Issues with my circuit design.

Discussion in 'The Projects Forum' started by Jemson, Sep 1, 2011.

  1. Jemson

    Jemson Thread Starter New Member

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    Hi all,
    I must point out that I am quite new to electronic design.
    I have designed my first circuit with the help of a friend (who is now overseas), and I am having some issues.

    It is a very small circuit for a micro controller to drive a relay for turning on/off an a fan in my computer room. The circuit works, but the transistors keeps burning out on the board. Once I replace it, it is ok for a short while then dies again. I am not sure if I got a dodgy batch of transistors or there is a problem with my circuit... If there is a problem in the circuit I am thinking it may be with the placement of D1, or R1 is not big enough but in reality, I am stumped!

    I have attached a small design i sketched up in paint, i am still learning my way around LTspliceIV and this is all I could whip up in the short term.
    I should also point out the ground is common between the 5v and 12v sources!

    Any help you can offer would be greatly appreciated. I will be only too happy to answer any questions you may have if additional info is required!

    Attached Files:

  2. strantor

    strantor Well-Known Member

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    What's the actual part# for D1? 1N4001? 1N4002? 1N4003? It should be 1N4004 or better. otherwise that may be your problem.
  3. Jemson

    Jemson Thread Starter New Member

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    Sorry, typo. Yes it's a 1N4004!
  4. strantor

    strantor Well-Known Member

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    well then in that case I'm out of ideas. Transistor & diode seems up to par. the way you've drawn it, seems it should work from what I see. Maybe check to make sure its actually wired the way you drew it (obligatory dummy check :)).
  5. Jemson

    Jemson Thread Starter New Member

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    Yeah definitely checked and rechecked that.
    The circuit actually works (for a while) then the transistor seems to die and I am stumped as to why... Replacing it fixes it for a while, before the next one goes :/
  6. strantor

    strantor Well-Known Member

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    you could alway try a different type of transistor. maybe the ones you have (or wherever you are getting them) are from a bad batch. I usually use TIP120 darlingtons for this application. you can get them at radioshack.
  7. praondevou

    praondevou Well-Known Member

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    As far as I remember, 1N400x are general purpose rectifiers, reverse recovery time could be too long, so the transistor would die from overvoltage, when the relay turns off.
    You could replace it with the UF4004 or a schottky diode.

    Or use a transistor with a higher collector-emitter breakdown voltage.
    strantor likes this.
  8. strantor

    strantor Well-Known Member

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    That's an attribute I never thought of. I usually just look at the max peak voltage when choosing one. (actually I never do any choosing, I just go straight to 1n4004) I have used 1n4004s all the time for this application (but never with this transistor) and never had a problem. I guess I should pay more attention to the whole picture.
  9. praondevou

    praondevou Well-Known Member

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    I'm not saying, this IS the problem, but the transistor dies either of overcurrent or overvoltage. Overcurrent is not likely, since it has 100mA continuous max. rating. To be sure if it's the turn-off voltage spike of the relay, one would have to measure it with an oscilloscope. The 549 has 45V, actually quite distant from 12V, but you never know..:)

    I remember that I used once the 1N4001 for the secondary flyback converter diodes, it didn't work at all, I had to change them to the UF type.
  10. Audioguru

    Audioguru New Member

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    A 1N914 or 1N4148 is a fast ordinary little diode rated for 75V and 300mA.
  11. SgtWookie

    SgtWookie Expert

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    The reverse recovery time of the 1N4004 really shouldn't be an issue, as it only conducts when the transistor shuts off. The diode should turn on pretty quickly.
  12. praondevou

    praondevou Well-Known Member

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    Sgt, I was thinking, if the diode doesn't change it's state from non-conducting to conducting fast enough, the voltage on the transistors collector could rise over it's breakdown voltage at relay turn-off and damage it even if it's only for a few ns/us.
    (I know reverse recovery time is actually defined as the other way around, from forward to reverse. I think I meant the diodes turn-on time)

    EDIT: Yes, I guess you are right, just read about it, turn-on time is extremely short, probably not enough to allow the voltage to rise too much.
    Last edited: Sep 1, 2011
  13. praondevou

    praondevou Well-Known Member

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    Well I just found this on google books... Now I'm not so sure anymore.
    See attached picture... Looks like the turn-on time for the 1n4004 can vary quite a bit.

    Attached Files:

  14. SgtWookie

    SgtWookie Expert

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    Well, if it were slow to turn on, that would definitely do it.

    Could replace the diode, or add a 220pF to 470pF cap across the coil terminals to "buy time" for the diode to turn on. Another possibility is to slow the turn-off time for the transistor.

    1N4148/1N914 diodes have extremely short recovery times; < 4nS. I don't remember if the turn-on times are measurable; but I haven't seen a spec for them.
  15. Audioguru

    Audioguru New Member

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    Like I said, use a 1N914 or 1N4148 fast ordinary signal diode, not a slow 50Hz rectifier.
  16. paulktreg

    paulktreg Active Member

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    I agree that a 1N4148 is the better choice and perhaps a resistor from the base of the transistor to GND, a 1K perhaps? You only need >0.6V or so and the maximum E-B voltage is 5V for the BC549. May help?
  17. praondevou

    praondevou Well-Known Member

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    Is there any way to estimate the forward current through the diode? I don't think so without having the relays characteristics...

    The 1N914 and 1N4148 diodes have both a much smaller peak forward current rating than the 1N4004. If I had to choose between the two I'd prefer the 1N914 because it can handle more peak current.
    If for any reason they can't handle the current, use the UF4004 or another more powerful diode.
  18. Audioguru

    Audioguru New Member

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    The peak current in the diode is the relay coil operating current (only 30mA).
  19. praondevou

    praondevou Well-Known Member

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    Do you mean because the coil resistance would limit the current? :confused:
    Yes, but isn't 30mA for 12V (in this case)? The inductance is generating a much higher voltage, wouldn't that increase the current spike too if the coil is clamped with a diode? I think I will have to test this one if I find the time...

    Edit: simulation software confirms what you are saying... I guess you are right then
    Last edited: Sep 1, 2011
  20. Audioguru

    Audioguru New Member

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    The diode keeps the current flowing (into the power supply) so that the inductance can discharge slowly.
    Without the diode then the inductance creates a rapidly reducing discharge current. The rapid change of current causes the inductance to try to generate a voltage high enough to keep the discharging current flowing slowly. The high voltage is created when the rapid change in the magnetic force cuts across the windings of the coil making a generator.
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