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| hamster_nz:
--- Quote from: electrodacus on May 08, 2022, 04:38:31 am --- --- Quote from: hamster_nz on May 08, 2022, 04:13:21 am --- Black cable is held against the negative of the 9V battery. --- End quote --- I see so you where charging and discharging the capacitors. LED's can glow at a few uA so it will take quite some time for them to dim as capacitors charge and those capacitors may have leakage current in that region. I just tested with a large 4700uF 35Vdc new capacitor and my small white LED had a small glow even after 30 to 50 seconds (did not count) the same level of glow it has if it conducts through my finger so I suspect a few nA I never used electrolytic capacitors in any of my projects and did a quick search and it seems they do have some leakage current due to the type of electrolyte they use. Water based ones have the largest leakage but all have even a few uA much more than nA modern LED's can glow at. So I learned something new and that is that electrolytics are not just capacitors but also a parallel resistor. There are good reason I never consider electrolytics for my projects (I do not like things with finite life). Edit: This is the exact one that I just used https://www.mouser.ca/ProductDetail/United-Chemi-Con/EGPA350ELL472MM40S?qs=beQ1fBGcmj2M%2Fui1vdONPg%3D%3D Looking at the spec it has 4uA of leakage current so fairly significant and this is not a non brand and purchased from mouser. Is just there on first page and that 4uA is best case can bemore https://www.mouser.ca/datasheet/2/420/GPALL_e-2509122.pdf In my particular case should be about 705uA based on use case 8V battery 3V on LED so 5V on the capacitor 5V * 4700uF * 0.03 but mine was no where near close to that probably is worse case and was more like 10uA max based on almost invisible glow. --- End quote --- It is a good glow - I put a meter in line with the 10k resistor - current starts out at well over half a mA, falling off over time (as expected). I just left it to settle down to steady-state, and there is no discernable glow, at least during the day, so it isn't leakage. Actually... I'll just go measure the leakage then update this post... leakage is less than a microamp. |
| electrodacus:
--- Quote from: hamster_nz on May 08, 2022, 05:10:29 am --- Actually... I'll just go measure the leakage then update this post. --- End quote --- Please do as there is sure a leakage. Those red LED's may not be as efficient as the modern white ones. The white one I have is visible at 1uA and that is about where it settles with that 4700uF cap after about one minute. You have two capacitors in series and seems like lower capacity not sure what voltage rating but there will be a leakage with any electrolytic. In any case they are perfectly fine for testing things like charge one cap with the other directly or with a small DC-DC with constant current control. |
| T3sl4co1l:
Yea nah, that's a mA or so. Clearly not leakage. It also clearly fades over time, as the capacitor(s) charge; exactly as conventional theory would predict. How odd... :) Tim |
| SandyCox:
Circuit theory isn't sufficient to solve the two-capacitor problem. We need to take the electrodynamic behavior of the system into account. So let's do the thought experiment of connecting a charged and a discharged capacitor in parallel at t=0s. We assume that all the conductors are perfect and that the two capacitors are of the parallel plate type with vacuum between the plates. The two capacitors can be modelled as two lossless transmission lines (see for instance House and Melcher example 14.2.1). After the two capacitors are connected in parallel, an electromagnetic wave will bounce back and forth between the two capacitors. The amount of energy stored in each capacitor will fluctuate, but the total amount of stored energy in the two capacitors will remain the same. If we go to the next level and take electromagnetic radiation into account, then al the energy will be radiated into space over time. |
| timenutgoblin:
--- Quote from: SandyCox on May 08, 2022, 09:49:28 am ---So let's do the thought experiment of connecting a charged and a discharged capacitor in parallel at t=0s. --- End quote --- May I suggest an alternative thought experiment? Assume lossless capacitors and conductors. Assume two capacitors of equal capacitance, C. Assume one of the capacitors is discharged. Assume that the other capacitor is charged to the exact value 1.602176634×10−19 Coulombs (the charge on an electron). Assume that the charged capacitor has a voltage of 1V, equivalent to 1eV Joules of energy. When the switch is closed, the electron migrates from the initially-charged capacitor to the initially-discharged capacitor. The electron then migrates back to the initially-charged capacitor. Oscillation occurs. Is my alternative thought experiment valid? --- Quote from: electrodacus on May 03, 2022, 12:40:07 am ---If those two capacitors are identical (same capacity) the voltage after switch is closed will be 0.707 * Vi --- End quote --- I was thinking further with this idea of 70.7% of Vi per capacitor after the switch is closed and noted a disparity regarding the issue of time relating to current flow between the two capacitors and the voltages across the capacitors. If both capacitors are 1F and the charged capacitor is charged to 3V then the initial charge is 3 Coulombs. If the initially-charged capacitor discharges from 3V to 2.121V then the change in voltage is -29.3% of Vi. If the initially-discharged capacitor charges from 0V to 2.121V then the change in voltage is +70.7% of Vi. If both capacitors have 70.7% of Vi after the same time period has past then how did the initially-discharged capacitor charge up faster from 0V to 70.7% of Vi than the initially-charged up capacitor discharge from 100% to 70.7% of Vi? This would violate Kirchhoff's Current Law where the current flowing into the node is not equal to the current flowing out of the node. In this case the node is located between the capacitors. The current flowing out of the initially-charged capacitor must be equal in magnitude to the current flowing into the initially-discharged capacitor. The initially-discharged capacitor would need to charge from 29.3% of Vi to 70.7% of Vi in a time period of zero seconds. This is impossible. I have seen a similiar problem in a YouTube video. https://youtu.be/q398AqtTEL8 \$I = C \frac{dV}{dt}\$ \$I.dt = C.dV\$ Time taken for capacitor to charge/dischage to target voltage: \$dt = C\frac{dV}{I}\$ For reference: +70.7% \$V_i \approx \frac{V_i}{\sqrt{2}}\$ -29.3% \$V_i \approx \frac{-V_i{(\sqrt{2}} - 1)}{\sqrt{2}}\$ |
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