Connect with us

Zap a volt regulator w/ no cap??

Discussion in 'Electronic Design' started by eromlignod, Oct 31, 2006.

Scroll to continue with content
  1. eromlignod

    eromlignod Guest

    Hi guys:

    First of all, forgive me for being a dumb ME meddling in electrical

    I have an application in a machine that has a 24-volt power supply. I
    have a tilt sensor (analog output) that needs 12 Vdc as its input
    supply. Until now this part of the machine used an embedded circuit
    that the sensor mounted directly to and that provided its own 12 V.

    I converted this part of the machine's control to be read from a PLC
    analog input module, so I had to come up with my own 12 V source. What
    I did was add an LM2940CT-12 voltage regulator (TO-220 package) to the
    circuit right before the analog sensor. This provided my 12V and
    worked great for about a day and a half, then my analog sensor fried.
    When I measured the voltage from the regulator I found that it was now
    the full 24V apparently the regulator failed first and then
    cooked the sensor with 24 V.

    When I took a look at the spec. sheet on the LM2940, I see that they
    recommend putting capacitors from the input and the output to
    ground...I didn't do this. Other than this mistake, I can't see
    anything else wrong with the circuit. Could the absence of these caps
    have caused the regulator to fry? If not, what else could have caused
    it? I'd like to know exactly what my problem is before I toast another
    sensor (they're $175 a pop).

    Thanks for any advice (or chastisement) that you can provide.

    Kansas City
  2. eromlignod wrote:
    Low drop out regulators (of which the LM2940 is one) are
    only stable with certain values of capacitance (and
    sometimes only with certain values of series resistance in
    that capacitance).
    The graph on page 11 shows how tricky this is.

    Since you have lots of excess voltage, a follower output
    type linear regulator like an LM7812 would be less
    problematical. It still works best with a small capacitor
    from both input and output to ground, close to the regulator
    (say, .1 uF) but the values and series resistance are not at
    all critical, compared to the LM2940.
    See page 22 of:
  3. Tim Wescott

    Tim Wescott Guest

    "Works best" is a mild way of saying "leave them off at your peril". I
    stopped in the middle of build a circuit with an embedded 78L05
    regulator; when powered up without an input cap it oscillated at 80MHz.

    You should also check on power dissipation. These regulators are
    supposed to have thermal shutdowns, but that doesn't mean they work
    right, and you don't want a shutdown event anyway. Check the input
    current to your 12V module; you'll be dissipating that times 12V (24V -
    12V), and more if your 24V supply goes higher.

    Make sure you have enough heat sinking to keep the regulator cool. The
    best rule of thumb I know is to run it for a while then put your thumb
    on it. If you pull back and say "ouch" then the heat sink isn't
    dissipating enough.


    Tim Wescott
    Wescott Design Services

    Posting from Google? See

    "Applied Control Theory for Embedded Systems" came out in April.
    See details at
  4. PeteS

    PeteS Guest

    Tim's caution about heat sinking should be heeded; I have worked with
    some industrial sensor modules (OPTO-22) and they can draw significant
    amounts of current. In your case, that would give a regulator power
    dissipation equal to the sensor (12V x load current) and even a TO-220
    without a heat sink can't handle too much if it's within a box (so
    ambient is pretty high).

    Do you know your load current?


  5. Eeyore

    Eeyore Guest

    The capacitors are rather important so you should fit them.

    Other than that I assume the current drawn @ 12V isn't greater than say 100mA.
    I'm thinking of the thermal limitation of the regulator here without a heatsink.

  6. Rich Grise

    Rich Grise Guest

    I've always used the caps, and never had a regulator problem. But, being
    obsessive/compulsive about capacitation, I soldered the caps right to
    the regulator leads right where the leads neck down.

  7. Tim Auton

    Tim Auton Guest

    The LM2940 has thermal overload protection, so given the symptoms
    described (overvoltage rather than shutdown) I doubt that was the
    problem in this instance.

    Shutdown isn't desirable though, so the thermals do need checking.
    eromlignod: If your sensor does draw more than around 100mA and/or the
    ambient temperature is much higher than room temperature you can let us
    know and we can show you how to calculate the required heatsink (or
    check your calculations if you've already figured that out).

  8. Yes.
    All regulators should have input and output caps, but "Low dropout"
    regulators like the LM2940 in particular are very touchy and require an
    output capacitance of a certain type and value to stop it oscillating,
    and hence stop the vital smoke from escaping the case.

    Unless you have a very low input voltage (<14V or so) then don't use
    the LM2940, use a bog standard LM7812 instead, they are much more
    stable. Yes, use an input and output cap on the LM7812 as well, value
    is not critical, 0.1uF should be fine.

    Dave :)
  9. eromlignod

    eromlignod Guest

    Thanks for the help so far, guys.

    The National spec. sheet (and my chip is a National) mentions reverse
    polarity protection and short circuit protection, but not overtemp
    protection, so I'm not sure if it was heat or not that made it fail.

    The sensor is all that the output drives and it's a very low
    load--maybe a few mA. The regulator is mounted perpendicular on a PC
    board with no heat sink and the ambient is normal indoor room
    temperature (maybe 75 F). It isn't enclosed in a box and gets good
    ventillation. I remember touching the regulator several times to see
    if it was red hot. It was hot, but not soldering-iron hot, like I've
    seen some power transistors get before they blow. I had experienced
    some low voltage problems on the 24 V supply due to a separate problem,
    but no lower than about 22 V or so. The regulator failure did occur
    during this low voltage period though...I'm not sure if it's related or

    I have ordered an LM7812 and the appropriate capacitors to go with it
    to try out (as well as a new sensor). Do you think I'll be all right
    without a heat sink?

    Kansas City
  10. Baron

    Baron Guest

    Based on this and previous posts, its quite likely that the LM2940
    simply took off ( Self Oscillation ). As others have pointed out, the
    capacitors are essential. They also need to be as close as practicable
    to the regulator and if there is more than a few inches of wire on the
    output, another one near the load.
  11. If it's too hot to keep your finger on the metal tab then add a
    dissipating say 0.5W or less you'll be fine without a heatsink. For a
    24V input and 12V output that's an output current of about 40mA

    Dave :)
  12. Tim Auton

    Tim Auton Guest

    The Feb 2006 National datasheet I have here mentions thermal overload
    protection, buried in the text at the start. The last sentence in the
    paragraph at the top of the second column of the first page:

    "Familiar regulator features such as short circuit and
    thermal overload protection are also provided."
    Anything from around 13V to 26V would be OK for the LM2940, so your 10%
    dip in input voltage won't have been a problem. It's only rated for
    brief pulses of over 26V input though - could your supply have gone much
    higher than the nominal 24V, as well as lower? Anyway, that question is
    academic now you're using the LM7812, which will be OK with anything
    from around 15V to 35V input.
    The LM7812 in a TO-220 has similar thermal specs to the LM2940 and will
    be dissipating the same power, so if the LM2940 didn't burn your finger
    the LM7812 should be fine.

    If you want to do the calculations, you'll need to find out how much
    current the sensor draws. Multiply that by the voltage drop across the
    regulator (input - output, 12V in this case) to get the power the
    regulator has to dissipate. Multiply that by the 'junction to ambient'
    or 'junction to air' thermal resistance (65 C/W according to the
    datasheet I have to hand). That'll give you the temperature rise, so add
    worst-case ambient ( hot summer day... ) to get the junction
    temperature. Specified absolute maximum junction temperature is 150 C,
    so you want to be well short of that to allow for accumulated dust,
    higher than normal input voltage etc. Anything less than, say, 110 C
    would be conservative.

  13. Guest

    I can't speak for all thermal shutdown schemes, but on the parts I
    designed, the shutdown was not very intelligent. That is, it would
    sense the die temperature, shut down, then cool due to the shutdown,
    power up, and of course get hot again enough to trip the shutdown
    circuit. You could make an oscillator using the shutdown. As part of
    the characterization of the part, a diode on the chip would be
    characterized over temperature, i.e. with the part in an oven and no
    load so the die temp equals ambient. Then if you short the output, you
    can monitor this diode to see the temperature trip points.

    There may be smarter regulators that stay shut down once tripped. It is
    a marketing issue since the smart design would need a power on reset
    from the input voltage. Not impossible, but POR circuits can be fooled.
  14. Guest

    I converted this part of the machine's control to be read from a PLC
    Guess it's the 00882206 diagram "Output Capacitor ESR".

    Makes me wonder if these LDO really are good. Seems more like liability :)
  15. I have used them successfully, but I always use a low esr
    cap and an external series resistor.
  16. John - KD5YI

    John - KD5YI Guest

    Since your load is only a few mills, I would suggest using a zener and
    resistor to set the 12V. Of course, the zener would not be as accurate in
    voltage as the regulator. But, the zener should be inherently safer from a
    failure standpoint as they usually fail shorted.

    If you insist on using a regulator, you could add some resistance between
    the 24V and the regulator input. The capacitors mentioned by others should
    be connected directly at the regulator pins. Then put a 13 or 14 V transient
    suppressor on the regulator's output. If you have another failure, the clamp
    should keep the voltage down to 13 or 14V.

    Just a thought. Good luck.

  17. They are essential when you actually need real LDO operation! e.g.
    dropping say 5V to 4.5V with low noise.

    Dave :)
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day