Skip to main content

Self and line commutation; minimum gate pulse width

Short answer

Self-commutation uses a resonant LC to create a current zero; line commutation relies on AC supply reversal. For the RL load, current rises as i(t)=1(1-e^{-1000t}) A. Setting i=0.2A gives required gate pulse width t = -ln(0.8)/1000 = 223.1µs.

Explain about self and line commutation with circuit and its waveform. A 10 V source is connected across RL load having R = 10 Ω and L = 10 mH. A thyristor having latching current of 200 mA is used as switch for the circuit. Calculate minimum gate pulse width to Turn ON the thyristor.

Commutation is the process of turning OFF a conducting thyristor by bringing its anode current below the holding current and applying reverse voltage for longer than its turn-off time.

Self commutation (load / resonant commutation)

   +Vs                 
    |     L      C
    +---[SCR]--LLLL--||---+
    |                     R (load)
    +---------------------+

The SCR is in series with an under-damped LL-CC (and load RR) circuit. When fired, the current is a damped sinusoid. After half a period the current tries to reverse and falls through zero; the SCR turns OFF by itself, and the charged capacitor applies reverse voltage across it. Used in series inverters and DC choppers.

 i  ^   .-.
    |  /   \
 ---+-'-----'-------> t
       SCR off at i = 0
 vC ^       ______
    |    .-'      (holds reverse
 ---+-.-'          voltage on SCR)

Line (natural) commutation

  ~ vs ---[SCR]---+
                  R
  ----------------+

In a.c. circuits the supply voltage reverses every half cycle. At the end of the positive half cycle the anode current falls to zero naturally and during the negative half cycle the SCR is reverse biased, so it turns OFF without any extra circuit. Used in controlled rectifiers, AC voltage controllers and cycloconverters.

 vs ^  .-.         .-.
    | /   \       /
 ---+-|----\-----/----> t
      a     \___/
 io ^  |\
    |  | \  current falls to 0
 ---+--+--\------------> t

Numerical: minimum gate pulse width

Given: V=10V = 10 V, R=10 ΩR = 10\ \Omega, L=10L = 10 mH, IL=200I_L = 200 mA. For an RL circuit the current after firing is

i(t)=VR(1−e−tR/L),VR=1 A,LR=1 msi(t) = \frac{V}{R}\left(1 - e^{-tR/L}\right), \qquad \frac{V}{R} = 1\ \text{A}, \quad \frac{L}{R} = 1\ \text{ms}

The gate pulse must last until ii reaches ILI_L:

0.2=1(1−e−t/10−3)e−1000t=0.8t=−10−3ln⁡(0.8)=0.2231 ms\begin{aligned} 0.2 &= 1\left(1 - e^{-t/10^{-3}}\right) \\ e^{-1000t} &= 0.8 \\ t &= -10^{-3}\ln(0.8) = 0.2231\ \text{ms} \end{aligned}

Answer: minimum gate pulse width =223.1 μ= 223.1\ \mus.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Power Electronics chapter 1 questions →