# How to calculate the effect of different voltages on an incandescent bulb?

How can you calculate the effects of running a 120 V, 7 W lamp on, say, 80 V, or any other voltage?

What will the current draw be when you reduce the voltage? Is it possible to calculate knowing the above, or do I have to measure?

• Well, I started to write out all the parameters I could imagine and decided to give up. Yeah, you have to make some measurements. And you will also have to measure for not one but many of the same bulb type and manufacturer and date of manufacture. And that will only work in the exact same circumstances (emissivity and temperature and shape factors of surrounding environment, etc.) If this were easy, the lifespan of electron beam IC mask lithography emitters would have been a lot less difficult to manage. (Something I got paid to help improve.) Nov 5, 2023 at 10:11
• Whoa, slow down there @periblepsis -- do they just want to know current and intensity, or every detail of the tungsten surface? Nov 5, 2023 at 11:31
• @TimWilliams I see a can of worms. That's all. Nov 5, 2023 at 11:45
• Looking at one of periblepsis woms :-). This MAY NOT be useful. I'm making up the method s I go based on assumptions that pop up as I type :-). Radiation occurs as the 4th power of the temperature difference. I assume that the filament & enevlope are two points at either end of a temperature gradient. The same power is handled by both. The envelope to ambient is another with the same power BUT more (probably) convection. You can probably find data on normal filament temperatures. For the two power levels you can probably produce an equation that balances the two paths. The envelope ... Nov 5, 2023 at 12:12
• ... temperature will shift far less than the filament temperature for reasonable part-power levels. So you get a rough set of equations that allow the lower filament temperature to be approximated. So the resistance. And then you probably have an iterative process. And then ... agh - no, I'd measure it at various voltages for several wattages of bulbs and types of bulbs and see if some obvious empirical rules because obvious. It seems very likely that this will be available on web already. Nov 5, 2023 at 12:15

If it's an incandescent bulb, then the following wikipedia page allows you to estimate how the parameters will change.

From that page, rated current goes as $$\voltage^{0.55}\$$. Note that it doesn't vary linearly, as the resistance increases with filament temperature.

As lamps vary in construction and materials, you'll get a more accurate answer from measurement.

These power laws are only estimates, and they are better in the region of the lamp rating, than far away from it.

• Nice reference. From that page this is useful. Note that that they say Power is proportional to V^1.6 BUT they also say " ... The relationships above are valid for only a few percent change of voltage around standard rated conditions, ...". Nov 5, 2023 at 12:22
• @RussellMcMahon - Note that the current exponent and the power exponent should differ by exactly 1. They are both approximations, and apparently two people came to slightly different conclusions. See my answer. Nov 6, 2023 at 14:32
• @Mattman944 Without vast thinking (rushing): Sounds right. Rear of brain niggles. Changing temperature of filament affects resistance (of course). Power loss through enevelope is 4th power of delta t related, and so temperature and so resistance vary with wattage with differing laws. Then there is envelope to ambient, which is less variable. (Convection also matters in m=both cases, almost certainly to a lesser extent). There are so many mechanisms with different exponents and laws that I suspect that "it's complex" :-). Maybe not. Nov 7, 2023 at 1:05
• @RussellMcMahon Lots of mechanisms, so measure it, and fit it. The main takeaways are that the current exponent is a long way from linear, and the lifetime exponent is huge. Nov 7, 2023 at 6:22
• @Neil_UK Yes. Even a few percent works woners. Lonnnnng ago people used to make startup surge reducers. Memory said that just those made a very great difference. || Meaurement is what I'd usually do :-) . Nov 7, 2023 at 8:06

Measured data from a 4 W bulb. The data fits the V1.6 curve remarkably well!

Constant K determined by trial and error.

Feel free to analyze this data further and post results yourself.

The resistance at zero volts is from an ohmmeter.

Edit1: I also measured data for a 60 W bulb. Results are almost as good.

Edit2: I plotted the current equation referenced in the Wikipedia article. I0.55. For the 4 W bulb, the data doesn't fit as well. This puzzled me until I realized that both exponents are approximations, and two people came to slightly different conclusions, according to math, the exponents should differ by exactly 1.

$$\ P_a = {\left(\frac{V_a}{V_d}\right)}^{1.6} P_d \$$

$$\ I_a = {\left(\frac{V_a}{V_d}\right)}^{0.55} I_d \$$

$$\ P_a = V_a I_a = V_a {\left(\frac{V_a}{V_d}\right)}^{0.55} I_d \$$

$$\ I_d = \frac{P_d}{V_d} \$$

$$\ P_a = {V_a}^{1}{V_a}^{0.55}{V_d}^{-0.55}{V_d}^{-1}{Pd} \$$

The equations for the 4 W bulb fit best when the exponents are 1.6 and 0.6. The equations for the 60 W bulb fit best when the exponents are 1.5 and 0.5.

Finally, note that you don't need to use trial and error for the constants, they are easily calculated (see final image).

The best fit from theory and experiment is:

$$P_2=P_1\cdot\left(\frac{V_2}{V_1}\right)^{1.54}$$ and $$I_2=I_1\cdot\left(\frac{V_2}{V_1}\right)^{0.54}$$

These equations hold for vacuum incandescent light bulbs made with tungsten filaments, and are typically true for bulbs that are designed to be operated at 25 W or less. Higher wattage bulbs have an argon nitrogen gas fill. The equations for these bulbs will be slightly different due to the heat convected by the gas. See Kykta: Incandescent lamp design and lifetime. I am the author of the article cited and linked at the AIP (American Institute of Physics).

Sample calculation:

$$P_2=7\cdot\left(\frac{80}{120}\right)^{1.54} = 3.75\ \mathrm{W}$$

$$I_1=\frac{P_1}{V_1}=\frac{7}{120} = 0.058\ \mathrm{A}$$

$$I_2=0.058\cdot\left(\frac{80}{120}\right)^{0.54} = 0.046\ \mathrm{A}$$