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Thread: Standlight circuits (Will this circuit work?)

  1. #31
    Flashaholic* znomit's Avatar
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    Default Re: Will this circuit work?

    Great thread here.
    I've just built up a 5 led bikelight with a 1F 5.5V supercap across one led.
    I'm using Martins circuit 10 which has served me well in the past.

    I just wanted enough light for traffic light stops, seems to work well enough on the bench. The standlight led is on an oval polymer optic and the rest are on a cutter narrow quad.

    I did wonder about putting all 5 on super caps but was a little worried about lag with the capacitors charging up. Coming over some hills and going from 7 to 50kph quickly I don't want to wait for my lights to catch up. SteveK what do you think, is this an issue?

    Bandgap send me the circuit pics and I'll host them.

  2. #32
    Flashaholic* Steve K's Avatar
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    Default Re: Will this circuit work?

    Quote Originally Posted by znomit View Post
    Great thread here.
    I've just built up a 5 led bikelight with a 1F 5.5V supercap across one led.
    ....<snip>.....

    I did wonder about putting all 5 on super caps but was a little worried about lag with the capacitors charging up. Coming over some hills and going from 7 to 50kph quickly I don't want to wait for my lights to catch up. SteveK what do you think, is this an issue?

    Bandgap send me the circuit pics and I'll host them.

    Hard for me to say if it will be a problem or not. The best estimate I can make is with the equation I = C x dv/dt, where I is capacitor current (amps), C is capacitor value (farads), and dv/dt is the change of capacitor voltage over time.

    Using I = 0.5A, C = 1F, and dv = 3.5v, then dt is 7 seconds. i.e. if you assume that all of the dynamo current is dedicated to charging up the supercap instead of flowing through the LED, then it will take 7 seconds to get the cap charged up. That means that you'll be riding for 7 seconds without any light from the LED.

    Of course, this is a simplification to allow a quick analysis. In reality, the supercap probably won't discharge all the way to 0v, and some dynamo current will flow through the LED even at low capacitor voltages. Still, it does indicate that the LED will be dim for a while.

    Based on this, I'd be reluctant to put supercaps across all of the LEDs. It would be nice to have at least one LED that's providing full output at low speeds right away.

    good luck,

    Steve K.

  3. #33

    Default Re: Will this circuit work?

    I've had no problems with supercaps across all the LEDs. On every light I've built this way they light pretty much immediately, even on the 6LED big guy. The video I just posted in my MC-E dyno light thread starts with a fully discharged capacitor bank (probably close to 0V because I have balancing resistors on the supercaps). You can see the light comes online right away once I'm rolling.

  4. #34
    Flashaholic* znomit's Avatar
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    Default Re: Will this circuit work?

    Thanks Steves. On the bench it takes a long while for the 5th LED to come on if the cap is fully discharged, but very quick subsequently so yeah I guess the lag wont be significant while rolling.

    Is there any reason not to use a similar setup on a battery light? Any driver-hates-supercap issues? For roadies a helmet light is not needed but you need something for redundancy in case the main light goes out. Supercaps would give you enough light to see while you stop and reconnect your battery, or when your bflex hits the battery protection limit (I'm sure we all have stories!) etc ...

  5. #35
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    Default Re: Will this circuit work?

    Here is my idea for a slightly better controlled standlight.


    Thanks to znomit for putting this image up for me

    The LM317L initially steals 100mA from the dynamo to charge the supercap, but once charged - to 5.3V (in 10s for a 1F 'Amp-type' super cap) - current consumption drops to 5mA (this is unavoidable as the 317 needs its voltage setting resistors to draw 5mA).
    This reduction in current once charged is a distinct advantage over zener-based charging schemes which keep drawing current after charging.
    There is also a simple one transistor circuit that can be used, except that it needs more supporting components.
    Other 5V 100mA three terminal regulators could be user, but they have to tolerate big capacitors on thier output when the input is open circuit. not all can cape with this like the 317 can.

    Charging the supercap to close to its maximum voltage means that it stores more energy than charging it to only 3.6V, so more standby time.

    EDIT - Following Alex Wetmore's work (see further down), I should point out that tolerance build-up can cause this circuit to produce more than 5.5V.
    So if you build it, check the output voltage and reduce the value of the 820 Ohm resistor if the output turns out to be too high.

    The resistor sets the initial discharge current.
    So for 20mA - which would last quite a while but not be very bright -
    for 20mA chose V/I=R = (5.3-3.6)/0.02=85ohms.

    If using a 'mA' type supercap, subtract its internal resistance (from its data sheet) from the resistance you get from the calculation. mA types will charge slower.

    The asterisk indicates where a diode would be required to protect the 317 if either the top led, or the bridge rectifier, leaked more than a few mA.
    But as they don't, no diode is needed.

    The 0.1uF input capacitor is required for stability on the input of the 317.

    This circuit is untesed, but the 317 is a pretty predictable beast.

    Steve
    Last edited by Bandgap; 11-19-2008 at 04:54 AM.

  6. #36

    Default Re: Will this circuit work?

    That's a nice idea! I toyed with the idea of IC regulated cap charging a few months ago but I kept coming up with overly complex ideas... This is as simple as it gets! I like the fact that you can control output current from the fully charged supercap, I bet this setup will yield some exceptional standlight runtimes! Also, it seems that the circuit can be scaled up for more LEDs. Check me if I'm wrong here, but with a few component swaps you could set it up to run 2LEDs from a 3LED array or 3LEDs from a 4 LED array. I like it!

  7. #37
    Flashaholic* Steve K's Avatar
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    Default Re: Will this circuit work?

    hey Mr. Bandgap,

    As a long-time user of the LM317, I can't help but notice that you haven't included a protection diode connected from the output to input of the regulator. The datasheet recommends one in the event of a shorted input, but the circuit should be protected from that.

    The datasheet also has an "absolute max" rating for the input-output voltage of -0.3v, and it seems possible that this might be exceeded. Of course, this usually assumes that the current could be significant. In this circuit, the current would be limited to the leakage current of the diode bridge in addition to whatever the upper LED would conduct at a very low Vf.

    Was the omission intentional, and if so, what was the justification?

    Overall, I think this circuit strikes a nice balance between "ultra simple but short run time" and "very complex but long run time".
    I'll still be using a nicad and the Zetex boost circuit, though. :-)

    good to hear from you again!

    Steve in Illinois

  8. #38
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    Default Re: Will this circuit work?

    Quote Originally Posted by Bandgap View Post
    The 0.1uF input capacitor is required for stability on the input of the 317.
    Many people use a "smoothing" cap that appears in parallel with the 0.1 uF input cap shown here, and it is typically 4 to 5 orders of magnitude larger than this input cap. I assume the 0.1 uF input cap forms the dominant pole of the loop? If so, retaining the "smoothing cap" should not be a stability problem. It will just take longer for the loop to settle.

  9. #39

    Default Re: Will this circuit work?

    Quote Originally Posted by Bandgap View Post
    Here is my idea for a slightly better controlled standlight.

    Cool! I've tried to work your mods into the original drawing. Please tell me if this drawing is accurate:



    I'm assuming that this standlight charging circuit could just be dropped in Martin's circuit 8 in between C1 and the LEDs in his design? If the goldcap were a pair of 1.5F supercaps in series instead of a single, how do the internal resistances of the caps in series fit into the calculation for the discharge rate resistor? Do you just subtract the sum of both internal resistances? I guess it might also be necessary to change the values of the control resistors for the LM317L so that it sets a different max voltage for the supercaps in series?

    Thanks again!
    Last edited by syc; 11-20-2008 at 12:41 AM.

  10. #40
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    Default Re: Will this circuit work?

    That's it.

    Except the left hand capacitor is not a smoothing cap, but an 0.1uF stability cap for the regulator.
    It has only a tiny smoothing effect on the main voltage waveform.
    It should be near to the pins of the regulator chip.

    As I said somewhere above, I am not a great fan of big smoothing capacitors on dynamos unless you have hub dynamo and you spend a lot of time cycling below 6mph (10km/h)
    If you must have a big smoothing capacitor, you must put Steve K's diode in - see below in this post - or put a diode in the asterisk position and keep the smoothing capacitor up-stream of this.

    FrontRanger
    It is years since I understood poles and zeros!
    I think it is the impedance of the capacitor at potential oscillation frequencies that is important.
    Big electrolytic capacitors tend to have quite high impedance in the MHz range where I suspect the 317 would oscillate if it has the chance.
    Mid-sized ceramic capacitors have low impedance at low-MHz frequencies.

    Overall
    A 0.1uF ceramic (I would use a modern multi-layer ceramic) right close to the pins of the LM317 will keep it happy. There things are only 5x5mm at the biggest.
    It is the price you pay for bringing in an electronic regulator.

    This is National Semiconductors take on that particular capacitor from the LM317L data sheet http://cache.national.com/ds/LM/LM317L.pdf :
    "An input bypass capacitor is recommended in case the regulator is more than 6 inches away from the usual large filter capacitor. A 0.1μF disc or 1μF solid tantalum on the input is
    suitable input bypassing for almost all applications. The device is more sensitive to the absence of input bypassing when adjustment or output capacitors are used, but the above values will eliminate the possibility of problems."

    Syc
    I am not sure having caps in series would gain you anything except the ability to have both leds on when stopped.
    You will have a tough job charging two 5.5V caps in series using a similar arrangement because to get them full, you would need at least 11V from somewhere to charge them.
    And the 317 needs a volt or two to work.
    And unless they are hand-selected for balance, supercaps in series need some kind of balancing or limiting circuit (like an led directly across each one) if they are to be reliable - not difficult, but more hassle.
    Aside from the cosmetic difference, if you need more light you might as well put twice the (small) current through the one led that is easy to drive.
    If you have two caps, put them in parallel and get a longer or brighter standby.
    For the record, the resistances of series capacitors add.
    The resitances of parallel capacitors is a bit more complex, but for identical capacitors you divide the series resistance of one by the number of capacitors.
    If integrating with Martin's circuit 8 http://www.candlepowerforums.com/vb/...post&p=2679207, feed the 317 and its little capacitor from the top of LED5, and connect the discharge resistor to the centre point of LED5 and LED6.

    Steve K
    Most excellent to hear from you again.
    My justification for missing out a diode from output to input is: AFAIK, the internal junction can take care of small leaks.
    Providing you do not have an external low impedance conduction path from input to anywhere important, you don't need an external diode across the output-input, or in series with the input.
    So long as there is no big smoothing capacitor, and you stick below the reverse breakdown voltage of the led, this design has no low-impedance paths from the input of the 317.
    For reference: the data sheet says: "The discharge current depends on the value of the capacitor, the output voltage of the regulator, and the rate of decrease of VIN. In the LM317L, this discharge path is through a large junction that is able to sustain a 2A surge
    with no problem. This is not true of other types of positive regulators."


    Phew.

    Steve
    Last edited by Bandgap; 10-29-2008 at 06:09 AM.

  11. #41
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    Default Re: Will this circuit work?

    Quote Originally Posted by Bandgap View Post
    FrontRanger
    It is years since I understood poles and zeros!
    I think it is the impedance of the capacitor at potential oscillation frequencies that is important.
    Thanks for the link to the data sheet. From what's written there, the 0.1 uF input cap is not for stabilizing the regulator loop but rather for bypassing noise on the input side of the regulator. (It's not even part of the feedback loop.) So if desired, you should be able to use a "smoothing cap" with this circuit.

  12. #42

    Default Re: Will this circuit work?

    Quote Originally Posted by syc View Post
    Cool! I've tried to work your mods into the original drawing. Please tell me if this drawing is accurate:



    I'm assuming that this standlight charging circuit could just be dropped in Martin's circuit 8 in between C1 and the LEDs in his design? If the goldcap were a pair of 1.5F supercaps in series instead of a single, how do the internal resistances of the caps in series fit into the calculation for the discharge rate resistor? Do you just subtract the sum of both internal resistances? I guess it might also be necessary to change the values of the control resistors for the LM317L so that it sets a different max voltage for the supercaps in series?

    Thanks again!
    I built this circuit and it is working nicely. I just have it on a breadboard now, I haven't finished my headlight.

    My only concern is getting better life out of the supercap. On my Lumotec IQ Fly headlight I get about 5 minutes of life at what appears to be half brightness. With this circuit and no discharge rate resistor I get about a minute at more like 1/4 brightness. 50ohm resistor increases the life, but makes it even dimmer.

    I'm wondering if I bought the right supercaps. The better of the ones that I have is a NEC/Tokin FYH series. 5.5V rating and 1.0F.

    What is the best mechanism to measure internal resistance of the capacitor? Is ESR on the data sheet (20 ohms) the internal resistance? That actually sounds quite reasonable, but I don't think it is accurate. I'm seeing it deliver 100ma for a very short period, and then it drops down to about 25ma quickly before levelling off a bit.

    What are the best goldcaps/supercaps for this type of project? I see that Panasonic makes a HW series goldcap designed for high ma projects, but they are limited to 2.1 volts...not enough.

  13. #43
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    Default Re: Will this circuit work?

    Quote Originally Posted by Alex Wetmore View Post
    My only concern is getting better life out of the supercap. On my Lumotec IQ Fly headlight I get about 5 minutes of life at what appears to be half brightness. With this circuit and no discharge rate resistor I get about a minute at more like 1/4 brightness. 50ohm resistor increases the life, but makes it even dimmer.
    I understand that the IQ Fly also uses an XR-E. Perhaps not the R2 bin, but other than that, a fair comparison to your light. Why the difference? It may simply store more charge on its cap, as you mention. Alternatively, the Fly may use something fancier than a resistor (more efficient) for the standlight. With that said, the simplest fix may well be to increase the resistance and tolerate the reduced brightness (doesn't take much to be seen) or increase the capacitance, if you have room for it.

  14. #44

    Default Re: Will this circuit work?

    Quote Originally Posted by FrontRanger View Post
    I understand that the IQ Fly also uses an XR-E. Perhaps not the R2 bin, but other than that, a fair comparison to your light. Why the difference? It may simply store more charge on its cap, as you mention. Alternatively, the Fly may use something fancier than a resistor (more efficient) for the standlight. With that said, the simplest fix may well be to increase the resistance and tolerate the reduced brightness (doesn't take much to be seen) or increase the capacitance, if you have room for it.
    FWIW, I picked up a pair of 1.5F 5.5V Panasonic Gold Caps spec'd at 30 ohms (P/N P11343-ND at Digi-Key) for the simpler circuit. I was only planning on using one at a time for my projects, but you were saying that putting them in parallel was an option on your smart standlight circuit, right?

  15. #45

    Default Re: Will this circuit work?

    Quote Originally Posted by Alex Wetmore View Post
    My only concern is getting better life out of the supercap. On my Lumotec IQ Fly headlight I get about 5 minutes of life at what appears to be half brightness. With this circuit and no discharge rate resistor I get about a minute at more like 1/4 brightness.
    ps.
    What's the difference in performance when you wire it up using the simple supercap charging circuit? How much of a difference in light intensity do you notice during the charging phase, and any difference in standlight duration?

  16. #46
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    Default Re: Will this circuit work?

    Quote Originally Posted by syc View Post
    ... you were saying that putting them in parallel was an option on your smart standlight circuit, right?
    syc, you must be thinking of another poster, as I have not smart standlight circuit. I was just replying to Alex that increasing the capacitance would help his standlight runtime. One way to do that is put more of the same capacitor element in parallel.

  17. #47

    Default Re: Will this circuit work?

    Quote Originally Posted by syc View Post
    ps.
    What's the difference in performance when you wire it up using the simple supercap charging circuit? How much of a difference in light intensity do you notice during the charging phase, and any difference in standlight duration?
    With the simplest circuit (just a supercap in parallel with one of the LEDs, no resistor) it takes a couple of seconds to charge the capacitor and the LED is ramping up in brightness as that occurs. The capacitor charges very quickly. With the "smart" circuit the primary LED doesn't dim noticably as the capacitor charges (so it is instantly bright), but it takes more like 20-30 seconds to charge the capacitor fully. Discharge seems similar in both cases...which makes sense since the LM317 is out of the way during discharge.

    alex

  18. #48

    Default Re: Will this circuit work?

    Quote Originally Posted by Alex Wetmore View Post
    I built this circuit and it is working nicely. I just have it on a breadboard now, I haven't finished my headlight.

    My only concern is getting better life out of the supercap. On my Lumotec IQ Fly headlight I get about 5 minutes of life at what appears to be half brightness. With this circuit and no discharge rate resistor I get about a minute at more like 1/4 brightness. 50ohm resistor increases the life, but makes it even dimmer.
    I think I've figured out one of the major problems with this circuit.

    The LM317 is setup to deliver around 6 volts (actually too much for the supercap). However the LED is in parallel with the supercap and seems to prevent more than 3.5 volts or so from being dropped across it. If I remove the LED from being in series to charge the supercap, then unplug my power supply and plug in the standlight LED I get both a brighter LED and a longer lifetime.

    Since the supercap never seems more than 3.5 volts it isn't charging fully. It is wasting 66% of the useful charge because the LED stops having useful light below about 2.5V. I'm getting 1V of useful potential out of the supercap instead of the ideal 3V of useful potential (3 + 2.5 = 5.5, the capacitor's maximum rating).

    I tested this by disconnecting the second LED while charging the capacitor, then unplugging my power source and plugging in the second LED. It had a longer life and was brighter than in the normal circuit.

    Any ideas on how to address this? Would it be easier to have a 3rd LED that only came on as a standlight?

    alex

  19. #49

    Default Re: Will this circuit work?

    Sorry, my mistake (shouldn't rush off posts just before bed). It was Bandgap's design.

    Quote Originally Posted by FrontRanger View Post
    syc, you must be thinking of another poster, as I have not smart standlight circuit. I was just replying to Alex that increasing the capacitance would help his standlight runtime. One way to do that is put more of the same capacitor element in parallel.

  20. #50
    Flashaholic* Steve K's Avatar
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    Default Re: Will this circuit work?

    hi Alex,

    You've stumbled across the fun and annoying aspect of standlight circuits..... the simple ones don't work that well. :-)

    The one suggested by Bandgap relies on some isolation between the led and the supercap in order to fully charge the supercap. A 500 ohm resistor is a partial way to achieve this, but involves a lot of I x R losses.

    A current source could also work, but involves more parts and will have some headroom requirements (i.e. will require some voltge to just run the circuit).

    Something that I'm planning to use in an upcoming standlight circuit is to use a small boost converter to drive the led. The Zetex ZXSC310 can run from voltages from 0.8v to 8v, so it could use quite a bit of the charge in the supercap.

    Plenty of options, so you have an excuse to buy parts and play around. :-)

    Steve K.
    (known on the i-bob list simply as Steve Kurt)

  21. #51

    Default Re: Will this circuit work?

    Quote Originally Posted by Steve K View Post
    Something that I'm planning to use in an upcoming standlight circuit is to use a small boost converter to drive the led. The Zetex ZXSC310 can run from voltages from 0.8v to 8v, so it could use quite a bit of the charge in the supercap.
    That looks like it'll do the job, but surface mount gives me a headache and inductors give me an even larger headache (unless I don't need to think about how they are working). Any ideas on similar devices that are designed for through soldering?

    I guess I also don't understand how you charge the capacitor to ~5 volts without a large resistor. The ZXSC310 will help you make use of the 2.5 down to .8 volt range of the capacitor, but it won't let you charge it over the 3.5 volt range clamped by the LED. Is that correct? If you put a large resistor between the capacitor and ZXSC310 then it will still restrict the current.

    Figuring this out and dealing with optics really makes me appreciate the IQ Fly that much more. It is expensive but a well dialed in solution. It's fun to build these things, but not as simple as it first appears.

    alex

  22. #52
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    Default Re: Will this circuit work?

    Quote Originally Posted by Alex Wetmore View Post
    I think I've figured out one of the major problems with this circuit.

    The LM317 is setup to deliver around 6 volts (actually too much for the supercap). However the LED is in parallel with the supercap and seems to prevent more than 3.5 volts or so from being dropped across it. If I remove the LED from being in series to charge the supercap, then unplug my power supply and plug in the standlight LED I get both a brighter LED and a longer lifetime.

    Since the supercap never seems more than 3.5 volts it isn't charging fully. It is wasting 66% of the useful charge because the LED stops having useful light below about 2.5V. I'm getting 1V of useful potential out of the supercap instead of the ideal 3V of useful potential (3 + 2.5 = 5.5, the capacitor's maximum rating).

    ...

    Any ideas on how to address this? Would it be easier to have a 3rd LED that only came on as a standlight?
    Hi Alex, your analysis looks correct to me. As you point out, during discharge, your capacitor is limited from discharging much below 2.5 V, because the LED is nearly off at that point. And during charging, the same LED clamps it to around 3.5 V or so. (It will actually be a bit higher than the LED voltage because of the resistor between the cap and the LED.)

    As Steve K. says below, the simple standlights don't work that well. Let's say that "simple" means only passive elements and diodes here. With that restriction, I don't think you can do much about the first problem (the 2.5 V min voltage). The boost regulator that Steve K. mentions sounds promising for extracting as much charge as possible out of that cap, but that's a bit more complicated. If you want to keep it simple, you might be able to attack the second problem (3.5 V max voltage). To maximize the amount of charge stored on the cap, I scribbled out this little circuit:




    If each LED is 3.6 V @ 500 mA, you'll get 7.2 V at the top. An ideal clamp would charge your cap to 5.5 V (assuming no margin). If standard Si diodes drop 0.7 V, and Schottky diodes drop 0.4 V, then two of the former plus one of the latter will clamp the cap to 7.2 - 1.8 = 5.4 V, just under the 5.5 V max. The resistor must be sized so that only a small current goes through it during normal steady-state operation. Let's say we choose 10 mA for that current. Then the resistor should be (5.4-3.6 V)/10mA = 180 ohms. The values of the diode drops are approximate, so you'd have to play with the resistor value a bit, or measure the diodes accurately and calculate exactly.

    I don't have the parts laying around to try this myself. This won't be as good as the boost regulator, but will nearly triple the usable stored charge if it works as planned (3.5-2.5) becomes (5.4-2.5). Hope it helps.

  23. #53

    Default Re: Will this circuit work?

    I didn't get all that the discharge rate resistor was doing for me. It isn't only slowing down the capacitor discharge, but it also allows the capacitor to charge to a higher voltage.

    With a 150 ohm discharge rate resistor I have a 10ma current flowing through it when power is applied to the circuit. The voltage drop across the resistor is 1.4 volts. That allows my capacitor to charge an extra 1.4 volts above the 3.6 volts, giving me 5 volts at the capacitor.

    During discharge I start with around 110ma going through the circuit at first, and it drops down to around 80ma where it stays for a long time. That works out pretty well.

    I had made the mistake of trying to maximize discharge rate, without realizing that it was also hurting my maximum possible charge with this circuit.

  24. #54
    Flashaholic* Steve K's Avatar
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    Default Re: Will this circuit work?

    Quote Originally Posted by Alex Wetmore View Post
    I didn't get all that the discharge rate resistor was doing for me. It isn't only slowing down the capacitor discharge, but it also allows the capacitor to charge to a higher voltage.

    With a 150 ohm discharge rate resistor I have a 10ma current flowing through it when power is applied to the circuit. The voltage drop across the resistor is 1.4 volts. That allows my capacitor to charge an extra 1.4 volts above the 3.6 volts, giving me 5 volts at the capacitor.

    During discharge I start with around 110ma going through the circuit at first, and it drops down to around 80ma where it stays for a long time. That works out pretty well.

    I had made the mistake of trying to maximize discharge rate, without realizing that it was also hurting my maximum possible charge with this circuit.

    hi Alex,

    The resistor does serve two functions, and optimizing one tends to degrade the other.

    I've got an idea that may help, but of course, it adds parts and eats up a little voltage. Not much tho.....
    The idea is to add a transistor switch between the LED and the cap/resistor. The transistor only turns on when the dynamo is stopped. The means that the original resistor only has to control the discharge current.



    I've left it to the student to calculate the resistor values. :-)

    The only thing that requires attention is the base resistor value. Set the base current to be about 0.03 of the collector current (roughly). Make sure that you do this calculation for the lowest desired operating voltage of the cap.

    I certainly like most surface mount parts, and it can work even with crude tools. I just hack up a copper clad board with a dremel tool to make most circuits. This is an example of the Zetex boost circuit as used in a rear standlight.



    That ought to provide some options.

    Steve K.

  25. #55

    Default Re: Will this circuit work?

    Quote Originally Posted by Steve K View Post
    The idea is to add a transistor switch between the LED and the cap/resistor. The transistor only turns on when the dynamo is stopped. The means that the original resistor only has to control the discharge current.
    Cool - I was wondering how you'd use a transistor switch to isolate the supercap during the charging - I was going to ask a friend with an ECE background about it, but looks like CPF came through first.

    By the way, my ECE friend also pointed out the wasted charge in the supercap, and said that he could sketch out a simple current pump circuit for me that could squeeze out more of the charge. Then he wandered down a path with adding a pic microcontroller...
    I'll try to get the current pump out of him before feature bloat takes over, and post some updated diagrams.

  26. #56

    Default Re: Will this circuit work?

    Quote Originally Posted by FrontRanger View Post
    As Steve K. says below, the simple standlights don't work that well. Let's say that "simple" means only passive elements and diodes here. With that restriction, I don't think you can do much about the first problem (the 2.5 V min voltage). The boost regulator that Steve K. mentions sounds promising for extracting as much charge as possible out of that cap, but that's a bit more complicated. If you want to keep it simple, you might be able to attack the second problem (3.5 V max voltage). To maximize the amount of charge stored on the cap, I scribbled out this little circuit...
    I was wondering if you could simplify things a little more with this circuit?


    (sorry if its nonsensical - I've been reading up a little on electronics, but have been slow about it)

    I wasn't crazy about the supercap soaking up current from both LED's while charging, so went back to the supercap parallel to only a single LED. What I'm trying to get is the benefit of the 150 ohm resistor that Alex reported, but on a very simple circuit. Would this circuit allow the supercap to charge up to around 5V, as well as throttle the discharge rate?

  27. #57
    Flashaholic* Steve K's Avatar
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    Default Re: Will this circuit work?

    Quote Originally Posted by syc View Post
    I was wondering if you could simplify things a little more with this circuit?
    <image removed... see above>

    (sorry if its nonsensical - I've been reading up a little on electronics, but have been slow about it)

    I wasn't crazy about the supercap soaking up current from both LED's while charging, so went back to the supercap parallel to only a single LED. What I'm trying to get is the benefit of the 150 ohm resistor that Alex reported, but on a very simple circuit. Would this circuit allow the supercap to charge up to around 5V, as well as throttle the discharge rate?
    the circuit does simplify things, but the performance is greatly reduced. The supercap will only charge up to the LED's Vf, which means you won't get much use of the stored energy.
    [to review: energy = 0.5 x C x V^2]

    The use of the diode and resistor does allow the supercap to charge up faster while producing a dimmer standlight with a longer run time.

    The circuit that uses the voltage regulator does divert current from the LEDs, but only while charging. Since it charges to a higher voltage, it stores more energy, but does reduce light output longer than when the supercap is only charged to Vf. This is a normal trade-off, and up to the user to decide if it is appropriate.

    Everyone has to decide what factors are important for their use. The factors include size, cost, complexity, efficiency, ease of assembly, reliability, run time, brightness, and how it will be used. (I may have missed some factors...)

    To me, the voltage regulator and boost regulator are pretty simple, and the supercap is expensive and big and limited in ability to store energy. As such, I tend to use a AA nicad with a boost regulator and am pretty happy. There are times when the supercap will be better than the nicad (i.e. if you expect that there won't be much time to charge, and the nicad or cap will completely discharge). You'll need to pick the best solution for your use.

    regards,
    Steve K.

  28. #58
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    Default Re: Will this circuit work?

    Quote Originally Posted by syc View Post
    I was wondering if you could simplify things a little more with this circuit?


    (sorry if its nonsensical - I've been reading up a little on electronics, but have been slow about it)

    I wasn't crazy about the supercap soaking up current from both LED's while charging, so went back to the supercap parallel to only a single LED. What I'm trying to get is the benefit of the 150 ohm resistor that Alex reported, but on a very simple circuit. Would this circuit allow the supercap to charge up to around 5V, as well as throttle the discharge rate?
    Quote Originally Posted by Steve K View Post
    the circuit does simplify things, but the performance is greatly reduced.
    Yes, and...

    Quote Originally Posted by Steve K View Post
    The supercap will only charge up to the LED's Vf
    ... it's not even that good. The cap will only charge up to one diode-drop below the LED's steady-state Vf. As Steve K showed that's a quadratic degradation in stored energy.

    The circuit I posted in #52 dramatically improves this while not electrically overstressing your cap. Increasing the stored energy while restricting design to only two-terminal devices was the idea there. The circuit in post #56 works on the same clamping principle as the one in #52, except that the clamp is referenced to only one Vf above ground instead of two, and the clamp voltage is smaller.

  29. #59
    Flashaholic* Calina's Avatar
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    Default Re: Will this circuit work?

    This is likely the most interesting thread that I've seen on CPFabout stand lights and super caps.

    I'm not sure it deserve a sticky status since stand lights are unfortunately not a very popular subject but it should at least be included in a "thread of interest" on the first page.

    Thanks to all for this very informative thread.
    Health is merely the slowest possible rate at which one can die.

  30. #60
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    Default Re: Will this circuit work?

    Quote Originally Posted by Alex Wetmore View Post
    I think I've figured out one of the major problems with this circuit.

    The LM317 is setup to deliver around 6 volts (actually too much for the supercap). However the LED is in parallel with the supercap and seems to prevent more than 3.5 volts or so from being dropped across it. If I remove the LED from being in series to charge the supercap, then unplug my power supply and plug in the standlight LED I get both a brighter LED and a longer lifetime.
    alex
    Ooops.
    I have been off-line for a few days.

    Circuit charging at 6V - which is indeed too much for the supercap

    My calculations show 5.3V for 240 and 820 Ohms.
    Did you use exactly those?
    If so, the 820 need to be a little lower.

    Also - if the capacitor voltage never gets to 5V under charge, too much current is being drawn by the led, the current limiting resistor needs to be higher.
    This is a simple circuit - only really good for 20 or 40mA discharge.

    EDIT
    Just noticed the the wise Steve K has already said this: The one suggested by Bandgap relies on some isolation between the led and the supercap in order to fully charge the supercap. A 500 ohm resistor is a partial way to achieve this, but involves a lot of I x R losses.


    With a 30 Ohm internal impedance, the supercap output voltage will immediately drop as soon as you draw current because of this resistance.

    Hope I am not missing the point.

    By the way: If you want to have a high-current (greater than 50mA ) standby, supercaps are the wrong technology in my view.

    Steve
    Last edited by Bandgap; 11-18-2008 at 07:51 AM.

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