Skip to main content

Single-phase AC voltage controller with R load

Short answer

Two anti-parallel thyristors (T1 positive half, T2 negative half) are fired at angle α. Output is zero until α then equals supply until π, where current naturally extinguishes. The rms output voltage is V_o = V_s √[(1/π)(π - α + sin2α/2)].

With the help of suitable circuit diagram and waveforms, explain the operation of single phase AC voltage controller with resistive load. Derive the expression for rms value of output voltage as a function of firing angle.

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 α\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<ωt<α0 < \omega t < \alpha: T1 is forward biased but not fired; no current; vo=0v_o = 0.
  2. ωt=α\omega t = \alpha: T1 is fired; it conducts and vo=vsv_o = v_s.
  3. π\pi: supply and load current fall to zero together (resistive load), so T1 turns off naturally (line commutation).
  4. π<ωt<π+α\pi < \omega t < \pi + \alpha: T2 is forward biased but not yet fired; vo=0v_o = 0.
  5. ωt=π+α\omega t = \pi + \alpha: T2 is fired; vo=vsv_o = v_s (negative) until 2π2\pi.

The current has the same shape as the voltage: io=vo/Ri_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 vs=Vmsin⁡ωtv_s = V_m \sin\omega t and by symmetry of the two halves:

Vo=[1π∫απVm2sin⁡2ωt  d(ωt)]1/2=[Vm22π∫απ(1−cos⁡2ωt)  d(ωt)]1/2=[Vm22π(π−α+sin⁡2α2)]1/2\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 Vs=Vm/2V_s = V_m/\sqrt{2}:

Vo=Vs1π(π−α+sin⁡2α2)V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}
  • α=0\alpha = 0: Vo=VsV_o = V_s (full voltage).
  • α=π\alpha = \pi: Vo=0V_o = 0.

So the rms output is controlled smoothly from VsV_s to 0 as α\alpha goes from 0 to π\pi. The input power factor is Vo/VsV_o/V_s, which falls as α\alpha increases.

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

✓ Answer checked by Claude Opus 5.5 (high effort)

All Power Electronics chapter 6 questions →