Why does an Ekasi oscillator oscillate?

From my research, I learned that when this is first plugged in, the capacitor is charging, so the LED doesn't shine. When the capacitor reaches 12v, then the current switches to the top loop (which is possible because the voltage of the battery equals to the breakdown voltage of the flipped transistor). Note, because we are using its breakdown voltage, the base doesn't need to be connected.

As for its oscillation, I read that when the current goes through the top loop (and the LED is shining), the capacitor begins to discharge. So, because the voltage across the capacitor goes down, the current stops flowing to the top loop and goes back to the capacitor to recharge it (so the LED stops shining). This repeats, creating an oscillation/blinking LED light.

My question is: why would the capacitor discharge after it reaches 12V? What causes it to begin to discharge? Is it because when it's fully charged, no current is flowing towards it, "preventing" it from discharging? So it's free to begin discharging, but as soon as it does, then it begins to slowly recharge once again, creating this repetition? Sorry if it's a bad question folks, I just recently got into electricity! :)

Here's the schematic:


simulate this circuit – Schematic created using CircuitLab

Edit: Added resistor to circuit.

  • \$\begingroup\$ What is the internal resistance of the battery? \$\endgroup\$ – analogsystemsrf Oct 15 '18 at 3:44
  • \$\begingroup\$ Your circuit does not work, the capacitor is simply across the battery and does nothing. Read this for the correct circuit: cappels.org/dproj/simplest_LED_flasher/… \$\endgroup\$ – Jack Creasey Oct 15 '18 at 5:13
  • \$\begingroup\$ If you'd like a detailed explanation, please fix the circuit diagram and explain a little about your current level of knowledge. \$\endgroup\$ – pjc50 Oct 15 '18 at 13:21

First, the circuit as you present it is a bit too simple -- the resistor in series with the battery as shown on this page is necessary, and must be about the right value.

Second, the reason the capacitor will discharge (in the correct circuit) is because the LED-transistor string is consuming more current than the battery (through that resistor you left out) is supplying -- so the capacitor discharges.

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  • \$\begingroup\$ Good catch, I forgot to add the resistor! However, I'm still confused why the capacitor would discharge. If you don't mind, can you elaborate on what happens conceptually? \$\endgroup\$ – F16Falcon Oct 16 '18 at 1:01
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    \$\begingroup\$ Conceptually, the transistor turns on and stays on as long as the current is above some low threshold. That lights the LED and discharges the cap, until the current falls below the low threshold. Then the transistor turns off and stays off until the voltage reaches some high threshold. The cap charges, the high voltage threshold is reached, the transistor turns on again, and the cycle repeats. \$\endgroup\$ – TimWescott Oct 16 '18 at 17:25

If you just got into electronics, skip this example and learn the fundamentals.

This 2 pin reverse-biased part is being over-stressed to create an abrupt discharge switch to cap while allowing charging via the series R to current limit the Cap charge rate Ic=CdV/dt

This mode puts the transistor into DIAC like mode but only a poor version . It is a Vthreshold triggered negative incremental resistance until the hold current threshold reverts back to high impedance.

In other words we call it a simple Relaxation Oscillator which is far better done with a simple Schmitt Logic Inverter. With R feedback and C to gnd.

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    \$\begingroup\$ "which is far better done with a simple Schmitt Logic Inverter". Oh, perish the thought. It's far better done with a cap, a resistor or two, and a neon bulb. At voltages that'll make you sit up and pay attention! \$\endgroup\$ – TimWescott Oct 15 '18 at 23:00
  • \$\begingroup\$ The hysteresis inverter is also useful for triangle + sine + square + variable f over 3 decades which I have answered here before. Add a reed relay coil to the square wave and now you have a high voltage buzzer. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 16 '18 at 0:22

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