# How does a magnetic ballast (large inductor) stabilize a negative resistive circuit (such as a fluorescent light)?

When reading about fluorescent lights, I noted many sources stating that a large inductor (the magnetic ballast) in series with a negative resistor (the light, in this case) will stabilize it by blocking large AC currents. (Sources that say this include: [Reactive_ballasts] or [Fluorescent lights])

However, the impedance of a negative resistor in series with an inductor is $$\frac1{j\omega L+R},$$ where R is negative, leading to an unstable pole. What am I missing? Are the sources wrong, or am I misreading them? If so, how does a magnetic ballast actually stabilize the light?

I don't think that the fluorescent light is actually a negative resistor. If something actually has negative resistance, putting a positive voltage across it causes negative current flow, in other words, (conventional) current flows from the lower voltage to the higher voltage. A fluorescent light can't just do that by itself. Negative resistors are powered circuits that, when taken as a whole, appear to have negative resistance. A fluorescent light just has a resistance that drops as the amount of current through it increases. Its resistance is always positive. The purpose of the ballast is to resist the current briefly--until the next AC phase change--in order to limit the current flowing through the light.

This page has a schematic for a negative resistor. The more voltage you apply to the input, the more current flows from the negative resistor to the input. What's actually happening is that the op amp is producing a voltage at node B that exceeds the input voltage, causing current to flow to the input.

Maybe the sources didn't exactly mean "negative resistance" per se. It is a fact that, in certain fluorescent lamps, lamp current decreases with increase in applied voltage. CFLs for example. This way you may say dV/dI is negative. Also, a lamp does not offer constant resistance. Resistance is a function of voltage applied.

It isn't really stabilizing in the sense of a linear feedback network. Rather, the inductor is simply limiting the current to some acceptable value.

The lamp is a non-linear hysteretic load. When there is no arc, the impedance is very high. Once the arc forms, the impedance drops to a very low value. So yes, over a certain range of operation increasing the current decreases the voltage, but I don't think calling it a 'negative resistance' load is helpful here -- that is a linear response / small signal analysis terminology, but we actually care about large signal behavior where a linear model is not appropriate.

The ballast is simply a current limiter. It limits the average current to approximately Vac/(w*L) when the arc is established. Without it, it would cause a destructively high current. In a traditional florescent bulb it is also part of the starter circuit to generate a high voltage pulse to ignite the arc.

Actually the resistance of the tube drops as it warms up. Kind of like a semiconductor in thermal run away. As the tube warms up mercury vaporizes and lowers the resistance and with any ionized plasma conductivity goes up with temperature. In an AC circuit an inductor stores energy during a portion of the cycle and releases it as the input voltage drops and changes polarity. This serves to limit current through the load allowing temperatures to stabilize and preventing thermal runaway.