author=ion link=topic=3649.msg68820#msg68820 date=1532449346]
OK,so lets get down to the nitty gritty here.
The hidden energy causes the supercap to supercap transfer voltage to be higher than the expected one-half of the original voltage.
How can it do this?
If the energy is hidden,then the voltage measured across the capacitor should represent a lower value of stored energy--if the hidden energy cannot be seen or accounted for.
This is not seen in real world good quality film capacitor tests where the transfer produces a very close to one-half.
We are not talking about film capacitors here--never have been.I have been dealing with supercaps only.
The makeup of the two are completely different. To compare a film capacitor to that of a super capacitor,is like comparing a lead acid battery to a lithium ion battery--no pun intended
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It is my belief that if the actual energy was tallied to charge the first supercap, it would be a bit more than the calculated value shows and why 1/2CV^2 is not an accurate method of calculating the energy of a hybrid cap/battery.
You simply cannot calculate the capacity of a supercap by how much energy went into charging it,due to heat loss.
And i see your reply to this below--
The dissipation factor and heating effect is miniscule for DC and becomes large at AC frequencies, which is why supercaps capacitance must be rated at a given frequency. At any rate the temperature rise can be measured and accounted for, which will be tiny for DC
Im afraid this is not correct,and a good chunk of the charging energy is lost to heat.
In the below video,i limited the charging current to 500mA,and it took over 20 minutes at that current to get the cap to 2.3 volts. The current then of course started to drop off as we got closer to the input voltage value of 2.7v
Even at this low current value,i could feel the cap was warm,and the others were at room temperature.
https://www.youtube.com/watch?v=6z_WLRtIBqESo,the only way to calculate the energy stored within a supercap,is by means of discharging it across a pure resistor over time--which i have done 3 times now,to the point where there is less than 10mV regaind after the circuit is disconnected. Those results were within 99.3% of the calculated capacitance value.
Then when you use the 1/2CV^2 formula to arrive at resultant energy you find you have a bit more than the start energy calculated of the first capacitor using the same formula.
I would think that the manufactures would know the value of there capacitors,through years of research,and take all this into account.
Of course the cheaper ones will some where in the ball park of there claimed value.
You know that terminal voltage alone cannot tell the actual energy of a battery, nor can it tell the actual energy of a supercap.
I do not agree with that.
A batteries makeup is different to that of a supercap.
A battery can have a surface charge,and a supercap cannot.
You can have a battery that has 1.5v across it,and as soon as you place a load across it,the voltage will drop maybe .5v in an instant--you cannot get that with supercaps.
So the supercap is "a little bit pregnant" with battery effect.
Very little.
Therefore your COP values are actually a little higher than expected, which would be 1.0 if measured correctly.
If measured correctly
How much more accurate can you get than the way i have done it.
We calculated the actual capacitance of each cap down to the third decimal point,by way of discharging over a resistance to the first time constant ,down to a single second. I then discharged the total stored energy over the same 2.2 ohm resistor,and calculated the energy dissipated from the resistor,which was within .7% of the first calculation,and that .7% very well could have been dissipated as heat from the cap it self.
As an example I took a fully charged NiMH battery with terminal voltage of 1.396 volts and dumped it into a fully discharged identical cell, and the terminal voltage on both was 1.394 after a couple of hours. This is the extreme case where you have 100% battery effect, the transfer voltage is well above one-half of the original terminal voltage, in fact it is close to the starting voltage. There are percentages of battery effect and normal ideal capacitor effect in supercaps.
That test actually proves how little battery effect supercaps have,as you would never be able to get those results with supercaps.If a rapid cap to cap transfer is done,where the resistive path between the two caps is very low,then the circuit is disconnected,you would see the supply caps voltage go up,and the receiving caps voltage go down. If the transfer is done slowly,say through a 10 ohm resistor,then when the transfer is complete,and the circuit is disconnected,you would not see hardly any voltage rise or fall on either cap.
Depending on the mix of the two will be the degree the voltage rises above the "one-half value" in both supercaps after the transfer.
And if the voltage remains stable,with the circuit still connected after the transfer,then what?.
There is also this that requires an answer.
In my tests,i use the same caps in the same location,where the supply cap is always charged to 2.7v exactly,and the receiving cap is always discharged until it's voltage reads no more than 10mV after 5 minutes sitting.
The first test is a transfer using the 2.2 ohm resistor in series,and the caps settle to 1.413v each.
The second test is using the small DC PM motor which has a small propellor as a load.
At the end of the transfer each cap has a voltage of 1.502v
The odd thing is,the motor and load(propellor) dissipated more energy than the 2.2 ohm resistor did in the same test--how do we explain this?.
p.s. I agree that there may be some other strange gains that occur in supercap to supercap transfers, but first we must weed out the obvious effects to get to the stranger data reliably. We need to exhaust the "knowns" properly then we can diligently proceed to the "unknowns".
This is going to take a while i feel,as it seems that the knowns are not so well known at this point in time.
Brad
Never let your schooling get in the way of your education.
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Topic: Generating System (Read 59655 times)
