# Why doesn't the leakage currents automatically turn on thyristor?

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?

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:

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).

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?