Go Back   DIY Chatroom Home Improvement Forum > Home Improvement > Electrical

CLICK HERE AND JOIN OUR COMMUNITY TODAY...IT'S FREE!

Reply
 
Thread Tools Search this Thread Display Modes
Old 07-25-2012, 10:01 AM   #1
Newbie
 
Kookie's Avatar
 
Join Date: Jul 2012
Posts: 3
Share |
Default

Light Bulb Trickery


I use a very large 300W incandescence light bulb 120VAC. (Its a huge bulb with the large socket.) Actually I have a whole crate of them. The problem I would like to address with the use of these bulbs is that they donít last very long. The bulb will burn out between 1 and 4 months, 4 months if Iím lucky.



I remember from a long time ago using equipment that would trickle a little current to the light bulbs to make them last longer. You could barely see that they were still on. In essence, keep the filament warm so that the shock of full on, from cold to blazing hot, would extend the filaments life time. It makes sense, but whether it works, or will work in my application is another story.


I am thinking about trying this for my big bulbs. There are many ways of doing this and Iím not really sure how much current would be enough.



I would like to keep it simple so Iím thinking just a resistor across the switch (in parallel, the switch shorts out the circuit). I could also include a diode to give me only the half cycle 60V.



So, on full on it should be drawing 2.5A. Iím not sure how much current to use to keep the filament warm.



Does anyone have any experience with this type of work? What would be my best approach and how much current do you thing I should use? Do you think it would work too?


Thanks--

Kookie is offline   Reply With Quote
Old 07-25-2012, 10:15 AM   #2
Master Electrician
 
Join Date: Mar 2010
Location: Toronto Ontario
Posts: 1,165
Default

Light Bulb Trickery


A diode isn't going to give you half of 120v its going to give you 120v 30 time a second instead of 60. To drop it to 60volts you would need a series resistor valued at exactly the same resistance as the lights have at the desired brightness. The thing with a setup like that is that even though your lights are down low your still going to be useing the same amount of electricity as you would if they were on full bright. Your just bleeding it off through a resistor. So your call. Buy more bulbs or have a higher hydro bill.

__________________
Sarcasm is my friend
I'm here to learn too, i do mostly commercial/industrial/new construction and this place is a great way to pick up tips on residential from some good electrical minds. Excuse the spelling, my phone has a mind of it's own.
andrew79 is offline   Reply With Quote
Old 07-25-2012, 10:26 AM   #3
JOATMON
 
ddawg16's Avatar
 
Join Date: Aug 2011
Location: S. California
Posts: 6,829
Default

Light Bulb Trickery


My first question....Why?

Why have a 300w bulb, but then want to run it at a lower current? Sort of defeats the purpose.

The diode would cut the RMS (actual work the electricty does) in about half...but you would need a really large diode to handle the forward current.

An easier solution....put two bulbs in series. You will get 60 Vac across one bulb and 60Vac across the second one.

Put 4 in series and you will have 30 Vac across each bulb....or about 1/4th the brightness.
__________________
"The dream is free but the hustle is sold separately."

My 2-Story Addition Build in Progress Link ... My Garage Build Link and My Jeep Build Link
ddawg16 is online now   Reply With Quote
Old 07-25-2012, 11:01 AM   #4
Newbie
 
Kookie's Avatar
 
Join Date: Jul 2012
Posts: 3
Default

Light Bulb Trickery


No you guys don't understand my question at all then. No, I want to extend the bulbs life by warming the filament. You've never seen that done? They do this in high end equipment mainly on the indicator lights. I was thinking it may work for my light.

I just want to bust out the ladder less frequently to change that bulb.

I don't want to dim the light, I want to have it be barely on so that when I do turn it on, full brightness, the filament isn't completely shocked from room temp to very hot. It goes from warm to very hot. That is the theory.

BTW, some of these answers are just plain wrong. LOL. And I assume you don't think it will work at all.

Last edited by Kookie; 07-25-2012 at 11:04 AM.
Kookie is offline   Reply With Quote
Old 07-25-2012, 11:52 AM   #5
Member
 
Join Date: Nov 2007
Location: Nashua, NH, USA
Posts: 6,864
Default

Light Bulb Trickery


Control it with an ordinary dimmer wall switch. This will apply the voltage more gradually (depending on how you turn the knob) and also not waste electricity during times it is off.

You can still move the dimmer to maximum when you turn it on, but the lamp will last much longer if you don't turn the dimmer quite to max. And you could leave the dimmer not quite at minimum if you want to keep the lamp filament a little warm to avoid the thermal shock. (Not needed if you don't turn the knob too fast.)

Caution: Some math follows:

By the way, if you use a diode or resistor, you can't use the little ones you would find inside a radio or television, even an old fashioned TV. Those handle only about one watt or 50 milliamperes each, respectively, if that.

To significantly increase the life of the lamp you would want to decrease the voltage by at least 7 percent or about 8 volts. Using a resistor in series with the lamp, dropping 8 volts at 2-1/2 amps within the resistor means 20 watts eaten up by the resistor, which in turn needs to be rated for 20 watts or more and not stuffed into a small enclosure like a junction box. A diode is rated in amperes and here you need the full 2-1/2 amps (or 2500 milliamps).

Old fashioned theater/stage lighting dimmers were indeed huge resistors. Modern dimmers of all sizes achieve the voltage reduction with far fewer watts dissipated in the dimmer unit and generating heat and wasting the energy (except in a location that could use the heat in winter).
__________________
Stop wasting time re-adjusting the pattern. Have several lawn sprinklers, one for each pattern.

Last edited by AllanJ; 07-25-2012 at 07:34 PM.
AllanJ is offline   Reply With Quote
Old 07-25-2012, 12:01 PM   #6
JOATMON
 
ddawg16's Avatar
 
Join Date: Aug 2011
Location: S. California
Posts: 6,829
Default

Light Bulb Trickery


Quote:
Originally Posted by Kookie View Post
No you guys don't understand my question at all then. No, I want to extend the bulbs life by warming the filament. You've never seen that done? They do this in high end equipment mainly on the indicator lights. I was thinking it may work for my light.

I just want to bust out the ladder less frequently to change that bulb.

I don't want to dim the light, I want to have it be barely on so that when I do turn it on, full brightness, the filament isn't completely shocked from room temp to very hot. It goes from warm to very hot. That is the theory.

BTW, some of these answers are just plain wrong. LOL. And I assume you don't think it will work at all.
Oh...ok.....

Not really going to make a big difference.

Basically, the limiting factor on an incandescent bulb is how long the tungston stays intact. When you light it up, it throughs out tungston atoms. That is that gray stuff you see on the inside of bulbs when they die. It's the tungston on the glass.

That is the reason for Halogen bulbs. The halogen gas helps to redeposit the tungston back onto the filiment.

Unless your cycling the lights a lot....having them on partially is not going to help.

If you want to increase the life, then try reducing the voltage a little. As much as a 10% drop will increase the life and the difference in light output may not be noticable.
__________________
"The dream is free but the hustle is sold separately."

My 2-Story Addition Build in Progress Link ... My Garage Build Link and My Jeep Build Link
ddawg16 is online now   Reply With Quote
Old 07-25-2012, 12:16 PM   #7
Member
 
Join Date: Jul 2008
Location: NW of D.C.
Posts: 5,990
Default

Light Bulb Trickery


My bad. Try this.

according to Wiki, P = kV^1.6 for an incand. Bulb
so k = P/(V^1.6)
and P = VI
so VI = kV^1.6
so I = kV^0.6

For 300W & 120v
I= 2.5 amps
R= 48 ohms
k= 0.141394091

V P I R
120 300 2.5 48
60 98.96 1.65 36.38
50 73.92 1.48 33.82
40 51.73 1.29 30.93
30 32.65 1.09 27.57
20 17.06 0.85 23.44
10 5.63 0.56 17.77

Just dial up your preferred bulb power level and pick a series impedance to get it.

Last edited by Yoyizit; 07-25-2012 at 12:58 PM.
Yoyizit is offline   Reply With Quote
Old 07-25-2012, 12:44 PM   #8
Member
 
Join Date: Jan 2012
Location: IL
Posts: 526
Default

Light Bulb Trickery


Get a dimmer switch without push on push off. The types that rotate and then click off at the end of the CCW rotation. That way every time you turn it on you are effectively soft starting the bulb by the mechanical movement. Forget the ideas with resistors. A dimmer switch makes more sense and resistors would need to be high wattage and probably a fire hazard depending how you mounted them.


http://www.amazon.com/GE-18021-Dimme...words=ge+18021

Last edited by curiousB; 07-25-2012 at 12:49 PM.
curiousB is offline   Reply With Quote
The Following User Says Thank You to curiousB For This Useful Post:
DangerMouse (07-25-2012)
Old 07-25-2012, 02:06 PM   #9
JOATMON
 
ddawg16's Avatar
 
Join Date: Aug 2011
Location: S. California
Posts: 6,829
Default

Light Bulb Trickery


Are you using just one bulb or serveral bulbs at a time?
__________________
"The dream is free but the hustle is sold separately."

My 2-Story Addition Build in Progress Link ... My Garage Build Link and My Jeep Build Link
ddawg16 is online now   Reply With Quote
Old 07-25-2012, 03:03 PM   #10
Master Electrician
 
Join Date: Mar 2010
Location: Toronto Ontario
Posts: 1,165
Default

Light Bulb Trickery


I want to know what was wrong with anything anyone said. Diodes don't limit voltage but in ac they only allow one half of the cycle to pass. As for the other posters everything that was said is true as well. Educate us sir, please. The rotary dimmer is by far the best solution I've heard so far. I think you forget that when you stick the resistor across the switch you've created a series parallell circuit at that point. That resistor is going to have to drop the bulk of 120v to make it so the lights aren't visibly on.
__________________
Sarcasm is my friend
I'm here to learn too, i do mostly commercial/industrial/new construction and this place is a great way to pick up tips on residential from some good electrical minds. Excuse the spelling, my phone has a mind of it's own.
andrew79 is offline   Reply With Quote
Old 07-25-2012, 04:28 PM   #11
Member
 
Join Date: Jan 2012
Location: IL
Posts: 526
Default

Light Bulb Trickery


Quote:
Originally Posted by Yoyizit View Post
My bad. Try this.

according to Wiki, P = kV^1.6 for an incand. Bulb
so k = P/(V^1.6)
and P = VI
so VI = kV^1.6
so I = kV^0.6

For 300W & 120v
I= 2.5 amps
R= 48 ohms
k= 0.141394091

V P I R
120 300 2.5 48
60 98.96 1.65 36.38
50 73.92 1.48 33.82
40 51.73 1.29 30.93
30 32.65 1.09 27.57
20 17.06 0.85 23.44
10 5.63 0.56 17.77

Just dial up your preferred bulb power level and pick a series impedance to get it.
Whoa not so fast. This is a very bad idea. You failed to mention the power dissipation of the resistor would be I^2*R. If you pick 30W (10%) of the bulb's full power as a good "idle power" then I is 1.09 and R is 27 Ohms or a 32W power dissipation. Add a safety margin to say 50W and now you are talking something very large and hot. Not only is that a large amount of power to burn (about $25/year in wasted power) it could literally burn down your house...

Light dimmers use solid state switching for this very reason. Inexpensive and they can operate in saturation mode (the Triac switching element) so their power dissipation is the lowest possible. Imagine the size of a 600W dimmer if it were a bank of (large) resistors and a rotary switch.

This is mutum plus the rear of a donkey

Last edited by curiousB; 07-25-2012 at 04:32 PM.
curiousB is offline   Reply With Quote
Old 07-25-2012, 05:07 PM   #12
Member
 
Join Date: Nov 2007
Location: Nashua, NH, USA
Posts: 6,864
Default

Light Bulb Trickery


Quote:
Originally Posted by Yoyizit View Post
... according to Wiki, P = kV^1.6 for an incand. Bulb
so k = P/(V^1.6)
and P = VI
so VI = kV^1.6
so I = kV^0.6

For 300W & 120v
I= 2.5 amps
R= 48 ohms
k= 0.141394091
.
What is k? (in more than 25 words)?

Also note that the resistance of the bulb filament varies with its temperature.
__________________
Stop wasting time re-adjusting the pattern. Have several lawn sprinklers, one for each pattern.
AllanJ is offline   Reply With Quote
Old 07-25-2012, 05:13 PM   #13
Member
 
Join Date: Jul 2008
Location: NW of D.C.
Posts: 5,990
Default

Light Bulb Trickery


Quote:
Originally Posted by AllanJ View Post
What is k? (in more than 25 words)?

Also note that the resistance of the bulb filament varies with its temperature.
K is an arbitrary constant adjusted for each bulb wattage.
This resistance/temp variation is included in the formula.

BTW, for formulas like this I prefer to see:
for P in the 10w to 75w range the error is +/-10% (for example)
for P in the 1w to 100w range the error is +/-20% (for example)
etc.

At some very low or very high wattage this formula probably becomes practically useless.

Mr B?
Would you like me to interpret your motives for your replies to my posts? I can do you and Mr. Port for the same low price. . .
Yoyizit is offline   Reply With Quote
Old 07-25-2012, 06:04 PM   #14
An old Tradesmen
 
Join Date: Oct 2008
Location: PA
Posts: 24,529
Default

Light Bulb Trickery


5000 ohms, 5 watts.
beenthere is offline   Reply With Quote
Old 07-26-2012, 01:46 AM   #15
Newbie
 
Kookie's Avatar
 
Join Date: Jul 2012
Posts: 3
Default

Light Bulb Trickery


Ah, now this talk is what I was looking for. Thank you very much for your inputs. I'm on an island over here and don't get to talk much about these things. Its been a long time since I've done anything creative, so to speak, in the electrical field.

I posted in haste earlier (ripped that off just before rushing off to work), and the wrong I was mainly speaking of seems to be gone now somehow. Although I didn't get the nick Kookie for nothing.

Yeah, this is why I brought this up because I was having a hard time justifying that resistive load on all the time. (Yes, for just one bulb.) I don't know why a simple dimmer just did not come to mind. I like that alot.

Although, beenthere's statement of 24mA just might work too.

I like ddawg16's statement on how the tungsten filament works. Makes me wonder if this will actually be worth the effort. Nonetheless, I want to baby my last stash of those bulbs I have.

Now, how to solve when other people Tom Bodett that light. (You know, "Leave the light on for ya'.")

And don't worry, I wont burn anything down. Although I do remember building a circuit in my youth that had a few indicator lamps. Turned it on and saw an extra light. I said to myself, "Hey, where did that light come from?" then immediately realizing that is no lamp, that is a resistor that is obviously miscalculated. Burned real nice the whole time I was befuddled.

Kookie is offline   Reply With Quote
Reply


Thread Tools Search this Thread
Search this Thread:

Advanced Search
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off


Similar Threads
Thread Thread Starter Forum Replies Last Post
Light Wiring Question mataguri Electrical 4 11-09-2011 02:15 AM
Light bulb in fan goes off and on ilyaz Electrical 2 07-19-2010 09:00 PM
Light Bulb Burst Mark Harvey Electrical 4 02-20-2010 12:54 PM
Do I need training to change a light bulb? Mechelle Electrical 11 08-13-2008 07:26 AM
Flourecent light bulb question shadango Electrical 7 06-28-2008 08:28 PM




Top of Page | View New Posts

Copyright © 2003-2014 Escalate Media. All Rights Reserved.