# Qustion about Inductor + Neon Lamp Example

#### SiegeX

Joined Jul 22, 2004
20
In the Inductor Chapter in the DC Volume, there is an example with a 6V battery and an inductor with a neon lamp attached across the inductor. The example says the following

In this circuit, a lamp is connected across the terminals of an inductor. A switch is used to control current in the circuit, and power is supplied by a 6 volt battery. When the switch is closed, the inductor will briefly oppose the change in current from zero to some magnitude, but will drop only a small amount of voltage. It takes about 70 volts to ionize the neon gas inside a neon bulb like this, so the bulb cannot be lit on the 6 volts produced by the battery, or the low voltage momentarily dropped by the inductor when the switch is closed
The part I question is highlighted in bold. When the switch is initially open, the current in the circuit is zero, however when the switch is closed, there is a very sudden current impulse making di/dt a large number (depending on intrinsic resistance in the wire). To the best of my knowledge this large current impulse should cause the inductor to induce a large voltage in opposition of this current, acting as a LOAD.

It makes no sense to me why the example states that going from zero to positive amount of current in a very short time causes the inductor to "drop only a small amount of voltage" but going from some positive amount of current to zero in a short time all of a sudden creates this huge voltage (as it should).

It very well may be that the neon lamp is polarized, and thus going from zero->postive current creates a voltage in the wrong direction not igniting the neon gas, but that still doesnt mean there isnt a large voltage present.

So am I right? If not what I am failing to consider. Thanks

#### pebe

Joined Oct 11, 2004
626
Originally posted by SiegeX@Jan 11 2005, 09:12 PM
In the Inductor Chapter in the DC Volume, there is an example with a 6V battery and an inductor with a neon lamp attached across the inductor. The example says the following
The part I question is highlighted in bold. When the switch is initially open, the current in the circuit is zero, however when the switch is closed, there is a very sudden current impulse making di/dt a large number (depending on intrinsic resistance in the wire). To the best of my knowledge this large current impulse should cause the inductor to induce a large voltage in opposition of this current, acting as a LOAD.

It makes no sense to me why the example states that going from zero to positive amount of current in a very short time causes the inductor to "drop only a small amount of voltage" but going from some positive amount of current to zero in a short time all of a sudden creates this huge voltage (as it should).

It very well may be that the neon lamp is polarized, and thus going from zero->postive current creates a voltage in the wrong direction not igniting the neon gas, but that still doesnt mean there isnt a large voltage present.

So am I right? If not what I am failing to consider. Thanks
[post=4535]Quoted post[/post]​
The bold quote is correct about the current rise, but the bit about '...but will drop only a small amount of voltage..' makes no sense. At the instant when the switch is closed and 6V is applied to the inductor, the current in the coil is zero because the current attempting to rise in the inductor will generate a back-EMF that will oppose the rising current. The current will increase (relatively) slowly until a maximum is reached where there is no further increase in current because the resistance of the inductor has limited it and Ohm's law has taken over.

That steady state current will continue to flow while the switch is closed.

When the switch is opened, current will stop abruptly. That rapid decrease in current will generate a high voltage across the inductor and that could exceed the striking voltage of the neon. BTW, a neon lamp is a non-polarized device.

#### Martin P Dembrowski

Joined Dec 3, 2004
10
HI...

PEBE's description is very good. Also, the voltage rise across the inductor as well as the slowing and eventual (virtual) cutoff of current flow is hyperbolic (I believe) and so the rate of change at first of these two elements is very fast at first, gradually changing to a almost flat curve in time. Opening the circuit causes the rapid (almost instantaneous) collapse of the magnetic field in the coil to produce a voltage in the reverse direction to the originally opposing EMF when charging the coil, which turns out to now be in the same polarity as the battery voltage and gives you a voltage burst many times higher than the battery alone making the neon bulb flash, a technique used in many old television pulse circuits to provide the 10,000 or more volt potential at the anode of the picture tube.