AC Soft Start Incandescent Lamp switch (*FINAL EXPERIMENTAL RESULTS*)

[hgg]

Not bad.  So can I use for example 3 thermistors in series to increase the ramp up time?
Can I also use a capacitor in parallel with the bulb to damp the off time as well?

[edavid]

eDavid,  Where is the faulty premise?

They main thing that burns the incandescent bulbs is inrush current.  When cold they have low
resistance and the sudden increase in voltage creates very large currents which raise the temperature of
the filament very fast.

Do you have any actual data which show that inrush current and thermal stress are not important factors
for the lifetime of incandescent lightbulbs?  If you have, you would save me time trying to build this circuit.

The failure mechanism of incandescent lamps is evaporation of the filament.  Don Klipstein's old page does a good job explaining the positive feedback mechanism.  Soft start doesn't change this.

Let me turn it around and ask if you have any actual data that supports your idea that soft start will significantly extend lamp life.

[Paul Moir]

Just attach the NTC to a big block of something to slow it down?

Circuit #2 won't work either I think; it only accounts for the postive going cycle of the AC waveform.  Imagine what happens when everything flips over.

My thinking is that you're trying to do two things at one time; open up the firing phase angle with an RC like a normal light dimmer (which must be done each half cycle, so fast) and second to do this slowly.  Two ways strike me as being possible.  One is to have two circuits essentially, one for the postive and one for the negative cycle.  Probably would be with two SCRs but maybe not.  Second would be brute-force and ignorance.  Take a bog standard dimmer circuit, change out the pot for a optocoupled LDR (so it works on both cycles) and build a slow R/C charge circuit to power the LED.  Stupid but would work!

PS - calculate your current through R1 and R2.  You've got a lot of power being dissipated there.  Don't forget there's a big difference between US/Canada 110v^2/R and 220v^2/R when looking at generic circuits on the web!  And big Rs mean small Cs which is good.  You just need enough current to turn on your TRIAC/SCR when you want it to.

PPS - I have no idea why everyone is getting all high and mighty on this.  I don't agree with the premise either but that's not the point of this discussion, as was covered some time ago. 
 

[edavid]

I have not seen a dimmable CFL on the shelves of any brick and mortar store, ever. Believe me, I look every time I go past the lighting section. Where are the dimmable ones to be found?

Here's one that's sold by Home Depot: http://www.homedepot.com/p/Feit-Electric-60W-Equivalent-Soft-White-2700K-GU24-Spiral-Dimmable-CFL-Light-Bulb-BPESL13T-GU24-DIM/203450789

[Kjelt]

Here's one that's sold by Home Depot: http://www.homedepot.com/p/Feit-Electric-60W-Equivalent-Soft-White-2700K-GU24-Spiral-Dimmable-CFL-Light-Bulb-BPESL13T-GU24-DIM/203450789 

And be amazed about the "ultra expensive" dimmable Led lamp they also sell: $13.-  that's cheap!

http://www.homedepot.com/p/Philips-13-Watt-65W-BR30-Indoor-Soft-White-2700K-Dimmable-LED-Flood-Light-Bulb-E-423798/203408017#.UpO3mklgWCg

[hgg]

The failure mechanism of incandescent lamps is evaporation of the filament.  Don Klipstein's old page does a good job explaining the positive feedback mechanism.  Soft start doesn't change this.

Don't get me wrong but where are the data? 
Neither you nor Don provide any actual data.

The theory says that evaporation is the cause and indeed it is a major factor,
but does anybody leave their lamps on all the time?  I say that another
major factor is thermal (hot & cold) stress.  Going from cold to extreme hot
and vice versa creates the small cracks on the filament which then will get
much worse with time and will decrease its lifetime considerably more.

Here is an example of an immortal bulb...   
http://sanfrancisco.cbslocal.com/video/8903225-worlds-longest-running-lightbulb-goes-out-in-livermore/

[Zero999]

It sounds like the problem is that you don't like 2700K "warm white" light.  You must  really hate daylight.

The 2700K phosphor doesn't produce exactly the same colour as a filament at 2700K. It's just an approximation and not a very good one at that. Rather than emitting a continuous spectrum, there are spikes and gaps so the colour rendering index is poor.

I like real day light but the yucky greyish blue light emitted from daylight simulation bulbs is nothing like the real thing.

[edavid]

It sounds like the problem is that you don't like 2700K "warm white" light.  You must  really hate daylight.

The 2700K phosphor doesn't produce exactly the same colour as a filament at 2700K. It's just an approximation and not a very good one at that. Rather than emitting a continuous spectrum, there are spikes and gaps so the colour rendering index is poor.

Not relevant since he doesn't like 2700K incandescents either.

[jahonen]

I recently changed most of the fluorescent tubes at home from 2700/3000K (827/830) to 4000K (940), thus ones with 4000K CRI > 90. I used Philips TL-D 90 De Luxe series, but they also have TL-D 90 Graphica which has better still CRI. I think that they make those up to 6500K if one prefers higher color temperature. There is some penalty of lumen output for having that improved CRI but not anything huge. lm/W FOM still competes quite well against LED bulbs, especially with electronic ballast (I did upgrade those too).

I think that those 940 ones look much nicer, almost pure white. Colors look much better with those. Only problem with those that they seem to be almost unobtanium locally, all you can find here is just that "piss" colored 2700K.

Regards,
Janne

[Kjelt]

Hi the 8xx series has a usual color rendering around 85 and the 9xx series has a color rendering around 91.
Difference is small and usually the 900 series are used mainly where printed material has to be checked on colors.
So for 4000K you could also look for the cheaper and more obtainable 840 

[BravoV]

I say that another major factor is thermal (hot & cold) stress.  Going from cold to extreme hot and vice versa creates the small cracks on the filament which then will get much worse with time and will decrease its ....

Just curious, do you have any data to back that up ? 

[hgg]

Just curious, do you have any data to back that up ? 

Maybe the following will convince you:
http://spinoff.nasa.gov/spinoff/spinitem?title=Bulp+Miser+%25281978%2529

At least 300% greater runtime! ...
... and they could have done better if they had taken into account the coolling stress as well...
with more than 300% increase, the filament evaporation becomes a secondary factor now.


wilfred, where did you find them ??   Very rare these days.

[BravoV]

Hmm .. not very convincing, besides that is an "advertisement" archive, not a research paper or report.

[dfmischler]

And be amazed about the "ultra expensive" dimmable Led lamp they also sell: $13.-  that's cheap!
http://www.homedepot.com/p/Philips-13-Watt-65W-BR30-Indoor-Soft-White-2700K-Dimmable-LED-Flood-Light-Bulb-E-423798/203408017#.UpO3mklgWCg

I prefer this one (5000K and 10 year warranty).

[Kjelt]

I prefer this one (5000K and 10 year warranty).

5000K that is very cold light to do a lot of work in and keeps you from sleeping or even sustain from creating sleephormones, naah too cold for me at home, nice for in the shed, garage or your job but that's it.

[hgg]

Hi,

Since I cannot (yet) build an AC ramp up and ramp-down circuit I will try to use an in series
thermistor to protect the bulb from inrush current.  Its a bit tricky to size the correct thermistor,
so maybe you can verify my choice.

A typical OSRAM 60 Watt incandescent lamp has a 61 Ohms resistance when cold and 808 Ohms
when hot, with 220V AC.

So the inrush current will be 3.6A which is 13 times more than its working current of 272mA.

The thermistor that I chose has the following specifications:
Resistance @ 25 degC        = 400R
Resistance @ Max Current  = 9.8R
Max Steady State Current  = 200mA
Thermal Time Constant       = 32 sec
Dissipation Factor               = 11mW/degC

I will use the thermistor in series with the bulb and these are the two states of the circuit:


Thermistor Circuit

So the inrush current during the cold state will be 477mA and the working current when the
thermistor is hot will be 269mA.  There will be no stress on the bulb but the initial current
will be 1.7 times more than the thermistor max value.   Do you know if the thermistor will be able
to handle that?  It will be for a short time.  I don't want to save the bulb and stress the thermistor...

Do you think that the above thermistor choice is correct?
How can you calculate the time it takes for the bulb to light up fully?

There is a way to short the thermistor when it gets hot with a latching relay but I not sure
yet how to do it.

Any help is appreciated.
Thank you!

[IanB]

So the inrush current during the cold state will be 477mA and the working current when the
thermistor is hot will be 269mA.  There will be no stress on the bulb but the initial current
will be 1.7 times more than the thermistor max value.   Do you know if the thermistor will be able
to handle that?  It will be for a short time.  I don't want to save the bulb and stress the thermistor...

I don't follow your logic here. You have indicated "Max Steady State Current  = 200 mA" and yet you have calculated a steady state current when the bulb is operating of 269 mA. Your steady state current exceeds the maximum current by a factor of 269/200 = 35%.

This is not for a short time, this is continuous. Exceeding the maximum current rating will mean the thermistor gets hotter than it is designed to get and I presume its life will be shortened as a result.

(You can do a simple calculation about the thermistor temperature. The power dissipated at maximum current will be I²R = 0.2² x 9.8 = 0.39 W = 390 mW. With a thermal coefficient of 11 mW/C the temperature rise above ambient will be 390 / 11 = 35 C. If ambient temperature is 25 C then the thermistor will heat up to about 60 C.)

It may not hurt to exceed the maximum recommended current by 35%, but it is not likely recommended. You might want to choose a part with a higher current rating.

[hgg]

I will not leave the thermistor in the circuit after the initial surge.  I will short it out with
a relay as I said.  (I did not make it clear).  That is why I need to know how long it will
take for the thermistor to reach its lowest resistance.  Your calculation if for the final
temperature of the thermistor, but I need to calculate the time it will take to reach
its lowest resistance.

[IanB]

The time is not something that can be reliably calculated with the data available. It would be far more accurate to put the device on the bench and measure it.

[hgg]

I thought that the time could be calculated from the specification values.

If I use two similar thermistors in parallel they will be able to handle 400mA continuous current.
I need to short them out afterwards, because now I will dissipate more energy than the bulb itself,
they will be running hot and in case of a sudden blackout the thermistor will not protect the bulb
because they need enough time to cool down.

That's why I need to use a latching relay anyway.
Do you know how to connect it so that when the thermistors reach a low resistance they will trip
the relay?

[IanB]

No, time can only be calculated from parameters that have time in them. There are no parameters with a time dimension in the values you listed.

However, good design is a combination of calculation and experiment. You do calculations to help set up the experiments, and then you do experiments to confirm those things that are hard to predict from calculations. The exact time for a thermistor to heat up is not something you can calculate accurately, but it will in any case only be a few seconds. What is a second or two here or there?

How can you say the thermistor will dissipate more energy than the bulb itself? We already did the calculation: thermistor 0.39 W, bulb 60 W. The thermistor is dissipating only 0.65% of the power in the circuit.

Also, please note you can't put two NTC thermistors in parallel because they have a negative temperature coefficient. One of them will simply take all the current and the other one will stay high resistance.

As for the relay, I'm not sure exactly how to do that without some thought and careful choice of the relay. It it easiest just to use the thermistor.

[hgg]

Also, please note you can't put two NTC thermistors in parallel because they have a negative temperature coefficient. One of them will simply take all the current and the other one will stay high resistance.

I didn't know that...

How can you say the thermistor will dissipate more energy than the bulb itself? We already did the calculation: thermistor 0.39 W, bulb 60 W. The thermistor is dissipating only 0.65% of the power in the circuit.

Yes you are right... my mistake.   When its hot its power dissipation will be very low.
P = I2R = 0.269A * 9.8 Ohms = 0.7 watts.

So if I find a way to short it out afterwards, the specific thermistor is not a bad choice, correct?

[SeanB]

If you are going to short it out anyway why use the thermistor? Just use a regular power resistor and a nonresettable thermal fuse in thermal contact with it in case of anything happening to the bypass circuit.

I did that using a 24V ultra low current relay that only had to handle the running current so wear on the contacts is not a factor. The relay was powered via a resistive dropper that fed a bridge rectifier that fed the relay in parallel with a 1000uf capacitor. Gave a delay of around a second on power up to not nuisance trip the breakers on inrush current, but still would trip on a short circuit. A 130c thermal fuse on the resistors was there for the relay failing open circuit. Resistors were 10W ceramic wire wound units of 10R, 2 in parallel for my application.

For your application using 100R 10W would work.

This setup ran for many years until I upgraded and installed separate NTC thermistors on each power transformer ( 2 300VA units to give 220VAC, 50VAC and 24VAC to run the machine) as well as a separate feed for the new inverter via a mains filter, as it has it's own inrush limiter.

[hgg]

Because the thermistor will give the bulb filament a very nice ramp up current with no abrupt steps.
After the lamp is fully on, I will short out the thermistor.  I am trying to think of a way to use the
property of the thermistor so that when it reaches its very low resistance the relay will short it out.
(I haven't found anything yet...)

Do you have any schematic of the resistor & relay circuit you used?

[Rufus]

I will short out the thermistor.  I am trying to think of a way to use the
property of the thermistor so that when it reaches its very low resistance the relay will short it out.
(I haven't found anything yet...)

The only way soft start circuits extend lamp life by more than a few percent is by leaving something in series with the lamp to reduce its operating voltage. Shorting out a soft start thermistor makes it even more pointless.