Re: Seeking low chemical reaction caps
- From: Nemo <Paul@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Wed, 9 Jan 2008 22:01:33 +0000
Aha, for once I can contribute something useful.
Boris' link is essentially correct. It is dielectric relaxation. I first came across this effect with "supercaps". I was attempting to figure how long my battery would power a micropower circuit, but the circuit's micro used current in pulses, so it wasn't easy to measure. I figured I could tell the energy required by hooking up a 5 volt, 1 farad supercapacitor to the circuit running voltage, then seeing how long it took to discharge by say 1 volt while the circuit ran. [See note 1.]
I found that if I charged the supercap to 5V, then short circuited it for say 30 seconds, its terminals would gradually spring apart to say 1V.
After a literature trawl I discovered that some dielectrics store energy for a long time - I think of it as them being twisted like an LCD molecule and only gradually returning to their rest position. After discharging them the dielectric continues to unsqueeze and charge is induced back into the capacitor plates. Which is why certain types of capacitor are recommended for sample-and-hold circuits. Supercaps are particularly bad for this because of their enormous internal resistance, up to 200 ohms, so they have a long RC time constant. However I found you CAN discharge them if you short them for long enough.
The tants which John Larkin checked, unused for years, may be showing a residual charge from all that time ago. I'm not saying you CAN'T get a battery effect, especially with a leaky cap, but I would think if you did, the capacitors would have corroded to a blob after 6 years. Fresh capacitors in a reel may have a residual charge from when they were made and tested.
- - -
Note 1: the practicalities of this technique for measuring average current drain
Of course it's not that simple. First, you need to determine what the capacitance of the beast really is, at the voltage you're going to use, as supercaps are not precision devices. Once you've figured that out:
Since Q = It = CV, then the average current of the circuit is CV/t
where V is the voltage drop in time t
Next real-world issue is, you actually want a voltage drop nearer 0.1V because the circuit current drain is going to vary considerably between a supply of 3.5V and 2.5V.
Then you need to charge the cap up on its own and leave it, say 10 minutes to find out what its self discharge is from the voltage drop(about 1uA as I recall)
Then you need to check the 100 - 200 ohm internal impedance of the supercap is not going to affect your circuit current drain, voltage supply etc
Around this point you realise that the DVM monitoring the supercap has a noticeable draining effect on the experiment and is best used intermittently rather than as a constant load 8)
Apart from that, it's pretty straightforward.
--
Nemo
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