capacitorcurrentcurrent measurementdischargetime constant
I observed the discharge of a capacitor on an energy harvester chip via an oscilloscope. The Output was used to power a BLE module.
I know the capacitor value, discharge time, and the top and bottom voltage. I don't think I know my load resistor because it is a bit complicated to measure.
Is there a way to measure the total discharged current?
The circuit you have included is essentially a high voltage source.
The switching frequency and capacitor impedance are not relevant for the discharge cycle. The oscillator will boost the voltage and charge the capacitor by virtue of the diode rectifier until it is charged to approximately 1.1kV. From that point on, the switching frequency of your oscillator is not relevant.
The discharge of the capacitor will take on the profile of Vcap=Vo*e^(-t/RC) The complex variable in this is the unknown discharge impedance R. You are using a gas discharge tube to discharge into the output multiplier transformer in a similar manner Tazer guns utilise spark gaps. The problem is that when the Xenon vapour/air gap ionises, its impedance is extremely dynamic and is only ionised as long as there is sufficient ionisation energy stored within the capacitor. The main limiting factor for the apparent lack of power is the ionisation characteristics of the Gas tube as it will only be conductive for a short amount of time and not deliver the full charge of the capacitor.
Also be aware, that a spark, whether in an air gap or within a ionisable vapour, is NOT a short circuit. It dissipated energy in the same manner as a resistor. All that light that comes out of the tube is energy which is NOT delivered to your prongs.
When a camera flash is fired, the storage capacitor is never fully depleted. There is always a residual hazardous voltage on the capacitor terminals because the tube vapor has a minimum ionisation energy.
It will discharge when the +Vcc is switched off. The left plate is then connected to the ground via the power supply. The RST input then becomes -Vcc with respect to the ground and discharges via the resistor. +Vcc present:
+Vcc not present:
When the capacitor is charged the left plate is +5V with respect to the right plate. Since the +5Vcc power supply is connected to the left plate of the capacitor with the opposite polarity, the RST pin is +5v-(+5V) = 0V. However when you cut the power supply the left plate is connected to the GND (0V). RST then becomes 0V - (+5V) = -5V.
Best Answer
If you know the capacitance C (assuming it is constant) then the slope of the curve is proportional to the discharge current.
i(t) = C \$dv\over dt\$