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Current source OK until load disconnected

Discussion in 'General Electronics Discussion' started by eem2am, Nov 16, 2012.

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  1. eem2am

    eem2am

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    Aug 3, 2009
    Hello,

    We have a lamp which runs off a current source when the mains is on, this current source drives the LED luminaire, and another current source drives current into the battery to charge it, (battery runs the lamp when mains fails)

    The problem comes when the lamp is disconnected when the mains is on, because the current source just keeps being a current source and pushes the rails up to very high voltages which blows it up.

    ...The same applies to the battery , when the battery is removed the current source keeps flowing but has nowhere to flow into.

    It would be good if we could short out the battery and lamp connections in the event of such dicsonnection......then there would be a path for the current to flow.

    As such, i wondered if their is such thing as a connector which when disconnected, it leaves a short across the terminals........then when re-connected, it breaks the short.

    do you know of such a thing.?
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    OK, reading on, you really do mean "current source".

    It sounds like a poor design. I'm not sure when I would ever want to change a battery from a current source, and neither would I switch the load on and off from a current source (I'd switch the current source on and off.

    OK, that's kinda reasonable, however unless you have a perfect current source (or at least a good one) it may not tolerate having the output shorted.

    And you'll have a problem with the battery because you can't short it before you remove the battery (or you'll short the battery) and yo can't disconnect the battery first or the voltage will spike.

    OK, I think this is a bad idea, but I'll give you my advice. I think the correct solution would be a different design.

    However... The solution for the battery is to have a SPDT switch (or relay, etc.) that in one position charges the battery, and in the other position disconnects the battery and shorts the current source. "But wait" (I hear you say) what about the brief time while neither the battery nor the short is connected? Here's the trick -- place a high power zener across the current source. This zener must have a voltage higher than seen with the load connected. For the short period of time while the relay switches, it clamps the voltage across the current source. It's only going to be conducting for a matter of milliseconds, so it's OK to overload it, but perhaps not more than 10 to 20 times (e.g. use a zener rated at 5W for a 50W load)

    For the LEDs, you can simply use a switch to short them out.

    Remember, this advice deals with the problem as you stated it. I think there would be several far better solutions that would come to light if we knew more about the problem.
     
  3. eem2am

    eem2am

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    Aug 3, 2009
    thanks, i cant really say too much about why a current source is used and the nature of it.

    ill think about what you said..when you mantioned switch , i am now thinking of a switch, whereby if the user is going to disconnect the lamp or battery, they have to move the switch lever to access the lamp or battery connector, and by moving this lever they short out the current source connection.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    What current are we talking about? And what voltage across the lamp under normal conditions?

    A simple solution may be to create a shunt regulator with a voltage a couple of volts higher than the lamp, placing it across the lamp. If you remove the lamp, the shunt regulator takes up the load. It would have to be rated for the full lamp power (and maybe a little more).
     
  5. eem2am

    eem2am

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    Aug 3, 2009
    the lamps can be 5V, 10A or 40V, 1.2A........or anything in between 5V and 40V , but 50W max.

    .....The battery charge current is 428mA DC.

    ..........its the 10A one that causes the most problems here, because where do we ship that 10A without needing a big heatsinked component.......
     
  6. CocaCola

    CocaCola

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    You don't "ship" it anywhere, you fault detect and shut it down...

    As Steve said "I think the correct solution would be a different design."
     
  7. eem2am

    eem2am

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    Aug 3, 2009
    shutting it down sounds fine....but the current source, is remote from the lamp.....we cannot simply shut it down......not only that, but the single current source supplies many lamps on the ring, and they would all go off if the current source is shut down.

    the whols advantage of our system is the fact tthat its a current source.and yet as you say, it has this weakness too......
     
  8. CocaCola

    CocaCola

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    Apr 7, 2012
    Sure you can, break that branch of the circuit... No circuit no current flow, just like when you flip a switch...
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Why do you want to replace bulbs when the power is on, and without turning the others off?

    Assuming they're LEDs, it's not going to be because they've failed.

    If you have control over the current source, you should be able to prevent it spiking in voltage. Perhaps you can limit the rate at which its output voltage increases (without limiting the rate at which it decreases)
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    At the moment I'm confused as to which of these is the real problem:

    1) Voltage spike from current source as load is removed (causing arcing perhaps?). The phrase "which blows it up" is used. Does this mean, the lamp, the battery, the constant current source, or something else? And how long does this take?

    2) Other lights going off as one in a string is removed

    Oh, I asked what current an voltage was normally required and I get an answer that the current varies. This means either a variable current source, several current sources, or something else. What else?

    The lack of an ability to give us detailed information is making this like pulling teeth.
     
  11. eem2am

    eem2am

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    Aug 3, 2009
    i cannot unfortunatley say too much as its a new product.

    Depending on the way we "tap" in to the current source, we get a 1A, or a 2A, up to 10A.

    An errant customer may disconnect a lamp (yes theyre led you are right) or the battery while the system is on.....we dont want them too , but they might.

    There really is no way of stopping the current source......it supplies lots of different lamps, so we cannot just switch it off.

    So we have to live with the fact that the taps that we attach to the "bus" will pull Current amount "X" , and we have to live with that.....we cannot shut it off......if the led lamp must be turned off, then we must simply short the output of the "tap", and have the current flowing through the short, instead of through the lamp.

    There is this probelm of what happens if the battery is removed, because currently all the control circuitry is supplied by the battery.......and the battery is charged by the charger "tap" which supplies 428mA DC of current, .........and when the battery is fullly charged, we simply short this tap so the 428ma dc no longer flows into the battery.

    .....if the battery is removed, the control circuitry looses its supply, we cannot supply the control circuitry from the charger "tap" as it gives 428ma, and the control circuitry only draws ~10mA.

    ....so i am thinking we need another "tap" for the control circuitry.
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Not sure what these "taps" are, but maybe you can arrange the fittings for the lights to short out when the bulb is removed.

    That an re-engineering the constant current drivers so they don't fail if driving an open circuit.
     
  13. CocaCola

    CocaCola

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    Apr 7, 2012
    If this is to be a commercial product, maybe it's time to step back and redesign it 'better' from the ground up with design thoughts for the issues you are worried about as it appears to riddled with serious potential issues that were not thought out and planned for...
     
  14. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    May 8, 2012
    Personally I don't think this problem can be resolved in this section of EP. Perhaps if it's moved to the "Secret" section ...?

    Since I find it fun to guess what secret function secret circuits are providing as well as the secret consumer targets, I cast my secret ballot on some sort of secret alternative power / secret emergency power.

    I think I have a simple solution but I can't securely post it until this is moved to the secret section. :cool:

    Chris
     
  15. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Chris, consider this a formal warning!

    What is the first rule of "the secret section"?
     
  16. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    Uhh, I don't know. I'm clueless! :confused: Do I get a prize if I get it right? I hope its not a secret.

    Chris
     
  17. eem2am

    eem2am

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    Aug 3, 2009
    sorry i can't reveal.

    I'd never heard of this till 3 weeks ago myself.

    ...anyway, ive thrown in a bjt latch for the open led situation...i had to put a hardware latch there because it was 10A......for the battery charger , (in case the battery is disconnected, i just put in a comparator, and filtered the trip signal such that it goes low long enough for the uC to catch it with it interrupt sensor.....the batt circuit is much less current so i didnt bother with a hardware latch there.

    i hate hardware (bjt) latchs because they can often trigger on transients...so ive put some filtering in there...........
     
  18. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    May 8, 2012
    You never did say what exactly "blows up" but here's some food for thought. Whichever component fails it did so from over voltage breakdown. Either replace the failing components with models that can handle the open loop voltage or investigate whether the unloaded voltage can be reduced while still maintaining the constant currents you require.

    Chris.
     
  19. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The first rule of "Secret Section" is "Do not mention the Secret Section"
     
  20. Harald Kapp

    Harald Kapp Moderator Moderator

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    I guess the "secret section" has a backdoor. Someone published this on the internet. This could be used in parallel to the LED. If the voltage rises due to the removal of the LED or battery, the crowbar will short circuit the ring.

    Only problem: how to turn off the crowbar after a new LED or battery has been inserted?

    Anyway: a design that doesn't shut off when the circuit is interrupted (or short circuited) doesn't look sound to me. I fully agree with that part of the discussion.

    Harald
     
    Last edited: Nov 18, 2012
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