Silicon controlled rectifier (SVR) question

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Silicon controlled rectifiers (SCRs) operate through a regenerative effect involving two transistors, where a small leakage current can trigger a significant increase in current flow. The gate voltage controls the SCR by allowing current to flow into the base of the NPN transistor, which subsequently turns on the PNP transistor, creating a feedback loop that keeps the SCR in the "on" state. Once activated, the SCR remains on until the current falls below a specific holding current, and it cannot be turned off simply by removing the gate current. The discussion also highlights the importance of understanding the equivalent circuit and the interaction between the components. Overall, SCRs are complex devices that require careful consideration of their operating principles and circuit design.
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Okay so I have to work with SVRs in a schematic I'm looking at and I've never seen them before

http://img269.imageshack.us/img269/6985/19892699.png

Uploaded with ImageShack.us


I know how BJTs work but I don't understand how the current is flowing

so far I see that a Vin at base gives a current from bottom base to emitter, which let's current flow from bottom collector to emitter I think

So does the current from the 12V go to the collector on the bottom AND the base at the top?

My boss says its for fixing some short circuit problem and that one BJT makes sure that its always "on"...

anyways I'm pretty confused and I've read some articles over SVRs but they don't explain them in terms of BJTs, but as a whole component

any help would be appreciated
 
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Helo Kr-etc, welcome to Physics Forums.

A good suggestion is to use the forum's search function. Some useful search keywords appear in the next paragraph.

The SCR (silicon controlled rectifier) and its close cousins an SCS (silcon controlled switch) DIAC and TRIAC are examples of four layer diodes also collectively called thyristors.

There are quite a few threads discussing these here eg

https://www.physicsforums.com/showthread.php?t=508643

The circuit you have displayed is and 'equivalent circuit' using close coupled transistors. However you should realize that part of the thyristor action depends upon bringing the junctions close enough to interact so you cannot exactly reproduce the action with discrete components.

What exactly do you wish to know about them?
 
Well since I asked this question not so long ago, it would be nice of me to provide my current hopefully correct understanding.

As Mr. Studiot said that picture that you provided us is equivalent circuit. That circuit is good for understanding what is going on in thyristor.

As this I found myself is very messy on the internet I will try to explain that circuit.

490px-Thyristor.svg.png


Here is a similar picture I will work with. Thyristor is a component that is controlled by a external signal that is sent through gate.

Gate of thyristor is the collector of the pnp transistor, and the base of the npn transistor. Let's review the case if gate voltage(current) is 0.

If you increase the voltage between anode and cathode, no current will flow. Only small leakage current will flow from np junction of the pnp transistor.

This small leakage current goes into base of npn transistor.

However if you increase the voltage high enough so that u have significant leakage current, thyristor will suddenly turn itself on.

How?

Well you see that leakage collector current, when its big enough it gets amplified by the npn and sent through collector to base of pnp. This current gets amplified also by the pnp and again sent through collector to npn etc. This is called the regenerative effect and voltage needed for this, breakdown voltage. When thyristor goes into "on" state, its voltage drop suddenly drops down to few tenths of volts.

You can see on the graph, that the voltage drops very fast, when the regenerative effect occurs. Regenerative effect would be between Vbo and Vh above IL.

374px-Thyristor_I-V_diagram.svg.png


This is marked as Ig=0 on the graph.

So basically in regenerative effect, you get this coupled amplification between 2 transistors.

In practice this is mostly unwanted turning on of the thyristors. So they are built with high breakdown voltages, depending on what one might need.

Now let's review if Ig is greater than 0. Now you are basically sending current into base of npn transistor, which gets amplified and gets sent into base of pnp... effect like before. But the thing here is, you turn on thyristor earlier, before breakdown voltage.

Important thing that you have to say is that thyristors will be turned on until you lower the current anode-cathode below holding current.

Holding current is current needed to "hold" the thyristor in "on" state. It cannot be turned off by removing current from the gate either. It can be turned off by applying negative voltage to gate.

I believe mr. Studiot can verify this.
 
You can also use the base of the npn transistor (the upper n layer) as a gate.

Gates are described as cathode gate or anode gate according to which is used.
 
Bassalisk said:
Now let's review if Ig is greater than 0. Now you are basically sending current into base of npn transistor, which gets amplified and gets sent into base of pnp... effect like before. But the thing here is, you turn on thyristor earlier, before breakdown voltage.

Important thing that you have to say is that thyristors will be turned on until you lower the current anode-cathode below holding current.

Holding current is current needed to "hold" the thyristor in "on" state. It cannot be turned off by removing current from the gate either. It can be turned off by applying negative voltage to gate.

I believe mr. Studiot can verify this.

So far I understand it as this:

1) Vin at gate provides current to go into the base of the NPN
2) This turns on the bottom BJT allowing C to E current to flow (the amplification)

now looking at my drawing,,does the current from the left Volt source split itself between the collector of NPN and the base of the PNP and how do I determine how much. I feel like I heard somewhere that it all goes into base maybe??

3) the PNP gets turned on through current flough between E and B turning on the PNP and amplifying the base current, (current from E to C) which ensures that B of the NPN is turned on.

Another dumb question...Lets say I put 1V at the gate before the resistor. Now after the Resistor the voltage should be 1V minus the turnon voltage Vbe correct?

thanks for the quick responses thus far!
 
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