I'm starting to work with simple circuits (learning how to measure electromagnetic variables). I have been looking up on the internet for a formula or deduction of a formula that relates the current $I$ and the voltage $V$ in a filament lamp, but haven't succeeded.

I did the experiment of measuring these variables and got something similar to a natural logarithm, but of course there should be some other constants in the equation $I(V)$. Do you know where can I find such an equation?


2 Answers 2


Resistive Filament - Voltage vs. Current

If you referring to measuring the $V(I)$ dependence of the heated filament I would derive an approximate equation as follows.

$T: \textrm{Temperature of the filament}$

$P_{dis} \equiv V \cdot I: \textrm{Power dissipated}$

$k: \textrm{Coefficient of temperature vs power}$

$R \equiv V/I : \textrm{Ohmic Resistance}$

$\alpha: \textrm{temperature coefficient of resistance}$


$$T = k P_{dis} = k I^2R$$

Now using a linear approximation for the resistance over temperature we have:

$$R = R_o[1+\alpha(T - T_o)] = R_o[1+\alpha k I^2 R - \alpha T_o]$$

Rearranging terms we have:

$$R[1 - R_o \alpha k I^2] = R_o[1 -\alpha T_o]$$

By definition we can substitute $R=V/I$ to get

$$V(I) = I \cdot R_o \cdot \left( \frac{ 1-\alpha T_o}{1-R_o \alpha k I^2} \right)$$

I have not tested this equation though. Note that you have one parameter that you can measure directly ($R_o$), one parameter you can look up ($\alpha$) and one that is unknown ($k$) because it depends on many factors including the shape of the wire. And I doubt that the equation will be accurate over a wide temperature range.

Thermionic Emmission of Electrons

If you are looking for the emission of electrons from a heated filament then I think you need the Richardson or Sommerfeld equations (http://en.wikipedia.org/wiki/Thermionic_emission) which are discussed more in depth here (http://www.tubebooks.org/Books/chaffee.pdf) on page 59. I don't think this is what you mean though.

  • $\begingroup$ I will try to fit this model to my data tomorrow, and will tell you how it goes $\endgroup$ Feb 21, 2015 at 5:16

If you go to the Oshino lamp catalog for tungsten filament lamps you find:

((new voltage)/(rated voltage))^0.55×rated current=new current

which means current is proportional to the voltage raised to the 0.55 power. Or You could say that voltage is proportional to the Current raised to the 1.81 power. The powers are approximate and depend on whether the lamp is vacuum filled or gas filled and the temperature of the filament. You can find this equation published in the article, "Mathematical and physical bases for incandescent lamp exponents" by David D. Van Horn, in Illuminating Engineering.

I took some data for a lamp off the web and plotted it in excel below.

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