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I have a 110 Vac incandescent light bulb but I need to use it in 220 Vac.

If I add a diode in series with the bulb, the RMS power is the same as before but the peak value of voltage and current is doubled.

The bulb is kind of special and costs about $200. So will this modification affect its lifetime?

Edit: I'm sorry that I had miscalculated the RMS power with diode. If we use dimmer so that the power is the same as 110Vac but peak value of voltage and current is still doubled, will it affect the lifetime?

-- I've tested series diode for 5 minutes and bulb is still alive!

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    \$\begingroup\$ @tonym I really appreciate your help. \$\endgroup\$
    – M Nasr
    Jan 16, 2022 at 19:14
  • \$\begingroup\$ Have you considered a dimmer switch? \$\endgroup\$
    – stretch
    Jan 16, 2022 at 19:28
  • \$\begingroup\$ Don't forget that some devices ( circuit breaker ) does not like "DC current" ... It would be perhaps interesting the use of a serial capacitor ... Check current necessary. Just perhaps use an electronic switch fired at the "right" place. \$\endgroup\$
    – Antonio51
    Jan 16, 2022 at 19:29
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    \$\begingroup\$ "What will happen now?" Why not include the lamp part number and a link to the datasheet? \$\endgroup\$
    – Transistor
    Jan 16, 2022 at 22:25
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    \$\begingroup\$ For a $200 lamp, I would probably invest in a small transformer to power it. \$\endgroup\$
    – user57037
    Jan 17, 2022 at 18:13

8 Answers 8

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The RMS voltage value of a half sine wave (rectified by putting one diode in series with the load) is the RMS value of the source multiplied by 0.707 (sqrt2/2). If you apply 220 V half wave rectified, you are applying 156 volts. If the nominal supply voltage is 220 volts, the actual voltage could be higher or lower. The bulb is likely designed to withstand a higher voltage than the nominal 110 volts. It has been many decades since a 110 volt nominal supply voltage has been a standard in the USA. Today the nominal residential lighting voltage is 120 volts in the USA. The few other places with 110 volts may or may not have also increased the nominal voltage. Bulbs for use in the USA are nominally rated 120 volts and will likely last a long time at 5% or 10% above that. At 156 volts, 30% high, lasting 5 minutes would not be too surprising, but even with all of the tolerances in you favor you can not expect a satisfactory lifetime.

If the bulb resistance increase due to temperature is ignored, you can calculate the power based on RMS voltage squared divided by resistance. Based on the stated 110 and 220 volt stated voltages, the power for half-wave 220 V would be doubled - (2 X sqrt2/2) squared.

Since there is no question stated in the post, I have assumed that the questions are why did the bulb survive the test? What power was the bulb subjected to? Is the half wave rectifier a good solution to using a 110V bulb on a 220V supply? What is a good solution? I leave it to others to answer the last question since there are already suggestions about that.

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  • \$\begingroup\$ Thanks. But could you please explain more about your first sentence? I thought it's half. \$\endgroup\$
    – M Nasr
    Jan 16, 2022 at 19:58
  • \$\begingroup\$ The RMS voltage is the way of quantifying AC voltage and waveforms other than pure DC so that the heating effect is equal. So the RMS value is what is what you must use. I recalled that the RMS value for half wave was something more than half and looked it up. I can not write out the derivation from memory. You can probably easily find on the internet all about RMS voltage, formulae for common waveforms and derivations. \$\endgroup\$
    – user80875
    Jan 16, 2022 at 20:20
  • \$\begingroup\$ @Tony Stewart EE57 point out that peak power is important for looking at what may happen instantaneously with a dimmer, particularly with cold start. That would also apply with half-wave cold start. Your two best options are probably a dimmer as described by Tony Stewart of an autotransformer. \$\endgroup\$
    – user80875
    Jan 16, 2022 at 20:40
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As others have correctly pointed out, you are running the bulb grossly overvoltage so it cannot be expected to last very long (life is inversely proportional to operating voltage \$V^N\$ where N can be in the range of 12). You could get a control transformer or autotransformer to knock the voltage down (the most expensive, heavy and, in many ways, the best solution). You don't mention how many watts you need, but there are very inexpensive adapter boxes as step up/step down (auto) transformers. They are generally in metal boxes, so safe enough most likely. 500W, 1000W, 1500W etc.

Also pointed out by others, DC current contributes to tungsten filament "notching" which reduces the life, particularly for high voltage (thin) filaments such as what you apparently have. But that's a moot point given the 41% higher than nominal voltage you are feeding it.

One possible solution is to replace the potentiometer in a triac dimmer with a fixed resistor or trimpot. A typical triac dimmer schematic looks like this (often with some EMI-mitigation components- image from here):

enter image description here

There are other ways, but it would be easiest to use an RMS-reading multimeter to easily set the pot to the correct value (a cheap average-reading meter will give a dangerously wrong answer).

Personally, I would use the step down transformer. Dimmers cause EMI and can result in audible noise from the filament.

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  • \$\begingroup\$ Your answer was very clear and helpful. Thanks. \$\endgroup\$
    – M Nasr
    Jan 16, 2022 at 20:09
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Edit 2 - 18/Jan/2022:

Summary: I continue to believe a Transformer - or Ballast as others suggested - is the most recommended approach to Preserve service life of the lamp.
Diode and Triac-dimmer reduce RMS voltage, but are not the most adequate to preserve service life, for the reasons presented below.

I have invested time to research Incandescent lamps and presented my views, but somehow was voted down, something I don’t truly understand the reason(s) for that, as no comment or criticism was shared. Maybe my suggestion to not use a Triac-based dimmer was not clear, so here are some reasons for Not using - or to be very cautious to use such dimmers to reduce the RMS value for an incandescent lamp.

Lamp Filament can show instantaneous temperature fluctuations as high as 20% - that also reflects as Lumens, or the MSCP (Maximum Spherical Candle Power). To avoid this brightness fluctuation or “sub-flicker”, the frequency should be much higher than 60 Hz - The article mentions frequencies as 400Hz, where 60Hz and 1000 Hz are illustrated here:

enter image description here

Life of Filament depends to Power 12 of (ratio of Voltages)
and Power 3.5 of (ratio of Lumens), as seen here: enter image description here

\$ \frac{Life_1}{Life_0} = [{\frac{Volt_0}{Volt_1}}]^{12} \$ ; and \$ \frac{Life_1}{Life_0} = [{\frac{Lumen_0}{Lumen_1}}]^{3.5} \$ , Exponents and Ratios are highlighted in color - see picture.

Article references are marked inside each picture, and the complete Power Ratios were posted as can be interesting to someone else.

As Triac-dimmed voltage behaves as a PWM control at 120Hz, and instantaneous voltage will reflect as an instantaneous higher MSCP, life will be significantly shorter, even if perceived average luminous power is the same. So, the same MSCP will have +10% of instantaneous MSCP and will reduce service life to 70% - obviously, using the above modeled & tested data.

I’m just saying a dimmer here is not enough, being an excellent method to lower the light of properly rated lamps. But, for an expensive (and special) incandescent lamp, the smoother and more sinusoidal the voltage is important for filament life. A Triac-dimmer mimicking the average light output could be ok for a common lamp, or when a redder color is acceptable and moreover, it is much more compact than a transformer. But a dimmer would just attenuate the shortening of Filament life that continues to be reduced if the dimming adjustment is made to the same perceived color temperature.

If I’m misguided, please share your findings & comments and we all would appreciated to know more.


Edit 1: Another post suggested the use of ballast/inductor to limit the current. If you know the proper current at rated voltage and find a ballast compatible with it, that’s a nice (and smooth) idea too.


Original:
About Lamp operating voltage: Most incandescent lamps have a thermal time-constant that is larger than one line cycle: 1/60s, but they “feel” and react to the instantaneous current - as some lamp filaments make acoustic noise when driven by TRIAC dimmers, at certain dimming levels. Resistance of a cold filament usually if 10% of an incandescent-hot one. Tony’s answer/posting provided a simulation to play with. So it seems the instantaneous current can vary from 100% (at rated voltage) to 150%~1000% depending on if driven by half-wave, dimmer, or during inrush current.

The bottom line is: Even if average current (as in 1sec) is the rated one and presenting the same brightness level, lamp filament will pass a higher peak current in half-wave AC than if the filament would operate under perfect sinusoidal AC.

As this lamp is special and expensive (~ US$200), I would prefer to provide the smoothest voltage to the filament, I would not use a dimmer (Triac), as its instantaneous voltage would still be higher than nominal. So, I suggest to use an auto-transformer, and not a too oversized one, as the start-up / inrush current of the incandescent lamp would be further limited by the resistance of the transformer winding. This inrush current limitation may save the lamp from burning its filament during turn-on, as frequently happens.

A sidetrack about 110V/220V: If you are in a country with homes that can be driven by 3-Phase system, as in Brazil, the “110V” is in fact 220/(sqrt 3) = 127Vac.rms, while “220V” is actually 220Vac.rms. In the USA the “110V” is nowadays 120VAC.rms and the “220V” voltage is actually 240V, being provided by a split-phase transformer winding - as illustrated here.

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  • \$\begingroup\$ IMHO, The ratio of 11 or 12 is the resistance ratio from hot/cold depends of convection thermal resistance, which might not be constant if turbulence increases with temperature reducing temp rise. Luminaire air flow restriction makes a huge difference in MTBF from the temp rise, yet intensity changes are not obvious as most of the change is thermal \$\endgroup\$ Jan 19, 2022 at 21:54
  • \$\begingroup\$ @Tony, there is another very interesting Paper about the Temperature prediction of coiled Coil Tungsten Filaments, which depends on some heavy-duty modeling. Author Agrawal, here: tinyurl.com/Tungsten-Temperature (See Eq. 24 of item II.G). There, on Table !, Power losses due to heat transfer with the inert gas are less than 10% for a 500W lamp to 20% for a 40W one. \$\endgroup\$
    – EJE
    Jan 19, 2022 at 22:58
  • \$\begingroup\$ @Tony, About Resistance versus Temperature, this paper (vcclite.com/wp-content/uploads/2017/03/…) on "Figure 8" (page 4) shows as linearly proportional the resistance along with temperature (color temperature). It shows as Tamb = 1R, raising at 2000K = 10R and at 2500K = 12R. This may be a somehow simplified (linear) version than the GE & Agrawal papers, but I would use it for everyday life. But to use more precisely, "Figure 10" shows light output (and dispersion) are not linear, probably caused by non-linear thermal convection. \$\endgroup\$
    – EJE
    Jan 19, 2022 at 23:09
  • \$\begingroup\$ +1for excellent resources I was interested in this topic 30 yrs ago when I noticed a swag lamp with a heat and light 5" rear reflective (chrome face) on the eating room table lasted a couple decades while others in small glass domes lasted a few months. \$\endgroup\$ Jan 20, 2022 at 0:00
  • \$\begingroup\$ So then at that time I designed and built a large Rec room with 19x deep parabolic reflectors worked out perfectly for illumination without glare and the reflector was barely warm. Using a dimmer on 60W bulbs. Longevity and illumination was improved , far better than the cheap PAR lamps that are usually used. \$\endgroup\$ Jan 20, 2022 at 0:09
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Yes, both voltage and current is doubled, thus power is quadrupled for the half-cycle when the diode is forward-biased. Since the other half-cycle has no current, in effect the power is just doubled. Frankly, I'm puzzled that the lamp is alive. But then again, it is a non-linear device, whose resistance increases with temperature, so likely the total luminous power is less than 2x of the nominal.

Why don't you get an autotransformer? This way, you'll be able to output the 110V it is made to handle. Or you can buy one of those bulky 220-to-110VAC fixed transformer. It may cost more than the bulb, but in the long run it may be worth it.

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  • \$\begingroup\$ Apart from an autotransformer,one can get the right power by letting just a part of each half-wave through using a simple circuit, perhaps based on a triac. There's even a COTS device for that called a "dimmer" ;-) That's going to be vastly cheaper and less bulky than a transformer. \$\endgroup\$
    – TooTea
    Jan 16, 2022 at 19:09
  • \$\begingroup\$ @anrieff thanks for answering my question. If we ignore my RMS power miscalculation and do what TooTea said, now power is the same but peak value of voltage and current is still doubled. Will it affect the lifetime? \$\endgroup\$
    – M Nasr
    Jan 16, 2022 at 19:27
  • \$\begingroup\$ The power is not quadrupled. See my answer. An autotransformer is a good recommendation. A dimmer would be ok, but could fail or be misadjusted in a way that would kill the bulb. \$\endgroup\$
    – user80875
    Jan 16, 2022 at 19:41
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The best-case is to use a dimmer that starts from 0 and ramps up to 90 deg max.

Transient failure is very likely

  • Using the model of 100 W bulbs for 120 V, 60 Hz and 240 V, 50 Hz with the worst-case turn-on Vac at 90º , the 240V bulb will draw 4 kW peak and the 120V bulb will draw about 16 kW peak. Typically a random phase start current is 10x but the worse-case is starting at peak voltage and cold resistance.
  • Best-case is 0 deg start the proper-rated bulb voltage the peak power = 6.8 x rated power. These are significant factors that shorten lifespan and your experience of seeing a bulb burnout at turn-on, more often, may validate these factors for you.

Steady-State will shorten lifespan and be hotter.

  • the 120 V bulb with a diode will be 4821 ºK , 173 W or 1.73 x rated
  • the 240 V bulb will be the normal warm 2400 ºK, 100 W.

The better solution is to use a dimmer that always starts at 0 Deg phase after a power interruption and ramps up to 90º limited by Rpot.

updated Simulation proof Added slider for resets 0 to 90 º phase.

Although bulb failure is a combination of heat, age and peak force=current, the 240 V Triac will produce the 6kW peak worst-case if not started from a lower power than 90 º.

enter image description here

DMM's use peak and scale to RMS badly on dimmers unless true RMS type.

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  • \$\begingroup\$ If the OP is going to use a dimmer then choosing one with soft-start may be an option, also trailing edge rather than leading edge may be kinder to filament lamps, although the OP doesn’t state the type. \$\endgroup\$
    – Frog
    Jan 18, 2022 at 19:40
  • \$\begingroup\$ I agree @Frog good advice but then inductive spikes might occur if it is a ballast type \$\endgroup\$ Jan 18, 2022 at 20:25
  • \$\begingroup\$ Tungsten Filament melts at 3695K. Regular lamps operate at 2300K-2800K, Photoflood lamps operate at 3200K-3400K range, lasting just few hours. Simulation results above melting point, as stated 4821K, shows the model of the component (lamp) is not adequate to be extrapolated, at least. \$\endgroup\$
    – EJE
    Jan 19, 2022 at 17:34
  • \$\begingroup\$ Good point @EJE I am surprised the Op didn't exclaim the obvious change in illumination or even the type of lamp. Hmmm \$\endgroup\$ Jan 19, 2022 at 21:42
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Although using an autotransformer is probably the best answer, you can also use:

  1. series inductor. Perhaps 1 (or even more) "ballasts" from old fluorescent lamps. Inductor must handle the maximum current the filament will draw.

Example calculation (remember, inductive reactance does NOT add to resistance of the bulb; it adds in square-root-of-sum-of-squares): Assume bulb is 100W; RMS current at rated/correct operation is therefore (100W)/120V = 0.83A. Effective resistance of bulb will be 120V/0.83A = 144Ω. Then X_L = 221Ω and L should be (at 60 Hz) = 0.587 H. Here I have used only Ohm's laws, the formula for inductive reactance, and the impedance formula Z^2 = R^2 + X^2.

  1. Series capacitor ( ! ). It can be done. The correct value (at 60 Hz) would be 12.0 µF (and rated at least 220V). Since it will have to pass 0.83A continuously, it will be large, expensive, and will probably fail soon. Okay, scratch that idea!

  2. 2nd identical bulb in series. Let there be light!

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If the bulb costs $200, it could be worthwhile to try to make it last.

As the others have said, the diode solution is problematic, especially at turn-on.

You could use a triac dimmer with a soft-start.

Or you could use a 110V DC switching power supply with soft-start. For example Meanwell ELG-200-C1750A. This is also dimmable, if necessary.

However, there is an extremely low tech and cheap solution: just put a "resistor" in series, for example another 100W 110V incandescent lightbulb. This will not solve the soft-start problem though, and it doubles the power consumption.

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What definitely will work is putting two of your light bulbs in series. I'm guessing you have spares on-hand since you took a chance with one. My earlier comment suggesting a dimmer switch could be hard to implement unless you have an RMS reading voltmeter as @Spehro Pefhany suggested, or (if you know how to do the calculations) an oscilloscope. Dimmer switches are cheap, though, and available in retail stores everywhere.

As for what happens next: it's up to you. If you found one that answered your question, please accept it. If not, you can ask for more information about anything that seemed to be on the right track but not quite what you need. I think the answers pretty well cover all the possibilities, but the effort required is more than you can expend.

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