Need help with DC power supply

Discussion in 'General Electronics Chat' started by acat800, Jan 13, 2012.

  1. acat800

    Thread Starter New Member

    Dec 28, 2011
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    0
    (newbie)
    Hi, I have the following circuit which I hope to use to power some accessories from a 12V AC power source. The problem I am having is that even with no load I am noticing that the rectifier GBL005 is getting really hot. Anyone have some ideas on what might be the cause? I'm afraid I might burn it up if left connected too long.
    [​IMG]
     
    Last edited: Jan 13, 2012
  2. jimkeith

    Active Member

    Oct 26, 2011
    539
    99
    Welcome to AAC

    1. check your wiring

    2. measure the current going into the 7812 regulator--if it is shorted or wired incorrectly, it will draw lots of current--it is easy to wire these things reverse--done it myself many times...

    3. not likely your wall-wart can source enough current to destroy your bridge rectifier if wired incorrectly
     
  3. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    Does the LM812 have a live mounting tab? (I can't find a datasheet for it). Bolting a live mounting tab to a grounded heat-sink may cause a short. If this is the problem, use an insulating mounting kit.

    Another thing to check is that the mains supply's output is not already grounded - another way of getting a short.

    Anything reversed - from the rectifier on in - can give problems. Any reversed electrolytic capacitors can blow up unpleasantly. Best avoided!

    I note also that your reservoir capacitor is only 2μF. That would not be big enough to smooth the supply for the LED alone. Try 470μF.
     
  4. acat800

    Thread Starter New Member

    Dec 28, 2011
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    Ok, thanks guys we may be getting somewhere....

    As suggested, I've gone through the connections and I find they are all good. One thing I will likely need to change is the LM7812, after a closer read of the datasheet, I find that it requires at least 14V input to operate correctly but this doesn't explain the heat up... however, I did check the mains, and both leads are grounded. The mains come from a small engine stator for reference so it's not your typical transformer output. So, I was grounding the DC side, but it seems the AC side shares the same ground. I wouldn't have thought to check this. Any thoughts on what to do regarding this?

    I'll get a larger cap to replace the 2.2uF. I have a 100uF available but may be able to go out and pick up something even bigger. Thanks again!
     
  5. acat800

    Thread Starter New Member

    Dec 28, 2011
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    0
    So, the answer was pretty simple. Remove the ground wire from the DC side to the chassis of the vehicle (which is shared by the mains). Now, I've got some nice DC power with no overheating! Thanks for the hints guys!!!
     
  6. Adjuster

    Well-Known Member

    Dec 26, 2010
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    What kind of engine is this (diesel? gasoline?), or do you mean this is an extra winding on an electric motor?

    Possibly an isolating transformer would be the best way forward, but if the motor has a big speed range this may not be so easy.

    Depending on what the waveform of the AC is like, a "12V"input rectified might actually produce over 14V. You would have to use a really big smoothing capacitor though if you are running close to the bone on input voltage, as you could not then afford much ripple droop.

    Edit: Yes, floating the output would do it, but then you will have to beware of short-circuits.
     
  7. acat800

    Thread Starter New Member

    Dec 28, 2011
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    0
    This is a 2 cycle gas engine for a snowmobile. There is a lighting coil powered by a stator (I may not be stating this correctly) that is then fed to an AC voltage regulator which limits the voltage to 12V to power lights etc. The voltage I really need is in the range of 7-12VDC so, I could get a smaller regulator (LM7809) for 9V which would give me some headroom. My only left concern is how it will operate at idle because the voltage on the AC side can drop to 7 volts when idling.
     
  8. Adjuster

    Well-Known Member

    Dec 26, 2010
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    You will definitely need a plenty big smoothing capacitor to help when the motor is slow, because the frequency will drop as well as the voltage. I think you should be thinking of thousands of μF rather than hundreds.

    An anti-surge fuse on the input of this contraption would be a wise move. You do not want to have a fire if things go wrong, to say nothing of an expensive wrecked generator.
     
  9. gootee

    Senior Member

    Apr 24, 2007
    447
    50
    Since the AC drops to 7 volts at idle, you might have a problem.

    First of all, I'm assuming that you are reporting RMS voltages, as might be measured with a common multimeter. If that is not correct, please set us straight.

    The peak voltage of a sinusoid would be 1.414 x RMS voltage. That would be 9.9 Volts 0-peak for 7 V RMS.

    Unless you add a switchmode type of converter, it seems like about all you can do with your existing circuit topology is shoot for 7 Volts DC output. And even with that, you'd probably need to use a low-dropout-voltage type of regulator. Remember that your rectifier bridge will subtract roughly two diode drops (e.g. 1.5 Volts, but you have to look at the datasheet for your particular bridge; it could be more).

    Maybe you could use a voltage-doubling full-wave rectifier topology, if there is enough current available (since the max available current would be half as much as with your existing type of rectifier circuit). Here's a random one that I found with Google:
    http://www.play-hookey.com/ac_theory/ps_v_multipliers.html

    You might then be able to get the 14V minimum that you needed, barely (but I would use a low-dropout regulator). But you would also see 24 V at the regulator input, at other times. If you don't need too much current, maybe the regulator would be able to handle it without getting too hot. It might need a large heatsink.

    NOTE that you would need to make sure that the smoothing capacitance was large-enough to ensure that the bottom "peaks" (troughs) of the ripple voltage waveform didn't come down far-enough to cause the regulator's input voltage to drop below the point where Vin-Vout ≤ Vdropout, or the output would get pretty ugly.

    The smoothing capacitance needed should be calculated based on the desired maximum ripple amplitude and the maximum load current (regulator input current, in this case). The desired maximum ripple amplitude should first be calculated for a worst-case-low input voltage and worst-case-high output current. The basic equations that you will need are at the following URL:
    http://www.zen22142.zen.co.uk/Design/dcpsu.htm .

    If I was doing this, I would probably just use a simple buck-boost (or SEPIC) type of converter circuit, after the smoothing cap in your original circuit (but with a much larger cap, of course), with a specified input voltage range of 6V to 14V DC, or so, and an output voltage of whatever I needed. I might even add a linear post-regulator, after that.

    If you go to national.com and use the Webench tool, it will design one for you. Make sure that you specify a through-hole DC-DC converter chip, if you don't want to use surface-mount.

    Linear.com is very well-known for DC-DC converter chips. They too have an automated software tool that will design power supplies for you. It used to be in their free LT-Spice (SwithcerCad III). But I don't know if it is still in the newer version of that software or not.

    And both of those companies have tons of great application notes, which I would bet already have a few SMPS (switch-mode power supply) circuits that would do what you need.

    Unfortunately, you didn't specify the maximum load current. So I couldn't be more specific. (You should always specify as MUCH as you can, when posting requests like this.)

    Cheers,

    Tom
     
    Last edited: Jan 13, 2012
  10. JMac3108

    Active Member

    Aug 16, 2010
    349
    66
    You don't have to guess if your regulator will get hot - its a simple calculation to determine the temperature rise in the regulator.

    Take the voltage across the regulator and multiply it by the load current to determine the power dissipated in the regulator, P=VI. In your case measure the voltage and current.

    Look at the datasheet for your part and find the thermal resistance, junction to ambient. It will have units of degrees C per watt (C/W).

    Multiply the power by the thermal resistance and the result is the temperature rise. Trise = P x (C/W). Remember this is temperature rise, so you add it to room temperature, about 25C, and that will tell you how hot your part is.

    This type of calculation should always be made for all semiconductors that dissipate power in any design: diodes, transistors, MOSFETS.

    Just yesterday I had to redesign a circuit because we found a zener diode that was reaching 100C! The original designer had not done the thermal calculation and did not expect this to happen!
     
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