Asymmetrical thyristor:

• The asymmetrical thyristor, also known as the asymmetrical silicon-controlled rectifier (ASCR) is a modified version of the thyristor.
• Its turn OFF time is much shorter. Therefore, it can be used for switching at a higher repetitive frequency than the ordinary thyristor.
• The shorter turnoff time is made possible at the cost of the ability to block reverse voltages.
• The junction structure is exactly the same pnpn four-layer structure of the thyristor, with one difference. The middle n layer now consists of low resistivity (high impurity) region labeled as n+, and the usual high resistivity (low impurity) region labeled as n¯.
• The reason for the long turn OFF time in the conventional thyristor is that, during the reverse recovery transient, the flow of reverse current causes holes to be injected across the junction J₂ from the p₂ to the n₁ layer.
• These holes have to disappear, mainly by recombination, before the junction J₂, which is the junction responsible for blocking forward voltages, recovers its blocking ability.
• In normal thyristors, this recombination process takes a long time because of the high purity level of the n₁ layer.
• In the asymmetrical thyristor, the presence of the higher impurity n+ layer speeds up the recombination process and so shortens the turn OFF time.

Holding current is the minimum anode current to maintain the thyristor in the on-state.

The holding current implies that the thyristor will turn OFF if the current tends to fall below the value of 150 mA.

The highest value of R possible with the thyristor ON is

$R=\frac{V}{I}=\frac{300}{150\times {10}^{-3}}=2k\mathrm{\Omega }$

Conduction angle (α) = 90°

For rectangular wave,

The average current, $I_{avg}=\frac{\alpha }{360}\times I=\frac{90}{360}I=\frac{I}{4}$

The rms value of current, $I_{rms}=\sqrt{\frac{\alpha }{360}}\times I=\sqrt{\frac{90}{360}}\times I=\frac{I}{2}$

Concept:

For the above circuit the waveform of the load current, capacitor voltage inductor voltage and SCR voltage are shown by  respectively.

• From the waveform of SCR voltage we can observe that the SCR is in on-state only for .
• That is, conduction time duration of the SCR is given by $t_0=\frac{\pi }{{\omega }_0}$

Calculation:

Resonant frequency for the above circuit is given by

${\omega }_0=\frac{1}{\sqrt{LC}}$

Therefore, conduction time duration of the SCR is

$t_0=\frac{\pi }{\frac{1}{\sqrt{LC}}}$

$t_0=\pi \sqrt{LC}$

The current in the circuit should reach to the latching current during the firing pulse for the thyristor to switch ON.

The equation for current is,

$i\left(t\right)=\frac{V}{R}\left(1-e^{-\frac{R}{L}t}\right)$

Given that R = 50 Ω, L = 0.5 H

At t = 50 μs,

$i\left(t\right)=\frac{250}{50}\left(1-e^{-100\times 50\times {10}^{-6}}\right)$

$=5\left(1-e^{-5\times {10}^{-3}}\right)$

= 5 (1 – (1 – 5 × 10^-3)) = 25 mA

$R_s=50\mathrm{\Omega }$

$\frac{di}{dt}=10A/\mu s$

$V_s=100V$

R_L = 50 Ω

$i\left(t\right)=\frac{V_s}{R}(1-e^{\frac{-Rt}{L}})$

$\frac{di\left(t\right)}{dt}=\frac{V_s}{R}\left(\frac{R}{L}\right)e^{\frac{-Rt}{L}}$

$att=0,{\left(\frac{di}{dt}\right)}_{max}=\frac{V_s}{L}$

$\Rightarrow L=\frac{V_s}{{\left(\frac{di}{dt}\right)}_{max}}=\frac{100}{10}=10\mu H$

Holding current (I_H) = 2 mA

Voltage across SCR (V_T) = 1 V

SCR will turn off when I_a ≤ holding current

$\Rightarrow I_a\le 2mA$

$\Rightarrow 2mA=\frac{V_A-V_T}{45}$

$\Rightarrow 2\times {10}^{-3}=\frac{V_A-1}{45}$

⇒ V_A = 1.09 V

Initially the inductor will be treated as open circuit and entire voltage appears across the inductor so the maximum value of rate of change of current

${\left(\frac{di}{dt}\right)}_{max}=\frac{V_{smax}}{L}=\frac{\sqrt{2}\times 220}{20\times {10}^{-6}}=15.56A/\mu sec$

If SCR is off then current flows through the resistor and capacitor

Let I and I_sb be the one – cycle and sub – cycle surge current ratings of the SCR respectively.

$I^{2}T=I_{Sb}^2.t$

t = 10 ms T = 20 ms

$\begin{array}{l}{I}^2\left(20\right)={\left(2400\right)}^2\left(10\right)\\ \Rightarrow I^2=\frac{{\left(2400\right)}^2\left(10\right)}{20}\\ \Rightarrow I=\frac{2400}{\sqrt{2}}=1200\sqrt{2}=1697A\end{array}$

I²t rating

$\begin{array}{l}=I^2\times \frac{1}{2f}\\ ={\left(\frac{2400}{\sqrt{2}}\right)}^2\times \frac{1}{100}=28,800Am{p}^2.sec\end{array}$

From the gate drive circuit of the thyristor,

E = Ri_g + V_g

The diode D clamps the gate voltage to zero when E goes negative.

Now for i_g = 0, V_g = E.

Since the maximum value of V_g = 10 V, E = 10 V

And $E=\frac{N_2}{N_1}\times 15=10$

Now, the turns ratio $\frac{N_1}{N_2}=\frac{15}{10}=1.5$