storing energy to a capacitor

[MrAl]

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.

Hi,

Ok you answered one question, but why do you want to save the energy in the first place, what purpose is this going to serve?

 

[bazramit]

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.

But please be aware that after the diode I will have a capacitor and a resistance so how should I make its impedance 4R?

What do you mean by a FWbridge with isolation?

thanks

 

[alec_t]

I need to calculate the energy available in the first place

You have shown only a voltage waveform. A voltage alone doesn't have energy. Energy is the product of voltage and current (which you haven't specified, but which flows into the capacitor as it charges).
If you charge a capacitance C to a voltage V the resulting energy stored is 1/2 * C * V^2.

 

[ericgibbs]

But please be aware that after the diode I will have a capacitor and a resistance so how should I make its impedance 4R?

What do you mean by a FWbridge with isolation?

thanks

Considering only the first half cycle of the waveforms.

When the load was 1R the Vpk was 5V,,, so thats W1=V^2/R = 25/1 =25Wpk

with a 4R load W2= 12.5^2/4 = 39Wpk , which is a +50% increase.

If you did use a step up transformer , you would design it so that the reflected impedance of the secondary load into the primary was 4R.

 

[MrAl]

Hi,

Ok you answered one question, but why do you want to save the energy in the first place, what purpose is this going to serve?

Hi again,

I was asking this because what you have specified doesnt make that much sense. I thought if we had more information we could figure out what it was that you really wanted to do here.

If you specify a voltage then if that supplies a 10 ohm resistor we would end up with a certain amount of energy, but if it supplied a 1 ohm resistor then it would mean a different amount of energy. It doesnt matter what the source resistance is.

If you specify that the voltage is across a resistor say 1 ohm and the power is V^2/R then that's fine, but that energy is already lost so you can not 'capture' it. Energy dissipated in a resistor is lost as heat and the only way to capture that is to use it to heat something else up, if that's good enough.

Also, using a capacitor to store energy from a voltage source is not such a good idea because there is always half the energy lost in the transfer. It seems like if you reduce the resistance to a lower value that there would be less loss and thus more energy stored in the capacitor, but that's not how it works. As you reduce the resistance, the peak current goes up and so does the dissipation, but it's over a shorter period of time so the overall energy loss stays the same.

The classic example is the two capacitor one resistor circuit where one capacitor is charged at t=0 and the other capacitor is not charged, so one cap has some voltage across it and the other one has zero voltage across it and they are connected together with a resistor (and the remaining leads connected together or to ground). The two caps are the same value.
With one cap charged and the other cap discharged, when they are first connected together the energy transfer starts to take place. As one cap gains energy the other cap looses energy. It appears tempting to state that one cap receives half the energy (minus some small loss by the resistor which we make a low value) and one loses half the energy, but that's not true at all. What actually happens is one cap gains one quarter (1/4) the energy and the other cap looses three quarters (3/4) the initial energy so it ends up with just one quarter of the original energy too. So the question then is since each cap only has 1/4 the energy and 1/4 times two is only 1/2, what happened to the other 1/2 of the initial energy stored in the first capacitor? The answer is that it got dissipated in the resistor.
And the next question is usually what if we reduce the value of the resistor? The answer to that is that the energy transfer takes place FASTER, but the peak current in the resistor increases so the energy dissipation stays the same so we still loose 1/2 the total energy.
So in real life terms, if the two caps were 1F each and if the first cap was charged to 10v and the second was discharged to 0v just before the start of the experiment, and then we connect them together with a 100 ohm resistor, after a period of time the first cap discharges to 5v and the second cap charges up to 5v. Now since the first cap initially had 10v across it that's 50 Joules, and now that they both have only 5v across them that's 12.5 Joules each, for a total of 25 Joules at the end of the experiment. That means we lost 25 Joules.
Reducing the resistor to 10 ohms, we see the same thing happen but it happens faster...that is, the two caps reach equilibrium faster but the energy lost is still 25 Joules.
Now do the same with an inductor in place of the second cap and we can get much more of the energy to transfer from the cap to the inductor, but we have to know when to cut the connection or else the inductor will give the energy back to the capacitor. It's still not 100 percent efficient because the resistor has to conduct current at least part of the time, but it ends up much much better. The real problem though is when to cut off the circuit.

So i have good reason for asking the questions i have asked.

 

[bazramit]

Hi again,


Also, using a capacitor to store energy from a voltage source is not such a good idea because there is always half the energy lost in the transfer. It seems like if you reduce the resistance to a lower value that there would be less loss and thus more energy stored in the capacitor, but that's not how it works. As you reduce the resistance, the peak current goes up and so does the dissipation, but it's over a shorter period of time so the overall energy loss stays the same.


Now do the same with an inductor in place of the second cap and we can get much more of the energy to transfer from the cap to the inductor, but we have to know when to cut the connection or else the inductor will give the energy back to the capacitor. It's still not 100 percent efficient because the resistor has to conduct current at least part of the time, but it ends up much much better. The real problem though is when to cut off the circuit.

So i have good reason for asking the questions i have asked.

Hi,

I am covinced that the cap will only store half the energy but this is if was charge from another cap. Is this true in case it was charged from a voltage source with 4R source impedance?

Please give me some more details on using an inductor instead of a cap and how should the switching be done?

 

[bazramit]

Considering only the first half cycle of the waveforms.

When the load was 1R the Vpk was 5V,,, so thats W1=V^2/R = 25/1 =25Wpk

with a 4R load W2= 12.5^2/4 = 39Wpk , which is a +50% increase.

If you did use a step up transformer , you would design it so that the reflected impedance of the secondary load into the primary was 4R.

Hi,
You are talking here about power but is this also true considering energy since time will change in the two cases?

 

[ericgibbs]

hi.
This is a LTSpice simulation using your measured results.

 

[alec_t]

Please give me some more details on using an inductor instead of a cap and how should the switching be done?

If you don't have another energy source you won't be able to do the switching

 

[MrAl]

Hi,

I am covinced that the cap will only store half the energy but this is if was charge from another cap. Is this true in case it was charged from a voltage source with 4R source impedance?

Please give me some more details on using an inductor instead of a cap and how should the switching be done?

Hi,

Well no the capacitor to capacitor energy transfer looses half the energy in the resistor which means each cap only has one quarter of the energy at the end of the day. So if the original cap had 2 Joules after the transfer each cap only has 0.5 Joules each.

With a voltage source and resistor (the resistor can be part of the voltage source series resistance) then one half of the total energy expended gets lost in the resistor and half stored into the capacitor. So if during the transfer the source puts out 2 Joules the resistor absorbs 1 Joule and the cap gets 1 Joule stored.

That inductor and switching example was for use with a constant voltage source. For a changing signal it would become a little more complicated. The switching would have to be done automatically at the peak with a constant voltage source.

It's still not clear what you are trying to do here. You need to clarify what you are trying to do and what the purpose is. I think this would help to determine how to do this or if this is even possible for your application.

 

[bazramit]

Hi,



It's still not clear what you are trying to do here. You need to clarify what you are trying to do and what the purpose is. I think this would help to determine how to do this or if this is even possible for your application.

I am trying to store the energy in this signal and then re-transmitting its equivalent amount of energy after a
certain delay time. So I should store this energy to be used later and this should be a self powered circuit.

 

[ericgibbs]

I am trying to store the energy in this signal and then re-transmitting its equivalent amount of energy after a
certain delay time. So I should store this energy to be used later and this should be a self powered circuit.

hi.
Two options.

 

[bazramit]

hi.
Two options.

Hi,
thanks for the simulations. Do I understand that the first option is much better than the second?
And please let me know what you mean by 20m and 40m for the transformer?
Sorry but I don't know PSpice....

 

[MrAl]

I am trying to store the energy in this signal and then re-transmitting its equivalent amount of energy after a
certain delay time. So I should store this energy to be used later and this should be a self powered circuit.

Hi,

Well what i need to know is what exactly is generating this energy or where does this energy come from. If you say 'a voltage source' that wont help so i'll caution you right up front. What kind of device or object is producing this energy.

As i've said before, a voltage source produces energy but the amount that gets used is dependent on the load. The maximum energy is dissipated within the source with a short circuit, but that doesnt allow any energy transfer. To allow energy transfer means to loose some of the energy that the source is developing.

There are theories about the energy contained in a signal but im not sure you are concerned so much about that because you say you want to 'capture' this energy.

What do you mean by transmitting an 'equivalent' amount of energy? Do you mean that you just need to know how much energy is available and then you will produce that energy some other way, or does the entire circuit have to do this without any additional power input?
Also, what do you mean by 'transmit', what form of transmitting the energy will you be using?

Feel free to explain exactly what you are trying to accomplish here being as detailed as you possibly can. Too little information will never get this working the way you want it to work

 

[bazramit]

Hi,

Well what i need to know is what exactly is generating this energy or where does this energy come from. If you say 'a voltage source' that wont help so i'll caution you right up front. What kind of device or object is producing this energy.

As i've said before, a voltage source produces energy but the amount that gets used is dependent on the load. The maximum energy is dissipated within the source with a short circuit, but that doesnt allow any energy transfer. To allow energy transfer means to loose some of the energy that the source is developing.

There are theories about the energy contained in a signal but im not sure you are concerned so much about that because you say you want to 'capture' this energy.

What do you mean by transmitting an 'equivalent' amount of energy? Do you mean that you just need to know how much energy is available and then you will produce that energy some other way, or does the entire circuit have to do this without any additional power input?
Also, what do you mean by 'transmit', what form of transmitting the energy will you be using?

Feel free to explain exactly what you are trying to accomplish here being as detailed as you possibly can. Too little information will never get this working the way you want it to work

This waveform is generated from an electromagnetic coil I have designed but as you see the waveform is in ugly shape. I want to transform the energy of this waveform into an rectangular pulse holding the same energy.

Then I will have a train of rectangular pulses . My main concern is that whenever I receive an electromagnetic pulse I should convert it to a rectangular pulse after a certain amount of delay.

In this I will have a train of rectangular pulses where I could also play with its period. This is my main task. I hope I have explained well....

 

[alec_t]

The switching device to create rectangular pulses would require a source of energy. Where is that coming from?

Edit: Are you trying to build a perpetual motion machine?

 

[MrAl]

This waveform is generated from an electromagnetic coil I have designed but as you see the waveform is in ugly shape. I want to transform the energy of this waveform into an rectangular pulse holding the same energy.

Then I will have a train of rectangular pulses . My main concern is that whenever I receive an electromagnetic pulse I should convert it to a rectangular pulse after a certain amount of delay.

In this I will have a train of rectangular pulses where I could also play with its period. This is my main task. I hope I have explained well....

Hi,

Ok but why do you need to do this without any additional input power? It would be possible with a circuit of some practical kind but that circuit would need input power. If it can not get input power for some reason, then the best you could do is use SOME of the power from the signal itself to power the circuit which would then create the rectangular pulses.

Perhaps you could explain *why* you dont want to provide any additional power to power the circuit that could accomplish this task.

 

[bazramit]

Hi,

Ok but why do you need to do this without any additional input power? It would be possible with a circuit of some practical kind but that circuit would need input power. If it can not get input power for some reason, then the best you could do is use SOME of the power from the signal itself to power the circuit which would then create the rectangular pulses.

Perhaps you could explain *why* you dont want to provide any additional power to power the circuit that could accomplish this task.

This is the aim of my proof of concept project.
" Generating a rectangular pulse with an electromagnetic coil and a self powered circuit"

The excitation of the electromagnetic coil would be with some natural mechanical motion. Nature is beautiful.

In this case I could make a practical example to show my self that energy could be transformed from one shape to another.

Now with small amount of energy stored in the cap, how could I create a rectangular pulse at a certain time which holds the same amount of energy as the original waveform?

 

[bazramit]

The switching device to create rectangular pulses would require a source of energy. Where is that coming from?

Edit: Are you trying to build a perpetual motion machine?

from the energy stored in the cap. Any practical circuit for this?

 

[Nigel Goodwin]

from the energy stored in the cap. Any practical circuit for this?

No, as it's absolutely impossible.