A plain 115VAC light bulb into 230VAC...

[Externet]

Will an incandescent light bulb rated for 115VAC work normally if plugged into 230VAC with a diode in series, so only half cycle powers it ?

 

[crutschow]

In theory, yes. But putting a half-wave signal into an incandescent bulb will likely cause noticeable flickering since the flicker frequency is now 60Hz rather than 120Hz. It also makes the power transformer unhappy since you are putting DC through it. Why do you want to do that when you should have 115VAC available in the U.S.?

 

[ronv]

I did that with a hard to change porch light. Made it last forever, but it was a bit yellow.

 

[dougy83]

In theory, yes.

I don't agree. The diode halves the duty cycle, but the voltage is still double the rated voltage of the lamp. If the filament resistance was constant, the power would be doubled. Because the resistance increases with temperature/power, the power to the lamp will be less than double, but more than its rated power output. It may not last long.

I did that with a hard to change porch light. Made it last forever, but it was a bit yellow.

That was for a 110V lamp on a 110V circuit though, wasn't it?

 

[ronv]

Yes it was still 110 - 110.

 

[audioguru]

It takes many AC cycles for an incandescent light bulb to reach its operating temperature so I think a series diode reduces its power to half pretty well. Then doubling the AC voltage causes a normal amount of power.

If the doubled voltage, diode and light bulb is powered at exactly the peak voltage then the huge cold current surge might physically break a worn out filament.

 

[crutschow]

I don't agree. The diode halves the duty cycle, but the voltage is still double the rated voltage of the lamp. If the filament resistance was constant, the power would be doubled. Because the resistance increases with temperature/power, the power to the lamp will be less than double, but more than its rated power output. It may not last long.

You are correct. My thinking cap was on crooked. For a given resistance the power would be 4 times for a full-wave signal and 2 times for a half-wave signal or the equivalent of applying about 162Vrms to the bulb for a diode in series with 230Vrms. So the bulb would likely burn out in short order, as you noted.

One way to operate the bulb at 230Vac would be to use a 230V light dimmer switch and adjust it to output no more than 115V using a True RMS meter.

 

[MrAl]

Hi,

Yes, looking at the bulb as a resistor the resistance of a 115vac, 115 watt bulb would be 115 ohms. At 230vac it would consume 4 times that power if it could survive which would be 460 watts. Half that is 230 watts, still too much. We actually need 1/4 of that not 1/2.

Solving for the RMS voltage with time, then solving for the time to produce 115 watts, we find that it would take a firing angle of 90 degrees but for only one half cycle (as with an SCR). With a triac firing every half cycle we'd need to fire at 113.827 degrees and 180+113.827 degrees. That would give us two partial sines of width about 66.173 degrees each.
The total average power with either of these would be 115 watts at 230vac.

The peak of this second scheme would be a max of about 297.5v so it would be much higher than the normal peak of 170v for a 120vac line, but the peak would not last as long.
The peak of the 90 degree scheme would be just as high as the line voltage peak which is 230*1.4142=325 volts, pretty high for a 170v max bulb, and the peak would last just as long as the normal operating peak.

The average heating power would be the same now but I've never tried this myself. If the filament mass is not enough to properly integrate the voltage peaks, it could see a temporary but repetitive rise in temperature that exceeds it's normal rating and thus limit the life to less than normal. Average power heating is one thing, but instantaneous power could cause instantaneous rises in temperature which although average out to the normal working temperature could be far above that for short time periods.

There was a bulb model around on the web a few years back. The time constant of the filament would be the important thing.

We see unusual side effects with LED's too when they are driven with pulses instead of a constant level. Sometimes the color changes a little bit simply because the peak current is not at the normal level.

 

[MikeMl]

How about:

 

[ronsimpson]

Some people put two identical 110 bulbs in series.
"how to use 110 bulbs at 220 volts"

Mike, good use of spice!

 

[crutschow]

Using a capacitor as shown in Mike's last simulation has the possible problem of generating a very large peak current if the switch is closed when the waveform happens to be at it's peak voltage. That would put double the peak starting current (and four times the power) through the bulb when cold as compared to starting on its rated voltage. It may not blow the filament but it would certainly be an added stress.

 

[MrAl]

Hi,

Yes there are some drawbacks to using a cap.

First, it takes a relatively large, quality cap to get the required power at 240vac.
Second, the line current is around 1.1 amps (apparent power is over 240 watts) while the power in the 'load' is only about 91 watts. A 91 watt resistive load would be around 633 ohms at 240vac (rms) and draw a current of around 0.38 amps. So the power factor is pretty low.
On the other hand the triac solution generates spikes on the line which are not super simple to filter out.

We could look at an inductor solution too (although it would take a 0.7 Henry inductor to get 91 watts ha ha), and also an LC solution to maybe keep the power factor high.

There is also the "chopped" solution which presents a PWM current to the load. The input is chopped up with PWM to reduce the total power that gets to the load. The higher frequency PWM makes it easier to filter.

I like the two bulb idea best because it doesnt involve any additional circuitry, as long as the application can accept two bulbs in a place where there is normally just one. To keep the power down, we could use bulbs at half the power of the original and that would keep the power and the lighting about equal.