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Well, I've recently started reading about thyristors and I have been stuck by something my instructor said about the turn-on process.

During the forward-biased state of the thyristor the junctions J2 is reverse-biased, which is essentially responsible for the forward "blocking" mode. However, after the break-over voltage, J2 starts conducting magically.

How does this change come about? Is this due to the breakdown or avalanche process? If so, how can a device recover from it and turn-off? I mean, we say diodes are destroyed if the reverse voltage exceeds the breakdown voltage but how come the same process not destroy the thyristor junction?

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    \$\begingroup\$ Where is J2? Show the diagram if you refer to one! \$\endgroup\$ – stevenvh May 29 '12 at 6:31
  • \$\begingroup\$ J2 is the middle junction of a (Anode)-pnpn-(Cathode) SCR. The one between n-p and is held at forward bias if a forward voltage is applied between Anode and Cathode. Sorry for the ambiguity. \$\endgroup\$ – Shrikant Giridhar May 29 '12 at 9:48
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    \$\begingroup\$ Look at the 2 transistor model which James Smith posted. Are you happy that that behaves logically without any magic required? Are you happy that that models what you are seeing? Are you happy that in an eg NPN transistor the CB junction conducts in the forward direction during the transistor's on state but the BE junction conducts in what is usually the reverse direction? If you are happy with the transistor action you can apply it to the SCR. If you are unhappy with the transistor action go and delve into it unto happy - at which stage the SCR will make sense as well. \$\endgroup\$ – Russell McMahon May 29 '12 at 11:24
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    \$\begingroup\$ "we say diodes are destroyed if the reverse voltage exceeds the breakdown voltage" -- This is often true in practice if breakdown occurs unintentionally and there is no current limiting device in a circuit. But if done correctly, it is possible to operate a diode in breakdown without destroying it; and some diodes (Zeners) are intended to be operated this way. \$\endgroup\$ – The Photon May 29 '12 at 21:20
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There are 4 quadrant sensitive gate type Thyristors with max voltage ratings and there are 3 quadrant types which are better for commutation and preventing back EMF from false triggering. In both cases, however dv/dt can trigger false latching and low R and medium sized plastic cap called a snubber circuit or LPF for high voltage are required to avoid that.

You can try two discrete transistors to make an SCR and examine the Absolute Maximum Ratings to fully appreciate how this latch works.

SCR layout from Wiki....

Thank you for letting me explain to assist your learning process..

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The mode which you had mentioned will starts the thyristor conducting but where as the thyristor will destroy when you use it in reverse conduction mode.

As you mentioned that the diode will destroy in reverse bias but not in forward bias. like that the thyristor will works fine but it will work fine when the reverse voltage exceeds the its withstanding voltage.

Typicallly this voltage will be much higher when compare with the diode.

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A high voltage between anode and cathode results in an avalanche breakdown. A voltage at the gate results in a breakdown at a lower voltage. The thyristor is conducting now. If the current flows stops or the voltage between anode and cathode is zero, the thyristor turns off (and recovers). The conception of an avalanche neccesarily overheating and destroying diodes is wrong. There are even avalance diodes, which work fine.

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  • \$\begingroup\$ That, precisely, is my question. A diode is destroyed after an electron avalanche due to overheat as it loses its bound electrons but the SCR junction recovers to its initial state. Is anything different about this breakdown process? \$\endgroup\$ – Shrikant Giridhar May 29 '12 at 9:43
  • \$\begingroup\$ The difference is that P=V*I for thyristor is say few watts only because voltage drop is 1-2 volts in open state, when ordinary diode will stay in zener mode with say 100-200v. Power difference is then 10000% \$\endgroup\$ – user924 May 30 '12 at 1:31
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From your question that says:

How does this change come about? Is this due to the breakdown or avalanche process? If so, how can a device recover from it and turn-off? I mean, we say diodes are destroyed if the reverse voltage exceeds the breakdown voltage but how come the same process not destroy the thyristor junction?

It is true that the thyristor will switch on due to zener or avalanche breakdown. The thing is, it is not the breakdown itself that damages the semiconductor crystal but the increased temperature in the material. Devices intended to work in the avalanche region are specifically designed to withstand high temperature and in fact while normal diodes have their breakdown voltages decrease with increase in temperature, the zeners are designed to have temperature coefficients that are independent of the rated breakdown voltage.

Of course the device will still get damaged if too much current is allowed to flow because then too much heat will damage the device.

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