3

I’m very new to assembling circuits. I don’t understand why the capacitor in an Esaki oscillator discharges, causing the LED to flash.

From my understanding of capacitors (I’m sure there is something I’m missing), the capacitor should charge up to V_EC+V_LED and stay charged at that since the source is supplying enough voltage to get across V_EC+V_LED, which in my case was around 10.06V. When I remove the capacitor, the LED will stay on, but when I leave the capacitor in the circuit the LED will flash. Why does this happen?

enter image description here

Please disregard the value of the source voltage on the image. My values are:

  • V_cc=12.80V
  • R1=440ohms
  • C=100uF
  • V_LED=1.83V
  • V_EC=8.23V

I am using a BC547 transistor.

I’ve been reading Colin Mitchell’s capacitor section in his Electronic Circuits book, and read about capacitors “jumping up and down." I don’t know if that has something to do with why this happens, and if so, I can’t figure out why.

Ethan Coyle
  • 33
  • 4

1 Answers1

7

This is a so-called relaxation oscillator which exploits the fact that a reverse-biased transistor exhibits negative differential resistance.

In simpler terms, the transistor starts to break down at a certain voltage, which means that it starts to pass current once the voltage across it gets high enough. Due to the negative differential resistance, the breakdown voltage drops as soon as current flows through it, which means that it'll only stop passing current once the voltage across it is much lower than its original breakdown voltage.

You can view it as a strange kind of Zener diode. To give you some numbers, the transistor might start passing current at 10V, and only stop again once the voltage is 9V or lower.

What happens in the circuit is this (using the example voltages I gave above): The capacitor slowly charges up to 10V; the transistor is off during that time, and the LED doesn't light. As soon as it hits 10V, the transistor goes into breakdown and passes a current through the LED, making it light up. It keeps passing current until the capacitor's voltage drops down to 9V (because it gets discharged through the LED), at which point it turns off again. Now the capacitor has to charge back up to 10V for the LED to light up again, which takes a while due to the limited charging current. This happens in an endless cycle as long as you supply a charging current to the capacitor, causing the LED to flash.

This also means that the LED will flash slower the larger the capacitor is, and faster as you supply more current (lower the value of the resistor, or increase the supply voltage).

Without the capacitor, the charge just slowly "trickles" through the LED and transistor as there's nowhere else for it to go, which means that there's no way for the LED to turn off.

Jonathan S.
  • 14,865
  • 28
  • 50
  • 1
    This makes perfect sense now! Is there a specific name for the lower limit voltage across the transistor when the current causes it to drop to that potential? – Ethan Coyle Apr 22 '23 at 18:02
  • @EthanCoyle I don't know if it has any particular name, unfortunately. – Jonathan S. Apr 22 '23 at 18:52
  • The datasheet for the BC547 and for most other little transistors lists its maximum allowed emitter-base voltage at 6V. Then a higher voltage causes it to have avalanche breakdown. Its emitter base is in series with the 0.7V base-collector in your circuit. – Audioguru Apr 22 '23 at 19:33