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I can't wrap my head around how the capacitor gets discharged in UJT relaxation oscillator. I understand when \$ V_E \$ reaches or exceeds a threshold (say \$ V_{ON} \$), current flows from emitter to \$ B_1 \$. But \$ V_S \$ supply the circuit with a constant DC voltage and eventually after an \$ \tau = RC \$, the capacitor locks in a charged up state with voltage of \$ V_{ON} \$, as \$ V_E \$ also holds onto a constant \$ V_{ON} \$ passing current from emitter to base.

The capacitor has no where to discharge the current unless \$ V_{S} \$ is removed and only then can the capacitor supply voltage to the UJT instead of \$ V_{S} \$.

I've read a few tutorials which all sum up, without much explanation, that once the capacitor charges up to a threshold voltage, it discharges through the UJT. And once the capacitor gets fully discharged it gets charged up again and the cycle goes on producing a pulsed waveform out of the base. I do not see how the capacitor can discharge unless \$ V_{S} \$ is taken out.

UJT oscillator circuit diagram

KMC
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    Try reading these: [UJT](https://electronics.stackexchange.com/a/358420/38098) and [PUJT](https://electronics.stackexchange.com/a/429550/38098). Then tell me what else is left over that wasn't covered and you still don't follow. – jonk Jun 24 '20 at 04:38

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The key point that you're missing is that R1 has a much lower value than R3. Before the UJT conducts, all of the current through R3 goes into charging C1. But once the UJT conducts, R3 and R1 form a voltage divider that produces a much lower voltage. This is why C1 discharges through the UJT and R1. When C1 reaches that voltage, then the UJT stops conducting and the cycle starts over again.

Dave Tweed
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  • It's way more technical than that. Details matter ***a lot***. There are so many ways to screw this up. And only a few ways to get it right. – jonk Jun 24 '20 at 04:59