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?

| 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.

Options:

Get a 24 volt, or 12 volt, or a 24/12 volt transformer, and replace the 18 volt transformer. You'd have some rather wimpy wiring if it can't handle either 24 volts or the 0.25 amps you get at 12 volts.

This is probably not an option if the transformer also powers the bell.

Put a 24 ohm 1.5 watt resistor in series with the 12 volt bulb and run that at 18 volts. The total resistance would be 72 ohms (24+48) giving you the correct 0.25 amp current for the 12 volt bulb. Just be sure the 18 volt transformer can handle the extra load as it will now be driving 4.5 watts at 0.25 amps instead of the previous 3 watts at 0.166667 amps.

In lieu of a 24 ohm 1.5 watt resistor, you can use lower resistances in series or higher resistances in parallel, and divide the wattage accordingly.

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.

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.

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".

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?

|> | 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.

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.

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.

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.

On Sat, 8 May 2004 07:58:41 -0500 Bill Shymanski wrote: | | |> |>

|> |> 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.

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?

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.

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

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?

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.

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?

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.

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.

Are you aware that automotive "12 volt" bulbs are usually rated at closer 15 volts due to the typical charging voltages found in automobiles. Lead acid cells run something over 2.25 volts (more when being charged)

6 times 2.25 is 13.5 volts. Trickle chargers are set to about 13.8 volts. You might find that an automotive bulb would be reasonably happy, albeit bright, on 18 volts.

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