N
Nemo
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.
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.
