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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

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    \$\begingroup\$ Try reading these: UJT and PUJT. Then tell me what else is left over that wasn't covered and you still don't follow. \$\endgroup\$
    – jonk
    Commented Jun 24, 2020 at 4:38

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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.

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  • \$\begingroup\$ 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. \$\endgroup\$
    – jonk
    Commented Jun 24, 2020 at 4:59

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