light bulb & adaptor

The current specification on those wall warts is not what you get, it is a warning not to draw more than that or the thing can overheat and fail.

So, using a lower voltage wart actually gives you lower current.

To answer your question, I don't have any real facts, but I've heard it rumored that a bulb will last longer at lower voltages.

So, I think you are ok, and probably better off (as far as bulb lifetime) than you would be at 6.3V.

Regards, Bob Monsen

Not clear what you mean by "it", but unlikely that the lamp OR the adapter will "wear out sooner".

Running incanescent lamps on voltages lower than their design rating virtually always results in a longer (or MUCH longer) lifetime. There are charts that show life expectancy vs. operating voltage. This information is almost certainly available online somewhere, Google is your Friend (TM).

OTOH running a lamp at lower voltage will also result in significantly lower light output, especially for a large (>25%) reduction such as you are proposing. There are graphs that show this relationship to voltage also.

The adapter current rating is the maximum capacity rating. The lamp draws whatever it is designed to draw at the given voltage. Operating the lamp at lower voltage will cause it to draw lower current.

OK, but presumably there are other factors such as how much light you are expecting, etc. OTOH, you would likely get more light vs. power input (and even longer lifetime) from a white LED, etc.

It's more than a rumor:

(It's more that the lamp was VERY conservatively built relative to the known voltage source.)

The whiteness of the light suffers at lower voltage, however. It's going to glow yellow or even orange.

In article , snipped-for-privacy@yahoo.com mentioned...

Most wall wart adapters are unregulated, which means they put out their rated voltage at the rated current, but with less load, the voltage will be higher. So supposing that the one you have puts out 6V at no load but 5V at 250 mA, then your lamp will be running at about 1.3V less than the lamp's design voltage. Your adapter's 500 mA rating is a maximum, so it will run at half its rating and not get hot. So the lamp will not be at full brightness, will take less power and last longer. If you're satisfied with the lamp's brightness, then everything seems to be fine.

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In article , snipped-for-privacy@xprt.net mentioned... [snip]

Not likely. Current LEDs are not a lot more efficient than light bulbs. Read Don K's LED main page for more info.

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"Wats> Not likely. Current LEDs are not a lot more efficient than light

I didn't word that the way I was thinking, but I wasn't refering exclusively to efficiency, but to the ability to generate a reasonable amount of WHITE light given the two conditions:

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First of all, I like to thank for all of your inputs. I think I might buy another adaptor, maybe a little higher volt, 5 or

7 volt (I have another light bulb takes 7 volt). The reason I was using the 4.3 volt adaptor is becuase I have two of those laying around, and I thought it probably won'b blow the bulb up.

A longer life time is necessary for this project. The brightness is not so crucial. Although the current brightness is a little too yellow.

Ok, here is the point that confuses me. I read an article a while ago, about installing an electric kiln. "The amp rating of a kiln also determines what size circuit breaker or fuses need to be installed for care-free firing. The circuit breaker or fuses nee to be rated 7% higher than the amps the kiln draws when on high.....The reason for the 7% is because the voltage provided can vary up or down. when less voltage is delivered, the kiln pulls more than the designated amps to make up for the loss of power" Maybe this has nothing to do with my "bulb and adaptor" project. But >::Operating the lamp at lower voltage will cause it to draw lower current.::>, will not make the bulb to work harder to draw more amps, and causing the bulb to retire earlier?

BMB

Look up "Ohms Law". It says that there is a fixed (and predictable) relationship between the voltage, the resistance (of the lamp in your case) and the amount of current it will draw.

The current (in Amps) will be the voltage (in Volts) divided by the resistance (in Ohms). Note, however, that with incandescent lamps, the resistance changes as it gets hot (lights up). The "cold" resistance is usually lower than the "operating" resistance.

For a given lamp and a particular voltage, the lamp will draw exactly (E/R) amps (where E=volts and R=resistance) no more, and no less (unless the power supply can't keep up.)

Your kiln example appears to have some sort of active loop mechanism to attempt to keep consant power/temp even if the voltage varies. It may be an mechanical and/or electrical or electronic device or circuit. Likely some sort of thermostat, etc. Same effect as filling your bathtub. If there is lower water flow, you will run it longer to get the same water level.

A simple lamp has no such feedback mechanism. Once it heats up (to whatever power is available) it maintains the same filament resistance and won't draw any more current than Ohm's Law predicts. If you want more light for a given voltage, you will have to select a different lamp (ie. one with a lower resistance that is designed for that lower voltage).