Burn out 120V lamp on 120V circuit?
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From: krw on 9 Aug 2010 18:49 On Mon, 9 Aug 2010 17:41:21 -0400, "Paul E. Schoen" <paul(a)pstech-inc.com> wrote: > ><krw(a)att.bizzzzzzzzzzzz> wrote in message >news:ajcs56h4aem5la2screarnk1pocf065ps6(a)4ax.com... >> On Sat, 07 Aug 2010 20:51:43 -0700, foo(a)bar.com wrote: >> >>>On Sat, 07 Aug 2010 22:15:34 -0500, "krw(a)att.bizzzzzzzzzzzz" >>><krw(a)att.bizzzzzzzzzzzz> wrote: >>>>>> >>>>>> http://www.cbo.gov/ftpdocs/100xx/doc10068/effective_tax_rates_2006.pdf >>>>> >>>>>Thank you. Another typical righteous wing lie exposed. Sad that they >>>>>actually believe what their extremist conservative deities pronounce as >>>>>gospel. >>>> >>>>Which "lie" is that? The one that states that the "rich" already are >>>>slaves >>>>of the state, who pay for everything? That one, loser? >>>> >>> >>>From the same document, table 1: >>> >>>Top 10% have 41.6% of all pretax income and they pay 55.4% of all >>>federal taxes. >>>Top 1% have 18.8% of all pretax income and pay 28.3% of all federal >>>taxes. >> >> Yes, I get it. Convince the loser and we'll have one fewer Obamanut. >> >> I was specifically answering PeterD's assertion that the middle class pays >> all >> the taxes. No, it's far worse than that. Close to half are net >> receivers. > >The money paid by people in taxes to the government either goes back to >other taxpayers (wages, purchases, and even social security, Medicare, and >welfare), who spend it locally, stimulating the economy, or it goes to those >who do not contribute to economic health. The only bad that results from tax >and spend is a slight reduction of discretionary income for the upper tax >brackets, and a net outpouring of wealth when the money finds its way into >the criminal element (such as the huge drug market), and foreign powers and >terrorist groups who use the money for destructive purposes. Even so-called >wasteful spending still (ideally) recirculates back into the economy and >promotes business growth and jobs. Tax and spend means it hurts a bit now. >Borrow and spend means it's gonna hurt more later. That was the result of >Bush's tax cuts. Restoring the previous structure stabilizes and perhaps >reduces the debt. Absolutely clueless drivel "Government largesse" <> productivity "slight reduction" - NOT! "discretionary income" <> "investment" It's not only the "upper income" that saves/invests There are more real gems in there, but that's enough for now.
From: markp on 9 Aug 2010 19:31 "Paul E. Schoen" <paul(a)pstech-inc.com> wrote in message news:O8%7o.60837$lS1.14657(a)newsfe12.iad... > > "Sylvia Else" <sylvia(a)not.here.invalid> wrote in message > news:8c6oelFig4U1(a)mid.individual.net... >> On 6/08/2010 4:03 AM, VWWall wrote: >>> A while back, in another Usenet ng, someone asked about using an >>> ordinary 120V incandescent lamp to slow down a fan motor. A number of >>> posters replied that they had successfully done this. The usual >>> discussion of the merits of doing this ensued. >>> >>> Then, one frequent poster replied that he had tried this with a small AC >>> motor, and the 120V bulb, in series with the motor, burned out when the >>> circuit was completed. >>> >>> There was much discussion, with many saying that it was impossible for >>> any two terminal passive device in series with a 120V incandescent bulb >>> on a 120V circuit to cause that bulb to burn out. >>> >>> Some even set up Spice simulations which were difficult because of the >>> large variation in the bulb's resistance from cold to fully "on". The OP >>> was asked to repeat the experiment, which he did several times, with the >>> same results of the bulb burning out. >>> >>> The final conclusion, perhaps not shared by all, was that it was >>> possible for a passive device to act in this way. One poster even showed >>> Spice results with an increase in line current due to motor inductance. >>> It was never proved that this increase was enough to cause the bulb to >>> fail. >>> >>> I haven't tried the experiment myself, since I don't have a suitable >>> small motor available, and with 120V incandescent bulbs on the >>> endangered species list, I don't care to sacrifice even one! Some time >>> ago, I did use a series 120V 100W bulb to slow down the compressor fan >>> motor in my refrigerator, when the proper replacement was not available. >>> >>> I have my own theory and can postulate a two terminal passive device >>> capable of behaving in this way. (It doesn't even need pre-"charged" >>> condensers.) >>> >>> What say ye all? >>> >> >> When the circuit's turned on the bulb filament has a lowish resistance, >> so a large current can build up in the inductance of the motor. Then the >> filament heats up, its resistance increases, and the inductor pushes the >> voltage up to keep the initial current flowing, with the result that the >> bulb filament has to handle an overcurrent for a period. >> >> Whether this will destroy the bulb would depend critically on the >> characterstics of the filament and the size of the inductor. >> >> So I'd have thought that the answer was that it's possible, but actual >> mileage will vary. > > That is food for thought (and thanks for replying on topic instead of > getting into politics). > > I don't think a very large current would build up in the motor, especially > a small fan motor which is relatively self-limiting even with locked > rotor. If the motor stalls, then the maximum current would be initially > limited by the bulb cold filamant resistance, and the locked rotor > characteristics of the motor. At worst it would be about the same as > connecting the bulb to the line through a short, and since this is AC > there is not a constant build-up of inductive energy. If the motor runs, > then there may be some energy build up from the flywheel action of the > load, but this would build very slowly in comparison to the heating of the > filamant and eventual stabilization at a lower speed. The motor itself > should only be able to generate the same amount of power that is involved > in running at full load, which should be less than 1/8 HP or less than the > 100W rating of the bulb. So even flywheel action should not be able to > account for enough power being pumped into the bulb to blow it out, > especially at turn-on. > > Paul I think it is actually due to the inductor slowing the rate of change of current down. I'm going to assume a simple inductor here, not a fan. If you had just a bulb you'd get a high current initially and it will rapidly fall to steady state. So there's a peak power at the beginning but only for a very short time, this energy is absorbed easily by the element due to it's thermal heat capacity. When you add an inductor the whole thing slows down, and instead of starting at peak power it starts at zero power. It still reaches the same steady state condition (no more rate of change of current = 0V across the inductor), but because things have slowed down by the inductor, the heat capacity doesn't come into play anymore (it already absorbed the heat early on), so there'll be a maximum steady state power that the bulb could handle before blowing and this is much lower than the normal peak power with no inductor. This steady state maximum could, for example, be 90W for a 60W bulb.The question is therefore whether the product of the voltage and current through the bulb, i.e. the power, would ever rise above this value before falling back down to steady state. If it does peak like that then the bulb could blow before steady state is reached. I've no idea what would really happen, but my gut feel is that the power curve could have some 2nd order, or even 3rd order, terms in it and may well have a maximum. BTW I found this model of a tungsten lamp in case anyone wants to have a go at converting it to LTspice and simulating it: www.intusoft.com/nlpdf/nl11.pdf Mark.
From: VWWall on 9 Aug 2010 21:12 markp wrote: > "Paul E. Schoen" <paul(a)pstech-inc.com> wrote in message > news:O8%7o.60837$lS1.14657(a)newsfe12.iad... >> "Sylvia Else" <sylvia(a)not.here.invalid> wrote in message >> news:8c6oelFig4U1(a)mid.individual.net... >>> On 6/08/2010 4:03 AM, VWWall wrote: >>>> A while back, in another Usenet ng, someone asked about using an >>>> ordinary 120V incandescent lamp to slow down a fan motor. A number of >>>> posters replied that they had successfully done this. The usual >>>> discussion of the merits of doing this ensued. >>>> >>>> Then, one frequent poster replied that he had tried this with a small AC >>>> motor, and the 120V bulb, in series with the motor, burned out when the >>>> circuit was completed. >>>> >>>> There was much discussion, with many saying that it was impossible for >>>> any two terminal passive device in series with a 120V incandescent bulb >>>> on a 120V circuit to cause that bulb to burn out. >>>> >>>> Some even set up Spice simulations which were difficult because of the >>>> large variation in the bulb's resistance from cold to fully "on". The OP >>>> was asked to repeat the experiment, which he did several times, with the >>>> same results of the bulb burning out. >>>> >>>> The final conclusion, perhaps not shared by all, was that it was >>>> possible for a passive device to act in this way. One poster even showed >>>> Spice results with an increase in line current due to motor inductance. >>>> It was never proved that this increase was enough to cause the bulb to >>>> fail. >>>> >>>> I haven't tried the experiment myself, since I don't have a suitable >>>> small motor available, and with 120V incandescent bulbs on the >>>> endangered species list, I don't care to sacrifice even one! Some time >>>> ago, I did use a series 120V 100W bulb to slow down the compressor fan >>>> motor in my refrigerator, when the proper replacement was not available. >>>> >>>> I have my own theory and can postulate a two terminal passive device >>>> capable of behaving in this way. (It doesn't even need pre-"charged" >>>> condensers.) >>>> >>>> What say ye all? >>>> >>> When the circuit's turned on the bulb filament has a lowish resistance, >>> so a large current can build up in the inductance of the motor. Then the >>> filament heats up, its resistance increases, and the inductor pushes the >>> voltage up to keep the initial current flowing, with the result that the >>> bulb filament has to handle an overcurrent for a period. >>> >>> Whether this will destroy the bulb would depend critically on the >>> characterstics of the filament and the size of the inductor. >>> >>> So I'd have thought that the answer was that it's possible, but actual >>> mileage will vary. >> That is food for thought (and thanks for replying on topic instead of >> getting into politics). >> >> I don't think a very large current would build up in the motor, especially >> a small fan motor which is relatively self-limiting even with locked >> rotor. If the motor stalls, then the maximum current would be initially >> limited by the bulb cold filamant resistance, and the locked rotor >> characteristics of the motor. At worst it would be about the same as >> connecting the bulb to the line through a short, and since this is AC >> there is not a constant build-up of inductive energy. If the motor runs, >> then there may be some energy build up from the flywheel action of the >> load, but this would build very slowly in comparison to the heating of the >> filamant and eventual stabilization at a lower speed. The motor itself >> should only be able to generate the same amount of power that is involved >> in running at full load, which should be less than 1/8 HP or less than the >> 100W rating of the bulb. So even flywheel action should not be able to >> account for enough power being pumped into the bulb to blow it out, >> especially at turn-on. >> >> Paul > > I think it is actually due to the inductor slowing the rate of change of > current down. > > I'm going to assume a simple inductor here, not a fan. > > If you had just a bulb you'd get a high current initially and it will > rapidly fall to steady state. So there's a peak power at the beginning but > only for a very short time, this energy is absorbed easily by the element > due to it's thermal heat capacity. > > When you add an inductor the whole thing slows down, and instead of starting > at peak power it starts at zero power. It still reaches the same steady > state condition (no more rate of change of current = 0V across the > inductor), but because things have slowed down by the inductor, the heat > capacity doesn't come into play anymore (it already absorbed the heat early > on), so there'll be a maximum steady state power that the bulb could handle > before blowing and this is much lower than the normal peak power with no > inductor. > If this were true powering an incandescent bulb from a Variac or other variable voltage source would result in the bulb's failure before its rated voltage was reached. Photo-flood lamps are over voltaged, so it takes quite a bit of excess voltage to instantly fail the lamp. > This steady state maximum could, for example, be 90W for a 60W bulb.The > question is therefore whether the product of the voltage and current through > the bulb, i.e. the power, would ever rise above this value before falling > back down to steady state. If it does peak like that then the bulb could > blow before steady state is reached. > > I've no idea what would really happen, but my gut feel is that the power > curve could have some 2nd order, or even 3rd order, terms in it and may well > have a maximum. > > BTW I found this model of a tungsten lamp in case anyone wants to have a go > at converting it to LTspice and simulating it: > www.intusoft.com/nlpdf/nl11.pdf > A couple of people have tried modeling the bulb in series with an inductance. The current does rise above the expected, but not enough to fail the bulb. Paul Hovnanian was on the right track. It's possible to conceive a two terminal device which could behave in this manner, but it would involve the storage of energy and the release of that energy in a way that would over-voltage the bulb. Any ways I could conceive involved some sort of switch, operated at a precise time. A hint: The motor which consistently failed the bulb was a small capacitor start device. My theory may be wrong. -- Virg Wall
From: Jim Thompson on 9 Aug 2010 21:35 On Mon, 09 Aug 2010 18:12:40 -0700, VWWall <vwall(a)large.invalid> wrote: [snip] A lab-tested Spice model for a lamp... **** ** AIRCRAFT LAMP SUBCIRCUIT *VO=NOMINAL OPERATING VOLTAGE *IO=NOMINAL STEADY STATE OPERATING CURRENT *RCOLD=FILAMEMT RESISTANCE MEASURED AT ROOM TEMP (300K) *TAU=CURRENT TIME CONSTANT AFTER A 0 (zero) TO VO STEP IS APPLIED ..SUBCKT LAMP 1 2 PARAMS: VO=28 IO=25m RCOLD=112 TAU=22m TAMB=300 H1 6 0 VML 1 RH1 6 0 1 GP 0 4 VALUE={V(6)*V(1,2)} *V(4,0) = FILAMENT TEMPERATURE IN KELVINS RT 4 5 {300*(VO-IO*RCOLD)/(IO*IO*VO*RCOLD)} CT 4 5 {TAU*IO*IO*VO*RCOLD/(300*(VO-IO*RCOLD))} VAMB 5 0 {TAMB} El 7 0 1 2 300 R1 7 0 1 E2 8 0 VALUE={V(4)*RCOLD} R2 8 0 1 E3 10 0 7 9 10MEG R3 10 0 1 E4 9 0 VALUE={V(8)*V(10)} R4 9 0 1 GR 1 3 10 2 1 VML 3 2 0 ..ENDS LAMP **** ...Jim Thompson -- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | Spice is like a sports car... Performance only as good as the person behind the wheel.
From: Sylvia Else on 9 Aug 2010 23:54
On 10/08/2010 8:18 AM, Paul E. Schoen wrote: > > "Sylvia Else" <sylvia(a)not.here.invalid> wrote in message > news:8c6oelFig4U1(a)mid.individual.net... >> On 6/08/2010 4:03 AM, VWWall wrote: >>> A while back, in another Usenet ng, someone asked about using an >>> ordinary 120V incandescent lamp to slow down a fan motor. A number of >>> posters replied that they had successfully done this. The usual >>> discussion of the merits of doing this ensued. >>> >>> Then, one frequent poster replied that he had tried this with a small AC >>> motor, and the 120V bulb, in series with the motor, burned out when the >>> circuit was completed. >>> >>> There was much discussion, with many saying that it was impossible for >>> any two terminal passive device in series with a 120V incandescent bulb >>> on a 120V circuit to cause that bulb to burn out. >>> >>> Some even set up Spice simulations which were difficult because of the >>> large variation in the bulb's resistance from cold to fully "on". The OP >>> was asked to repeat the experiment, which he did several times, with the >>> same results of the bulb burning out. >>> >>> The final conclusion, perhaps not shared by all, was that it was >>> possible for a passive device to act in this way. One poster even showed >>> Spice results with an increase in line current due to motor inductance. >>> It was never proved that this increase was enough to cause the bulb to >>> fail. >>> >>> I haven't tried the experiment myself, since I don't have a suitable >>> small motor available, and with 120V incandescent bulbs on the >>> endangered species list, I don't care to sacrifice even one! Some time >>> ago, I did use a series 120V 100W bulb to slow down the compressor fan >>> motor in my refrigerator, when the proper replacement was not available. >>> >>> I have my own theory and can postulate a two terminal passive device >>> capable of behaving in this way. (It doesn't even need pre-"charged" >>> condensers.) >>> >>> What say ye all? >>> >> >> When the circuit's turned on the bulb filament has a lowish >> resistance, so a large current can build up in the inductance of the >> motor. Then the filament heats up, its resistance increases, and the >> inductor pushes the voltage up to keep the initial current flowing, >> with the result that the bulb filament has to handle an overcurrent >> for a period. >> >> Whether this will destroy the bulb would depend critically on the >> characterstics of the filament and the size of the inductor. >> >> So I'd have thought that the answer was that it's possible, but actual >> mileage will vary. > > That is food for thought (and thanks for replying on topic instead of > getting into politics). > > I don't think a very large current would build up in the motor, > especially a small fan motor which is relatively self-limiting even with > locked rotor. If the motor stalls, then the maximum current would be > initially limited by the bulb cold filamant resistance, and the locked > rotor characteristics of the motor. At worst it would be about the same > as connecting the bulb to the line through a short, and since this is AC > there is not a constant build-up of inductive energy. If the motor runs, > then there may be some energy build up from the flywheel action of the > load, but this would build very slowly in comparison to the heating of > the filamant and eventual stabilization at a lower speed. The motor > itself should only be able to generate the same amount of power that is > involved in running at full load, which should be less than 1/8 HP or > less than the 100W rating of the bulb. So even flywheel action should > not be able to account for enough power being pumped into the bulb to > blow it out, especially at turn-on. A signficant question, to which I don't know the answer, is how long it takes a filament to approach its working temperature. If it's a short time compared with half the AC period, then the mechanism I suggested could work in the right circumstances. It it's an AC period or more, then my proposal looks rather suspect. Sylvia. |