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I'm curious as to why?
An analogue approach would be to use op-amps to precision-rectify the signal and integrate it, using the capacitor as part of the integrator.
Or the signal could be applied to an A/D converter in a microcontroller programmed to integrate the converter output, the integration result being applied via a D/A converter to the capacitor.
shot in the dark here........ diode?
I'm curious as to why?
An analogue approach would be to use op-amps to precision-rectify the signal and integrate it, using the capacitor as part of the integrator.
Or the signal could be applied to an A/D converter in a microcontroller programmed to integrate the converter output, the integration result being applied via a D/A converter to the capacitor.
I thought perhaps you were simply trying to measure the energy. So scrub the analogue and digital approaches. In the absence of any other power source the most efficient way of storing for re-use would, IMO, be as Crosslakeguy suggested in post #4: use a diode.This is a self powered circuit so I need to store the energy found in this waveform to be used later
What you are trying to do is called energy harvesting. Gathering little bits of power parasitically from other circuits and operations,
How much energy do you need?
How much energy is there in the waveform you show? (You need to show the time scale to know how much power is there) How often does that pulse occur?
Only when you have answered those questions, can you know how efficient your harvesting circuit needs to be. Or if it is even practical.
I thought perhaps you were simply trying to measure the energy. So scrub the analogue and digital approaches. In the absence of any other power source the most efficient way of storing for re-use would, IMO, be as Crosslakeguy suggested in post #4: use a diode.
No, you didn't. Some energy was dissipated in the rectifier diodes.
Of course if the 1Ω resistor remains in-circuit energy is dissipated in the resistor, too, so there won't be much left to store
That's not necessarily true. Depending on the source resistance the first pulse may not charge the cap to a voltage above the second pulse.since the second pulse has a less voltage than the first pulse it will not charge the cap since the cap has already a higher voltage.
That's not necessarily true. Depending on the source resistance the first pulse may not charge the cap to a voltage above the second pulse.
The total time of the pulse shown is 13 msec. There is only one pulse I will benefir from. How can I calculate the energy in this waveform?
Do you recommend any harvesting energy circuit?
How could i know if this is happening or not. I think that the source impedance is 4 ohms. The capacitance is 880uF.
Hello,
You seem to be asking two different questions:
1. How to store or 'save' the energy in the wave to be used later.
2. How to 'calculate' the energy in the wave.
So which is it, do you really need to save the energy for some efficiency requirement or do you just need to calculate it?
Why do you want to do this in the first place, what is the benefit going to be?
Rerun your initial waveform tests with a 4R load resistor in place of the 1R, post your new waveform.
Then
Wind a simple 10:1 step up transformer using a piece of ferrite core, add a fullwave bridge across the secondary and charge the capacitor [880uF sounds much too high a value].
Measure the waveforms and voltages, post your results.
I have tried placing a 4R load resistor before and the waveform was as the previous one with the 5 volts increasing to 12.5 volts and the rest of the waveform proportionally increased.
Can I please know why you want to place a 10:1 step up transformer?