In this example
simulate this circuit – Schematic created using CircuitLab
After the initial charging of the cap to 3V, current gets blocked, but over time does it consume any energy from the batteries? Is this safe to make?
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\$\begingroup\$ I thought you were asking another (related) question ... in this circuit the energy given out by the battery (theoretically) is CV but energy stored in the capacitor is only half of that. The rest of the energy goes away in the form of heat in the battery, and in EM radiation. So even theoretically an ideal capacitor also wastes some energy. \$\endgroup\$– KartikSep 4, 2016 at 16:01
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5 Answers
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Leakage current will drain the battery, most likely not that significantly compared to the internal self-discharge of the battery.
An aluminum electrolytic might leak 100nA long term, which is not much compared to the self-discharge of even a button cell. The guaranteed maximum of a typical e-cap of this size is 0.002CV or 400nA (whichever is greater) after 3 minutes. Most parts will beat that significantly. Some SMD parts are not nearly as good.
Your second question was whether this safe to make. Generally, yes, however there are almost always exceptions in engineering. If your 3V battery has a large current capacity (perhaps an unprotected 18650 Li cell) and your capacitor is something like a 6.3V tantalum capacitor there is a significant risk of an 'ignition' event upon connecting the capacitor to the battery (picture flames shooting out, a bright light and some noxious fumes). The risk can be considerably reduced by adding some series resistance of some tens of ohms.
answered Sep 3, 2016 at 11:21
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\$\begingroup\$ "The guaranteed maximum of a typical e-cap of this size is 0.002CV or 400nA (whichever is greater) after 3 minutes": interesting, what is your source? \$\endgroup\$ Sep 5, 2016 at 11:57
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\$\begingroup\$ Capacitor datasheets, eg. Nichicon. \$\endgroup\$ Sep 5, 2016 at 12:22
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\$\begingroup\$ @SpehroPefhany Do you recall which series? I only ask because a short while ago I was looking for a low leakage electrolytic and the best I could find was 0.01CV or 3uA (whichever is greater). \$\endgroup\$– bitshiftSep 15, 2016 at 13:17
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\$\begingroup\$ @bitshift Try the UKL series, available in small quantities from Mouser. 0.002CV or 200nA. But many of the makers have low leakage types in their portfolio since they're pretty popular in Asia, just harder to find in distribution. Some lines of regular parts actually perform pretty well (without the guarantees), but some don't- higher ESR larger parts tend to be less leaky for whatever reason(s). \$\endgroup\$ Sep 15, 2016 at 14:13
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In steady state (after a long time) an ideal capacitor does not draw significant current from a battery. A real capacitor will draw some small leakage current. The amount of leakage current will depend on the type of the capacitor, electrolytics will have higher leakage than films and ceramics.
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2\$\begingroup\$ An "ideal capacitor" charge instantly from an ideal battery (with ideal zero-inductance wiring), in a spike of infinite current. I guess you're talking about a real battery with non-zero internal resistance, and the RC time constant for current to drop to zero. \$\endgroup\$ Sep 3, 2016 at 13:24
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1\$\begingroup\$ @PeterCordes you're correct, I was considering an ideal capacitor but a real battery and wiring, in which case the current starts out large and decays exponentially to zero. But if they are both ideal as you point out you would get an impulse of current and the cap would charge instantly. \$\endgroup\$– John DSep 3, 2016 at 16:33
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\$\begingroup\$ Yes, I meant to say "RC time constant for the current to drop toward zero", not "to zero". That's what I get for nit-picking :P \$\endgroup\$ Sep 3, 2016 at 16:55
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An ideal capacitor would be open circuit to DC, so no current would flow, and no energy would be consumed after the capacitor is fully charged.
However, real capacitors do have some small leakage current, so, in Real Life, energy would be consumed from the battery very slowly after the initial charging.
answered Sep 3, 2016 at 3:17
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You should check something called "insulation resistance"
I quote from Murata:
The insulation resistance of a monolithic ceramic capacitor represents the ratio between the applied voltage and the leakage current after a set time (ex. 60 seconds) while applying DC voltage without ripple between the capacitor terminals. While the theoretical value of a capacitor's insulation resistance is infinite, since there is less current flow between insulated electrodes of an actual capacitor, the actual resistance value is finite. This resistance value is called "insulation resistance" and denoted with units such as Meg Ohms [MΩ] and Ohm Farads [ΩF].
I checked a datasheet I had (part number: GRM32ER71H106KA12) for an example to approximate how much leakage gets to pass. Check the image below:
To fully understand the behavior of the capacitor at steady state (as in directly connecting a capacitor to a battery) I highly recommend reading the following article: http://www.murata.com/support/faqs/products/capacitor/mlcc/char/0003
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If the polarity of the battery is reversed in this scienario, then even an ideal capacitor will consume current to change it's polarity in tune with the battery. But in this case only a real capacitor will be able to consume energy due to springing effect i.e leakage of charge from the edges of the capacitor. However it'll depend on the type of capacitor and the material used in making the capacitor.
