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In this figure, the battery charges the capacitor up to 70V. After reaching 70V, the neon lamp turns on and basically becomes short and the capacitor is discharged through the neon lamp. After discharging, the capacitor then charges again, repeating the cycle.

Why does the capacitor discharge at 70V if the switch is still closed? Shouldn't the battery continue charging the capacitor until 110V and, in turn, turn on the neon lamp after reaching 70V?

enter image description here

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To understand this, you need to know the neon lamp impedance.

If the load is purely capacitive, then when the switch closes it charges both capacitors at the same time. Voltage on both capacitors is the same. The neon lamp cannot be purely capacitive of course.

If the load is purely resistive, then the current starts flowing through the load as soon as switch closes. Voltage on the load will increase at a finite rate due to the capacitor, but when it reaches steady state it will remain at that level. The steady state voltage level will depend on the voltage divider determined by R1, R2, RLOAD, and the voltage increase rate will depend on R1, R2, RLOAD and C.

This is all simple with linear elements such as resistors and capacitors (and inductors). But now consider a load that blocks current below some voltage level, and conducts current above the voltage level. Diode shows behavior similar to this. In this case, voltage increases on the capacitor at finite rate determined by R1, R2, and C. If it was not for the load, the voltage on the capacitor would reach 110 V and the current would stop flowing at that moment. But when the load "conduction voltage" is reached, the load is "suddenly" turned on. What happens now largely depends on how load behaves.

The voltage level which triggers neon lamp glowing is called a breakdown voltage. It is higher than (normal) operating voltage. When breakdown occurs, some energy from the capacitor is transferred to the neon lamp at the rate determined by resistance between capacitor and neon lamp. In this case, wire resistance which is really small and this is why energy is transferred really fast, hence almost instant voltage drop on the capacitor.

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  • \$\begingroup\$ Oh, so capacitors cannot charge and discharge at the same time? Once the capacitor reaches 70V, why would the battery not supply the needed load voltage of 70V 'continuously'? Why does the load prevent the capacitor from going over the 70V breakdown voltage? I am just filling the gaps here. \$\endgroup\$
    – Flash
    Oct 24, 2021 at 9:03
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    \$\begingroup\$ What do you mean by "charge and discharge at the same time"? Current direction determines if capacitor charges or discharges - current that flows to the capacitor will charge the capacitor (voltage increases), and vice-versa. Once everything settles (steady-state), the battery will supply needed current and capacitor current will be zero. Resistances (and inductances) determine how fast can energy (or voltage if you want to simplify) be transferred from one place to the other, which is why in transients capacitor will supply some of the energy. This is why we use capacitors in the first place! \$\endgroup\$ Oct 24, 2021 at 10:33
  • \$\begingroup\$ @PatrickGabuna you see those resistors in between the battery and the lamp? Those are what stop the lamp getting all the current it "wants". If the resistance is too low then the battery will simply power the lamp and there will be no flashing, as Spehro Pefhany pointed out. \$\endgroup\$
    – user253751
    Oct 25, 2021 at 8:57
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You said the lamp becomes a short circuit which is the key part.

The lamp being a short circuit will have much lower resistance discharging the capacitor than the resistance charging the capacitor, so the capacitor will get discharged quickly.

The lamp will turn off when voltage has dropped to low enough to not sustain the arc, and then the capacitor voltage will slowly start to rise via the charging resistor.

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If the resistor R2+R1 is too low (or too high) then the lamp will remain 'on' (or barely conducting) and it will not flash.

enter image description here

The R2 load line illustrates the necessary condition for oscillation- the load line must intersect the negative resistance region of the discharge tube.

Image and more information in the 1966 GE Glow Lamp Manual

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Well, the charging voltage of the capacitor is given by:

$$\text{V}_\text{C}\left(t\right)=\hat{\text{u}}_\text{i}\left(1-\exp\left(-\frac{t}{\text{RC}}\right)\right)\tag1$$

So, when we can solve for \$t\$:

$$t=-\text{RC}\ln\left(1-\frac{1}{\hat{\text{u}}_\text{i}}\cdot\text{V}_\text{C}\left(t\right)\right)\tag2$$

So, using your values we get:

$$t=-\left(\text{R}_1+\text{R}_2\right)\ln\left(1-\frac{1}{110}\cdot70\right)=\left(\text{R}_1+\text{R}_2\right)\ln\left(\frac{11}{4}\right)\approx\text{R}_1+\text{R}_2\tag3$$

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