Replacement for 18v 3w bulb

[BobbyK]

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?

 

[[email protected]]

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?

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.

 

[John G]

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?

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

 

[BobbyK]

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.

 

[Bill Shymanski]

| 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?

Running at 18 volts, the 12 volt 3 watt bulb (48 ohms) would burn out
rather quickly. It would be dissipating 6.75 watts.

Running at 18 volts, the 24 volt 3 watt bulb (192 ohms) would just be
very dim at 1.6875 watts.

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

 

[[email protected]]

You *do* know that incandescent lamp bulbs are not even approximately
ohmic?

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.

The resistance of a bulb can vary by an order of magnitude between cold
and working temperature.

True, but so what? Do you see a specific problem with
what was posted?

 

[Bill Shymanski]

|>
|> Running at 18 volts, the 12 volt 3 watt bulb (48 ohms) would burn out
|> rather quickly. It would be dissipating 6.75 watts.
|>
|> Running at 18 volts, the 24 volt 3 watt bulb (192 ohms) would just be
|> very dim at 1.6875 watts.
|>
|
|
| 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.

That's going to have an inrush current when you turn it on, of course.
So what is your point?

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

 

[Bill Shymanski]

< responses interleaved below>

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.

It's my understanding that "not linear" = "not ohmic". Sure, V=IR for
all I till the lamp burns out
(and even then), but if you were to plot V and I for an incandescent
lamp you would find that
there is not a straight line curve. I don't have my "Standard Handbook
for Electrical Engineers" handy, it's at work, but I believe for an
incandescent lamp V is some non-integer power of I. R is decidedly not
constant.

True, but so what? Do you see a specific problem with
what was posted?

Why...yes... yes I do. It is....misleading. So misleading as to be of
doubtful utility. It assumes that the power dissipated in the lamp is
directly. proportional to the square of the voltage applied to its
terminals. This is not the case. A lamp bulb does not have a constant
resistance so the four-significant-figure calculations of power that
will be dissipated in the lamp are spurious.

The name of the news group has "engineering" in it....we should try to
get the numbers right.

Bill

 

[Poop Moderator]

BobbyK


I have indeed tried a doorbell supplier, as well as a couple of harware
stores,.

The Ace hardware near me has these. Also we went online to get a new solenoid
for a chime and they had all the lamps, accessorie etc.

Bob AZ

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.

 

[BobbyK]

BobbyK


I have indeed tried a doorbell supplier, as well as a couple of harware
stores,.

The Ace hardware near me has these. Also we went online to get a new solenoid
for a chime and they had all the lamps, accessorie etc.

Bob AZ

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

 

[Bill Shymanski]

Ahem.

|
|
|> |>
|> |> Running at 18 volts, the 12 volt 3 watt bulb (48 ohms) would burn
| out
|> |> rather quickly. It would be dissipating 6.75 watts.
|> |>
|> |> Running at 18 volts, the 24 volt 3 watt bulb (192 ohms) would just
| be
|> |> very dim at 1.6875 watts.
|> |>

|>
|> That's going to have an inrush current when you turn it on, of course.
|> So what is your point?
|
|
| 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.

But it will get that voltage with the correct dropping resistor. Sure,
if the resistor is wrong, the temperature is different and the resistance
is different. The cases of what it will do off-voltage are just examples;
they are not designs. Adding the dropping resistor to get the voltage on
the bulb to its design voltage is a design. Can you show how that is not
correct?

Yes, with the proper dropping resistor you can make 12 volts appear
across the bulb.
But the off-voltage calculations of power in the lamp are wildly off -
which was my point.
The examples aren't correct.

I've just realized we've used thousands of dollars worth of Net
resources to debate replacment of a 75-cent doorbell lamp. Isn't this
amazing?

Bill

 

[[email protected]]

< responses interleaved below>

It's my understanding that "not linear" = "not ohmic". Sure, V=IR for
all I till the lamp burns out
(and even then), but if you were to plot V and I for an incandescent
lamp you would find that
there is not a straight line curve.

The fact that the resistance is not linear does
not mean it is not resistive. It doesn't magically
lose that property. Ohm's law applies, as you stated.

I don't have my "Standard Handbook
for Electrical Engineers" handy, it's at work, but I believe for an
incandescent lamp V is some non-integer power of I. R is decidedly not
constant.

You harp on R not being constant. Where do you see anyone
claiming it is?

Why...yes... yes I do. It is....misleading. So misleading as to be of
doubtful utility.

You call it misleading because you think it says
something that it doesn't. As to dubious utility -
Do you maintain that the computed 24 ohm resistance
is so far off that the bulb will either burn out or
be very dim?

It assumes that the power dissipated in the lamp is
directly.
proportional to the square of the voltage applied to its
terminals.

No it does not. It gives the resistance of the bulb at
its rating. The series resistor value is computed based
on that, such the the bulb, once hot, will "see" its
rated voltage. A 12 volt bulb rated at 3 watts in a
12 volt circuit will draw .25 amps. We want to increase
the voltage to 18, so we need to drop 6 volts. .25 amps
times 24 ohms = 6 volts. The bulb's resistance is not
linear. When it is cold, assume it to very low. The 24 ohm
resistor fed by 18 volts will try to draw to .75 amps.
The bulb will begin to heat, and its resistance will rise,
and the current will decrease. Pick a point where the
bulb's resistance equals 24 ohms, and compute again.
The current will be .375 amps, and 9 volts will be dropped
across both the resistor and the lamp. That's 3.375 watts
in the bulb - and that causes the resistance to continue to
climb. It will climb until the wattage dissipated equals
the wattage rating of the bulb, which occurs when the
circuit current is .25 amps, and the voltage across the
bulb is 12 volts.

This is not the case. A lamp bulb does not have a constant
resistance so the four-significant-figure calculations of power that
will be dissipated in the lamp are spurious.

The computations were based on the bulb running at
its ratings. Perhaps you would have no objection
if he had said "about 1.6 watts" instead of posting
the 4 significant figure? Maybe that's the whole source
of your complaint?

The name of the news group has "engineering" in it....we should try to
get the numbers right.

I note that you say "we" should get the numbers right,
yet provide no numbers yourself. Since you object to Phil's
numbers, propose your own.

 

[Bill Shymanski]

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.

The fact that the resistance is not linear does
not mean it is not resistive. It doesn't magically
lose that property. Ohm's law applies, as you stated.

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.

You harp on R not being constant. Where do you see anyone
claiming it is?

The power values calculated above assume R is the same at both the rated
and the applied voltages.

You call it misleading because you think it says
something that it doesn't.

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.

As to dubious utility -
Do you maintain that the computed 24 ohm resistance
is so far off that the bulb will either burn out or
be very dim?

No it does not. It gives the resistance of the bulb at
its rating. The series resistor value is computed based
on that, such the the bulb, once hot, will "see" its
rated voltage. ....
It <lamp resistance> will climb until the wattage dissipated equals
the wattage rating of the bulb, which occurs when the
circuit current is .25 amps, and the voltage across the
bulb is 12 volts.

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 computations were based on the bulb running at
its ratings. Perhaps you would have no objection
if he had said "about 1.6 watts" instead of posting
the 4 significant figure? Maybe that's the whole source
of your complaint?

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 note that you say "we" should get the numbers right,
yet provide no numbers yourself. Since you object to Phil's
numbers, propose your own.

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

 

[ns]

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

 

[[email protected]]

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

news:[email protected]...

Yup - deep quotes do get mismatched. I did not
write the section below.

This part is clearly wrong.

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.

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.

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.

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.

Right - Phil stated that a 24 ohm series resistor was needed.

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.)

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.

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

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.