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Silicon controlled rectifier (SVR) question

  1. Jul 17, 2011 #1
    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 [Broken]

    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 lets 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
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 17, 2011 #2
    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


    The circuit you have displayed is and 'equivalent circuit' using close coupled transistors. However you should realise 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?
  4. Jul 17, 2011 #3
    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.


    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. Lets 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.


    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.


    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 lets 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.
  5. Jul 17, 2011 #4
    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.
  6. Jul 17, 2011 #5
    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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