storing energy to a capacitor

[bazramit]

Hi Guys,

I want to store the energy stored in this waveform to a capacitor. in the attachment is the voltage waveform across a 1 ohm resistor.

thanks for your advice.

 

[alec_t]

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.

 

[bazramit]

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.

 

[crosslakeguy]

shot in the dark here........ diode?

 

[bazramit]

shot in the dark here........ diode?

what do you mean

 

[bazramit]

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't use an active rectifier since the opamps need there own power supply and this is not found since this is a self powered circuit

 

[ChrisP58]

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.

 

[crosslakeguy]

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/

 

[alec_t]

This is a self powered circuit so I need to store the energy found in this waveform to be used later

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.

 

[bazramit]

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.

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?

 

[bazramit]

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?

 

[alec_t]

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

 

[bazramit]

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?

 

[alec_t]

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.

 

[bazramit]

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.

How could i know if this is happening or not. I think that the source impedance is 4 ohms. The capacitance is 880uF.

 

[MrAl]

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?

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?

 

[ericgibbs]

How could i know if this is happening or not. I think that the source impedance is 4 ohms. The capacitance is 880uF.

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.

 

[bazramit]

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?

I want to save the energy but in order to be sure that i saved all the energy possible I need to calculate the energy available in the first place.

 

[bazramit]

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?

 

[ericgibbs]

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?

If you think the source impedance is 4R you should have a matched 4R load for maximum power transfer.

If the Vpeak is now 12.5V, then you have should have 11.5Vpk after a diode

For maximum energy capture you need to utilise the -12.5V swing, so a simple transformer would enable a FWBridge with isolation.

As the voltage is now much higher than the first test, you could use a lower 1:n step transformer.