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enter image description here

I understand that when \$TR_1\$ and \$TR_2\$ are ideal transistors, we need an initial \$I_G\$ current pulse signal in \$G\$ to turn on the thyristor. This \$I_G\$ will turn on \$TR_2\$ and the collector will current of \$TR_2\$ will turn on \$TR_1\$ and the overall system (the thyristor) will conduct.

However, when we work with realistic electronic parts, there will be a a tiny leakage current through \$TR_1\$ (from emitter to collector). This tiny current will create a collector current through \$TR_2\$ which is \$h_{FE}\$ times multiplied. And this collector current will further increase the current through \$TR_1\$. So on, this loop of events will cause an avalanche effect and finally, in a very short time, both transistors will saturate and the thyristor structure will conduct.

But it doesn't happen so.

Once I tested a thyristor on bread board. Initially I didn't connect anything to its gate; I left it open. The thyristor didn't conduct until I touched its gate with a cable which had some positive voltage on it.

Why doesn't leakage currents turn on thyristor automatically?

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One must be very cautious when comparing complex semiconductor devices to their "equivalents" which employ less complex semiconductor devices. See for example my answer to this question: "Why can't two series-connected diodes act as a BJT?"

However, in this case I got a feeling that if you'll connect two BJTs in the suggested configuration they will not saturate (therefore acting similarly to thyristor). You're welcome to test this guess by either assembling a real circuit or simulating in Spice. Please let me know if the transistors do saturate.

As for the thyristor, then the fact that it does not turn-on due to leakage currents is quite intuitive (unless you are seeking for a complete explanation involving semiconductor physics)

Let us assume that the Gate is floating:

enter image description here
(source: radio-electronics.com)

When \$V_{AK}>0\$:

  • J1 is "forward-biased" (note the quotes)
  • J2 is reverse -biased
  • J3 is "forward-biased"

Why did I put quotes around "forward-bias"? The junctions are forward-biased, but the voltage is much lower than the usual voltage associated with forward-biased PN diode. In fact, voltages across the forward-biased junctions are very close to 0 - the most of the externally applied voltage is dropped across reverse-biased junction (J2).

In order to turn the thyristor ON, one of the following must occur:

  • Breakdown inside reverse-biased junction
  • Forward-biasing the bottom PN junction (J3) by a significant voltage, thus causing the "bottom NPN sandwich" to become active.

The first condition can occur for very high \$V_{AK}\$ (turn-on without Gate-Drive).

The second condition can not be satisfied by \$V_{AK}\$, because the majority of the voltage is "eaten" by J2. However, applying bias "below" J2 (bypassing J2) can help because this voltage will not see any reverse-biased PN junctions. This is exactly what happens when Gate is driven with voltage pulse.

Summary:

Thyristor won't be turned-on by leakage currents because the reverse-biased PN junction (J2) consumes most of \$V_{AK}\$, thus leaving forward-biased PN junctions (J1 and especially J3) with insignificant forward-bias. Thyristor will be turned-on when either J2 undergoes breakdown, or there is applied bias bypassing J2 (at Gate electrode).

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It's just a representation

Your two-transistor model of a thyristor is just a circuit representation to explain how these devices work. If you were to make a thyristor from components you'd probably add a 10k resistor from the base to emitter on both transistors to prevent ultra-sensitivity issues.

Another way of looking at this

You mention leakage current but this leakage current (through a switched-off transistor or two back-to-back diodes) is going to be less than the leakage current into the base-emitter junction (single diode) so which one wins?

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