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linear regulator crapping out

Discussion in 'Electronic Design' started by Jimbo, Apr 20, 2006.

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  1. Jimbo

    Jimbo Guest


    I'm using an UA7805C linear voltage regulator. It's a three terminal
    device with an input, common, and an output. Bascially I'm using a
    center tapped transformer that is connected to the wall socket and has
    an output of 14 Volts......0 volts.......14 Volts. Each 14 volt is the
    two terminals and the 0 volts is the center tap of the transformer.

    I'm using two diodes on the two 14 volts to rectifiy the AC and it is
    connected to 680uf elec. cap and a .01uf elec. cap that is connected to
    the input via a 50uH inductor. On the output side there is another
    608uf and a .01uf elec cap. The input is drawing .09 amps common .06
    amps and the output is drawing anwhere between .01 to .03 amp. It keep
    crapping out on me. I'm not too sure why but it's rated at 1.5 amps.
    I have put a fairly big sized heat sink on it. when it craps out it
    seems to output voltage around 10 Volts +. Does anyone know why it's
    doing this? is it a surge from plugging the thing in? Is it the
    inductor spiking the voltage when the output load changes? All the
    grounds are common to the center tap. Any help is appreciated.
  2. Jimbo

    Jimbo Guest

    Could it be a short on the common terminal? Would that cause on of
    these things to quit?
  3. Jimbo

    Jimbo Guest

    Could it be a short on the common terminal? Would that cause one of
    these things to quit?
  4. What, you are using a inductor on the input pin of the regulator?
    That isnt exactly what its supposed to be and likely oscillates.
    Standard connection is as follows:
    transformer-rectifier-large electrolytic-100nF-input
    output-100nF-10-100µF electrolytic
    the 100nF can be ceramic or polyester film.
    Inductors usually are no good idea.
    Grounds should be connected with short wires, use a ground plane or
    single point gnd.
    Also, a diode is recommended from out to GND with cathode on out (on
    dual supply systems this is necesary, on single supply it doesnt hurt)
    and one from in to out with cathode on in.
    The large electrolytic should be calculated based on load current and
    other requirements, last time i did a 5v supply i came up with 2200µ 16v
    (input from 9v Transformer, load current 0,6A).
  5. Jimbo

    Jimbo Guest

    isn't a 100nf a bit small to keep the ripples low? is there a reason
    the capicatance is so much higher on the output side? those cap I
    mentioned on the input is to convert the rectified circuit from
    rippling. Right now I'm looking at between .01 and .03 amps on the
    output side at 5 Volts. What size cap do you recommend? Thanks!
  6. On the output, theres no ripple (well, almost no) since the regulator
    reduces the input ripple (around 10-20%) by its ripple rejection ratio
    of around 70db.
    The output capacitors only serve to keep the regulator stable and
    improve transistent response.10µ usually is enough, if it has low ESR
    (tantalum) or 10µ in paralell with 100n ceramic/film if the 10µ is
    electrolytic.100µ only improves transistent response slightly, probably
    not needed.

    On the input side, you need a capacitor sized to get the ripple voltage
    you need (ensuring the voltage never drops below Uout+3v under all
    circumstances), a generally accepted value is around 20% maximum.
    Since your current is so low 220µ (rough guess) do nicely.
    The 100n capacitor needs to be close to the regulator, again to keep it

    Inductors if any sort should not be used unless one understands exactly
    what they do and why it is needed.

    Both 100n capacitors should be right at the regulator pins.
    The electrolytic caps can be at some distance.

    So thats it all again,to be viewed with fixed font:

    | ______ |
    | | ------ | | |
    === === | === === K
    | | | | | A
    | | | | | |
    220µ 100n 100n 10µ Diode
  7. Jimbo

    Jimbo Guest

    Sorry, the reason I had the inductor in there was because the load will
    be plugged in and unplugged. Also, the load may be doubled at any
    moment as well. THANKS FOR THE ADVICE!!!! it is very helpful.
  8. Guest

    How about dumping the linear part and go for a TI switching module like
    the PT5105A? That's a 6.5V output module, but you can find a 5V part.
    Best thing is you can get a free sample from TI.
    Read the datasheet carefully and see if it fits your needs, the big
    plus here is very much reduced power dissipation.
  9. No problem, the inductor will probably cause problems and you dont need
    it.The 78xx regulates quickly and the output capacitor keeps the voltage
    up during that time.With your low load currents the drop probably is in
    the microvolts.
    One last Q: is the 7805 connected the wrong way round?Input is on the
    left side if the heatsink tab points away from you.
    If it still doesnt do, try a new regulator.
  10. Jimbo

    Jimbo Guest

    It's properly connected. This is the 4th one that has crapped out.
  11. History.
    Been in this movie long time ago. Try a small resistor on input line
    ~10Ohm value to reduce inrush current.(Check values according to
    manufacturers data taking into consideration a dead short presented by
    the electrolytic capacitor(s) on power-up.) It helped in my/our case.

    Have fun

    Slack user from Ulladulla.
  12. Boris Mohar

    Boris Mohar Guest

    Did you remove the inductor yet? Is the input cap reasonably close to the
    regulator? Put a scope on it. It could be oscillating.
  13. Is it possible the output is even briefly seeing > 7V with the input
    off? If that's even a remote possibility, put a diode such as 1N4001
    from input to output (reverse biased, obviously). It usually is
    unnecessary for 7805 parts, but you've got funny things going on.

    Best regards,
    Spehro Pefhany
  14. Fred Bloggs

    Fred Bloggs Guest

    Get rid of the inductor- it is worthless at this frequency and does
    nothing for ripple reduction. Insert a series 47R at 1/2W rating between
    the common cathode junction of the rectifiers and the input to the first
    680u || 0.1u and the regulator.
  15. Jimbo,
    Now, you have a centertapped transformer, and you are full wave
    rectifying it. What is the actual voltage at teh input to your
    regulator- 14 or 28 volts? A ways back, I tried this with 28 volts, but
    forgot that this is RMS voltage, not peak. The actual measured voltage
    was 38 VDO at the input of the 7805, which was above the operating
    point. Regulators usually lasted about a day, at best, before giving up
    magic smoke...

  16. I never needed the diode on the output pins to ground. and in your diagram
    the one across the regulator is backwards. other then that it is fine by me.
    Your real problem is most likely the input voltage is too high. 7805 ic's
    are hardy little bugs. Jtt
  17. The problem can come if the input has a heavy load on it or can be
    shorted, in which case current can flow back from output capacitors
    through (IIRC) some B-E junction breakdown. But that won't happen on a
    7805 *under normal conditions* because the output voltage is only 5V
    (unless something else is causing the voltage to rise out of
    regulation), not enough to break down a B-E junction. For 7808 and
    above it's a concern.
    Well inside.

    Best regards,
    Spehro Pefhany
  18. Yes, sorry, the one diode is reversed.
    Im using these diodes from out to gnd for some time now but they are
    only necessary in dual-supply applications, where overloadung one rail
    could lead to reversed voltage across the regulator and damage it.This
    blows both regulators most of the time and also damages the parts
    connected to it because the regulators shorted input/output, resulting
    in increased output voltage with loss of current limiting.

    On single supply applications that cant happen but during testing stupid
    events happen, like connecting wires to a power supply instead of a meter...

    For the input voltage, if its too high this can cause problems, although
    i dont think this happens here, since the transformer is mentioned to be
    14v with 2-pulse-rectification giving around 25v worst case (no-load
    with high input V) which is still inside the specifications of a 7805.
  19. Another problem (which is why i put in the second diode from out to gnd)
    only happens in dual supply systems and will cause extensive damage
    without this diode.
    This happens if a load connected between both rails draws excessive
    currents and the regulators limit current.Usually one regulator has a
    little lower current limit and then its output gets negative, blowing
    the regulator and making it short from in to out, putting unregulated
    input to the output, which causes high current to flow and does the same
    to the opposing regulator.
    Then one has both regulators shorted and until the fuse blows, damage
    will occur and expensive chips fail if present and no cheap part shorted
    out already.
    The diodes avoid this condition by keeping high reverse currents from
    flowing into the outputs.
    This always served me well in the past, and i did a few power supplies
    by now.
    As always, one can often get away without these diodes but then dont cry
    around if it blows up.A 4001 costs around 1 cent and can save rather
    expensive loads connected to the regulators from blowing.
  20. Guest

    Sorry; I just saw this... nobody has posted the correct answer yet, so
    I'll chime in. Your problem is the 680 uF cap on the *output* of the
    regulator, and your failures are occuring when you *unplug* your power
    supply, not when you plug it in.

    Basically, you don't want current flowing from the regulator's output
    back to its input. That lets the smoke out. Lose the 680-uF output
    cap, or install a diode across the regulator with the cathode towards
    its input pin, and you should be OK.

    -- john, KE5FX
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