A current flow question

Discussion in 'General Electronics Chat' started by Paradoxine, Aug 12, 2012.

  1. Paradoxine

    Thread Starter New Member

    Aug 12, 2012
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    So I've been designing a circuit and I've simplified it slightly for both mine and your convenience. I'm new to electronics but in essence here is what I propose for an active switch: (The phototransistor etc have been omitted but you'll have to trust that the switching process itself is fine).

    The NPN transistor you see below is actually an IGBT and the capacitor used is fully charged to 500V @ 100uF. When the IGBT is activated, the capacitor will discharge through the inductor. My question is this. Is there a problem with putting the IGBT in the circuit like this? My only concern with the below circuit is that the 500V from the capacitor will flow down to the set of two parallel resistors in the circuit and maybe even to the 12V source battery. My initial reaction would be no it won't because it won't form a proper circuit, but I'm not certain and as such want some confirmation. In other word's I want the two circuits more or less isolated apart from the interaction with the IGBT.

    [​IMG]
    Direct Link: Here

    TL;DR: I've made a circuit where I'm genuinely not certain whether I'll have a problem with the multiple power sources. In the future, is there an easy way I can ascertain whether I'll get any unwanted current flow?

    Note: You may notice this was modelled in LTSpice, but I've literally just started to use it and I'd rather get some advice here first.

    Hope that made sense. Thanks.
     
  2. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    What are the two parallel resistors for? Measuring load current? In this case the 500V negative must be connected to the 12V negative.

    Anyway, there will be no current from the 500V resistor through the 12V supply.
     
  3. Paradoxine

    Thread Starter New Member

    Aug 12, 2012
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    Essentially, one of the parallel resistors is connected in series with a photo-transistor such that as the photo-transistor switches 'on', the effective resistance of the two resistors is decreased, thus increasing the voltage drop across the single resistor above the IGBT, reducing the gate voltage and switching the IGBT off. At least, that's how it works in theory.

    By 'the 500V negative must be connected to the 12V negative', I assume you mean that both need to be connected to ground? If that's the case then I would have to put the IGBT in parallel with the resistors, right?

    Also, could you tell me how you know there will be no current from the 500V cap? Thanks for the response.
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Perhaps you've drawn the schematic incorrectly in relation to the actual circuit topology but your proposed gate control won't function correctly [per your explanation] as shown.

    Also with the inclusion of diode D1, the capacitor discharge may not be all that you anticipate. D1 alters the circuit behavior quite dramatically when compared to the case when it is not included.

    Also you haven't indicated whether your IGBT has an internal reverse [anti-parallel] C-E diode.

    What is the value of L?
     
    Last edited: Aug 13, 2012
  5. Paradoxine

    Thread Starter New Member

    Aug 12, 2012
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    1. Could you explain why it won't work? I'm guessing I've misjudged how the voltage will be divided due to the parallel resistors. Thinking now, you may be right in that the voltage over the two resistors will be the same regardless.

    2. D1 was there just to prevent any possible back current from the inductor as the stored charge is released as the capacitor discharges less and less current. I have read that putting a diode in anti-parallel will prevent this and it's important particularly for electrolytic capacitors.

    3. I'm not sure really, this will be my first venture into custom circuits, let alone semiconductor's. Here's the datasheet regardless: Here

    4. It's not really important. The inductor was more or less just a placeholder. For all intents and purposes it could be a resistor and the anti-parallel diode could disappear. Read below

    My entire goal here is to build a circuit where, when the photo-transistor is 'off' I.E low-light conditions, the IGBT will activate, the capacitor will safely discharge into X load and continue to do so until the light conditions change. After thinking some more about what you said I'm convinced that my circuit won't work, but I still don't know how I could make this work. This should be simple in theory but in practice I'm having some difficulty.

    I really am very new to all this but I dare-say I'll be able to improve my skillset fairly quickly after getting to grips with some of this stuff. I haven't got on hand any equipment yet, not even a multimeter and I wholly expect to learn alot just by trying different configurations, but I'd like to get at least some theory down first. I plan to become an electrical engineer in the future but I'm still in college(UK) and know next to nothing.

    If you had to do what I intend to do (detailed above), how would you go about it? What would be the simplest possible way to achieve what I want? It'd be far easier if the photo-transistor in question turned OFF when under light. :rolleyes:
     
  6. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    Remove the two Rs in parallel. Connect the "-" of the 12V to the emitter of the IGBT.

    The simpliest is then to connect the IGBTs gate to a npn switch similar to this:
    [​IMG]
    You need of course recalculate all the values. Q2 and the LED correspond to the IGBT and your load.
     
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  7. Paradoxine

    Thread Starter New Member

    Aug 12, 2012
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    Thanks for the response. Just checking If I understand what's happening here. When the photo-transistor turns off, does the voltage being provided to Q1 become negative or near 0, switching off Q1, but switching on Q2? (What is the actual voltage at Q1 when the photo-transistor is off?)
     
  8. ramancini8

    Member

    Jul 18, 2012
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    I would be very careful about using a photodarlington because they are unusually sensitive to leakage currents, both junction and light generated. They often turn on at high temperature because the CB leakage current gets too high. R3 must be sized to suck up the leakage current without turning the transistor on.
     
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