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'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.
This is a self powered circuit so I need to store the energy found in this waveform to be used later. This waveform is generated from an electromagnetic coil. I like the analog approach so I hope you could give some more details.
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 can think of a CRYSTAL RADIO, are you trying to set up a series of capacitors and amplifiers, or draw an inductive load? I think Tesla and perhaps Faraday research may be in order?
google Faraday's law of induction, and research Teslas work on wireless electricity. these men were brilliant/
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.
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.
I used a bridge circuit to rectify the waveform and I stored the energy in a capacitor but I don't know if I stored all the energy in the waveform or not?
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
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
As you can see that the first pulse in the waveform will charge the capacitor to a certain voltage.
Now 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.
So how could I benefit from the secon pulse in charging the cap?
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.
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?
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?