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I understand that the capacitor is like a secondary DC source when it's almost fully charged. What I don't understand is the way it behaves in a circuit.

It discharges itself when it's full or when it is the only DC source in a closed circuit?

Consider this slow-speed oscillation example:-  Slow-Speed Oscillation

The Programmable Unijunction Transistor (PUT) and the capacitor are connected in parallel, thereby the voltage is the same for both of them. But I don't see why the PUT depends on the capacitor. Even if the capacitor acquired the entire current till it gets full, wouldn't the electrons go from the DC to the PUT directly?

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    \$\begingroup\$ Looks like the PUT will be off until the capacitor has charged through the 470K resistor. The size of the capacitor and the resistance of the resistor will set the time constant. Then the PUT will conduct and dump the capacitor through the LED until the voltage falls below the cutoff voltage. Then the cycle starts over. \$\endgroup\$ – zeta-band Jul 8 '19 at 18:42
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    \$\begingroup\$ Do read: circuitstoday.com/programmable-ujt section " to understand how a PUT works. The section relaxation oscillator describes your circuit \$\endgroup\$ – Huisman Jul 8 '19 at 19:02
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There are two things going on, which is no doubt part of what is confusing you.

The PUT stays off until some threshold voltage between it's anode and cathode is reached. Then it turns on and it stays on until the current through it drops below some threshold. So the PUT is a 2-state device.

A capacitor tries to hold its voltage, and the bigger the capacitor, the better it does. The rate of change of voltage on the capacitor is equal to the current into or out of it, divided by the capacitance.

So here's what happens in that circuit. I'll start with the PUT off (not conducting current) and the capacitor discharged.

  • The capacitor charges up, through the 470\$\mathrm{k}\Omega\$ resistor. No current flows through the PUT, because it's off. So, no current flows through the LED, either.
  • Because the current through the capacitor is small, its voltage grows, but slowly.
  • Eventually, the capacitor reaches the threshold voltage to turn on the PUT. It turns on. This creates essentially a short circuit from the capacitor to the LED*, and the LED emits light.
  • The PUT and LED in series discharge the capacitor. There's lots of current, so the capacitor voltage falls off quickly.
  • Because the voltage is falling, the current through the PUT falls. Eventually, the PUT turns off.
  • Now the PUT is off and the capacitor is discharged, and the process starts over again.

* Which is pretty hard on the LED -- it would be better to put a resistor in series with the poor thing. This circuit was probably originally designed for an incandescent lamp, which would have little trouble with being used this way. But I digress.

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  • \$\begingroup\$ So the capacitor is the only DC supplier to the PUT and LED, and it keeps pushing till it has enough power to surpass the threshold! \$\endgroup\$ – Kepler 186 Jul 8 '19 at 19:01
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    \$\begingroup\$ Until it has enough voltage to surpass the threshold, yes. \$\endgroup\$ – TimWescott Jul 8 '19 at 19:23
  • \$\begingroup\$ RIght! Just so I can make sense of the capacitor. The capacitor will always eat the entire current, right? Because it's in parallel? \$\endgroup\$ – Kepler 186 Jul 9 '19 at 8:44
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    \$\begingroup\$ If you're confused about parallel and series, yes. The cap is in series with the resistor, and in parallel to the PUT and LED. When the PUT is off, it may as well not be there -- this leaves the series connection of the cap and resistor as the only active thing, and yes, in the circumstance that the PUT is off the cap sees the same current as the resistor. \$\endgroup\$ – TimWescott Jul 9 '19 at 21:52

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