Power consumption with a diode is about 60% of full power rather than
50% since the filament has less resistance at that RMS voltage. But I think that 1/3 of full light output sounds about right.
According to a mental composite in my mind of the "1-size-fits-all" exponential rules for predicting performance of incandescent lamps at various voltages, the current is roughly proportional to the square root of applied voltage. I have seen the "current exponent" as low as .42 somewhere, and worked out as high as .57 from a "properties of tungsten" (in vacuum) table in the CRC Handbook. (The curve set mentioned below indicates that the "current exponent" is more than .5) According to this, a 120V lightbulb draws half power at anywhere from
73.6 to 77.2 volts, probably usually between 75.7 and 77.2 volts. For light output, I have heard often enough that being proportional to voltage to the 3.4 power - although this is a "1-size-fits-all attempt". At 77.2 volts, this means about 21.6% of full output while drawing at least 50% of full power.
Lumens/watt drops a bit more than watts does as voltage decreases. This would have light output at half power being a little less than 25% of full output.
(how much lightbulbs run less efficiently when dimmed)
I remember one night working with the properties of tungsten table and a few voltage and current reading I got from a 120V 100 watt "A19" 750 hour lamp, and working out a "rough 1-size-fits-all" rule for color temperature as a function of applied voltage. I figured color temperature being roughly proportional to voltage to the .375 power (square root 3 times and cube the result). Somehow I think a 750 hour 100W 120V bulb at full voltage has a color temperature of about 2850 Kelvin. At 71% of full voltage (when it would draw about 60 watts) its color temp. would then be about 2500 Kelvin.
Power consumption with a diode is about 56.5-60% of full power rather than 50% since the filament has less resistance at that RMS voltage. But I think that 1/3 of full light output sounds about right.
According to a mental composite in my mind of the "1-size-fits-all" exponential rules for predicting performance of incandescent lamps at various voltages, the current is roughly proportional to the square root of applied voltage. I have seen the "current exponent" as low as .42 somewhere, as high as .6 in the page TKM mentions below, and worked out as .57 (in vacuum, less for gas-filled bulbs) from a "properties of tungsten" table in the CRC Handbook. According to this, a 120V lightbulb draws half power at anywhere from
73.6 to 77.8 volts. For light output, I have heard often enough (including in the table TKM mentions below) that this is proportional to voltage to the 3.4 power
- although this is a "1-size-fits-all attempt". At 77.8 volts, this means about 23% of full output while drawing at least 50% of full power.
Lumens/watt drops a bit more than watts does as voltage decreases. This would have light output at half power being a little less than 25% of full output.
(how much lightbulbs run less efficiently when dimmed)
I remember one night working with the properties of tungsten table and a few voltage and current reading I got from a 120V 100 watt "A19" 750 hour lamp, and working out a "rough 1-size-fits-all" rule for color temperature as a function of applied voltage. I figured color temperature being roughly proportional to voltage to the .375 power (square root 3 times and cube the result). Somehow I think a 750 hour 100W 120V bulb at full voltage has a color temperature of about 2850 Kelvin. At 71% of full voltage (when it would draw about 60 watts) its color temp. would then be about 2506 Kelvin.
And in a chart in the page pointed out by Terry McGowan,
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temperature is said to be proportional to voltage to the .42 power. This means that a 100 watt 120V 750 hour lightbulb, if 2850K at full voltage, would have a color temperature of about 2468 K at 71% of full voltage (where it would draw nearly 60 watts and achieve about 31% of full light output).
I've often wondered what the temperature of my '51 Chevy Carryall engine was when I noticed it was running rough and pulled into a PA Turnpike rest stop one night and opened the hood and saw the whole block glowing a dull red... 1292 F, after it ran out of water and oil? :-)
Somehow at this hour my mind likes to pick out of Fahrenheit 1,000 degrees as dimly glowing and 1500 degrees as a medium-brightish-orangish "red-hot"/"cherry-red-hot" that has a hue more a reddish orange than really red...
1,000 F is to the nearest degree 811 Kelvin. I think that is visible in an otherwise dark area with some dark adaptation, not requiring full dark adaptation. I would expect seeing color of this, an orangish shade of dim red. Closer to 700 Kelvin, my experience is that light output is so low as to be only/mainly seen by human eyes through scotopic vision despite this light having wavelengths less favorable to scotopic vision - so the color at temperature that low gets more grayish or brownish-gray.
811 Kelvin blackbody has surface brightness of about 1.2E-6 candela per square centimeter, and color close to CIE 1931 x=.68 y=.32 z=0, or dominant wavelength (roughly but not perfectly exactly a specification of hue) about 614-615 nm (orangish orange-red).
I wonder how you determined 1292 F... That's 973 Kelvin. Blackbody achieves about 1.4E-4 candela per square centimeter, and chromaticity (using 1931 CIE system) is about x=.657, y=.341, z=.002, which has dominant wavelength about 608 nm - usually seen more as redish orange than orangish red, but usually called "red-hot" since 1.4E-4 candela per square centimeter from an ideal emitting surface is only about 4.4 lux, which is dimmer than most normal room illumination in homes.
1500 F is about 1089 Kelvin - blackbody achieves about .00186 candela per square centimeter, from ideal radiator or about 58 lux, a "brighter red-hot" or "good-and-hot-cherry-red-hot". I suspect such terms could be used to about 1600 F, or 1144 Kelvin (achieving .0053 candela per square centimeter, about 166 lux from an ideal radiator). As for color: 1500F / 1089 K has 1935 CIE system chromaticity of x=.6402, y=.3553, z=.0045, dominant wavelength close to 604 nm - usually considered orange or a slightly reddish shade of orange. 1600F /1144 K gets about x=.632 y=.362 z=.006, which has dominant wavelength about 602 nm, pretty much plain orange to maybe slightly reddish. Achieving a more-highly-arguably non-reddish orange (dominant wavelength by most of various means of 600 nm) requires from a blackbody 1250-1255 or so Kelvin, about 980 C, about 1795-1800 F. Heck, I believe that can get a little bit on the yellowish side! Oxidized copper at its melting point (1083 C, 1356 K, 1981 F) surely appears to me a rather yellowish orange, and has 1931 CIE chromaticity coordinates about x=.604, y=.382, z=.014. That is yellowish orange with dominant wavelength about 598 nm, maybe 597 nm. Compare to sodium-tinted flame with dominant wavelength 589.something, often seen as orange-yellow and sometimes more orange than yellow.
Maybe I need a vacation...
Also wondering what was the end of the story of a car whose engine block incandesced at 1292 degrees F!
Why 700? One artist at The Mattress Factory in Pittsburgh uses such low light in a room that it takes about 15 minutes to become barely visible. How many W/m^2 do we need to be able to see an object with no resolution at the our most sensitive frequency? Can we see a single photon?
A 3 AM math error... 700x1.8+32.
That early SUV never ran again. Too bad. Serious Detroit Iron. The engine even had a socket for a hand crank. I was nursing it home from Ithaca NY with a slightly cracked block, feeding a slow water leak on a very cold night and the gauges didn't work and I didn't realize it was overheating until too late.
Dimmed bulbs will burn out SLOWER because the filament does not get as hot and does not degrade as much. Normally the bulb is filled with argon or other inert gas so the fillament lasts as long as possible.
Yes, there is a minimum starting voltage for incandescent bulbs. Get a voltmeter and clamp on ammeter. Wire a dimmer in series with a light bulb. Measure voltage across the bulb and measure the current through the bulb as you turn the dimmer up from the OFF position. See how much voltage and power you need before you can see any visible light (there is infrared output at a much lower voltage). Years ago I did that using an osilliscope instead of a volt meter. You will learn a lot that way. You may not find that in a book, but you can test it for yourself, and the meters you need do not cost all that much.
If you choose to continue using incandescent bulbs, use one high-wattage bulb in place of multiple low-wattage bulbs wherever possible. A 100-watt light bulb produces as much light as two
60-watt bulbs.
Wattage is often used as a substitute for the brightness of a lamp, but it is a poor substitute. Usually, a 100 Watt lamp will burn brighter than a 50 Watt lamp. But just because the larger lamp is using twice as much power, doesn't mean it is putting out twice as much light. It could be that the 100 Watt lamp is very inefficient at turning electricity into light, and the 50 Watt lamp is much more efficient.
Reducing the voltage applied to a light bulb will reduce the filament temperature, resulting in a dramatic increase in life expectancy. One device sold to do this is an ordinary silicon diode built into a cap that is made to stick to the base of a light bulb. A diode lets current through in only one direction, causing the bulb to get power only 50 percent of the time if it is operated on AC. This effectively reduces the applied voltage by about 30 percent. (Reducing the voltage to its original value times the square root of .5 results in the same power consumption as applying full voltage half the time.) The life expectancy is increased very dramatically. However, the power consumption is reduced by about 40 percent (not 50 since the cooler filament has less resistance) and light output is reduced by reduced by about
: >> D. : >
: >Mine has a "click" for the off position. Based on the following, I think it : >still takes some power when turned down past visible light. : >
: >Perhaps someone knows for sure. : : If turned down enough that when turning back up, the light suddenly : comes on, then no power was being sent to the light. If the light fades : on in a manner like it faded off (except reverse), then the light probably : was being sent power. : : As for how much power a usual dimmer consumes when it is not sending : power to anything: Not zero, but generally a small fraction of a watt. : : - Don Klipstein ( snipped-for-privacy@misty.com)
I wouldn't call that a "starting voltage" in the sense of fluorescents and other types which give no light at all until an arc suddenly starts, altho the dimmer might act like that, passing no current until it reaches some minimum thyristor triggering threshold.
Hi, Sure Triac once triggerd to conduct, only way to stop current flow is switching the load off. I have quite a few dimmers in the house. I always turn it off when not in use, don't leave it at minimum setting. Tony
Hi, Trim the voltage? Or trim the conduction angle? Result is lower voltage on meter but Triac does not tirm the voltage. On theory at least. Rheostat does. Am I wrong? Tony
I agree to turn it off when not in use due to leakage and whatever load the trigger circuit creates.
But it's NOT because the triac is always on once triggered. It does indeed stay conducting but FYI it remains on only until the half of the power cycle is completed and the voltage is at zero, so 1/120th of a second.
I think we need a little discussion here of how these dimmers work. The power passes through the triac (which is the essentially the same as a pair of SCR's in parallel facing the opposite with, with combined triggers). The triac, when triggered, will remain on through the remainder of the half power cycle when it switches off and stays off until triggered again.
The trigger in this case is coming from the incoming power line but passes through the potentiometer (small variable resistance...the knob or slider you turn or move). It takes a certain tiny voltage to trigger the triac to conduct.
If you have the knob turned all the way up (zero resistance) you are getting full power line voltage as trigger. As the voltage rises with the sine wave wave form the trigger voltage is reached almost instantly so the triac starts to conduct and the full half power cycle is passed through. Then when the next half cycle happens (opposite polarity) the same thing happens again.
Now let's say you turn the knob back. Greater resistance means you've reduced the trigger voltage...imagine the sine wave but lower overall in height (voltage).
With the lower voltage trigger the voltage necessary to switch on the triac doesn't happen until slightly later in the half power cycle. At this point the triac does turn on (output voltage jumps from near zero to whatever voltage the power line is at that moment, minus the fractional volt drop going through the triac).
Because the voltage stayed turned off for part of the cycle the load isn't getting powered part of the time and thus for a light bulb it will be dimmer.
Turn the knob back farther (even greater resistance) and you delay the triggering even farther.
If you turn it far enough back it won't trigger until the waveform is at it's peak and you are effectively only giving the load half power.
Turn it back farther still and it won't trigger on at all.
There would still be leakage through the triac and also whatever current the trigger circuit passes.
The key here is the switching on and off. You could make a dimmer using transistors to have voltage partially on / partially off but then you're subject to the same Ohms's Law as a pure resistive dimmer (rheostat). Because the triac is always full on or full off Ohm's Law is avoided (except for the tiny voltage drop through the device).
But that is also a drawback because of the funny waveform. That's why these dimmers are not suitable for many things and why some light bulbs will "sing" at some dimming levels.
And also accounts for the only other components you will find if you crack open a dimmer, besides the triac and the potentiometer: Typically a small inductor (coil) and a capacitor. This is to reduce RF since that rapid rise in voltage when a partly dimmed dimmer triggers will create a lot of it.
I think you're right as long as the dimmer isn't set below 50%. It just reduces the duty cycle (the 50% is because at that point, cutoff would be before the voltage reaches peak).
Perhaps, but one that works very poorly. The trigger voltage across the triac (Vac) is a function of the gate voltage, but is rather sensitive. It's far better to use a diac or other negative resistance widget (e.g. neon bulb) to set the timing independent of the anode-cathode voltage.
Pot and triac alone can sometimes make an unreliable one variable from close to full, down to half of each half cycle going through, and needs a higher wattage pot. That would work by varying what point in the waveform the pot lets through enough current to trigger the triac. They don't do it that way. For one thing, tolerances in trigger current in the triacs would be a big issue.
The usual dimmer also has a capacitor and a diac. The pot (used as a rheostat) adjusts the amount of time it takes the capacitor to charge up enough to cause the diac to become conductive. Then the charge in the capacitor discharges through the diac and the gate of the triac. This arrangement does limit the range of the waveform at which the triac starts conducting to the range where the instantaneous voltage exceeds the breakover voltage of the diac. Most dimmers without a "positive on" do not let lamps reach full brightness.
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