# Replacement for 18v 3w bulb

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

1. ### BobbyKGuest

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 GGuest

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

4. ### BobbyKGuest

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. ### Bill ShymanskiGuest

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. ### Bill ShymanskiGuest

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. ### Bill ShymanskiGuest

< responses interleaved below>

9. ### Poop ModeratorGuest

Guess my net searching skills need improvement - still couldn't find it.

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

10. ### BobbyKGuest

Oops. Sorry about that signature - was using another computer where a family

Ahem.

13. ### Bill ShymanskiGuest

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. ### nsGuest

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
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.  