Question about Pearson-Anson Relaxation Circuit

[Infernorage]

As seen in the circuit diagram picture below, the Pearson-Anson relaxation circuit is often used to blink neon lamps. According to what I have read, the lamp does not permit a current until the capacitor reaches the voltage threshold, causing the gases in the lamp to ionize and a sudden current. Once the capacitor discharges enough, the voltage falls below the lower threshold, the current drops to zero, and the power source begins charging the capacitor again to repeat the process.

My question is, why doesn't the battery just turn on the lamp and keep it lit while charging the capacitor. The voltage of the capacitor will never exceed the battery so the battery must have enough voltage to turn the lamp on. Also, since the lamp is in parallel, isn't the voltage difference across the lamp the same as that of the capacitor? I don't see why the battery wouldn't just light the lamp right away. Can anyone clarify this for me? Thanks in advance.

 

[gnurf]

My question is, why doesn't the battery just turn on the lamp and keep it lit while charging the capacitor.

Because, once the voltage across the neon lamp increases beyond its threshold voltage and begins to conduct, the lamp acts like a short circuit which discharges the capacitor. Re-analyze the circuit treating the neon bulb as a short circuit when the lamp is on.

 

[Infernorage]

Because, once the voltage across the neon lamp increases beyond its threshold voltage and begins to conduct, the lamp acts like a short circuit which discharges the capacitor. Re-analyze the circuit treating the neon bulb as a short circuit when the lamp is on.

I understand why the capacitor discharges, but I suppose my problem is with what happens before the voltage threshold. The lamp is parallel to the capacitor, so I figured the voltage difference of the power source would also be across the lamp, but that would immediately light it. Why isn't the batteries voltage difference initially applied to the neon lamp?

 

[Bobbywhy]

Infernorage, It seems you’ve copied the schematic drawing from the wiki page http://en.wikipedia.org/wiki/Pearson–Anson_effect

If you have not done so yet, may I suggest you read the two paragraphs just below the diagram labeled “Charging” and “Discharging” where the operation of the circuit is discussed with elegant clarity.

If then you have some doubts or questions about the operation, please return here and post them. Members here are always willing to assist any true searcher for understanding of the mysteries of our natural world.

Cheers,
Bobbywhy

 

[gnurf]

I suppose my problem is with what happens before the voltage threshold. The lamp is parallel to the capacitor, so I figured the voltage difference of the power source would also be across the lamp, but that would immediately light it. Why isn't the batteries voltage difference initially applied to the neon lamp?

The lamp is basically an open circuit below its threshold, so that part of your problem is reduced to understanding the transient behavior of a simple RC circuit. You wouldn't expect a capacitor that is charged with a current limited (as in the resistor) voltage source to instantly reach the supply voltage, would you?

 

[vk6kro]

The lamp also has two thresholds.
Before it fires, it draws no current and produces no light.
Once it fires, (that is, when the voltage across it reaches the first threshold), it draws current and produces light as it gets current mostly from the capacitor and some from the resistor.
It then reaches the second threshold when the voltage across the lamp is not enough to maintain conduction. So the lamp switches off and the capacitor starts to charge up via the resistor.

These devices usually operate at a relatively high voltage of about 60 volts and this circuit produces a crude but large sawtooth waveform across the capacitor.

 

[AlephZero]

The value of the resistor is important for this to work. If R is too small, there will be a delay while C charges up to the threshold voltage, and then the lamp will be permanently on, as the OP suggested.

R needs to be big enough so the battery can't maintain the lamp voltage on its own, because of the voltage dropped across R.

C acts like a "second battery" to keep the lamp voltage high, but the voltage falls as C discharges, until the lamp goes off and the battery can then recharge C to repeat the process.

Incidentally, you can make a low voltage version of this using an Schmidt trigger IC instead of the neon lamp, and you can make a similar oscillator with the common "555" IC. Experimenting at 5V is much safer than messing around with 60V or 90V neon lamps!