Squeezing the power out of a Supercap

[Bobblehat]

Oh! Dear! 4mA max? ..... Does this mean SimonInd and ejoyu are stressing their 4F supercaps? They say their circuits have been running for sometime? Are the 1F 5.5V types able to run nearer the 100mA mark? Maplin do not mention their supercaps max current. What type do you use znomit?

[jdp298]

I use the Maplin 1F 5.5v caps with their 1W white star LEDs, no problems after the first 1500 miles.

C2 dicharges to ground, where ever that is. If you think of it as a column of water trying to reach ground then it's easier to imagine. The charge will exit C2 positive terminal and pass through the red Luxeon, but then it still has to get to a ground plane, which is either through the white LEDs or through R2 and a diode. All the voltage on C2 will be spread between these components, not just used up on the red LED, such is the way of things.

[Bobblehat]

Good to know that I could "get away with" caps of lower spec than ideal! I think you are not the only one who has practical experience with using these supercaps for a purpose they were not designed for. Anyone know the max current spec of the Maplin 1F 5.5V cap?

I still just couldn't get my head around C2 discharging through the 2 white LEDS ...... so bare with me on this. I've redrawn the circuit and included the bridge rectifier for completeness. Original on the left ....... on the right hand side I've tried to make C2 look like "a battery" in this stationary situation. The only route I can see C2 discharging is through the red LED and R2. If I got the redraw right, all other discharge routes for C2 are blocked by a reversed diode(s) ..... aren't they?



Help me out here folks .... have I got it wrong?

[ianfield]

Vf for LEDs is rarely presented as a "headline specification" because LEDs are normally specified by optimum forward current, with a 3V LDO driving a red LED you will still need to use a current limiting resistor - there are plenty of pages on the web presenting the formula.

Red LED Vf vary from about 1.7V to around 2.2V depending on material, if Vf isn't specified you can measure it by setting up a simple circuit with a battery and current limiting resistor (or better still a constant current source) set to give the specified If into a short circuit. The higher the battery voltage; the higher the resistor will have to be and the less difference there will be in current between the S/C & the LED Vf when you insert it.

[Bobblehat]

Vf for LEDs is rarely presented as a "headline specification" because LEDs are normally specified by optimum forward current, with a 3V LDO driving a red LED you will still need to use a current limiting resistor - there are plenty of pages on the web presenting the formula.

Red LED Vf vary from about 1.7V to around 2.2V depending on material, if Vf isn't specified you can measure it by setting up a simple circuit with a battery and current limiting resistor (or better still a constant current source) set to give the specified If into a short circuit. The higher the battery voltage; the higher the resistor will have to be and the less difference there will be in current between the S/C & the LED Vf when you insert it.

Sorry mate, you've lost me ....... whose message were you replying to?

[ianfield]

Sorry mate, you've lost me ....... whose message were you replying to?

Its a mystery! - navigation in this forum is almost as bad as facebook!!!

[Bobblehat]

Keep taking the tablets, mate! .......... I find it helps!

Now ..... anyone help me with the revised diagram?

[jdp298]

There's a fwd path through the red, then through the 2 XP-G white LEDs (what's going to stop that?)

Also, the re-draw has helped, kind of. Something I just spotted is that C2 can discharge into C1 and charge that up too, although the bank of 4 diodes will mean it won't be much, but it will be another inefficiency.

[Bobblehat]

Appreciate the response jpd ........

There's a fwd path through the red, then through the 2 XP-G white LEDs (what's going to stop that?)

As I understand it .... the 4 reversed D2 diodes will block any flow. The "reverse" leakage current through D2's is absolutely miniscule compared to what flows through R2.

Also, the re-draw has helped, kind of. Something I just spotted is that C2 can discharge into C1 and charge that up too, although the bank of 4 diodes will mean it won't be much, but it will be another inefficiency.

That's how I see it too....... but the 4 reversed C2 diodes and also the reversed white XP-G's "reverse" leakage currents are insignificant in giving charge to C1 compared to the current allowed through R2.

If I got it right, 1N4001 reverse current < 50uA (~600 times smaller than initial red LED current @ 100 Ohm) and I think Cree use 100uA as the reverse current failure criteria for their power leds (~300 times smaller than the initial red LED current @ 100 Ohm)

All in all, until I build the circuit, I'm only trying to postulate what should happen!

[ Sticks neck out for chopping by real experts ...... which I am not! ]

[jdp298]

Appreciate the response jpd ........

As I understand it .... the 4 reversed D2 diodes will block any flow.

They won't, because they aren't reversed. On the right hand diagram, you can consider the far right vertical line to be the effective ground plane. Everything drains down to this point, so to speak, and ends up back at the bridge rectifier.

So what happens is the charge in C2 can drain down via any of 3 routes.

After the red LED:
1. Through R2 and then the D2s
2. Through both white LEDs

Or

Through the D1s, at which point it can charge up C1 and/or go through R1 and the thence the second white LED.

I don't think I'm using terms you're happy with; can someone else try and translate what's in my head for him?

[Bobblehat]

I don't think I'm using terms you're happy with;

No problem jdp ...... I'm easily pleased

So .... If I removed D1's, R1, C1 and D3 ...... I can still light the red LED and the two white LED's from the C2 standlight components? How bright roughly would the white LEDs be, compared to the red LED?

[SimonInd]

My over stressed Super Caps are still working fine, where does the 4mA figure come from? From the light output when stopped they're discharging at well over that rate, although I haven't measured the current.

Regarding adding a second Front LED, why not put it it series with the standlight circuit? The dynamo should still generate the necessary Voltage to drive all three, after all once the caps are charged you're effectively left with 3 LEDs in series which a dynamo shoul be fine to power. You'll get a standlight front and rear and the extra front LED when you're on the move.

[SimonInd]

Apologies, didn't read the eBay listing where it states 4mA.

However I'm not sure the measures ring true. The ESR of 16 ohms, or even the DC impedence of 25, suggests that each cap should take up to 200mA when discharged, i.e. pretty much everything coming out of the dynamo. My experience of the lights coming on straight away doesn't tally with that. Without a manufacturer's data sheet it is difficult to know what the actual specification for these is.

I guess we'll see how long they last for me and ejoyu, I'm commuting > 100 miles a week (7 hours+) at the moment so they're getting a reasonable workout. I have no switch so they're getting fully discharged overnight and during the day.

The ideal is probably something that has a voltage rating of 5.5v and an ESR of 50-100 ohms allowing charging/discharging at 50-100mA

[SimonInd]

So here's the question? would charging with more current damage the SuperCap? Most of what I've read states that the charging/discharging current is limited by the ESR. I've read nothing that states the capacitor can be damaged other than if you try and chage it above it's rated voltage and the LED clampling the voltage prevents this.

[Bandgap]

I still just couldn't get my head around C2 discharging through the 2 white LEDS ...... so bare with me on this. I've redrawn the circuit and included the bridge rectifier for completeness. Original on the left ....... on the right hand side I've tried to make C2 look like "a battery" in this stationary situation. The only route I can see C2 discharging is through the red LED and R2. If I got the redraw right, all other discharge routes for C2 are blocked by a reversed diode(s) ..... aren't they?

Forgive me if I am way too late here, but isn't C2 feeding only the red led, and C1 feeding only one of the white leds?

Steve

[Bobblehat]

My over stressed Super Caps are still working fine, where does the 4mA figure come from? From the light output when stopped they're discharging at well over that rate, although I haven't measured the current.

Hi Simon ... I'm kinda hoping that practical experience wins out over theory .... otherwise it's a search for a less common cap and no doubt more expensive, probably bigger cap too! Commercial units seem to be using caps whose spec seems lower than the use they are being put to ...... looks like they have checked it out in practice and went with the more common caps. Not an ideal situation, but one that works, I guess.

It's good to hear that you can vouch for a decent test period for the cap you are using.

Regarding adding a second Front LED, why not put it it series with the standlight circuit? The dynamo should still generate the necessary Voltage to drive all three, after all once the caps are charged you're effectively left with 3 LEDs in series which a dynamo shoul be fine to power. You'll get a standlight front and rear and the extra front LED when you're on the move.

I thought that's what my diagram (msg#82 & msg#93) has suggested ... 2 white XP-G's in series, one of the White LEDs on the standlight, each supercap charged to close to, but below 5.5V, the other supercap powers the red LED? If you are suggesting an alternative to the way I drew the circuit, I'd love to see a sketch.

[Bobblehat]

Forgive me if I am way too late here, but isn't C2 feeding only the red led, and C1 feeding only one of the white leds?

Steve

Thanks Steve .... not too late

That's what I thought too ...... but jdp has put a little doubt in my mind ..... am I right in thinking the reverse diodes do the necessary blocking of more complicated discharge route(s)?

[Bandgap]

As I remember it, white leds have a reverse breakdown of about 5V.
For 1N400x it is humdreds of volts.

And neither leak much.

most of the odds-and-sods diodes are acting to subtract 0.6V approx during charging.

Steve

[SimonInd]

Sorry I'm at work and the diagrams are blocked, I'll take a look at home later on.

The extra 1N4001 diodes are to prevent overvoltage on the capacitor, the Schottky's are to ensure that each cap only discharges through one LED and don't cross charge each other.

[SimonInd]

I've had a look at the pictures and I think it should work, however just putting one of the XP-Gs in series with the original circuit, i.e. between +vcc from the bridge rectifier and +vcc on the circuit removes the need for all of those extra diodes and has the same effect of one XP-G being a standlight and one not. I've even considered adding a second light this way although I don't really need it.

[ianfield]

Cavus gluteus maximus.

[Bobblehat]

I've had a look at the pictures and I think it should work, however just putting one of the XP-Gs in series with the original circuit, i.e. between +vcc from the bridge rectifier and +vcc on the circuit removes the need for all of those extra diodes and has the same effect of one XP-G being a standlight and one not. I've even considered adding a second light this way although I don't really need it.

GENIUS! So simple! Doh!

I guess it would even work like this ..... just a slight variation to make re-wiring my 2 x XP-G easier .....

[Bobblehat]

Bandgap ... tried to reply to your PM ... not sure if it worked? Seemed to go a bit squiffy when I hit send!

[Bandgap]

Bandgap ... tried to reply to your PM ... not sure if it worked? Seemed to go a bit squiffy when I hit send!

Thanks Mr BobbleHat

Have emailed it to you

Steve

[Bobblehat]

Thanks Mr BobbleHat

Have emailed it to you

Steve

Thanks Bandgap

One note (thanks to Bandgap) .... using XP-G's with their higher Vf than XM-L's, I might need to adjust the supercaps diodes to make sure the caps don't get more than 5.5v. This will also depend on the Vf of the red LED I get too. I could add a schottky or maybe even another 1N4001 to the 1N4001's already shown to do the adjustment.

Thanks everyone!

[Bandgap]

Don't forget, that with these simple circuits that include a tail light, the headlight will go out if the long wires to the rear led break.
- but at least you know the back light is out though.

By putting a bunch of 1A diodes (with a drop greater then the led) across the circuit board by the rear led terminals, you can keep the front light on, although now you no longer know that it has blown.

And with an extra transistor, you can indicate if the rear light has gone too. ( I say this confidently, then start to wonder...)

Steve

[SimonInd]

Nice idea with the diodes Steve, I hadn't thought of it.

I believe if you lose the Red LED in the diagram on post #57 you will still have a front light, just at lower brightness as the current will flow through D1, R1 and D3. The problem will be that with the reduced current the dynamo output voltage will rise, either cooking R1 (more voltage results in more current and therefore power to dissipate) or putting too much voltage accross C1. So I'm talking myself into adding some diodes, not too many though otherwise they won't clamp the voltage low enough to avoid supercap damage...

Simon

[ianfield]

The 1N4001 has a PIV of only 50V - an unloaded 'dynamo' (alternator) can easily exceed this value.

Shottky barrier diodes have about 0.2 - 0.3V forward drop, but PIV starts at 20V for the most common types - 90V types can be had (at a price), they also have higher leakage.

If using SB rectifiers, a shunt regulator is advisable.

As I remember it, white leds have a reverse breakdown of about 5V.
For 1N400x it is humdreds of volts.

And neither leak much.

most of the odds-and-sods diodes are acting to subtract 0.6V approx during charging.

Steve

[Bandgap]

The 1N4001 has a PIV of only 50V - an unloaded 'dynamo' (alternator) can easily exceed this value.

Shottky barrier diodes have about 0.2 - 0.3V forward drop, but PIV starts at 20V for the most common types - 90V types can be had (at a price), they also have higher leakage.

If using SB rectifiers, a shunt regulator is advisable.

Thanks IanField
Sorry everyone. In4001=50V, 02=100V, so go for : 1N4003=200V 04=400V, 05=600V, 06=800V, or 07=1,000V
All about the same voltage drop, and all about the same price.
Leakage is a few uA
Steve

[Bandgap]

By the way, one of the LEDs is not going to have any smoothing across it and will flash like crazy at slow speed.
- anctually, the other two will flash almost as much.

This is a bit anoying when you are behind that handlebars, but must be awful when you are in front of them.

Having such a flashy system on my utility bike, I get emarassed when I have to follow a car in traffic - the light is right in theor back window and must be horribly distracting.

I have a small addition to the circuit proposed that will cut the flash - using a small manual push button - it can be one of the tiny ones.
Anyone interested?

The other Steve