How does a capacitor affect the flow of the current?

Question

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:-

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?

Answer 1

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.