Under controlled voltage (bench supply source):
Small 13x7x5 mm ferite ring: starts lighting at:
- without capacitor: - 0.85 V
- With capacitor 10 uF : 0,83 V
- Adding 47 uF to existing 10 uf, in parallel -> the led dimmed and started blinking!
With the capacitor, at the same value of voltage, the LED is brighter.
The source displayed current of 0 - 0,002 A, changing every second.
Certainly, the capacitor helps by lowering the supplied voltage to which the LED opens and to providing more current.
There is also a clear link between the capacitance, the inductance and the rise in the voltage to the LED pins.
Because the values of the parameters are close to the source and multimeter accuracy, I cant fine tune the voltage – in the range of 0.001 V.
The practical consequence of these findings is that there is a value of the capacitor that grants best efficiency of the circuit. But how to calculate?