= \tan^{-1}\frac{X_L}{R} = \tan^{-1}1 = 45° \end{aligned}$$

Load current and extinction condition

For $\alpha \le \omega t \le \beta$:

$$i(\omega t) = \frac{V_m}{Z}\left[\sin(\omega t - \phi) - \sin(\alpha - \phi)\,e^{-(\omega t - \alpha)/\tan\phi}\right]$$

At $\omega t = \beta$, $i = 0$:

$$\sin(\beta - \phi) = \sin(\alpha - \phi)\,e^{-(\beta - \alpha)/\tan\phi}$$

(i) Extinction angle

Here $\alpha = \phi = 45°$, so $\sin(\alpha - \phi) = 0$ and the transient term vanishes:

$$\sin(\beta - 45°) = 0 \;\Rightarrow\; \beta = 180° + 45° = 225°$$

(ii) Conduction angle

$$\gamma = \beta - \alpha = 225° - 45° = 180°$$

(iii) RMS output voltage

$$\begin{aligned} V_o &= V_s\sqrt{\frac{1}{\pi}\left[\beta - \alpha + \frac{\sin 2\alpha - \sin 2\beta}{2}\right]} \\ &= 230\sqrt{\frac{1}{\pi}\left[\pi + \frac{\sin 90° - \sin 450°}{2}\right]} = 230\sqrt{1} = 230\ \text{V} \end{aligned}$$

Load current $I_o = 230/2.828 = 81.3$ A (pure sinusoid).

Note: $\alpha = \phi$ is the boundary between discontinuous ($\alpha > \phi$) and continuous conduction. Each SCR conducts exactly 180°, the current is a pure sine wave lagging by 45°, and the controller gives full output; any $\alpha < 45°$ also gives full output and loses control.

(iv) Waveforms

 vs,vo  /\        /\
       /  \      /  \
 -----/----\----/----\---
     0  45  \  / 225
  io    __   \/
       /  \        /
 -----/----\------/------
     45    225   405
 T1: 45 -> 225   T2: 225 -> 405

Answer: $\beta = 225°$, $\gamma = 180°$, $V_o = 230$ V.

A step-up cycloconverter gives an output frequency higher than the input ($f_o = n f_i$). Since the thyristors must turn off before the supply current reaches zero, forced commutation is needed. Mid-point circuit:

          a o----+---P1->|---+
            |    +---|<-N1---+
  ~ vs    Tr|                |
  (input) O o----[ Load ]----+  (common point K)
            |                |
          b o----+---P2->|---+
                 +---|<-N2---+

• In the positive input half cycle ($a$ positive): P1 gives positive load voltage $+v_{aO}$; N2 gives negative load voltage.
• In the negative input half cycle ($b$ positive): P2 gives positive load voltage; N1 gives negative load voltage.

Operation for $f_o = 4f_i$ (50 Hz to 200 Hz)

Each input half cycle (10 ms) must contain 4 output half cycles, i.e. two full output cycles. So each input half cycle is divided into 4 equal intervals of 2.5 ms:

At the end of each interval the conducting SCR is turned off by a forced-commutation circuit and the next one is fired.

 vs   _____
     /     \
 ---/-------\---------/--
             \_______/
 vo  _   _      _   _
    | | | |    | | | |
 ---+-+-+-+----+-+-+-+---
      |_| |_|    |_| |_|
    P1 N2 P1 N2 P2 N1 P2 N1

The output consists of slices of the input sine wave with alternating sign; its fundamental is 200 Hz. The output amplitude follows the input envelope, so the harmonic content is high. Step-up cycloconverters are used less often than step-down types (e.g. induction heating, high-speed drives).

With an inductive load, the load current lags the voltage, so the outgoing thyristor is still conducting when the incoming thyristor is fired.

• Suppose the firing angle $\alpha$ is less than the load angle $\phi$. When the incoming SCR (say P2 or T2) is fired, the outgoing SCR is still carrying current until the extinction angle $\beta > \pi + \alpha$.
• The conducting SCR clamps the voltage, so the incoming SCR is reverse biased (or has no forward voltage) at the instant of the gate pulse.
• A short pulse ends before the outgoing SCR stops conducting, so the incoming SCR never turns on. It then waits for the next pulse, one full cycle later.
• Result: only one SCR (one group) conducts, the output becomes unsymmetrical, a DC component appears in the load and transformer, and frequency control is lost.

Remedy: use a long (wide) gate pulse lasting up to $\pi$ (or a high-frequency pulse train) so the incoming thyristor turns on as soon as the outgoing one stops and it becomes forward biased.

Doubling the time period means $T_o = 2T_i$, i.e. $f_o = f_i/2$ (50 Hz to 25 Hz). This is done by a single-phase step-down cycloconverter in which each output half cycle is made from two input half cycles.

Circuit (centre-tapped transformer type)

          a o----+---P1->|---+
            |    +---|<-N1---+
  ~ vs    Tr|                |
  (input) O o----[ Load ]----+  (common point K)
            |                |
          b o----+---P2->|---+
                 +---|<-N2---+

P1, P2 form the positive group (load current K to O) and N1, N2 the negative group (load current O to K).

Working (R load, firing angle $\alpha$)

• In half cycles 1 and 2 the positive group acts as a full-wave rectifier, so two positive humps appear.
• In half cycles 3 and 4 the negative group gives two negative humps.
• Each SCR turns off naturally at the supply zero (R load).

Waveforms

 vs  /\    /\
    /  \  /  \
 --/----\/----\/----\/---
             (alternate humps -ve)
 vo  /\  /\
    /  \/  \
 --/--------\--------/---
             \  /\  /
              \/  \/
    | P1 | P2 | N2 | N1 |
    |<--- To = 2 Ti --->|

With firing delay, each hump starts at $\alpha$ after the zero crossing.

RMS output voltage

Every output half cycle is made of two identical chopped half-sine humps, so the rms over one hump equals the rms over the whole output. With $v_s = \sqrt{2}V_s\sin\omega t$ (per half-winding):

$$\begin{aligned} V_o &= \left[\frac{1}{\pi}\int_{\alpha}^{\pi} 2V_s^2\sin^2\omega t\; d(\omega t)\right]^{1/2} \\ &= V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} \end{aligned}$$

• For $\alpha = 0$: $V_o = V_s$.
• Example (assumed values): $V_s = 230$ V, $\alpha = 30°$:

$$V_o = 230\sqrt{\frac{1}{\pi}\left(\pi - \frac{\pi}{6} + \frac{\sin 60°}{2}\right)} = 230 \times 0.9855 = 226.66\ \text{V}$$

Answer: $V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}$, e.g. 226.66 V for 230 V and $\alpha = 30°$; output frequency 25 Hz from 50 Hz.

A single-phase AC voltage controller varies the rms value of the AC voltage applied to a load, at the supply frequency, by delaying the firing of two anti-parallel thyristors (phase angle control).

Circuit and operation

        T1 --->|---
   +---|          |----+
   |    ---|<---  |    |
   |       T2          |
 ( ~ ) vs              R   vo
   |                   |
   +-------------------+

1. $0 < \omega t < \alpha$: T1 not yet fired, $v_o = 0$.
2. At $\alpha$: T1 fired, $v_o = v_s$ until $\pi$, where the current falls to zero and T1 turns off (natural commutation).
3. $\pi < \omega t < \pi + \alpha$: $v_o = 0$.
4. At $\pi + \alpha$: T2 fired, $v_o = v_s$ until $2\pi$.

 vo     .
        | \
 -------|--\--------.------
 0      a   pi      |    / 2pi
                    |  /
                    |/
                   pi+a

RMS output voltage

$$\begin{aligned} V_o &= \left[\frac{2}{2\pi}\int_{\alpha}^{\pi} V_m^2\sin^2\omega t\; d(\omega t)\right]^{1/2} \\ &= \left[\frac{V_m^2}{2\pi}\left\{\omega t - \frac{\sin 2\omega t}{2}\right\}_{\alpha}^{\pi}\right]^{1/2} \\ &= \frac{V_m}{\sqrt{2}}\left[\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)\right]^{1/2} \end{aligned}$$ $$V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}$$

As $\alpha$ goes from 0 to $\pi$, $V_o$ goes from $V_s$ to 0, and load power $V_o^2/R$ from full to zero.

Use in the ELC of a micro hydro (MHP) scheme

Micro hydro plants usually run with a fixed water input and no mechanical governor. The electronic load controller (ELC) keeps the total electrical load constant:

$$P_{gen} = P_{consumer} + P_{dump} = \text{constant}$$

 Gen ---+-------> Consumers
        |
        +--[ AC voltage controller ]--> Dump heater
        |            ^ alpha
        +-> sense f/V -> controller

• The controller senses generator frequency (or voltage).
• Consumer load decreases → frequency tends to rise → $\alpha$ is reduced → more power to the dump (ballast) heater.
• Consumer load increases → $\alpha$ is increased → less power to the dump load.
• So speed, frequency and voltage stay nearly constant without a governor.
• The dump load is usually water or air heaters, and the heat can be used.

Drawbacks of phase control here are harmonics and poorer power factor at mid-range $\alpha$; binary-weighted dump loads or integral cycle control reduce this.

A three-phase cycloconverter converts a fixed-frequency three-phase supply directly into a three-phase output of lower, variable frequency and variable voltage, without a DC link. Each output phase is a dual (positive and negative) phase-controlled converter.

Types

Circuit (3-phase to 3-phase, 3-pulse, one phase shown)

 Supply a b c
  |  |  |      Phase A of output
  +--+--+--> P-group (3 SCR) --+
  |  |  |                      |--> A (load)
  +--+--+--> N-group (3 SCR) --+
 Same for output phases B and C
 (output phases share a neutral)

Operation

1. Each output phase has a positive group (P) that carries positive load current and a negative group (N) that carries negative load current.
2. The firing angle of the conducting group is modulated during the output cycle. With cosine-wave crossing control:

$$\alpha_P = \cos^{-1}\left(r\sin\omega_o t\right), \qquad \alpha_N = \pi - \alpha_P$$

so the mean output voltage of that phase is

$$v_{A}(t) = V_{do}\, r\sin\omega_o t, \qquad V_{do} = \frac{3\sqrt{3}V_m}{2\pi}\ \text{(3-pulse)}$$

3. Output phases B and C use the same law with $\omega_o t$ shifted by $-120°$ and $-240°$, giving a balanced three-phase output.
4. $r$ (0 to 1) sets the output amplitude; the rate of modulation sets $f_o$.
5. With R-L or motor loads, the current lags the voltage, so each group works as a rectifier for part of its conduction and as an inverter for the rest; the cycloconverter is naturally four-quadrant and regenerative.

Waveform (one output phase)

 vA (pieces of supply phase voltages)
        _/\/\_
      /\      /\
 ----/----------\-----------/--
                 \/\_  _/\/
                     \/
   |<--P group-->|<--N group-->|

Features

• Output frequency limited to about 1/3 (3-pulse) to 1/2 of input frequency for acceptable distortion.
• Natural (line) commutation, high efficiency, power flow in both directions.
• Poor input power factor and complex harmonic content; many thyristors.
• Applications: large low-speed AC drives (cement and ball mills, mine winders, ship propulsion), variable-speed constant-frequency (VSCF) aircraft supplies.

Take the input frequency as $f_i = 50$ Hz. For one-quarter frequency the output is $f_o = 50/4 = 12.5$ Hz, so the output period is $T_o = 80$ ms = 4 input cycles, and each output half cycle contains 4 input half cycles.

Circuit (single-phase mid-point cycloconverter)

          a o----+---P1->|---+
            |    +---|<-N1---+
  ~ vs    Tr|                |
  (input) O o----[ Load ]----+  (common point K)
            |                |
          b o----+---P2->|---+
                 +---|<-N2---+

• Positive group P1, P2: load current K to O, positive output.
• Negative group N1, N2: load current O to K, negative output.
• With an R load, every SCR turns off naturally at the supply zero (line commutation), so no forced commutation is needed.

Operation

Output positive half (0–40 ms): P1 conducts in each positive supply half (a positive), and P2 in each negative half (b positive). Load voltage is positive in all 4 half cycles.

Output negative half (40–80 ms): N2 conducts in positive supply halves, N1 in negative halves. Load voltage is negative in all 4 half cycles.

Output waveform for $f_o = f_i/4$

 vs  /\  /\  /\  /\  /\  /\  /\  /\
    /  \/  \/  \/  \/  \/  \/  \/  \
 (every second hump is negative)

 vo  /\/\/\/\
    /        \
 --/----------\----------/---
               \        /
                \/\/\/\/
   0          40        80 ms
   |<-- 12.5 Hz period -->|

Output voltage

With firing angle $\alpha$ for all SCRs and R load:

$$V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}$$

To make the output closer to a sine wave, the firing angle can be varied: large for the first and last humps, small for the middle humps of each output half cycle.

Answer: 50 Hz input gives 12.5 Hz output; each output half cycle uses 4 input half cycles (P1, P2, P1, P2 then N2, N1, N2, N1).

With an R-L load, current lags; since $\alpha$ is greater than the load angle $\phi$, conduction is discontinuous and each SCR conducts from $\alpha$ to the extinction angle $\beta$.

Given: $V_s = 230$ V, 50 Hz, $R = 20\ \Omega$, $L = 50$ mH, $\alpha = 90°$.

Load parameters

$$\begin{aligned} X_L &= 2\pi f L = 2\pi(50)(0.05) = 15.708\ \Omega \\ Z &= \sqrt{20^2 + 15.708^2} = 25.431\ \Omega \\ \phi &= \tan^{-1}\frac{15.708}{20} = 38.15°, \quad \tan\phi = 0.7854 \end{aligned}$$

Since $\alpha = 90° > \phi = 38.15°$, conduction is discontinuous.

Extinction angle

$$i(\omega t) = \frac{V_m}{Z}\left[\sin(\omega t - \phi) - \sin(\alpha - \phi)\,e^{-(\omega t - \alpha)/\tan\phi}\right]$$

Setting $i(\beta) = 0$:

$$\sin(\beta - 38.15°) = \sin(51.85°)\,e^{-(\beta - 90°)\pi/(180°\times 0.7854)}$$

Solving numerically (trial and error):

$$\beta = 215.37°, \qquad \gamma = \beta - \alpha = 125.37°$$

Check: $\beta < 180° + \alpha = 270°$, so T1 stops before T2 is fired (discontinuous), as expected.

RMS load voltage

$$\begin{aligned} V_o &= V_s\sqrt{\frac{1}{\pi}\left[\beta - \alpha + \frac{\sin 2\alpha - \sin 2\beta}{2}\right]} \\ &= 230\sqrt{\frac{1}{\pi}\left[2.1880 + \frac{0 - 0.9440}{2}\right]} = 230\sqrt{\frac{1.7160}{\pi}} \\ &= 230 \times 0.7391 = 169.99\ \text{V} \end{aligned}$$

(Here $\beta - \alpha = 125.37° = 2.1880$ rad, $\sin 2\alpha = \sin 180° = 0$ and $\sin 2\beta = \sin 430.73° = 0.9440$.)

RMS load current

$$I_o = \left[\frac{1}{\pi}\int_{\alpha}^{\beta} i^2(\omega t)\, d(\omega t)\right]^{1/2}$$

Evaluating with $V_m/Z = 325.27/25.431 = 12.79$ A gives

$$I_o = 5.575\ \text{A}$$

Check by power balance: $P = I_o^2 R = 5.575^2 \times 20 = 621.6$ W, equal to the average of $v_o i_o$.

Answer: $\beta = 215.37°$, $V_{o,rms} = 169.99$ V, $I_{o,rms} = 5.575$ A.

A cycloconverter is a direct frequency changer that converts AC power at one frequency to AC power at another (usually lower) frequency in one stage, without an intermediate DC link. A single-phase cycloconverter does this for a single-phase supply and load, using two groups of thyristors: a positive group for positive load current and a negative group for negative load current.

Circuit (mid-point type)

          a o----+---P1->|---+
            |    +---|<-N1---+
  ~ vs    Tr|                |
  (input) O o----[ Load ]----+  (common point K)
            |                |
          b o----+---P2->|---+
                 +---|<-N2---+

• Positive group: P1, P2 (current from K to O through the load).
• Negative group: N1, N2 (current from O to K).

Operation as a step-down cycloconverter ($f_o = f_i/2$, R load)

1. Input half 1 (a positive): P1 is fired; load voltage $= +v_{aO}$.
2. Input half 2 (b positive): P2 is fired; load voltage $= +v_{bO}$, still positive. Two positive humps form the positive output half cycle.
3. Input half 3 (a positive): N2 is fired; current flows O → load → K → N2 → b; load voltage negative.
4. Input half 4 (a negative, b positive): N1 is fired; load voltage negative.

 vs  /\    /\
    /  \  /  \
 --/----\/----\/----\/---
       (alternate humps -ve)
 vo  /\  /\
    /  \/  \
 --/--------\--------/---
             \  /\  /
              \/  \/
    | P1 | P2 | N2 | N1 |
    |<--- To = 2 Ti --->|

So the output has half the input frequency. Using 3 or 4 humps per half cycle gives $f_i/3$ or $f_i/4$. The SCRs turn off at natural current zeros (line commutation).

Voltage control: firing each SCR at a delay $\alpha$ gives

$$V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}$$

Step-up operation ($f_o > f_i$) is also possible, by switching between P and N groups several times within each input half cycle, but then forced commutation is needed.

Inductive load: current lags, so a group continues to conduct after the voltage reverses (inversion mode); long gate pulses are used and a group is switched only after its current reaches zero.

Applications

• Speed control of large, low-speed AC motors: gearless cement and ball mills, rolling mills, mine hoists, ship propulsion.
• Variable-speed constant-frequency (VSCF) aircraft power: variable-frequency alternator output converted to fixed 400 Hz.
• Induction heating supplies (step-up type).
• Static VAR / slip power recovery schemes (Scherbius drives) of wound-rotor induction motors.
• Traction: conversion of 50 Hz to $16\frac{2}{3}$ Hz for railway supply.

Merits and limits

A single-phase cycloconverter converts single-phase AC at frequency $f_i$ directly to single-phase AC at a different frequency $f_o$ (usually $f_o = f_i/n$), with no DC link. It uses a positive group of thyristors for positive load current and a negative group for negative load current.

Circuit (centre-tapped transformer type)

          a o----+---P1->|---+
            |    +---|<-N1---+
  ~ vs    Tr|                |
  (input) O o----[ Load ]----+  (common point K)
            |                |
          b o----+---P2->|---+
                 +---|<-N2---+

Operation ($f_o = f_i/2$, resistive load)

• In halves 1 and 2 the positive group works as a full-wave rectifier, producing two positive humps (positive output half cycle).
• In halves 3 and 4 the negative group gives two negative humps.
• Each thyristor turns off naturally when the supply (and R-load current) reaches zero.

 vs  /\    /\
    /  \  /  \
 --/----\/----\/----\/---
       (alternate humps -ve)
 vo  /\  /\
    /  \/  \
 --/--------\--------/---
             \  /\  /
              \/  \/
    | P1 | P2 | N2 | N1 |
    |<--- To = 2 Ti --->|

Output period = 2 × input period, so 50 Hz gives 25 Hz. For $f_i/3$ or $f_i/4$, each group is kept on for 3 or 4 input half cycles.

Output voltage control

With firing delay $\alpha$ for every SCR (R load):

$$V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}$$

Varying $\alpha$ through each output half cycle (larger at the ends, smaller in the middle) makes the output closer to a sine wave.

In a full-wave AC voltage controller with resistive load, each of the two anti-parallel thyristors conducts from $\alpha$ to $\pi$ in its half cycle.

Given: $V_s = 230$ V, 50 Hz, $R = 10\ \Omega$, $\alpha = 45° = \pi/4$, $V_m = \sqrt{2}\times 230 = 325.27$ V.

RMS output voltage

$$\begin{aligned} V_o &= V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} \\ &= 230\sqrt{\frac{1}{\pi}\left(\pi - \frac{\pi}{4} + \frac{\sin 90°}{2}\right)} \\ &= 230\sqrt{\frac{2.3562 + 0.5}{\pi}} = 230\sqrt{0.9092} \\ &= 230 \times 0.9535 = 219.30\ \text{V} \end{aligned}$$

RMS load current

$$I_o = \frac{V_o}{R} = \frac{219.30}{10} = 21.93\ \text{A}$$

Input power factor

$$\begin{aligned} P_o &= I_o^2 R = 21.93^2 \times 10 = 4809.4\ \text{W} \\ pf &= \frac{P_o}{V_s I_o} = \frac{V_o}{V_s} = \frac{219.30}{230} = 0.9535\ \text{(lagging)} \end{aligned}$$

Average thyristor current

Each thyristor conducts only in its own half cycle:

$$\begin{aligned} I_{T,avg} &= \frac{1}{2\pi}\int_{\alpha}^{\pi}\frac{V_m}{R}\sin\omega t\; d(\omega t) = \frac{V_m}{2\pi R}(1 + \cos\alpha) \\ &= \frac{325.27}{2\pi \times 10}(1 + 0.7071) = 8.84\ \text{A} \end{aligned}$$

(For reference, rms thyristor current $= I_o/\sqrt{2} = 15.51$ A.)

 vo   (alpha = 45 deg)
        .--.
       |    \
 ------|-----\-----.------
 0    45    180    |    / 360
                   |  /
                   '-'
                  225

Answer: $V_o = 219.30$ V, $I_o = 21.93$ A, pf = 0.9535 lagging, $I_{T,avg} = 8.84$ A.

In a single-phase full-wave AC voltage controller with resistive load, T1 conducts from $\alpha$ to $\pi$ and T2 from $\pi + \alpha$ to $2\pi$.

Given: $V_s = 230$ V, 50 Hz, $R = 20\ \Omega$, $\alpha = 45° = \pi/4$, $V_m = \sqrt{2}\times 230 = 325.27$ V.

i) RMS output voltage

$$\begin{aligned} V_o &= V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} \\ &= 230\sqrt{\frac{1}{\pi}\left(\frac{3\pi}{4} + \frac{1}{2}\right)} = 230\sqrt{0.9092} \\ &= 230 \times 0.9535 = 219.30\ \text{V} \end{aligned}$$

ii) Load power and input power factor

$$\begin{aligned} I_o &= \frac{V_o}{R} = \frac{219.30}{20} = 10.97\ \text{A} \\ P_o &= \frac{V_o^2}{R} = \frac{219.30^2}{20} = 2404.7\ \text{W} \\ pf &= \frac{P_o}{V_s I_o} = \frac{2404.7}{230 \times 10.965} = \frac{V_o}{V_s} = 0.9535\ \text{(lagging)} \end{aligned}$$

iii) Thyristor currents

Each thyristor carries the load current only in its own half cycle.

$$\begin{aligned} I_{T,avg} &= \frac{V_m}{2\pi R}(1 + \cos\alpha) = \frac{325.27}{2\pi \times 20}(1 + 0.7071) = 4.42\ \text{A} \\ I_{T,rms} &= \frac{I_o}{\sqrt{2}} = \frac{10.965}{\sqrt{2}} = 7.75\ \text{A} \end{aligned}$$

The rms thyristor current formula follows from

$$I_{T,rms} = \left[\frac{1}{2\pi}\int_{\alpha}^{\pi}\left(\frac{V_m}{R}\sin\omega t\right)^2 d(\omega t)\right]^{1/2} = \frac{I_o}{\sqrt{2}}$$

Answer: $V_o = 219.30$ V; $P_o = 2404.7$ W; pf = 0.9535; $I_{T,avg} = 4.42$ A, $I_{T,rms} = 7.75$ A.

In a bridge-type single-phase cycloconverter, the positive (P) bridge supplies the positive half of the output and the negative (N) bridge the negative half. For $f_o = f_i/3$, each bridge works for 3 input half cycles in turn.

   P-bridge (io > 0)     N-bridge (io < 0)
   +--P1----P3--+        +--N1----N3--+
 ~ |            |        |            |
 vs+            +-A [R] B+            +
   |            |        |            |
   +--P4----P2--+        +--N4----N2--+
 P: current A -> B     N: current B -> A

Given: $V_s = 230$ V, 50 Hz, $R = 10\ \Omega$, $f_o = 50/3 = 16.67$ Hz, $\alpha = 30° = \pi/6$.

i) RMS output voltage

Each output half cycle consists of 3 identical half-sine pieces from $\alpha$ to $\pi$, so the rms value is the same as for one piece:

$$\begin{aligned} V_o &= V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} \\ &= 230\sqrt{\frac{1}{\pi}\left(\pi - \frac{\pi}{6} + \frac{\sin 60°}{2}\right)} \\ &= 230\sqrt{\frac{2.6180 + 0.4330}{\pi}} = 230\sqrt{0.9712} \\ &= 230 \times 0.9855 = 226.66\ \text{V} \end{aligned}$$

ii) RMS output current

$$I_o = \frac{V_o}{R} = \frac{226.66}{10} = 22.67\ \text{A}$$

iii) RMS current of each converter

Each bridge carries the load current for one half of the output period:

$$I_{conv} = \frac{I_o}{\sqrt{2}} = \frac{22.67}{\sqrt{2}} = 16.03\ \text{A}$$

iv) RMS current of each thyristor

Within its bridge, each thyristor pair (P1-P2 or P3-P4) conducts on alternate input half cycles, i.e. half of the bridge's conduction time:

$$I_T = \frac{I_{conv}}{\sqrt{2}} = \frac{I_o}{2} = \frac{22.67}{2} = 11.33\ \text{A}$$

Conduction pattern ($f_o = f_i/3$)

(Over a long time each thyristor pair conducts very nearly half of its bridge's half cycles, so $I_T = I_o/2$ is the usual textbook result.)

Answer: $V_o = 226.66$ V, $I_o = 22.67$ A, $I_{conv} = 16.03$ A, $I_T = 11.33$ A.

A single-phase cycloconverter is a direct AC-to-AC frequency changer: it builds a lower-frequency output from selected portions of the input sine wave, using a positive thyristor group (for positive load current) and a negative group (for negative load current).

Circuit

          a o----+---P1->|---+
            |    +---|<-N1---+
  ~ vs    Tr|                |
  (input) O o----[ Load ]----+  (common point K)
            |                |
          b o----+---P2->|---+
                 +---|<-N2---+

Operation ($f_o = f_i/2$, R load)

 vo  /\  /\
    /  \/  \
 --/--------\--------/---
             \  /\  /
              \/  \/
    | P1 | P2 | N2 | N1 |
    |<--- To = 2 Ti --->|

Mathematical expressions

1. Frequency relation. If each group conducts for $n$ input half cycles,

$$T_o = n\,T_i \quad\Rightarrow\quad f_o = \frac{f_i}{n}$$

2. Output voltage (R load, constant firing angle $\alpha$). With $v_s = V_m\sin\omega t$, in every input half cycle the load sees $\pm V_m\sin\omega t$ from $\alpha$ to $\pi$:

$$\begin{aligned} V_o &= \left[\frac{1}{\pi}\int_{\alpha}^{\pi} V_m^2\sin^2\omega t\; d(\omega t)\right]^{1/2} \\ &= \frac{V_m}{\sqrt{2}}\left[\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)\right]^{1/2} \end{aligned}$$

3. Average voltage of a group. Each group is a phase-controlled converter; for continuous current its mean voltage is

$$V_{d} = V_{do}\cos\alpha, \qquad V_{do} = \frac{2V_m}{\pi}$$

4. Sinusoidal output (firing-angle modulation). To get a sinusoidal fundamental $v_o = V_{om}\sin\omega_o t$, the firing angle is changed during the output cycle so that

$$V_{do}\cos\alpha_P = r V_{do}\sin\omega_o t \;\Rightarrow\; \alpha_P = \cos^{-1}(r\sin\omega_o t)$$

and for the negative group $\alpha_N = \pi - \alpha_P$, so that $V_{do}\cos\alpha_P = -V_{do}\cos\alpha_N$. Here $r = V_{om}/V_{do}$ ($0 \le r \le 1$) is the voltage ratio. This is called cosine-wave crossing control.

Notes

• With R load the SCRs turn off at supply zeros (natural commutation).
• With inductive load, current lags; each group works as rectifier while $v_o$ and $i_o$ have the same sign and as inverter while they differ; groups are switched only at current zero (or a reactor is used for circulating-current mode).
• Step-up operation ($f_o > f_i$) needs forced commutation.

A single-phase cycloconverter with a centre-tapped transformer uses four thyristors: P1, P2 (positive group) and N1, N2 (negative group). For one-third frequency ($f_o = f_i/3$, e.g. 50 Hz to 16.67 Hz), each output half cycle is built from three input half cycles.

Circuit

          a o----+---P1->|---+
            |    +---|<-N1---+
  ~ vs    Tr|                |
            O o----[ R ]-----+ K
            |                |
          b o----+---P2->|---+
                 +---|<-N2---+
 vo = v_KO; positive when current flows K -> O

• $v_{aO}$ positive = positive supply half cycle (then $b$ is negative w.r.t. O).
• P1 conducts when $a$ is positive; P2 when $b$ is positive.
• N1 conducts when $a$ is negative; N2 when $b$ is negative.

Conduction pattern and current paths ($f_o = f_i/3$, R load)

The pattern repeats every 6 input half cycles (3 input cycles = 1 output cycle).

Waveforms

 vs  /\    /\    /\
    /  \  /  \  /  \
 --/----\/----\/----\/----\/----\/--
     1   2   3   4   5   6  (halves)
 vo  /\  /\  /\
    /  \/  \/  \
 --/------------\------------/--
                 \  /\  /\  /
                  \/  \/  \/
    P1  P2  P1   N1  N2  N1
    |<----- To = 3 Ti ----->|

Remarks

• Each SCR turns off naturally at the supply zero because the load is resistive.
• The output frequency is $f_i/3$ and, with firing angle $\alpha$, the rms output is $V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}$, where $V_s$ is the rms voltage of each half winding.
• For an inductive load, longer gate pulses and a dead time at changeover between P and N groups are required.

An AC voltage controller varies the rms AC voltage applied to a load at the supply frequency by phase-delayed firing of two anti-parallel thyristors, T1 (positive half) and T2 (negative half).

        T1 --->|---
   +---|          |----+
   |    ---|<---  |    |
   |       T2         R
 ( ~ ) vs = Vm sin wt  |
   |                 (L)  vo
   +-------------------+

Operation with purely resistive load

• $0$ to $\alpha$: no SCR conducts, $v_o = 0$.
• At $\alpha$, T1 is fired and $v_o = v_s$ till $\pi$, where current falls to zero and T1 turns off naturally.
• At $\pi + \alpha$, T2 is fired and $v_o = v_s$ till $2\pi$.
• Load current $i_o = v_o/R$ has the same shape as $v_o$.

 vo     .
        | \
 -------|--\--------.------
 0      a   pi      |    / 2pi
                    |  /
                    |/
                   pi+a

RMS output voltage (R load)

$$\begin{aligned} V_o &= \left[\frac{1}{\pi}\int_{\alpha}^{\pi} V_m^2\sin^2\omega t\; d(\omega t)\right]^{1/2} = \left[\frac{V_m^2}{2\pi}\left(\omega t - \frac{\sin 2\omega t}{2}\right)\Big|_{\alpha}^{\pi}\right]^{1/2} \\ &= \frac{V_m}{\sqrt{2}}\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} \end{aligned}$$

Load current with R-L load

When T1 is fired at $\omega t = \alpha$, the circuit equation is

$$L\frac{di}{dt} + Ri = V_m\sin\omega t, \qquad i(\alpha) = 0$$

The solution has a steady-state part and a transient part:

$$i(t) = i_{ss} + i_{tr} = \frac{V_m}{Z}\sin(\omega t - \phi) + A\,e^{-\frac{R}{L}t}$$

where $Z = \sqrt{R^2 + (\omega L)^2}$ and $\phi = \tan^{-1}(\omega L/R)$.

Applying $i = 0$ at $\omega t = \alpha$ (i.e. $t = \alpha/\omega$):

$$A = -\frac{V_m}{Z}\sin(\alpha - \phi)\,e^{\frac{R\alpha}{\omega L}}$$

Since $\dfrac{R}{\omega L} = \dfrac{1}{\tan\phi}$:

$$i(\omega t) = \frac{V_m}{Z}\left[\sin(\omega t - \phi) - \sin(\alpha - \phi)\,e^{-\frac{(\omega t - \alpha)}{\tan\phi}}\right], \qquad \alpha \le \omega t \le \beta$$

Extinction angle $\beta$ (where $i = 0$ again):

$$\sin(\beta - \phi) = \sin(\alpha - \phi)\,e^{-\frac{(\beta - \alpha)}{\tan\phi}}$$

• If $\alpha > \phi$: $\beta < \pi + \alpha$, current is discontinuous, and output voltage is controlled.
• If $\alpha \le \phi$: current becomes continuous and sinusoidal, $V_o = V_s$, and control is lost. So the control range is $\phi \le \alpha \le \pi$.

The rms output voltage for R-L load is then $V_o = V_s\sqrt{\frac{1}{\pi}\left[\beta - \alpha + \frac{\sin 2\alpha - \sin 2\beta}{2}\right]}$.