I'm going to get into it this weekend to check the line voltage and make
sure all of the connections at the light and switch box are secure and
insulated but I can't imagine a loose connection or even a short would
blow a bulb (am I wrong on this?).
What else should I be looking for?
Thanks in advance!
-Dave
You can get a "rough service bulb". These are made to withstand vibrations.
And being next to the door, when the door closes, this may be vibrating the
bulb. Especially if you have kids running in/out.
Or might try compact fluorescent.
LED bulbs are quite expensive and not as bright, but will last forever.
"Dave" wrote in message
I was more curious if it was possible there was some electrical anomaly
causing this and if so, how I could find it.
I'm not opposed to CFL, and if my investigation turns up nothing I'll
probably switch to CFL.
Would this cause a bulb to blow?
I had no kids and didnt slam doors either and my porch light was
always blowing out too. It was due to vibration, but strangely, the
rough service lamps didnt fare much better than the regular bulbs.
What fixed it for me was going to the high voltage 130V bulbs. Ace
Hardware sells them. They have a beefier filament. After switching to
them, I never lost another bulb in that fixture. They are less
efficient though.
You could also try CFLs. If you live up north, you need to check the
label and get one rated for -20 degrees.
might be caused by bad socket that gets overheated because of poor
connection and causes bulbs to burn out. I have had this
happen..........
Probably not right when turned on. A drip on a hot bulb will break it.
--
Christopher A. Young
Learn more about Jesus
www.lds.org
.
"Dave" <dave@???.com> wrote in message
news:ELednfNM2rWMnu7X...@giganews.com...
Had same problin. Cured by adding a 5 amp diode in seris with the hot side
of line voltage. Installed a 300 watt bulb. Diode dropped wattageto about
75 watts. That was 24 years ago, WW
Except for most white ones. Some keep 70% of their output for 50,000
hours if not overheated, some I have seen lose half their output in 4,000
hours. I am aware of ones only rated to last 10,000 hours even at
"characterization current" and not overheated. I can cite one by a top
manufacturer rated for 15,000 hours at charcterization current.
If you really want fade resistance without verifying what LED is in it
and its honest life expectancy, get a bright green or blue LED light. The
spectrum of those stimulates night vision well anyway.
- Don Klipstein (d...@misty.com)
I have at a few times seen hot bulbs get cracked and last at least
several hours afterwards. But when the bulb is alowed to cool, it sucks
in air. The next time it is turned on, it burns out over a few seconds.
The telltale symptom is a slightly greenish gray-white smoky coating on
the inside surface of the bulb. This coating may be hardly visible in
white frosted bulbs.
- Don Klipstein (d...@misty.com)
<EDIT FOR SPACE>
>Had same problin. Cured by adding a 5 amp diode in seris with the hot side
>of line voltage. Installed a 300 watt bulb. Diode dropped wattageto about
>75 watts. That was 24 years ago, WW
A 300 watt incandescent with a diode dims to brightness of about a 75 or
100 watt incandescent, but consumes about 175-180 watts. (Not 150 because
the cooler filament has lower resistance.)
- Don Klipstein (d...@misty.com)
An electrical appliance, motor, etc on the same circuit can cause a
voltage variance. Have you tried other brands of light bulbs?
IIRC they used to make things for the bulb socket. One was a diode, the
other was a thermistor, I think. Are those still around and useful?
<SNIP to here to edit for space on making lightbulbs last longer>
>IIRC they used to make things for the bulb socket. One was a diode, the
>other was a thermistor, I think. Are those still around and useful?
I don't know how available these are, but I don't find them useful.
I tested a thermistor that claimed to double the life of light bulbs,
for dimming that accounts for life extension. I found enough dimming to
account for a 50% life extension. This reinforces what I heard from
experts about most lightbulbs not actually suffering significant wear from
cold starts, but merely becoming unable to survive a cold start a little
before they become unable to survive steady operation.
It appears to me that use of a 130V version (available at Lowes last
time I checked) works at least as well, without wasting that small amount
of power creating heat in the thermistor to keep it fairly conductive.
The diode types do indeed extend filament life greatly - but I consider
them energy wasters. They only reduce power consumption by about 40-41%
typically, while typically reducing light output by generally at least
70%. Energy efficiency is roughly halved. To restore original light
output, you would need to nearly double the amount of electricity used -
costing much more than what is usually saved by saving lightbulbs.
If you can be satisfied with light output being only about half that of
an undimmed incandescent of the same actual power consumption, then there
is a fair chance that an LED light on the market will be acceptable to
you even if a CFL is not.
- Don Klipstein (d...@misty.com)
Thank you. I will add that the same holds true for dimmer switches.
They waste a hell of a lot of electric due to running lights at less
than their rated wattage which makes them very inefficient.
Three way bulbs in table lamps save more $$ than any dimmer switch.
I'm sure they reduce power by less then 40%, I thought it was closer to
30%. I did a test in my shop some time ago but don't know where I wrote
down the numbers.
Don't overlook the possibility that a loose neutral (read higher
resistance than a "solid" connection) combined with current loads on the
opposite side of the 230 volt service could be causing higher than
normal voltage across that bulb. It's a long shot, but far from unheard of.
My own solution for increasing the life of incandescent bulbs is to
install a solid state dimmer in a non accessable location and crank it
slightly down from full on. I did that for all the 25 and 40 watt
bathroom mirror surround lights in our home when we built it 23 years
ago and its amazing how few times in all those years I've had to replace
a bulb in those bathrooms. I willingly accept the slight lumen per watt
inefficiency in return for the savings in bulb costs.
Jeff
--
Jeffry Wisnia
(W1BSV + Brass Rat '57 EE)
The speed of light is 1.8*10e12 furlongs per fortnight.
Using 130 volt bulbs when your voltage is 115 - 120 also makes them run
less efficiently also.
I wouldn't call a dimmer inefficient. If you mean it uses more than 1/2
the power to put out 1/2 the lumens, so what? You're still saving money
when you dim the lights. Almost every light in my house is on a dimmer,
including the closet. It's all about the mood.
Actually it's close to only a 30% reduction in power gives you a 70%
reduction in lumens. If that isn't inefficient *watt* ;-) is?
Anyway, I'm really am glad you have the right "mood" in your closets!
;-) (just watt are you doing in there anyway?)
All right, I'll give you the lumens per watt inefficiency, but I still
use far less electricity by dimming my lights than I would by leaving
them on full. I'd much rather have three softly glowing lights in a
room, than two lights off and one on full. People tend to light their
homes much too harshly, IME. Closets included.
Unless you have lots of lights the price differance will not be that great.
To get the best retrurn for the dollar, use some 25 watt or so bulbs instead
of using 60 watt or larger bulbs and dimming them. It only takes a small
reduction of power usge in a bulb to make a big differance in the ammount of
light output.
You need to run the bulbs at full power , but less wattage bulbs to save the
money.
Then you have to check to see how long a higher wattage bulb will last at
reduced voltage to come out in the long run.
So I'm supposed to have a dozen different wattage bulbs hanging around
for each fixture, and keep changing them out 20 times every day as my
lighting needs change? Nuts.
Dimmers are infinitely variable, and I fine tune the lighting
differently for dining, watching TV, working on the computer, reading,
visiting the closet dungeon, etc. ad infinitum.
<SNIP to here>
>> > A 300 watt incandescent with a diode dims to brightness of about a 75 or
>> > 100 watt incandescent, but consumes about 175-180 watts. (Not 150 because
>> > the cooler filament has lower resistance.)
>>
>> Thank you. I will add that the same holds true for dimmer switches.
>> They waste a hell of a lot of electric due to running lights at less
>> than their rated wattage which makes them very inefficient.
>>
>> Three way bulbs in table lamps save more $$ than any dimmer switch.
>
>I wouldn't call a dimmer inefficient. If you mean it uses more than 1/2
>the power to put out 1/2 the lumens, so what? You're still saving money
>when you dim the lights. Almost every light in my house is on a dimmer,
>including the closet. It's all about the mood.
If you are usually dimming the lights and they are incandescent
(including halogen and HIR), then chances are it is better to do one
of these:
1. Use fewer bulbs if doing so does not cause light to be unacceptably
unevenly distributed (such as causing dark spots).
2. Use lower wattage bulbs, so that you usually don't dim them or at
least make a lot less use of dimming - provided you don't run into
trouble from lack of brighter light when you need it.
3. Two lighting systems, a bright one and a lower wattage one, to rely
less on use of dimmed bulbs and their lower energy efficiency.
- Don Klipstein (d...@misty.com)
<I snip to what I said about diodes to edit for space>
>> The diode types do indeed extend filament life greatly - but I consider
>> them energy wasters. They only reduce power consumption by about 40-41%
>> typically, while typically reducing light output by generally at least
>> 70%. Energy efficiency is roughly halved. To restore original light
>> output, you would need to nearly double the amount of electricity used -
>> costing much more than what is usually saved by saving lightbulbs.
>
>I'm sure they reduce power by less then 40%, I thought it was closer to
>30%. I did a test in my shop some time ago but don't know where I wrote
>down the numbers.
A silicon diode combined with 120 volts AC means RMS voltage
delivered to the load very close to 84 volts, 70% of full voltage.
I did some testing in this area, which I publish in:
http://members.misty.com/don/incchart.html
For the "USA-usual" 120V 100W "A19" lightbulb with rated life
expectancy of 750 hours and light output 1710-1750 lumens (1670-1710
for "Soft White", at 84 volts:
RMS current is 82.5% of "full". .825 times .7 is .5775, or close
enough to 58 watts. Add to this power consumption of the diode, which
in this specific case is about .45 watt - total is about 58.5 watts,
58.5% of "full power".
Though I do report 28.4% of "full" light output.
120V incandescents with vibration resistant filaments and/or of lower
wattage will have wattage decreased a little less and lumens decreased a
little more than is the case of the example above by use of a diode, as
long as design watts and design amps don't get so low that an economy of
scale makes a vaccum fill gets better than a gas fill. (Vaccum in 120V
incandescents is used in one extreme example of a 60 watt one, most
tubular 40 and 25 watt ones, and most of lower wattages with
multi-supported filaments such as C-7 filament style.)
In the above web article of mine, I do state that a specific 25 watt
120V gas-filled incandescent at 84 volts has current consumption 83.5% of
"full", meaning power consumption about 58.5% of "full". Add another half
percent for the diode, and this data point becomes 59%.
I have yet to determine light output here, but I would guesstimate about
24-25% of "full".
- Don Klipstein (d...@misty.com)
<SNIP to here to edit for space>
>Don't overlook the possibility that a loose neutral (read higher
>resistance than a "solid" connection) combined with current loads on the
>opposite side of the 230 volt service could be causing higher than
>normal voltage across that bulb. It's a long shot, but far from unheard of.
>
>My own solution for increasing the life of incandescent bulbs is to
>install a solid state dimmer in a non accessable location and crank it
>slightly down from full on. I did that for all the 25 and 40 watt
>bathroom mirror surround lights in our home when we built it 23 years
>ago and its amazing how few times in all those years I've had to replace
>a bulb in those bathrooms. I willingly accept the slight lumen per watt
>inefficiency in return for the savings in bulb costs.
If by bulb costs you mean only the price of the bulbs, I expect that
your dimming is causing you to lose money unless the next lower available
wattage is unacceptably dim and you have a cosmetic or light distribution
problem with removing one bulb instead.
If you attribute value of time and effort expended to replace bulbs,
then it gets reasonable to dim them a bit, maybe to extent of achieving
average operating life around 3,000-5,000 hours. Use of 130V bulbs
available from Lowes, with the same CC-6 or CC-8 filament used in
"standard" incandescents 25 watts and up (more efficient with maybe less
vibration resistance), has a fair chance of achieving this to extent that
many desire.
- Don Klipstein (d...@misty.com)
This is true, but avoids the small amount of power waste that occurs
with a thermistor and a thermistor also dims the bulb.
Meanwhile, I do like to take the opportunity to repeat that Lowes has
130V bulbs with CC-6 / CC-8 filament design typical of 120V incandescents
that achieve less-bad lumens/watt. This means avoiding the additional
efficiency loss by using the less-efficient design C-9 filament used in
light bulbs of "vibration resistant" design.
- Don Klipstein (d...@misty.com)
Wouldn't a diode reduce RMS voltage by 50%? If that's true, your chart
says current would be 68.5%. Power would be 34%. Your chart indicates
that the lumens of a 100W bulb would be 125/1750, or 7%, an 80%
reduction in lighting efficiency.
By formula, power consumption is proportional to voltage^1.6, and light
output is proportional to voltage^3.4. That would mean 9% of the light
for 33% of the power or a 73% reduction in lighting efficiency.
Disregarding forward voltage drop, wouldn't a half-wave rectifier reduce
RMS voltage by 50%?
Rats! It would reduce RMS voltage by 29.3%!
So, by formula, power would be 57% and lumens would be 31%, a 46%
reduction in lighting efficiency.
Did you also figure the filament having a much lower resistance?
The answer is no, an "ideal diode" (with zero voltage drop) reduces RMS
voltage by about 29.3%, to original voltage times the square root of 1/2.
RMS means "square root of mean square".
This means that supply voltage being blocked half the time has the
average square or any other average-of-function-of input voltage being
halved.
Square root of mean square of 50% "duty cycle" is about .707.
======================
As for another way to analyze this: Adding an "ideal diode" in series
with an "ideally resistive load" powered by AC means that power
consumption is halved.
Since Ohm's Law says that power cosumption by an ideal linear resistor
is proportional to square of voltage, the "representative voltage" as a
result of adding an ideal diode to halve the duty cycle is not halved but
reduced by 29.3%.
> If that's true, your chart
>says current would be 68.5%. Power would be 34%. Your chart indicates
>that the lumens of a 100W bulb would be 125/1750, or 7%, an 80%
>reduction in lighting efficiency.
Which is true if RMS voltage is halved. Adding a diode to block half of
the AC cycle ideally does not halve RMS voltage, but ideally reduces RMS
voltage by a factor of square root of 2.
>By formula, power consumption is proportional to voltage^1.6, and light
>output is proportional to voltage^3.4. That would mean 9% of the light
>for 33% of the power or a 73% reduction in lighting efficiency.
By one somewhat-published "one-size-fits-all" sort of formula, this is
true, assuming effective/RMS voltage is halved.
>Disregarding forward voltage drop, wouldn't a half-wave rectifier reduce
>RMS voltage by 50%?
No, the "mean square" of input voltage is halved. Half the squares of
input voltage are unchanged, and half the squares of input voltage are
changed to zero.
Square root of mean square of input voltages becomes 70.7% of original.
70.7% of original effective voltage applied to an "ideal resistor"
(resistance not varying with input voltage or temperature) cuts
its power consumption and dissipation (heat output) by half.
70.7% of original effective voltage applied to a typical incandescent
lightbulb means power consumption about 59% rather than 50% because the
filament's resistance varies directly with its temperature.
(And the lower filament temperature here roughly halves the already-low
percentage of its 'outgo' being in the form of visible light.)
Please feel free to followup or to e-mail me (d...@misty.com) if I have
been insufficient so far in explanation here.
- Don Klipstein (d...@misty.com)
<SNIP to here>
>> Disregarding forward voltage drop, wouldn't a half-wave rectifier reduce
>> RMS voltage by 50%?
>
>Rats! It would reduce RMS voltage by 29.3%!
>
>So, by formula, power would be 57% and lumens would be 31%, a 46%
>reduction in lighting efficiency.
Rats - I responded to your post prior to the one above before I read the
one above! Double Rats or Rats squared!
- Don Klipstein (d...@misty.com)
Don's data shows current at various voltages, from which the changing
resistance could be calculated.
Saying power is proportional to voltage^1.6 implies that resistance is
proportional to voltage^0.4 (unless I've jumbled the math). I don't
know how accurate the formula is over a wide voltage range.
It says service life is proportional to voltage^(-16).
http://en.wikipedia.org/wiki/Incandescent_light_bulb
I'm not the brightest bulb in the box. I was thinking RMS voltage:
indicator of power: half the power = half the RMS voltage.
I'll seek solace in the adage that the only stupid question is the one
that isn't asked. :)
Before I discard a bulb that looks good, I check it in another fixture.
Fixtures and switches can have intermittent problems.
>Don's data shows current at various voltages, from which the changing
>resistance could be calculated.
>
>Saying power is proportional to voltage^1.6 implies that resistance is
>proportional to voltage^0.4 (unless I've jumbled the math). I don't
>know how accurate the formula is over a wide voltage range.
>
>It says service life is proportional to voltage^(-16).
>http://en.wikipedia.org/wiki/Incandescent_light_bulb
Power being proportional to voltage to the 1.6 power is something I
remember (I hope accurately) from something from GE on "miniature lamps",
and I found it largely accurate for vacuum-containing incandescent lamps,
at least ones of higher design voltage (14 volts and up maybe?).
This would mean current being proportional to voltage to the .6 power,
and resistance being proportional to the .4 power. It appears to me that
you did not jumble your math.
Meanwhile, I publish the "current exponent" for 3 specific incandescent
lamps, for application to a large number of voltage ranges where one end
of the range is 120 volts. I refer to this as log(i/I)/log(v/120) in
http://members.misty.com/don/incchart.html
For the vacuum one that I mention there, this figure is .58 to .61 for
all voltage ranges (effective average over the range) where one end is
120 volts and the other is anywhere from .5 to 144 volts.
For gas filled ones, this figure is lower. For 100 watt 120V 750 hour
lamps rated 1670-1750 lumens, this figure tends to be about .54. For
lower wattage 120V gas-filled lamps and 120V gas-filled lamps with
multisupported filaments, this figure will be lower. I seem to think that
a "1-size-fits-all" figure for household 120V gas-filled lamps is .53.
(Current is proportional to voltage to the .53 power, and power is
proportional to voltage to the 1.53 power.)
As for light output: There is not that good a "one size fits all"
exponent to describe what power of input voltage that light output is
proportional to. I have seen 3.4 and 3.5 published, and find that to be
reasonable. I have known this figure to be as low as 3.2, as high as
6-plus for extreme dimming, though it appears to me to usually be
between 3.3 and 4.
As for life expectancy: That exponent of -16 appears to me to be on
the extreme side. I usually figure life expectancy being inversely
proportional to voltage to the 12th power, although I find 13th fairly
credible. At one time I mathematicaly worked out -11 from relationship
of filament temperature with voltage and life expectancy of lamps with
different color temperatures, but I seem to think the truth is slightly
more extreme than -11.
I take a page or two of screen space below with cites for this exponent
being anywhere from -10.43 to -14.55.
A Google search has its description of one hit as a result of my
chosen search terms showing 12.86 power for this, not visible in the
link itself (describing a PDF that has to be paid for),
"Physics of Incandescent Lamp Burnout",
The Physics Teacher -- January 2008 -- Volume 46, Issue 1, pp. 29-35
Another statement repeated a bit is that voltage 10% above design
voltage will reduce life to 30% of that expected at design voltage.
I work out the exponent from this to be -12.6.
Two places saying this are:
http://www.candelacorp.com/products/lamps/miniatures/
http://www.sunraylighting.com/technicalinfo.shtm
These also say that life will be multiplied by approx. 3 by use of
90% of rated voltage. From that, I work out the exponent to be -10.43.
Although I saw a statement of reduction of life to 25% by use of 110% of
design voltage, in:
http://www.trft.org/b3/B3Pix/CHRSDialLamps.pdf
I work out an exponent of -14.55 from that statement.
In that document is also a statement that use of 90% of design voltage
results in life expectancy 400% of original. From that I work out the
exponent to be -13.2.
As far as I remember, I hope accurately, packages of 130V lightbulbs
with statements of life expectancy at both 130 and 120 volts have the 120V
life being about 2.5 times the 130V life. I work out the exponent from
this to be -11.45.
I also expect this exponent to vary somewhat with filament temperature,
presence/absence of gas fill and with percentage of input power becoming
heat conducted by the fill gas (which varies widely with design current
and filament style).
- Don Klipstein (d...@misty.com)
Wow! Thanks. Wikipedia was a starting point to help me comprehend your
stuff. Yesterday when I looked at your charts, I didn't make sense of
your logs involving current.
I think that's true of some materials. As tungsten gets hotter,
resistance increases.
From Don's data on a 100W gas-filled bulb,
it appears that
at 120V the resistance is 144 ohms and the temperature is 2865K.
It appears that
at 132V the resistance is 150 ohms and the temperature is 2975K.