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  1. #1
    Senior Member BikeManDan's Avatar
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    Knowledge of electronics needed

    Here's what I'd like to do but do not know how:

    When a line goes from unpowered to powered, I want to briefly connect two leads (as to mimic a momentary pushbutton) and then have them disconnected.


    Actual scenario is this:
    I'm going to rig my PB Superflash to be supplied by my 14.4V li-ion battery using a 3 pin voltage regulator to 3V. When I supply power to the Superflash, I want it to automatically turn on. In order to this, I'd have to simulate a push of its momentary pushbutton. I will solder directly to the pushbutton contacts but need the timer circuit or whatever mechanism to simulate the press.


    How would this be done? Schematic?

  2. #2
    Cries on hills supton's Avatar
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    I'm not familar with the batteries in the Superflash--but I know that using a common 3 pin linear regulator is lousy in regards to efficency. You're tossing quite a bit of energy into the air with that approach.

    I'm thinking a one shot. LM555 or 74HC123; been long time since I messed with the '555 but have used the '123 a number of times. Interfacing to the light would be the harder part--how is the switch on that wired?
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  3. #3
    Senior Member BikeManDan's Avatar
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    Don't know yet but my assumption is that its simply two leads connected to the pushbutton

  4. #4
    Cries on hills supton's Avatar
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    But is it active high or active low? Gotta pull the line one way or the other. Most likely, active low. With that, it should be easy to drive with an NPN or NFET. One shot -> transistor -> switch, plus a few support components.
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    Stupid question: why do you want to power your superflash with the 14.4 V battery? Wouldn't going with a basic set of NiMH rechargeables be far, far simpler?
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    Senior Member BikeManDan's Avatar
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    Quote Originally Posted by Ngchen View Post
    Stupid question: why do you want to power your superflash with the 14.4 V battery? Wouldn't going with a basic set of NiMH rechargeables be far, far simpler?
    My main goal is to be able to turn on both my tail lights (Superflash and xenon flasher) with a single on/off pushbutton. But I figured while I'm at it, I might as well never have to worry about batteries and just utilize the power already on hand


    Someone else found this for me utilizing a relay. Seems to do what I want. Solid state would be nice but this is simple and I can relatively understand it
    http://www.the12volt.com/relays/page5.asp#ctm

  7. #7
    Scott n4zou's Avatar
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    Instead of butchering your PB Superflash just build yourself a new light and use it in addition to the PB Superflash. You can just use resistors to set the current for an LED. This would be a constant on taillight which is best for after dark riding. You could blink your PB Superflash in addition to this light. Here is an on line resistance calculator so you know how much resistance for the LED and voltage you use.
    http://led.linear1.org/1led.wiz
    I would use this Cree high power RED LED.
    http://www.dealextreme.com/details.dx/sku.1776

    If you want it to blink just use an LED flasher can.

    http://autolumination.com/equalizers.htm

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  8. #8
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    The pushbutton is a momentary that cycles the light through the 3 modes, blink, steady, off. The unit is glued together, so I didn't check to see if it grounds or goes high to toggle.

    But I tried a little experiment. Set the light to blink mode, then took the batteries out. Left it overnight. Put the batteries back in, and it was still in the same mode.

    If yours does this, too, then all you have to do is get more efficient with your voltage regulation. The Superflash use 2 AAA cells in series for 3v. Like supton says, a 3 terminal device won't be too efficient, but may not be a deal killer at these currents. A DC-DC buck converter might work. More expensive, might be fun to try.
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  9. #9
    Thread Killer evblazer's Avatar
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    Would something like this work? For a while I was thinking of getting all the battery lights on my bike working off the same pack or a pack charged by a hub generator and using these for the 2 battery ones

    Switching voltage regulator Up to 30V input range
    83% typical efficiency, up to 87%
    <2% ripple
    1A output (continuous)


  10. #10
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    Yes, something like that converter should work. I'd probably go for the adjustable version to keep the output under 3v, unless it's proven that the Superflash can handle more. I suppose it's possible that the funky pulsing load might cause some problem with a converter, but I'm no expert. For $15, I'd probably just try it.
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  11. #11
    Cries on hills supton's Avatar
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    Just make sure that the efficency is still high at the actual current draw--sometimes the efficency drops dramatically, leading you to not really gain all that much. A decent sized bypass capacitor may be required to replace the batteries in the unit also, as the unit expects such and such battery impedance; and if the convertor is a distance away, the unit may act erratically. I'd think it'd be easy to add 10-100uF to "replace" the battery in the unit, allowing any convertor to work properly, regardless of wiring distance.
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  12. #12
    Senior Member BikeManDan's Avatar
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    Quote Originally Posted by Litespeedlouie View Post
    The pushbutton is a momentary that cycles the light through the 3 modes, blink, steady, off. The unit is glued together, so I didn't check to see if it grounds or goes high to toggle.

    But I tried a little experiment. Set the light to blink mode, then took the batteries out. Left it overnight. Put the batteries back in, and it was still in the same mode.

    If yours does this, too, then all you have to do is get more efficient with your voltage regulation. The Superflash use 2 AAA cells in series for 3v. Like supton says, a 3 terminal device won't be too efficient, but may not be a deal killer at these currents. A DC-DC buck converter might work. More expensive, might be fun to try.
    Well thats an interesting note. I briefly considered this as a possibility but never tested it! I will give it a try. If thats true then this whole thing is much simpler than I imagined

  13. #13
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    Quote Originally Posted by dwoloz View Post
    Well thats an interesting note. I briefly considered this as a possibility but never tested it! I will give it a try. If thats true then this whole thing is much simpler than I imagined
    The unit memorizes last selected mode just as long as there is some voltage left. Usually there is a capacitor in the unit so this really works but you have to use a diode in the power line to keep the capacitor from discharging.
    To make it 100% reliable it would be better to use one AAA cell as a back-up. If you replaced the second cell by a 1N4148 diode, the voltage would be under 1V and no current would be drawn from the cell. Then the last selected mode will be memorized forever, at least I hope it should work this way.

    It's a good idea to power it by a DC-DC converter. The cheapest IC I know for that use is the good old MC34063 used in car CL adaptors for cell-phones. It's very suitable. You can even use the 12V CL adaptor as it is only with some output voltage adjustment done.
    With a 3-pin regulator you woud have wasted 3/4 of the input power. Instead of that you could use 4 Superflashes in series with the some power consumption!

  14. #14
    Senior Member Zero_Enigma's Avatar
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    Question about resisters.

    I'm curious if I have 2 x Cree Q5 @ 3.7v vF for ~273lm (if I understand it correctly at 1A) and wish to power those two LED's with a 12v powersupply will resisters be more like a bottle neck and only give the LED's what they need and not waste and of the extra power of the 12v?

    Think of it like a water barrel full of water but you install a size down hose adaptor then run a smaller hose to say your plants. Because the hose is narrow and you have a dripline for the plants the plants only get the little water they need but the water barrel only loses like 0.0001% water to that plant and nothing else is wasted or lost.

    I'm not sure if I'm explaining myself properly. I'm not the best explainer of things. Basically I'd like to know if the 12v battery with oh lets say 5Ah capacity can I bottle neck that battery in a way with resisters so I can extend the juice I have in the battery to the LED's?
    Zero_Enigma

  15. #15
    Cries on hills supton's Avatar
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    I'm not so good at explaining these things, but here goes.

    Using your barrel of water, voltage is like height, and current is the flow of water. You're right: if you are looking just for water, it doesn't matter how fast or how slow you pull water, you'll have water. Problem is, you need more than just water for this analogy to work. This is why analogies can be so hard...

    It is more like you need to instead create shaft power. You need such and such torque, but you also need it at such and such rpm. You could use a high source of water, and a large diameter water wheel, and a slow flow rate of water. Or you could use a low source of water, a smaller diameter wheel, and a large amount of water flowing. In end, the shaft needs to turn at the same speed, and with the same torque applied.

    I probably didn't do it justice, but that's the problem: you are dealing with power, not just current or voltage. Power is the product of voltage and current.

    Resistors, in this usage of limiting voltage to an LED, work by basically changing power from electrical current and voltage into heat. Heat is basically wasted power, unless if you are intentionally heating something. LED's are nice devices in that they are nearly constant in voltage drop. A 2V LED will be approximately 2V whether the current through it is 1mA or 20mA. A resistor will set the current, but if the applied voltage varies, then the current through the LED will vary. Not a problem for most batteries, as the voltage droop isn't excessive before the battery conks out.

    Now, as to your question: a 12V battery being regulated down to 3.7V (I think that's a typical white LED): the resistor must drop the voltage from 12V to 3.7V. Power delivered by the battery is always 12V times the current draw; but power delivered to the LED will only be 3.7V times the current draw. As you can see, the efficency would be right around 30%, using a resistor.

    Let's say you had a 4V battery to power the LED. The efficency goes up over 90%, as very little power is wasted in the resistor.

    Long story short: if you used a fancy DC-DC convertor, your 12V battery powering a 3.7V LED would last perhaps 3 times longer than if you just used a resistor to set the current.
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