Connect with us

Replacement for 18v 3w bulb

Discussion in 'Electrical Engineering' started by BobbyK, May 7, 2004.

Scroll to continue with content
  1. BobbyK

    BobbyK Guest

    I need to replace a miniature bulb that illuminates my doorbell nameplate.
    The original is an 18 volt, 3 watt bulb, but I can't find any similar to it
    (it's a "festoon" style 8x31mm bulb). The closest I can find of this size
    and type is either a 12v3w bulb or a 24v3w bulb. Would either of these be
    OK?
     
  2. Guest

    The 24v bulb will probably be too dim. If you use the 12 v
    bulb, you need to add a resistance of about 24 ohms at 3 watts
    in series. Get a 25 ohm 5 watt resistor, and put it in series
    with the wire from the transformer to the button. 5 watts
    and 25 (vs 24) ohms gives you a very nice safety margin -
    the resistor will last forever, and bulb life will be extended.
     
  3. John G

    John G Guest

    Have you tried a shop that sells door bells and may have
    spares to fit.
     
  4. BobbyK

    BobbyK Guest

    Thanks very much for the informative answers and helpful suggestions.

    John G: I have indeed tried a doorbell supplier, as well as a couple of
    harware stores, two electrical supply businesses, a large lighting retailer,
    and tons of internet searching including the manufacturer's site (Siedle).
    Haven't yet tried automotive supply places.
     

  5. You *do* know that incandescent lamp bulbs are not even approximately
    ohmic?
    The resistance of a bulb can vary by an order of magnitude between cold
    and working temperature.

    For example, the spare 100 watt 120 volt bulb that I just picked up off
    the shelf has a cold resistance of about 12.5 ohms. At rated operating
    temperature it would have a resistance of about 120*120/100 = 144 ohms.

    One of my favorite inteview questions is "Tell me about Ohm's Law".

    Bill
     
  6. Guest

    They're ohmic, just not linear. Nothing wrong with Phil's
    math. But he should have specified a higher wattage than
    the math yielded. That's not due to the non-linearity -
    it's just bad practice to run a resistor right at its
    wattage rating.
    True, but so what? Do you see a specific problem with
    what was posted?
     

  7. It's not just "inrush" - the resistance is not constant.
    Characterization of the lamp as "12 volts 3 watts 48 ohms" is not a good
    way to calculate power dissipation at other than 12 volts.

    Bill
     
  8. < responses interleaved below>
     
  9. Guess my net searching skills need improvement - still couldn't find it.
    Could you post a link, please?

    BTW, I installed a 24v3w bulb and it's working OK for the moment, though a
    bit too dim as posters here predicted.
     
  10. BobbyK

    BobbyK Guest

    Oops. Sorry about that signature - was using another computer where a family
    member had been fooling around.
     
  11. Ahem.
     
  12. Guest

     
  13. Responses interleaved - deep quotes may not be matched up to the right
    source....


    This part is clearly wrong. The powers calculated here assume that the
    resistance of the lamp is the same at all applied voltages. A 48 ohm
    resistor would dissipate 18*18/48 = 6.75 watts with 18 volts applied. A
    12 volt lamp bulb will not have a resistance of 48 ohms at 18 volts
    applied.
    Yes, but we nearly always assume R is constant over some range of I and
    V - this is not even approximately true for an incandescent lamp. An
    ohmic resistance has constant R over some range of applied I or V.
    The power values calculated above assume R is the same at both the rated
    and the applied voltages.
    The paragraph quoted above says that a 3 watt 12 volt lamp will
    dissapate 6.75 watts if 18 volts is applied. The paragraph above also
    says that a 24 volt 3 watt lamp
    will dissipate 1.675 watts at 12 V applied. This is not correct.
    Yes, of course, this is like "load line" analysis. 48 ohms is not the
    series
    resistor value to get the 12 volt lamp to operate at 18 volts. 24 ohms
    is correct.

    If you had decided after all this trouble that you wanted the lamp to
    run a lot longer (so as to
    not touch off another lengthy Internet discussion), and say wanted only
    11 volts to appear across your 12 volt 3 watt lamp, what series
    resistance would you use in the 18 volt circuit?
    It's NOT going to be (18-11)/0.25 = 28 ohms, it will be something else
    best determined empirically.
    The bulb only runs at its rating at its rated voltage. Using the
    effective resistance calculated at
    rated voltage and trying to apply that to other than rated voltage is
    wrong. (The extra significan figures are also in some sense wrong but
    that wasn't my primary objection.)
    I bought a couple of No. 89 bulbs at Canadian Tire earlier this week - I
    will run some measurements and report them to the group. These are a
    little larger than 3 watts but the physics will be similar.


    Bill
     
  14. ns

    ns Guest

    And putting a few diodes in series will drop the voltage further...
     
  15. Guest

    Yup - deep quotes do get mismatched. I did not
    write the section below.
    To me it was clearly right. It assumes you know the resistance
    will change, but specifies the resistance used in the calculation
    at 48 ohms. But at least I finally understand your objection.
    It would have been better if he had said that the resistance
    would change, but that he was using the only known value we have.
    As already stated, I took it differently - that the
    calculation was based on the only known resistance value.
    The math is correct using that value, and supports the
    correct conclusions that a 12 volt bulb will burn out if
    powered by 18 volts, and that a 24 volt bulb will be dim
    if run at 18 volts. The fact that the resistance changes
    is true - and not at all important in the answer to the OP,
    who doesn't know enough in any event to have that factor
    explained in the answer. Phil took a shortcut in the
    explanation, leaving out parts based on them not being
    needed, just as I did in stating that "a 12 volt bulb will
    burn out if powered by 18 volts". The statement does not
    specify an 18 volt source capable of sufficient current.

    Right - Phil stated that a 24 ohm series resistor was needed.
    Well, then we disagree, and you disagree with yourself.
    You stated that 24 ohms is correct. That figure was
    computed based on the effective resistance calculated at
    the rated voltage.
    Great. If you expose the bulb to a voltage other than its
    rating, the resistance will change from the calculated
    value - that's a given. Build your voltage divider to keep
    the voltage that the bulb "sees" to its rating, based on the
    calculated effective resistance and let us know what you find.
     
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

-