Skip to main content

Extinction angle control for power factor

Short answer

Switch turns on at zero crossing and off at π−β, so output voltage is zero for the final β interval and input current leads voltage by β/2. Average output is (Vm/π)(1+cosβ). RMS output is (Vm/√2)[(π−β+0.5 sin2β)/π]^1/2.

With the help of suitable circuit diagram and waveforms, explain the operation of extinction angle control for power factor improvement in rectifier circuit. Derive the expression for average value and rms value of output voltage.

Extinction angle control is a power factor improvement method in which a forced-commutated switch is turned ON at the zero crossing of the supply voltage (ωt=0\omega t = 0) and turned OFF at ωt=π−β\omega t = \pi-\beta, where β\beta is the extinction angle. The output voltage is controlled by varying β\beta. Because the current pulse is advanced, the fundamental input current leads the voltage by β/2\beta/2.

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, io=Iai_o = I_a constant)

  1. At ωt=0\omega t = 0, S1 is turned ON; current path a→a \to S1 →\to load →\to D2 →b\to b. vo=vsv_o = v_s, is=+Iai_s = +I_a.
  2. At ωt=π−β\omega t = \pi-\beta, S1 is turned OFF by its gate (forced commutation, since vsv_s is still positive). Load current freewheels through Dm: vo=0v_o = 0, is=0i_s = 0.
  3. At ωt=π\omega t = \pi, S2 is turned ON; path b→b \to S2 →\to load →\to D1 →a\to a; vo=−vsv_o = -v_s, is=−Iai_s = -I_a.
  4. At 2π−β2\pi-\beta, S2 is turned OFF and Dm conducts until 2π2\pi.
 vs   ___
     /   \             /
 ---/-----\-----------/---
           \___/
 vo  ___       ___
    /  |      /  |        (sine from 0 to 180-b,
 --/---+-----/---+----     zero from 180-b to 180)
 is
 +Ia +------+
     |      |
 ----+      +--+      +---
               |      |
 -Ia           +------+
     0  180-b 180  360-b 360

Average output voltage

Vdc=1π∫0π−βVmsin⁡ωt d(ωt)=Vmπ[−cos⁡ωt]0π−β=Vmπ(1+cos⁡β)\begin{aligned} V_{dc} &= \frac{1}{\pi}\int_0^{\pi-\beta}V_m\sin\omega t\,d(\omega t) = \frac{V_m}{\pi}\big[-\cos\omega t\big]_0^{\pi-\beta} \\ &= \frac{V_m}{\pi}(1+\cos\beta) \end{aligned}

VdcV_{dc} varies from 2Vm/π2V_m/\pi (β=0\beta = 0) to 00 (β=π\beta = \pi).

RMS output voltage

Vrms2=1π∫0π−βVm2sin⁡2ωt d(ωt)=Vm22π[ωt−sin⁡2ωt2]0π−β=Vm22π[π−β−sin⁡(2π−2β)2]=Vm22π[π−β+sin⁡2β2]\begin{aligned} V_{rms}^2 &= \frac{1}{\pi}\int_0^{\pi-\beta}V_m^2\sin^2\omega t\,d(\omega t) = \frac{V_m^2}{2\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_0^{\pi-\beta} \\ &= \frac{V_m^2}{2\pi}\left[\pi-\beta - \frac{\sin(2\pi-2\beta)}{2}\right] = \frac{V_m^2}{2\pi}\left[\pi-\beta + \frac{\sin2\beta}{2}\right] \end{aligned} Vrms=Vm2[1π(π−β+sin⁡2β2)]1/2V_{rms} = \frac{V_m}{\sqrt2}\left[\frac{1}{\pi}\left(\pi-\beta+\frac{\sin2\beta}{2}\right)\right]^{1/2}

Power factor improvement

  • Fundamental input current: Is1=22Iaπcos⁡β2I_{s1} = \frac{2\sqrt2 I_a}{\pi}\cos\frac{\beta}{2}, leading vsv_s by ϕ1=β/2\phi_1 = \beta/2.
  • Displacement factor =cos⁡(β/2)= \cos(\beta/2) leading, compared with cos⁡α\cos\alpha lagging for a phase-controlled converter.
  • The converter can appear as a capacitive load and can partly compensate the lagging reactive power of other loads.
  • Self-commutated devices are needed, which adds cost.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Power Electronics chapter 2 questions →