Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Path: utzoo!linus!philabs!cmcl2!harvard!seismo!rochester!rocksanne!sunybcs!kitty!larry From: larry@kitty.UUCP (Larry Lippman) Newsgroups: net.consumers,net.sci Subject: Re: lightbulbs Message-ID: <1034@kitty.UUCP> Date: Sun, 18-May-86 01:29:46 EDT Article-I.D.: kitty.1034 Posted: Sun May 18 01:29:46 1986 Date-Received: Tue, 20-May-86 05:46:41 EDT References: <446@hropus.UUCP> Organization: Recognition Research Corp., Clarence, NY Lines: 69 Xref: linus net.consumers:4794 net.sci:577 Summary: How many lightbulbs does it take to change a ... In article <446@hropus.UUCP>, jin@hropus.UUCP (Jerry Natowitz) writes: > > 1) What is the spectrum (at least as filtered by the bulb) of the > blue flash that a dying light bulb (or the filament thereof) produces? > As a child I used to think that the bulbs gave off X-rays when they > died (I have no idea where that belief came from). The blue flash from a "dying" lightbulb is actually the momentary ionization of the argon in the nitrogen-argon fill gas mixture. When the filament fails, a momentary arc is struck as the pieces fall apart; this arc results in the ionization. No significant ionization takes place in a conventional incandescent lamp during normal operation. There could be a slight grain of truth to the x-ray story. X-rays are given off when electrons strike a HARD target material - like the tungsten lamp filament. Most x-ray tubes use anodes (i.e., the target) which are made of tungsten. It is quite possible that a momentary, miniscule amount of very _soft_ x-rays would be given off during the brief ionization which accompanies an incandescent lamp failure. I cannot overemphasize how brief and miniscule such x-ray emanation might be. Without doing some calculations, I am not certain if in fact there is enough energy in a 120 volt lamp to create x-rays all - but I would guess there is. Such x-ray radiation would be modulated at 60 Hz, and chances are only a few cycles would exist. > 2) I use Bulbsavers on some of my light bulbs. Bulbsavers are stick-on > thermisters, a device that changes from a high electrical resistance > to a very low resistance when they heat up. > The theory (well proven by my experience) is that the bulbs last longer > because the thermister reduces thermal shock to the filament caused by > the rush of current that normally occurs when turning on a light. > I have read that the light output of "thermistered" bulbs is decreased > even at normal operating temperature and have noticed a "yellow shift" > in the spectrum. > > My question is: how much of a decrease in light output do these things > cause and what change in the light spectrum does occur? There is always some resistance to the thermistor - even when hot - so the lamp will never see the full line voltage. If you have noticed a "yellow shift", then the lamps must be operating at a REALLY LOW voltage. An incandescent lamp behaves like a black body, and hence lamp energy is measured as "color temperature" - the temperature of an equivalent black body in degrees Kelvin. Normal household incandescent lamps have color temperature ratings of 2700 to 3100 degrees K. Lamp operating current determines color temperature (of a given lamp). It might surprise you to learn that while operating a _given_ lamp from 2700 to 3100 degrees K. (by varying current) causes a change in OVERALL radiated energy of almost 50%, the spectral distribution change in the VISIBLE portion (400 to 700 nanometers) is almost nil. Notice that I have not said what happens when an incandescent lamp is run much below 2700 degrees K.; below that value, an incandescent lamp becomes extremely inefficient, and will, of course, show a spectral shift toward red. In very rough terms - for small values of less than 20% from the rated value of the lamp - a change in lamp current results in an exact corresponding change in lamp energy output; i.e., decreasing lamp current by 10% results in roughly 10% less lamp energy (watts). Also for small values of less than 20%, a change in lamp current results in TWICE the amount of visible light output; i.e., decreasing lamp current by 10% results in 20% less visible light output (lumens). Also, notice that I have been referring to lamp current rather than lamp voltage. While decreasing lamp voltage decreases lamp current, it is a non-linear function, and the current is really the parameter which determines energy output. ==> Larry Lippman @ Recognition Research Corp., Clarence, New York ==> UUCP {bbncca|decvax|rocksanne|rocksvax|watmath}!sunybcs!kitty!larry ==> VOICE 716/688-1231 {rice|shell}!baylor!/ ==> FAX 716/741-9635 {G1, G2, G3 modes} seismo!/ ==> "Have you hugged your cat today?" ihnp4!/