SCR forward breakover

Hi

This may be an elementary question, but I haven't found a convincing answer in any reference book.
Why does the voltage across an SCR suddenly decrease after the forward break-over voltage?
As far as I understand, the middle junction breaks down at that voltage. Why does the voltage across it drop to zero after it breaks down?

Because the SCR is triggered. It behaves the same as triggered by a gate signal. So the question refers to SCR operation principle in general.

What exactly does happen when an SCR is triggered? Books explain the process by saying that the reverse biased middle junction breaks down. What I don't understand is how the voltage across it decreases to zero when an avalanche breakdown has occurred.

It's a simple positive feedback effect, that's very visually modelled by the well-known two-transistor equivalent circuit. One the holding current threshold is exceeded, the device switches to saturated state with low residual voltage.

Thanks for the explanation!

I was trying to visualise the triggering operation considering the behaviour of the three junctions, but I think I'd better stick to the two transistor analogy

Of course, the behaviour can be explained from a charge-carrier generation and recombination viewpoint as well. But I prefer an equivalent circuit comprised of components with known behaviour. For a real SCR. you should add a resistor in parallel to the gate-cathode terminals to model holding current respectively trigger threshold.

The main difference between the simple two-transistor model and a real SCR is in the effect of lateral extension of the active zone and respective bulk resistivity. This explains, why a regular SCR can't be turned-off from the gate. This works only with a special designed (lower intrinsic gain) GTO and by feeding rather large currents to the gate.

it is quite simple. if you consider/visualize an scr as a switch then before forward breakdown it behaves like an open switch and an open switch have total voltage drop across it. and after forward breakdown it behave like a close switch or short circuit so the voltage across it must drop down to zero or approaches to zero.

you can perform this experiment with switches of your home appliances; take a multimeter set on ac voltage range and measure the voltages across the switch in open state (appliance OFF) it will read 220V now when you close the switch the the reading of the meter drops down to zero and appliance ON.

Now same in the case of all Thyristers and Transistors when they are used as a switch they will give you full voltage applied across them( just like open switch with mains ) but after forward breakdown when they start conduction (i;e. it shorts off) the potential across it drops to zero and it starts behaving like a close switch or rightly say simply it behave like a wire/conductor; if you measure the voltage on a single conductor it will always read zero as it has no potential difference across it.

hope you got the point.

Well, I did understand that part, but in that case, why doesn't the same thing happen when the device is reverse biased?

In reverse bias, there's simple junction breakdown, but no internal feedback and thus no latching effect. The SCR will behave similar to a diode in reverse breakdown.