An old GE lamp catalog gives light output as proportional to the 3.6 power. As you noted, Dean's equation is missing the caret after the quotient of the voltages.
The same reference gives lifetime as being proportional to the inverse 12 power. Excessive voltage on an incandescent bulb really hurts. If you can, try to be a few percent low in lamp voltage. The light output drops slightly, but the life increases substantially.
light output = (applied voltage/design voltage)^3.6 x MSCP at design volts
lifetime = (design voltage/applied voltage)^12 x lifetime at design volts
The impending death of incandescent bulbs
Thank you Mr. Hayes, always having the equation.
My interest is peaked with the curve of the resistance of tungsten filaments. I have previously used a small instrumentation bulb when I designed a sine wave oscillator for our TV broadcast station (1KHz tone over color bars) because the resistance rises with temperature and stops the osc from clipping.
I have searched for a voltage/current curve on the internet and found various curves, some obviously wrong. Can you believe it, there is a University Physics department in Newfoundland that thinks the VI for a light bulb curve is linear like a resistor!
One of the curves has the current rising with voltage until a plateau is reached where the filament becomes constant current in nature, not changing with voltage across the filament. This agrees with the Tektronix paper that a hot filament is a constant current device.
However, I seem to remember way back when I was about 13, I saw a couple of 12V bulbs in series and driven by about 12v total. One bulb would be lit and the other dark. If the lit one was shorted for a moment, the dark one would light and the previously lit wone would be dark. The circuit was like a rudimentary bistable circuit.
Since the two identical bulbs were in series, this means that the current through both bulbs was the same, equal at all times but the voltage was not. This means that there are TWO voltage points on a V/I curve where the current is the same. So, I'm looking for a V/I curve for hot tungsten filaments that has a hysteresis loop.
I checked my Eshbach Engineering handbook and found a polynomial equation with a cube in it, possibly having multiple roots but I can't decrypt some some of the greek variable symbols and my math is getting rusty.
Bob
My interest is peaked with the curve of the resistance of tungsten filaments. I have previously used a small instrumentation bulb when I designed a sine wave oscillator for our TV broadcast station (1KHz tone over color bars) because the resistance rises with temperature and stops the osc from clipping.
I have searched for a voltage/current curve on the internet and found various curves, some obviously wrong. Can you believe it, there is a University Physics department in Newfoundland that thinks the VI for a light bulb curve is linear like a resistor!
One of the curves has the current rising with voltage until a plateau is reached where the filament becomes constant current in nature, not changing with voltage across the filament. This agrees with the Tektronix paper that a hot filament is a constant current device.
However, I seem to remember way back when I was about 13, I saw a couple of 12V bulbs in series and driven by about 12v total. One bulb would be lit and the other dark. If the lit one was shorted for a moment, the dark one would light and the previously lit wone would be dark. The circuit was like a rudimentary bistable circuit.
Since the two identical bulbs were in series, this means that the current through both bulbs was the same, equal at all times but the voltage was not. This means that there are TWO voltage points on a V/I curve where the current is the same. So, I'm looking for a V/I curve for hot tungsten filaments that has a hysteresis loop.
I checked my Eshbach Engineering handbook and found a polynomial equation with a cube in it, possibly having multiple roots but I can't decrypt some some of the greek variable symbols and my math is getting rusty.
Bob
Hi there Bob,Bob Scott wrote:Thank you Mr. Hayes, always having the equation.
My interest is peaked with the curve of the resistance of tungsten filaments. I have previously used a small instrumentation bulb when I designed a sine wave oscillator for our TV broadcast station (1KHz tone over color bars) because the resistance rises with temperature and stops the osc from clipping.
I have searched for a voltage/current curve on the internet and found various curves, some obviously wrong. Can you believe it, there is a University Physics department in Newfoundland that thinks the VI for a light bulb curve is linear like a resistor!
One of the curves has the current rising with voltage until a plateau is reached where the filament becomes constant current in nature, not changing with voltage across the filament. This agrees with the Tektronix paper that a hot filament is a constant current device.
However, I seem to remember way back when I was about 13, I saw a couple of 12V bulbs in series and driven by about 12v total. One bulb would be lit and the other dark. If the lit one was shorted for a moment, the dark one would light and the previously lit wone would be dark. The circuit was like a rudimentary bistable circuit.
Since the two identical bulbs were in series, this means that the current through both bulbs was the same, equal at all times but the voltage was not. This means that there are TWO voltage points on a V/I curve where the current is the same. So, I'm looking for a V/I curve for hot tungsten filaments that has a hysteresis loop.
I checked my Eshbach Engineering handbook and found a polynomial equation with a cube in it, possibly having multiple roots but I can't decrypt some some of the greek variable symbols and my math is getting rusty.
Bob
Very interesting about the two bulb experiment.
Are you sure you remember that long ago experience correctly?
Perhaps they were using neon bulbs, which are very well known
to have the kind of effects you are talking about?
The other possibility was that the bulbs were some special type, or
that there was something else in the circuit.
Since most bulb data is based on driving the bulb with a voltage
source, another possibility is that the behaviour is different
when driven with a current source, and that behaviour is more
complex to model then anyone cares to do, considering that there
may be little use for this kind of knowledge in modern design.
To see just how complex analog hysteresis can be, take a look at
a typical magnetic BH loop where B and H are traced out over just
a few cycles as exciting current I is increased and decreased.
A possible test on a particular bulb might be to energize it with
12vdc (assuming a 12v bulb) and measure the current Imax. Then,
disconnect and now
energize it with a variable current source, whos range goes from
0 to Imax, and record the voltage, then back the current down
to zero again (slowly) taking measurements again. If this is
done for say three or more steps in current you could see if
the voltage while turning the current up matches the voltage
while turning the current down.
The current source should be a true current source however
and not a resistor in series with a voltage source.
In the mean time, if you can take a scan of your polynomial
equation and post it somewhere on the web or simply email it
to me i would be glad to take a look. It sounds like it could
be interesting.
LEDs vs Bulbs, LEDs are winning.
I haven't seen very many collections of V-I curves for incandescent lamps. Sometimes you will see two or three types plotted when someone is building a wein bridge oscillator.
I think that the trick to the bistable circuit is the resistance change in the bulb as it warms up. A cold bulb usually has lower resistance than a hot one by a factor in the range of 7 to 10 (my guess, this isn't exactly data sheet information). Tungsten bulbs are noted for a high inrush current when they are first turned on.
When you short one of the two bulbs, you reduce its voltage and current to zero, and it cools off, lowering its resistance. The other bulb stays hot, with its resistance possibly ten times that of the cold bulb. When the short is removed, the total current is mainly determined by the hot bulb. It will drop by possibly 10 percent due to the slight resistance of the cold bulb. Though the current in the two bulbs is indentical, the power is not. The hot bulb will receive about 90 percent of the power and the cold bulb about 10 percent. This might be a stable condition, but I don't know how to analyze it.
I think that the trick to the bistable circuit is the resistance change in the bulb as it warms up. A cold bulb usually has lower resistance than a hot one by a factor in the range of 7 to 10 (my guess, this isn't exactly data sheet information). Tungsten bulbs are noted for a high inrush current when they are first turned on.
When you short one of the two bulbs, you reduce its voltage and current to zero, and it cools off, lowering its resistance. The other bulb stays hot, with its resistance possibly ten times that of the cold bulb. When the short is removed, the total current is mainly determined by the hot bulb. It will drop by possibly 10 percent due to the slight resistance of the cold bulb. Though the current in the two bulbs is indentical, the power is not. The hot bulb will receive about 90 percent of the power and the cold bulb about 10 percent. This might be a stable condition, but I don't know how to analyze it.
-
ecerfoglio
- Posts: 108
- Joined: Thu Nov 25, 2004 1:01 am
- Location: Buenos Aires Argentina
- Contact:
The cold bulb, receiving some power will probably heat up slowly.The hot bulb will receive about 90 percent of the power and the cold bulb about 10 percent. This might be a stable condition, but I don't know how to analyze it.
As its filament't temperature raises so does its resistance and the power that it is receiving. The other bulb will have less current and it will cool a little.
I think that, if you allow enough time, both bulbs will glow equally, dividing the supply's voltage in two halves (of course, assuming the two bulbs are identical ones).
Low voltage, high current bulbs, having thicker filaments, will take more time to heat up or cold down. It is more likely to work with say 12 V 50 W bulbs than with 120 V 40 W ones.
E. Cerfoglio
Buenos Aires
Argentina
Buenos Aires
Argentina
Hi again rshayes and ecer,
I seem to come to the same conclusion as ecer, in that the two bulbs
should come up to equal voltage once the short is removed.
rshayes, it seems that the 'cold' filament would start to heat up
anyway because some rather large current would flow through it,
and i dont see what could keep it from heating up because it will
have the same current as the other (hot) bulb as soon as the short
is removed. The current will develop a higher voltage, which in turn
develops more power heating, which again in turn heats the filament
up more and this process keeps going until some steady state is
reached.
Here is some data for an actual real life 1.5v, 25ma bulb:
The pairs are in millivolts and milliamps (102mv, 7.8ma for example)
{102, 7.8},
{202, 11.7},
{307, 13.3},
{400, 14.2},
{500, 15.1},
{599, 16.1},
{700, 17.1},
{745, 17.5},
{800, 18.1},
{902, 19.1},
{1000, 20.0},
{1050, 20.5},
{1101, 21.0},
{1150, 21.4},
{1201, 21.9},
{1248, 22.3},
{1304, 22.8},
{1350, 23.2},
{1394, 23.5},
{1451, 24.0},
{1504, 24.4}
This data was extracted from an actual real bulb using an adjustable
voltage supply.
For analysis, a rough approximation was found to be:
i=((1000*v)^0.435)/1000
with v in volts and i in amps.
Although this is approximate, it still mimics the behaviour quite well
and the numbers come out close.
For some reason i neglected to measure the 0 current resistance, which
i am estimating to be 10 ohms.
The following table was produced using the above equation...
V is in volts, I is in amps, R is in ohms, P in watts
Looking at the above numbers, it is interesting to think about
what happens when the short is removed. First, the total
resistance jumps up a little and this reduces the current
immediately. After a short delay, the hot bulb starts to cool
down a little and the cold bulb starts to heat up a little.
The cold bulb starts to heat up a little because now it has
both a rather large current through it and that large current
means it has a voltage across it, and this combination means
power heating takes place in the filament. The resistance
of the cold bulb would then have to rise which would mean
the current would be reduced, but the voltage would rise
meaning power heating would still take place.
This process can be modeled to some degree.
The main issue i think is how the power heats the filament.
I would think that if there were x watts being dissipated then
there would be T temperature (heating characteristics of the
filament would not change over small periods of time) and
this would produce a resistance R. In other words, if there
were 0.010 amps through a bulb and 1 volt across it
(which gives rise to 0.010 watts of heating power) the resistance
would be the same if there was 1.1 volts across it and
0.0090909 amps through it, because the power heating would
be the same and that is what determines temperature and
temperature is what determines resistance.
If anyone agrees or disagrees let me know because i am
in the process of trying this out on paper...that is, putting
together a model of the bulb and then connect them in series
and see what happens.
Also, if anyone cares to try with real bulbs all you need is two bulbs the
same and a voltage supply. I think 12v bulbs and a nice 12v supply
would be good.
I seem to come to the same conclusion as ecer, in that the two bulbs
should come up to equal voltage once the short is removed.
rshayes, it seems that the 'cold' filament would start to heat up
anyway because some rather large current would flow through it,
and i dont see what could keep it from heating up because it will
have the same current as the other (hot) bulb as soon as the short
is removed. The current will develop a higher voltage, which in turn
develops more power heating, which again in turn heats the filament
up more and this process keeps going until some steady state is
reached.
Here is some data for an actual real life 1.5v, 25ma bulb:
The pairs are in millivolts and milliamps (102mv, 7.8ma for example)
{102, 7.8},
{202, 11.7},
{307, 13.3},
{400, 14.2},
{500, 15.1},
{599, 16.1},
{700, 17.1},
{745, 17.5},
{800, 18.1},
{902, 19.1},
{1000, 20.0},
{1050, 20.5},
{1101, 21.0},
{1150, 21.4},
{1201, 21.9},
{1248, 22.3},
{1304, 22.8},
{1350, 23.2},
{1394, 23.5},
{1451, 24.0},
{1504, 24.4}
This data was extracted from an actual real bulb using an adjustable
voltage supply.
For analysis, a rough approximation was found to be:
i=((1000*v)^0.435)/1000
with v in volts and i in amps.
Although this is approximate, it still mimics the behaviour quite well
and the numbers come out close.
For some reason i neglected to measure the 0 current resistance, which
i am estimating to be 10 ohms.
The following table was produced using the above equation...
Code: Select all
V I R P
0.102000 0.007477 13.641405 0.000763
0.202000 0.010065 20.068898 0.002033
0.307000 0.012076 25.423374 0.003707
0.400000 0.013549 29.523232 0.005419
0.500000 0.014930 33.490224 0.007465
0.599000 0.016150 37.089088 0.009674
0.700000 0.017283 40.502366 0.012098
0.745000 0.017758 41.953509 0.013230
0.800000 0.018317 43.676300 0.014653
0.902000 0.019298 46.740321 0.017407
1.000000 0.020184 49.545019 0.020184
1.050000 0.020617 50.929798 0.021647
1.101000 0.021046 52.313025 0.023172
1.150000 0.021449 53.615984 0.024666
1.201000 0.021858 54.946723 0.026251
1.248000 0.022226 56.151484 0.027738
1.304000 0.022654 57.561465 0.029541
1.350000 0.022998 58.700065 0.031048
1.394000 0.023321 59.773470 0.032510
1.451000 0.023732 61.142345 0.034434
1.504000 0.024105 62.394316 0.036254
Looking at the above numbers, it is interesting to think about
what happens when the short is removed. First, the total
resistance jumps up a little and this reduces the current
immediately. After a short delay, the hot bulb starts to cool
down a little and the cold bulb starts to heat up a little.
The cold bulb starts to heat up a little because now it has
both a rather large current through it and that large current
means it has a voltage across it, and this combination means
power heating takes place in the filament. The resistance
of the cold bulb would then have to rise which would mean
the current would be reduced, but the voltage would rise
meaning power heating would still take place.
This process can be modeled to some degree.
The main issue i think is how the power heats the filament.
I would think that if there were x watts being dissipated then
there would be T temperature (heating characteristics of the
filament would not change over small periods of time) and
this would produce a resistance R. In other words, if there
were 0.010 amps through a bulb and 1 volt across it
(which gives rise to 0.010 watts of heating power) the resistance
would be the same if there was 1.1 volts across it and
0.0090909 amps through it, because the power heating would
be the same and that is what determines temperature and
temperature is what determines resistance.
If anyone agrees or disagrees let me know because i am
in the process of trying this out on paper...that is, putting
together a model of the bulb and then connect them in series
and see what happens.
Also, if anyone cares to try with real bulbs all you need is two bulbs the
same and a voltage supply. I think 12v bulbs and a nice 12v supply
would be good.
LEDs vs Bulbs, LEDs are winning.
This one is getting closer to my specialty which is curve tracing. I hadn't curve traced a bulb before so this was interesting. My instrument tops out at 15V and just over an Amp (though I squeezed almost 2A for this test). Its a digital curve tracer so I can adjust the time it takes to ramp a curve by adjusting the number of samples.
I used a 12V, 25W Tungsten Microscope bulb with the two pronged base. See the curves in my image. There are three groups on this one graph, so pay attention.
Most of the curves exibited a linear shape at lower current (cold filiment R). Then as the filiment heated up, the resistance changed and you can see the curve slope down for a bit then regain a new semi linear track of higher resistance than the original cold slope (Hot Filiment R).
Starting at the top is a set of 5 curves taken with 200 samples so that works out to about 300ms for the entire sampling of the curve (1.5ms per step).
The next set down has a similar shape but the crest of the curve happens at a much lower current. These curves were taken at 2000 samples so the curve took about 3 seconds to acquire thus the filiment had a chance to heat up to critical temp sooner (lower V andf I because T was larger)
The last group of 4-6 curves shows 2000 samples as well but with no cooldown time between curves. The filiment remained hot between measurements and thus started out close to the hot resistance.
As you can see, the curve trace of an incandescent bulb is not only non linear WRT the V/I plot but it is also time dependent. Deciding on how to publish a single characteristic for a given bulb would be tricky and you may end up with a situation where each manufacturer does it differently making it hard to compare specs. I am not aware of a standard practice or method regarding curve tracing of bulbs. Most characterizations are done steady state at the operating point.
I used a 12V, 25W Tungsten Microscope bulb with the two pronged base. See the curves in my image. There are three groups on this one graph, so pay attention.
Most of the curves exibited a linear shape at lower current (cold filiment R). Then as the filiment heated up, the resistance changed and you can see the curve slope down for a bit then regain a new semi linear track of higher resistance than the original cold slope (Hot Filiment R).
Starting at the top is a set of 5 curves taken with 200 samples so that works out to about 300ms for the entire sampling of the curve (1.5ms per step).
The next set down has a similar shape but the crest of the curve happens at a much lower current. These curves were taken at 2000 samples so the curve took about 3 seconds to acquire thus the filiment had a chance to heat up to critical temp sooner (lower V andf I because T was larger)
The last group of 4-6 curves shows 2000 samples as well but with no cooldown time between curves. The filiment remained hot between measurements and thus started out close to the hot resistance.
As you can see, the curve trace of an incandescent bulb is not only non linear WRT the V/I plot but it is also time dependent. Deciding on how to publish a single characteristic for a given bulb would be tricky and you may end up with a situation where each manufacturer does it differently making it hard to compare specs. I am not aware of a standard practice or method regarding curve tracing of bulbs. Most characterizations are done steady state at the operating point.
-
dyarker
- Posts: 1919
- Joined: Fri Aug 22, 2003 1:01 am
- Location: Izmir, Turkiye; from Rochester, NY
- Contact:
Well done haklesup!
The "no cool down" wouldn't vary between manufacturers. I guess that the manufacturers expect their bulbs to used near rated voltage to produce light, so the marketing dept. only publishes expected life.
I would not have guessed there would be a negative resistance region in the curve. But now that you've shown it, it makes sense. It shows the thermal time constant for the bulb. For this bulb, if the application involved frequencies from 0.1Hz to 1Hz, thermal time constant effect would be important.
If someone asked me out-of-the-blue about the bistable effect mentioned by Bob, my first EWAG would be the same as ecerfoglio's. The curve in the middle with straight lines to left and right would make it happen; and more likely with a sharper curve. The cold bulb will heat up some and the hot bulb will cool down some. But the bulbs can't creap to equal temperature because of the curve in the middle. The line to the right is less steep than the left, so hot and cold bulbs will settle at stable points on either side of the curve. With equal current through both bulbs because they're in series the hot bulb will continue to have more than half the power like rshayes said. (Steve, I just hope my rephrasing makes it clearer.)
For comparison; the temperature vs. resistance graph of semiconductors curves in the opposite direction of metals. From less steep to more steep. So while lamps a slightly self regulating, transistor curcuits go into thermal run-away (especially emitter followers).
The current vs. voltage graph for LEDs curves in the opposite direction from the incandescent bulb graph. That is why LED circuits must have current limiting somewhere.
Cheers,
The "no cool down" wouldn't vary between manufacturers. I guess that the manufacturers expect their bulbs to used near rated voltage to produce light, so the marketing dept. only publishes expected life.
I would not have guessed there would be a negative resistance region in the curve. But now that you've shown it, it makes sense. It shows the thermal time constant for the bulb. For this bulb, if the application involved frequencies from 0.1Hz to 1Hz, thermal time constant effect would be important.
If someone asked me out-of-the-blue about the bistable effect mentioned by Bob, my first EWAG would be the same as ecerfoglio's. The curve in the middle with straight lines to left and right would make it happen; and more likely with a sharper curve. The cold bulb will heat up some and the hot bulb will cool down some. But the bulbs can't creap to equal temperature because of the curve in the middle. The line to the right is less steep than the left, so hot and cold bulbs will settle at stable points on either side of the curve. With equal current through both bulbs because they're in series the hot bulb will continue to have more than half the power like rshayes said. (Steve, I just hope my rephrasing makes it clearer.)
For comparison; the temperature vs. resistance graph of semiconductors curves in the opposite direction of metals. From less steep to more steep. So while lamps a slightly self regulating, transistor curcuits go into thermal run-away (especially emitter followers).
The current vs. voltage graph for LEDs curves in the opposite direction from the incandescent bulb graph. That is why LED circuits must have current limiting somewhere.
Cheers,
Dale Y
Hi again,
Hackle, thanks for taking and posting that data...very interesting.
The 'snap' is due to the heat capacity of the filament.
The graphs also show that the filament does in fact heat up even
when driven with small voltages. Apparently the slower the voltage
rises the less 'snap' there is.
What the graphs dont show:
1. What happens when the voltage is reduced at various rates.
Presently the curves only show what happens when the voltage
is being *increased*. In order to compare the rising thermal
time constant to the falling thermal time constant we would need
to see this data too. This would be interesting too and help with
the modeling. I had suspected that the falling ttc is longer than the
rising ttc because of the insulating properties of the bulb package,
but did not make any attempt to measure just how different it is.
2. Also interesting would be that the device could be driven with a
constant current source rather than a constant voltage source.
In other words, set the current and let the voltage go to whatever
it may be instead of setting the voltage and letting the current go
to whatever it may be.
3. The time frame i am assuming to be in proportion to the length
of each line, but if possible, it would be good to see some tiny 'dots'
that indicate either each sample, or a group of say 10 (or even 100
in some cases) drawn along each line. This would make estimating
the relative time periods between various regions a lot easier.
What else would be interesting is to see the same stepping only
up to 1/2 the normal operating voltage (not half the normal
current, but up to 1/2 the voltage).
One other thing, do you happen to have two of these bulbs to actually
try the experiment?
Thanks again for posting this interesting data.
Hackle, thanks for taking and posting that data...very interesting.
The 'snap' is due to the heat capacity of the filament.
The graphs also show that the filament does in fact heat up even
when driven with small voltages. Apparently the slower the voltage
rises the less 'snap' there is.
What the graphs dont show:
1. What happens when the voltage is reduced at various rates.
Presently the curves only show what happens when the voltage
is being *increased*. In order to compare the rising thermal
time constant to the falling thermal time constant we would need
to see this data too. This would be interesting too and help with
the modeling. I had suspected that the falling ttc is longer than the
rising ttc because of the insulating properties of the bulb package,
but did not make any attempt to measure just how different it is.
2. Also interesting would be that the device could be driven with a
constant current source rather than a constant voltage source.
In other words, set the current and let the voltage go to whatever
it may be instead of setting the voltage and letting the current go
to whatever it may be.
3. The time frame i am assuming to be in proportion to the length
of each line, but if possible, it would be good to see some tiny 'dots'
that indicate either each sample, or a group of say 10 (or even 100
in some cases) drawn along each line. This would make estimating
the relative time periods between various regions a lot easier.
What else would be interesting is to see the same stepping only
up to 1/2 the normal operating voltage (not half the normal
current, but up to 1/2 the voltage).
One other thing, do you happen to have two of these bulbs to actually
try the experiment?
Thanks again for posting this interesting data.
LEDs vs Bulbs, LEDs are winning.
Hi again,
Ok, i got the preliminary model going and i put two of them in series
and checked out the waveforms for the center voltage and the two
bulbs resistance after the switch was opened.
Here are the results of that test...
The two bulbs were modeled after the 1.5v bulb posted previously
in this thread.
The two bulbs are connected in series and the center voltage is monitored,
along with the resistance of each bulb. The lower bulb is shorted out
with a switch that has close to zero resistance (1m ohm).
At t=0, 1.5 volts is applied across the two bulbs.
The current jumps up to 150ma.
The upper bulb starts to heat up, and its resistance rises,
which causes the current to ramp down exponentially.
After 39ms its resistance stabilizes to about 62 ohms and the current
stabilizes to 24.1ma . This is typical for a single bulb running at 1.5 volts.
The lower bulbs resistance stays at 10 ohms, having no current through it
because it is shorted out by the switch.
At t=40ms, the switch across the lower bulb is opened.
The current jumps down immediately to 20.7ma, and the center
voltage jumps up to 208mv. Note at this exact point in time
the resistance of both bulbs does not change, only the current
changes and the voltage across both bulbs changes.
Just after t=40ms, both bulbs resistance *begins* to change
and the total current begins to drop slightly (all exponential curves).
At t=60ms, the resistance of the upper bulb is 46 ohms, and the
resistance of the lower bulb is 39.6 ohms; the total current
has dropped to 17.5ma, but that's after it had dropped to a valley
of 17.45ma around 55ms, so it went down and came back up ever so
slightly.
At t=90ms, both bulbs resistances are very nearly equal, being
around 42.2 ohms each. The center voltage is very nearly 750mv,
the current is 17.8ma . The actual upper bulb resistance is
42.24 ohms and the lower bulb is 42.19 ohms.
After 150ms everything is still about the same and nothing
else changes after that.
To add a bit more, the rising thermal time constant was made equal to
the falling thermal time constant for this test, but later the falling
time constant (cooling) was made twice as large as the rising
(to mimic the insulation properties of the bulb casing) but the only
difference was that the settling time extended out by something like
another 20ms, after which the resistance of both bulbs still became
equal.
CONCLUSION
The two bulbs resistances became equal after several time constants.
In order to prevent this from happening (keep the lower bulb cold
while passing *almost* the rated current) once the
switch is opened, there would have to be some mechanism that
could prevent the heating of the filament even with this much current
flowing through it. Perhaps a very heavy set of lead wires? But then
again this would simply add another time constant. Apparently the
bulb modeled did not have this property, if it really does exist.
Ok, i got the preliminary model going and i put two of them in series
and checked out the waveforms for the center voltage and the two
bulbs resistance after the switch was opened.
Here are the results of that test...
The two bulbs were modeled after the 1.5v bulb posted previously
in this thread.
The two bulbs are connected in series and the center voltage is monitored,
along with the resistance of each bulb. The lower bulb is shorted out
with a switch that has close to zero resistance (1m ohm).
At t=0, 1.5 volts is applied across the two bulbs.
The current jumps up to 150ma.
The upper bulb starts to heat up, and its resistance rises,
which causes the current to ramp down exponentially.
After 39ms its resistance stabilizes to about 62 ohms and the current
stabilizes to 24.1ma . This is typical for a single bulb running at 1.5 volts.
The lower bulbs resistance stays at 10 ohms, having no current through it
because it is shorted out by the switch.
At t=40ms, the switch across the lower bulb is opened.
The current jumps down immediately to 20.7ma, and the center
voltage jumps up to 208mv. Note at this exact point in time
the resistance of both bulbs does not change, only the current
changes and the voltage across both bulbs changes.
Just after t=40ms, both bulbs resistance *begins* to change
and the total current begins to drop slightly (all exponential curves).
At t=60ms, the resistance of the upper bulb is 46 ohms, and the
resistance of the lower bulb is 39.6 ohms; the total current
has dropped to 17.5ma, but that's after it had dropped to a valley
of 17.45ma around 55ms, so it went down and came back up ever so
slightly.
At t=90ms, both bulbs resistances are very nearly equal, being
around 42.2 ohms each. The center voltage is very nearly 750mv,
the current is 17.8ma . The actual upper bulb resistance is
42.24 ohms and the lower bulb is 42.19 ohms.
After 150ms everything is still about the same and nothing
else changes after that.
To add a bit more, the rising thermal time constant was made equal to
the falling thermal time constant for this test, but later the falling
time constant (cooling) was made twice as large as the rising
(to mimic the insulation properties of the bulb casing) but the only
difference was that the settling time extended out by something like
another 20ms, after which the resistance of both bulbs still became
equal.
CONCLUSION
The two bulbs resistances became equal after several time constants.
In order to prevent this from happening (keep the lower bulb cold
while passing *almost* the rated current) once the
switch is opened, there would have to be some mechanism that
could prevent the heating of the filament even with this much current
flowing through it. Perhaps a very heavy set of lead wires? But then
again this would simply add another time constant. Apparently the
bulb modeled did not have this property, if it really does exist.
LEDs vs Bulbs, LEDs are winning.
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dacflyer
- Posts: 4751
- Joined: Fri Feb 08, 2002 1:01 am
- Location: USA / North Carolina / Fayetteville
- Contact:
i seriously doubt that filiment bulbs will be fazed out of existence
because of the color tone. also i haven't seen a light yet that can replace a halogen lamp, for crispness and brightness..
even now, i still hate it that the 1,500 watt incadescent lamps
(mogual base) are no longer available.. i work in my yard a lot in the evenings, (too hot during the day, especially insummer) and i hate to have to wait for metal halide light to warm up,,i liked it when i had instant light..
oh, well.. i am thinking to switch over to 3-500 watt halogen quartz lamps
because of the color tone. also i haven't seen a light yet that can replace a halogen lamp, for crispness and brightness..
even now, i still hate it that the 1,500 watt incadescent lamps
(mogual base) are no longer available.. i work in my yard a lot in the evenings, (too hot during the day, especially insummer) and i hate to have to wait for metal halide light to warm up,,i liked it when i had instant light..
oh, well.. i am thinking to switch over to 3-500 watt halogen quartz lamps
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Chris Smith
- Posts: 4325
- Joined: Tue Dec 04, 2001 1:01 am
- Location: Bieber Ca.
Hi there,
dacflyer, why do you think the color tone has anything to do with it?
LEDs and fluorescent bulbs can be obtained in 'warm' colors so why
not use one of those instead of a bulb?
I see filament bulbs getting phased out eventually, just dont know the time
line exactly. CA has already passed some laws however, banning
filament bulbs in new building constructions (dont know if this is the entire
state or just the high population areas). This ought to say something
about what the trend for the future is. Filament bulbs are the worst
on efficiency, and with the current global warming concerns starting
to peak the CO2 'footprint' of the filament bulb is larger than other
lighting types, so eventually it's got to go.
One other interesting fact is that the life of a bulb isnt all that long
either. I hate having to keep replacing bulbs, and in a flashlight
if you drop it once the bulbs usually breaks rendering it useless.
Many Mag-Lites keep a spare bulb in the tail end of the flashlight
just in case this happens, but with properly designed LED flashlights
this is not necessary.
dacflyer, why do you think the color tone has anything to do with it?
LEDs and fluorescent bulbs can be obtained in 'warm' colors so why
not use one of those instead of a bulb?
I see filament bulbs getting phased out eventually, just dont know the time
line exactly. CA has already passed some laws however, banning
filament bulbs in new building constructions (dont know if this is the entire
state or just the high population areas). This ought to say something
about what the trend for the future is. Filament bulbs are the worst
on efficiency, and with the current global warming concerns starting
to peak the CO2 'footprint' of the filament bulb is larger than other
lighting types, so eventually it's got to go.
One other interesting fact is that the life of a bulb isnt all that long
either. I hate having to keep replacing bulbs, and in a flashlight
if you drop it once the bulbs usually breaks rendering it useless.
Many Mag-Lites keep a spare bulb in the tail end of the flashlight
just in case this happens, but with properly designed LED flashlights
this is not necessary.
LEDs vs Bulbs, LEDs are winning.
-
dacflyer
- Posts: 4751
- Joined: Fri Feb 08, 2002 1:01 am
- Location: USA / North Carolina / Fayetteville
- Contact:
mr al >> i say that i think filiment bulbs will be around for a long time, maybe not in wide spread use, but in some appliactions that the others might be able to handle..example.. in hi temp. use like your oven..in mine the bulb is only protected by a glass cover, but you know its over 450F in there... and also not every bulb is readily replaceable by a led or CFL
and. as you mentioned about the mag lites, last night i seen the new 3 watt led conversion kits..they were 18.00 just for the led at wal mart, but 1 thing i noticed was,,once you change over to the led..the adjustable focus on it becomes useless..flood/spot adjuster.
Hmm another example i think would be..stage spot lights i don't see them becoming leds no time soon, altho i know there are some led colored lights out there, but not the high power spot light they follow you with.
i am glad for all the CFL and led technology out there, but i just don't see it making the filiment lamp becoming totally obselete.. i cannot wait to see the so called led headlights on cars soon. i hope they are focused like the HID ones
and. as you mentioned about the mag lites, last night i seen the new 3 watt led conversion kits..they were 18.00 just for the led at wal mart, but 1 thing i noticed was,,once you change over to the led..the adjustable focus on it becomes useless..flood/spot adjuster.
Hmm another example i think would be..stage spot lights i don't see them becoming leds no time soon, altho i know there are some led colored lights out there, but not the high power spot light they follow you with.
i am glad for all the CFL and led technology out there, but i just don't see it making the filiment lamp becoming totally obselete.. i cannot wait to see the so called led headlights on cars soon. i hope they are focused like the HID ones
-
Chris Smith
- Posts: 4325
- Joined: Tue Dec 04, 2001 1:01 am
- Location: Bieber Ca.
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