Squeezing the power out of a Supercap

[Bobblehat]

I'm getting slightly unexpected results... I'm running the lamp + supercap from a battery and regulator to test things and while connected to the batter the circuit sits about 3.2v once the power is disconnected the voltage drops to 2.7v, is this normal or is the supercap just taking a very long time to charge?

What is the LED doing just before, and shortly after you disconnect your voltage source? Does it suddenly dim at disconnection then stay relatively stable for a few or 10's of seconds?

[steveo_mcg]

Yep thats exactly what its doing, as soon as the power is reconnected it brightens back up again.

[Bobblehat]

Yep thats exactly what its doing, as soon as the power is reconnected it brightens back up again.

Looks like normal exponential decay set by the internal resistance of the supercap. Rapid drop off at first, but as the voltage of the cap drops closer to the LED cut-off point the LED only dims quite slowly.

As Frontranger predicts at the bottom of msg#52, that small voltage difference between the forward Vf of the LED at the normal running current and the Vf when the LED nears cut-off leaves little room to provide a decent standlight current for a decent time. The Vf of the LED at normal running current is what is limiting the effectiveness of the supercap. The cap is charging to 3.2V instead of potentially charging it to near 5.5V.

Depending on the LED cut-off point (2.5V ish), you have just a 3.2 - 2.5 = 0.7V(ish) differential compared to a potential 5.5- 2.5 = 3.0V(ish) differential. This can make quite a difference in either initial brightness and/or run time of the standlight.

If you could add a resistor to lower the standlight current (how?), you would see an initial large step down in the light levels when the test voltage is disconnected but the light level would stay steady for longer.

[steveo_mcg]

As far as i can tell, the supercap is only charging to 2.7v then rapidly dropping off as you observe. I've got a MM connected across the supercap and as soon as the battery is disconnected the voltage reads 2.7v, I would expect that if the supercap was charging properly it would read about 3.2v then go south, no?

[znomit]

Problem is i've only got one emitter so the circuit Vf is only ever ~3.2v.

Frontrangers circuit below is based on two LEDs but you can replace the top LED with a diode or two to up the super cap voltage a little bit. Even 1V should make a reasonable difference to stand light runtime. You'll lose a little in low speed light and a tiny bit of efficiency.

[Bobblehat]

Frontrangers circuit below is based on two LEDs but you can replace the top LED with a diode or two to up the super cap voltage a little bit. Even 1V should make a reasonable difference to stand light runtime. You'll lose a little in low speed light and a tiny bit of efficiency.

I thought about that ..... initially that's pretty obvious ..... but couldn't figure out how to optimise a current limiting resistor!

[znomit]

... but couldn't figure out how to optimise a current limiting resistor!

Oh thats easy.
Just need a selection of resistors and lots of time spent siting in a dark room with a stopwatch.

[Bobblehat]

Oh thats easy.
Just need a selection of resistors and lots of time spent siting in a dark room with a stopwatch.

Now, you're teasing ....... I think you might need at least one diode left in the feed-line to the supercap, just to block the cap discharging directly through the diodes used to "up" the volts!

So ...... would this work for steveo?

[steveo_mcg]

Yup that worked nicely thank you!

I didn't need the diode in line with the super cap it only seems to discharge through the emitter or put another way it didn't make any difference with or with out it.

Just for reference with three diodes I'm losing 3x0.7v @ 500ma, or about 1w?

[Bobblehat]

Yup that worked nicely thank you!

I didn't need the diode in line with the super cap it only seems to discharge through the emitter or put another way it didn't make any difference with or with out it.

Just for reference with three diodes I'm losing 3x0.7v @ 500ma, or about 1w?

Wow! Quick testing! Good to know it made a difference. Could you describe what effect the new circuit has on the brightness and time the standlight is lit compare to what you had before?

I'm a little surprised the diode in line with the supercap has no effect .... but then I'm not an expert on these matters, I had it in mind that the cap would discharge swiftly through the diodes in series with the LED with only the internal resistance of the cap to slow it .... i.e. still relatively quickly.

[steveo_mcg]

The old circuit discharged in about two minutes to <10ma and started about 40ma dropping to 20ma after about 30s. The new circuit starts ~100ma and takes about 5 minutes to drop to <10ma. I've got a 10r resistor in there so its discharging too quickly, I'm going to try and find one to keep it around 30-50ma as long as possible.

When i've missed out the link between the resistor and the super cap the current running out the supercap to the emitter via the diodes is very low, the resistor must be the path of least resistance.

I ended up with two silicon and 1 schottky which charges to about 5.3v.

[Bobblehat]

The old circuit discharged in about two minutes to <10ma and started about 40ma dropping to 20ma after about 30s. The new circuit starts ~100ma and takes about 5 minutes to drop to <10ma. I've got a 10r resistor in there so its discharging too quickly, I'm going to try and find one to keep it around 30-50ma as long as possible.

That's good data for anyone wanting to build any of these circuits.

When i've missed out the link between the resistor and the super cap the current running out the supercap to the emitter via the diodes is very low, the resistor must be the path of least resistance.

Yep! I got that wrong by the looks of it (I said I wasn't an expert). I had a feeling the diode route might have had more influence ..... good to know that the major control of the current is the resistor at the sort of currents we are likely to want.

I ended up with two silicon and 1 schottky which charges to about 5.3v.

A useful rise over 3.2V (or 2.7V, can't figure out why!) ...... and more energy to power the standlight. When you try a higher ohm resistor (50ohm ... 100ohm?) also try the diode in the feed line to the supercap ....... it might just make a difference to extending the run time ...... but then I was wrong last time

I can't wait to get all my bits and start building the more complicated 2 front + 1 rear circuit Diodes arrived today ...... still gathering bits.

[Savvas]

A useful rise over 3.2V (or 2.7V, can't figure out why!) ...... and more energy to power the standlight. When you try a higher ohm resistor (50ohm ... 100ohm?) also try the diode in the feed line to the supercap ....... it might just make a difference to extending the run time ...... but then I was wrong last time

So Steveo and Bobblehat, just to clarify things a bit, are we left with a circuit that has a string of diodes in series with the standlight led with a supercap and series resistor paralleling this same led?

No need for a separate diode (or diode series) to charge up the supercap directly from the +ve rail?

BTW, with the 'standard' 2 led/ 1 standlight set up (using XR-Es) a 100ohm resistor and 1F supercap gives me 5-10 mins quite noticeable light.

ta,

Savvas.

[znomit]

No need for a separate diode (or diode series) to charge up the supercap directly from the +ve rail?

The point of the diodes in the original circuit is to drop the +ve rail 2x Vf (~6 or 7V) to something the super cap can handle (below 5.5V).
Here the power rail is below the super cap rating so no worries.

From testing I noted that the super cap voltage quickly dropped to around 4V... so getting the full 5.5V into it doesn't increase runtime significantly, it is nice and bright for short stops though.

[Bandgap]

The sudden voltage steps - rather than gradual changes - could easily be due to the internal resistance of the super caps.

the smaller types - sometimes called uA types - have a lot of resistance inside.

mA types less so. Large amp types have very little internal resistance and brightness would hardly change whe the dynamo stops.

so from your initial voltage drops, you could calculate internal resistance - which also by the way slows charging.

internal resistance will also be on the capacitor's data sheet.

Steve

[steveo_mcg]

A useful rise over 3.2V (or 2.7V, can't figure out why!) ...... and more energy to power the standlight. When you try a higher ohm resistor (50ohm ... 100ohm?) also try the diode in the feed line to the supercap ....... it might just make a difference to extending the run time ...... but then I was wrong last time

As above, the 2.7v is the voltage sag with the (relatively) high drain from the emitter, the actual unloaded V was about 3-3.3v.

@znomit, I think for turning right that short bright stand light won't hurt, at lights I don't suppose it really matters once you've stopped in a traffic queue.