Banggood Network
Results 1 to 26 of 26

Thread: Electrical help....

  1. #1

    Sigh Electrical help....

    EDIT: See post number 3

    I'm currently designing a human powered LED flashlight. It will be powered by a Mabuchi 4.5v Generator (RC-280RA-2865) (1.36A, 3.47W)

    https://uwm.courses.wisconsin.edu/d2...topicId=534837

    I'm trying to find a LED Cluster reflector housing to give about 1-2 watts of light. I'm done a lot of searching but haven't had much luck. Does anyone have any ideas of what I should use to accomplish this. Keep in mind i'm going to have to do all the wiring and model the casing for the flashlight. Thanks for your help!
    Last edited by goucho; 03-06-2007 at 03:01 PM. Reason: Changed my title to something more appropriate

  2. #2

    Default Re: LED Reflector...

    You could probably accomplish your goal with a single Luxeon or Cree or Nichia Jupiter or any one of a number of one-watt LEDs on the market along with a single reflector (an IMS 27mm would be great for this, or you could hit Dat2Zip's online store and snag one of the aluminum reflectors there to suit your specific LED selection) and a boost driver to drive the LED properly.

    Powering the driver would then be the issue, and you could do that with the generator and a few large capacitors to act as supply stabilizers slash current sources for the driver.

    Or, you could direct-drive the LED and make do with variable output based on the speed the genny's turning.

    oO
    O d d O n e
    Webdude of "OddOne's White LED Information"
    http://www.e-f-w.com/community/

  3. #3

    Default Re: LED Reflector...

    ^-- Thanks for your help. After some research I've found a suitable LED. It's a Luxeon LED that comes complete with collimating lens (10 degrees) and aluminum submount.

    Luxeon White Star/O - Batwing
    Specs:
    Max Continuous Current 350 mA
    Forward Voltage 3.42 Vf

    Knowing that the generator I'm using gives the following:
    Nominal Voltage = 4.5v constant
    @ Max Efficiency; Current = 1.36A , Output = 3.47W

    Here are my calculations for Resistance-->
    V = (4.5v - 3.42v) = 1.08v
    R = V/I = 1.08v / .35A = 3.0857 ohms
    P = IV = (.35A)*(1.08v) = 0.378 Watts

    I don't know a whole lot about resistors but I figure I would need a 1/2watt 10 ohm resistor (do they make a 5 ohm resistor?)...or maybe use four 1 ohm resistors in series... Am I correct thinking this?

    As for the Capacitor(s), how would I go about finding a correct one for my application? Also, if I needed multiple capacitors i would want to add then in parallel right? I am going to have them hooked up to an on/off switch to either charge or discharge the current generated.

    If someone could check over my work to make sure i'm on the right track that would be great. I am new to all of this so it's a great learning experience. Thanks for your patience and help in the matter!
    Last edited by goucho; 03-06-2007 at 03:04 PM.

  4. #4
    Flashaholic* mudman cj's Avatar
    Join Date
    Dec 2005
    Location
    Where corn and pigs are grown unimpeded by trees
    Posts
    1,827

    Default Re: Electrical help....

    This is an interesting project you have here. I hope you can share the results with us when it is finished.

    While that LED will work, you will definately get higher output for the same amount of energy consumed using one of the new high efficiency LEDs, namely Cree or Seoul. Direct driven with a 5 Ohm resistor (yes they make them in 1/2 Watt) would limit the current a bit more than in your calculations, so you might want to use three 1 Ohm resistors in series.

    Multiple capacitors would be connected in parallel to increase capacitance. Of course, most capacitors don't hold a charge for very long, so realize that the user of the device would have to recharge it if it sat for long. As far as what type of capacitor to use, that depends on how much energy you want to be able to store. You definately would want to find a capacitor with a high capacitance and a voltage limit not much above your input voltage. The reason for this is that a capacitor with a high voltage capability will have a lower capacitance for a given size because to sustain a higher voltage requires a thicker dielectric (or a better one).

    If you can afford them, the new ultracapacitors would give you the best energy storage capacity among capacitor choices. NessCap makes a 5.4V, 2.5 Farad capacitor selling for about $6 ea. That will store 11.25 Coulombs of charge at 4.5V. As the capacitor discharges, the voltage available to drive the circuit decreases exponentially, so while it will start out driving the LED to spec, the LED current and light output will die down. How quickly? Well, after about 32 seconds (the time constant for this circuit) there will be only 7 Coulombs of charge left in the capacitor, which will be able to apply 2.8V to the circuit. This will still drive the LED at about 220 mA. At 64 seconds, you only have 1.7 Volts driving the LED at 130 mA. You could gang a bunch of these capacitors in parallel to increase runtime, but this is not a very good solution IMHO.

    So, while you could use a capacitor for this application, you only get a couple of minutes of usable light. I think a better way to store energy in this application is with a NiMH battery. The circuit needs to charge the battery at a rate not to exceed specifications (depends on the battery size) and is then free to discharge it at a much more consistent voltage. You will also be able to store more energy per unit volume and at a lower cost than if using capacitors. Capacitors have an advantage when you need to extract the stored energy quickly, but this application does not require it.

    Edit: Actually, you would want to use 3 NiMH batteries in series, not just one. Furthermore, a protected Li-ion rechargeable might be an even better choice, since you wouldn't have to worry about over discharge as much due to the built in protection. But, NiMH is more tolerant of overcharging, so each has its advantage. If choosing Li-ion, I recommend AW's protected batteries. You will want to avoid charging them beyond 4.2 V.
    Last edited by mudman cj; 03-07-2007 at 06:34 AM.


  5. #5
    *Flashaholic*
    Join Date
    Apr 2004
    Location
    Lost In Space
    Posts
    5,431

    Default Re: LED Reflector...

    Quote Originally Posted by goucho
    ^-- Thanks for your help. After some research I've found a suitable LED. It's a Luxeon LED that comes complete with collimating lens (10 degrees) and aluminum submount.

    Luxeon White Star/O - Batwing
    Specs:
    Max Continuous Current 350 mA
    Forward Voltage 3.42 Vf

    !
    when given the choice of Luxeon leds, EVEN if your going to run it at 1W, prefer to use the 3W version of them, they are more durable, and work about as well (more bin dependant) as a 1W when run at the same rate.

  6. #6

    Default Re: Electrical help....

    Quote Originally Posted by mudman cj
    I think a better way to store energy in this application is with a NiMH battery. The circuit needs to charge the battery at a rate not to exceed specifications (depends on the battery size) and is then free to discharge it at a much more consistent voltage.
    Thank you for your help! I have decided to go with the 1 Watt Luxeon LED due to cost and my application. After doing some calculations I found that it would not be very cost or runtime efficient to use the supercapacitors. I decided to take your advice and use the NiMH rechargable batteries. Using the 1.5v AAA batteries in series to give 4.5v would be enough to run the 3.42v LED.

    I know that overcharging a battery will ruin it. So is there a way to limit the total charge amount (ie.. a Zener Diode)? Is this the same as you were talking about above when you said "The circuit needs to charge the battery at a rate not to exceed specifications"?

    So here is what I think I am going to need (electrically):
    -1 watt Luxeon LED
    -Three 1 ohm (1/2 watt) resistors (in series)
    -3 AAA NiMH Rechargable Batteries (4.5v in series)
    -On/Off Toggle Switch

    ?-Limiting Diode (Zener)

    If you could please verify this list I would appreciate it! Seriously, thanks again for your help. It's hard to do this with my limited knowledge of circuits.

  7. #7
    Flashaholic* mudman cj's Avatar
    Join Date
    Dec 2005
    Location
    Where corn and pigs are grown unimpeded by trees
    Posts
    1,827

    Default Re: Electrical help....

    NiMH batteries provide 1.2V nominally under load (provided the load is not too large), so three would only total 3.6V, not 4.5. Using four of them in series would then give 4.8, but they would not fully charge with a 4.5V source.

    What I meant by charging rate is not the charging cutoff, but the current through the batteries during charging. This would be set by the internal resistance of the cells and possibly with the addition of a resistor. Manufucturers typically specify a charging current upper limit, but for AA size cells and a continuous (not pulsed) current it would be somewhere around 500 mA. It is tricky to sense when to stop charging NiMH and switch to a low current trickle charge because the cell voltage changes so little with state of charge. This typically requires a sensitive charge termination circuit. They can tolerate overcharging, but really only at low currents (trickle). You would not want to keep 500 mA on them after they are already fully charged.

    For these reasons I really think you should consider Li-ion instead. It is easier to make a dumb charger for them because their voltage changes a lot as they charge (from 3.3V or so up to 4.2 when charged). I would recommend using the electronics from the DSD charger to control charging, which runs typically from a 5.2V power supply. It may also work from 4.5V. They are really cheap on Dealextreme - $4 or so. Don't charge a Li-ion with a current exceeding 0.7C, where C is the capacity in mAh. The current into the DSD will determine the charging current.

    As far as going with the 1 Watt Luxeon for cost, check this out. It's the Seoul P4 U bin LED for $7.18 each, and its already mounted to a star.


  8. #8
    Enlightened
    Join Date
    Feb 2007
    Location
    Finland, Oulu, beyond 65N
    Posts
    97

    Default Re: Electrical help....

    Quote Originally Posted by goucho
    EDIT: See post number 3

    I'm currently designing a human powered LED flashlight. It will be powered by a Mabuchi 4.5v Generator (RC-280RA-2865) (1.36A, 3.47W)
    How do you plan to get enough RPM for that vibrator motor?

    I have been planning on something similar, one possibility is to take motor from deceased CD-ROM drive. (these require also pretty high rpm) or a diy version large-diameter axial flux machine with rare-earth magnets.
    Last edited by mzzj; 03-07-2007 at 01:44 PM.

  9. #9

    Default Re: Electrical help....

    Ok, So I think i've narrowed everything down to what I am going to be using. Since the parts have to be order by next tuesday, I am not going to be changing anything (unless it's absolutely vital). I've decided to go with 4 AA MAHA NiMH Rechargable Batteries (2700 mAh). Now, I'm trying to figure out my charge/discharge times. Here are the following specs.

    Generator:
    Nominal Voltage: 4.5v Constant
    No Load --> Current = .27A
    Max. Efficiency --> Current = 1.36A, Output = 3.47W
    Stall --> Current = 6.90A

    LED:
    Max Continuous Current = 350 mA
    Forward Voltage = 3.42v
    Voltage Range = 2.79v - 3.99v

    For discharge time would I simply take the 2700mAh rating for the fully charged batteries and divide it by the 350 mA current of the LED? (which would give 7.71 hours)

    As far as Charge up time, I was told that I should use half of the stall current (3.45A).
    Would I take the 2700 mAh divided by the 3450 mA? (which would give .783 hours) If I made these assumptions then that would give about 9.86 mins of light for 1 min of charge (obviously the charging is not going to be consistant, but this would be a rough estimate). The only resources I have to go off of is the web and the basic P=IV & V=IR equations so I am pretty overwhelmed trying to get things to work out. Also, all of this is completely new to me so all my knowledge in the subject I have gained just by reading on the internet.

    Also, for the resistor calculations, here's what i'm thinking. The Vsupply is the 4.8v from the batteries. The Vled is 2.79v. So [Vsupply - Vled]/I = 5.74 ohms. So would I be able to use a 6.2 ohm resistor (that is a standard value resistor right?) Also, according to my P=IV equation, this would give [.35A]*[4.8v-2.79v] = .7035 watts. So would that mean I would need a 1 watt resistor?

    Right now, you are basically my only guidance. And it seems like you really know what you are doing. So I want to thank you for actually taking the time to mentor me!

  10. #10

    Default Re: Electrical help....

    Quote Originally Posted by mzzj
    How do you plan to get enough RPM for that vibrator motor?

    I have been planning on something similar, one possibility is to take motor from deceased CD-ROM drive. (these require also pretty high rpm) or a diy version large-diameter axial flux machine with rare-earth magnets.
    Unfotunately, this is the motor I am given to work with... So I have to make due. I plan on using a 6 gear setup from a hand crank to obtain the desired speed of 11350 RPM's. This is a whole nother calculation set, so i'll worry about that later.....

  11. #11
    Silver Moderator
    SilverFox's Avatar
    Join Date
    Jan 2003
    Location
    Bellingham WA
    Posts
    11,641

    Default Re: Electrical help....

    Hello Goucho.

    Welcome to CPF.

    To clear up a couple of battery issues...

    Your 2700 mAh cells will usually only be actually capable of about 2400 mAh. This means your runtime will be a little less.

    Charging at 1.725 amps is a good charge rate. You can figure that you will need to put in a little more because the charge efficiency is not perfect. I would guess that you should be fully charged in approximately 1.5 hours when charging at 1.725 amps starting with discharged cells.

    There may be a problem with the maximum voltage. Fully charged cells end up at 1.44 volts. This means that 4 cells in series would be at 5.76 volts when fully charged, and to charge you need to have a higher voltage than that. I believe chargers usually charge at roughly 2 volts per cell, so in order to get a full charge, you may need to supply up to 8 volts.

    There is no immediate issue with under charging, except reduced runtime, however there can be long term effects.

    I do know that if you hook up a fully charged pack (4 cells at 1.44 volts = 5.76 volts) with a pack that is completely discharged, there will be roughly less than a 10% charge of the empty pack in 24 hours. In order to charge, you need a higher voltage. By the way, this is why parallel chargers don't balance batteries.

    Is there a way to get a higher voltage out of your generator?

    Tom
    Behind every Great man there's always a woman rolling her eyes...

    Most batteries don't die - they are tortured to near death, then murdered...

  12. #12

    Default Re: Electrical help....

    Why not use the RayoVac hybrids as they retain their charge much better than the regular NiMh's and are just a little less storage if you say the 2700s actually have about 2400. As this appears to be hard earned energy and you don't want to waste it. I'm not 100% sure what youre trying to do because the link doesn't work but I see someone said hand cranked so I assume you have to be putting labor into this machine.

  13. #13
    Flashaholic* mudman cj's Avatar
    Join Date
    Dec 2005
    Location
    Where corn and pigs are grown unimpeded by trees
    Posts
    1,827

    Default Re: Electrical help....

    As I posted before: "Using four of them in series would then give 4.8, but they would not fully charge with a 4.5V source." Silverfox was able to be much more specific with respect to this issue, and I defer to him on such matters.

    Seems to me Li-ion is looking better all the time since you would be greatly undercharging NiMH cells with only 4.5 V supplied, and it would not be trivial to step up that voltage to 8 V. Does anyone else think he should use Li-ion, or am I missing something? And you don't have to use the DSD to charge the Li-ion either, I just thought it would be quick and easy if it works.


  14. #14

    Default Re: Electrical help....

    Ok, i've been doing some more thinking about this. I agree with you that the Li-Ion batteries would work better than the NiMH batteries because they will probably never be fully charged. Also, when i was using the 4 AA batteries before it was giving me 2700 mAh. This is impractical for my application because nobody is going to sit there and charge the batteries for about an hour to get them fully charged. So I was thinking that the typical desired runtime for the LED should be about 10-20 mins. Therefore it would lead me to believe that I should use some smaller coin cell batteries (maybe three 1.2v cells). Or would i be able to use just one 3.7v Li-Ion battery? Would i want to find the Li-Ion coin batterie with about a 150 mAh rating? Also, to protect the batteries from being over charged, could I use a Zener Diode? This should also work with the motor i'm using because it can give up to 6 volts. (4.5v nominally) Am I thinking about this the right way?
    Last edited by goucho; 03-09-2007 at 10:10 AM.

  15. #15
    Silver Moderator
    SilverFox's Avatar
    Join Date
    Jan 2003
    Location
    Bellingham WA
    Posts
    11,641

    Default Re: Electrical help....

    Hello Goucho,

    Lithium chemistry is wonderful as long as it is used within its limits. When you get outside of its limits, it has a tendency to "rapidly vent with flame," more commonly called an explosion accompanied with a fireball.

    Li-Ion cells need to have the maximum voltage to them limited to 4.2 volts. When you exceed this, you run the risk of an explosion.

    Li-Ion cells need to have a low voltage cutoff. Running them below roughly 3.0 volts will can ruin the cell, and it has the possibility of creating problems during the next charge cycle.

    The current draw from a Li-Ion cells should be limited to twice the capacity of the cell. For example, if you have a 1200 mAh Li-Ion cell, the maximum current draw should be limited to 2400 mA. Exceeding this causes the cell to heat up and results in a reduced cycle life, and if it gets hot enough, an explosion.

    The charging current for a Li-Ion cell is usually recommended at 0.7 - 0.8 times the capacity of the cell, with a maximum charge rate of 1C. If you have a 1200 mAh Li-Ion cell, the ideal charging rate would be 840 - 960 mA, with a maximum charge rate of 1200 mA. Faster charging rates cause the cell to heat up, resulting in cell damage. If the cell get hot enough, you end up with an explosion.

    Now, if you want to go with Li-Ion cells, make sure you have all the protection circuits in place and make sure they all work.

    On the other hand...

    There are many flashlights that have been built running off of 3 NiMh cells. These are often direct drive, but some have a small amount of resistance (around 1 ohm if I remember correctly) in line with the LED.

    AAA NiMh cells can be charged at 3.5 amps and AA cells at 7.5 amps in the pulsed 15 minute chargers. NiCd cells are regularly charged at 2C constant current, and charge to higher voltages than NiMh cells do.

    Pulsed current charging results in cooler cell temperatures than constant current charging. However, NiCd cells do not heat up until they are nearly full. During the bulk charge period, they are nearly 100% efficient in accepting the charge.

    You may want to consider running 3 AA NiCd 1000 - 1200 mAh cells. If you can manage to get 3 amps charging current from your set up, it would only take around 22 minutes to go to a full charge, and 5 - 6 minutes at 3 amps charging would probably give you your 20 minute run time with a 350 mA draw.

    Tom
    Behind every Great man there's always a woman rolling her eyes...

    Most batteries don't die - they are tortured to near death, then murdered...

  16. #16
    Flashaholic* mudman cj's Avatar
    Join Date
    Dec 2005
    Location
    Where corn and pigs are grown unimpeded by trees
    Posts
    1,827

    Default Re: Electrical help....

    @ SilverFox - Couldn't he just use a protected Li-ion cell to prevent overdischarge and a simple charging circuit to prevent overcharge? (I know the protected Li-ion cells tend to limit charging closer to 4.35 V rather than 4.20 V) I have a bookmark on my computer at home for a Li-ion charging circuit, which I thought was on CPF Wiki, but I can't seem to get to it right now. I will post it later tonight. You probably know the circuit I am referring to anyway. Perhaps that would work even better for him than the DSD circuit? I can't recall the input voltage requirements for said circuit.


  17. #17
    Silver Moderator
    SilverFox's Avatar
    Join Date
    Jan 2003
    Location
    Bellingham WA
    Posts
    11,641

    Default Re: Electrical help....

    Hello Mudman cj,

    That is an option, but he would also have to be able to limit the charging current to 1C. Charging at 1C means that you need to spend more time charging.

    However, if he was able to charge at a steady rate of 2200 and was using 2200 mAh cells, he could get enough charge to run his 350 mA load for 20 minutes in roughly 12 minutes of charging.

    Tom
    Behind every Great man there's always a woman rolling her eyes...

    Most batteries don't die - they are tortured to near death, then murdered...

  18. #18
    Flashaholic* mudman cj's Avatar
    Join Date
    Dec 2005
    Location
    Where corn and pigs are grown unimpeded by trees
    Posts
    1,827

    Default Re: Electrical help....

    He seems to be limited to 1.36 Amps, otherwise I understand that other battery chemistries would allow him to charge more quickly.

    Is there a simple way to detect charge termination? It seems to me that if you just monitor voltage across the cells you will detect the charging voltage instead of the cell open circuit voltage. I imagine that chargers have to briefly interrupt charge to detect open circuit voltage, therefore excluding the use of a Zener diode to end charging, but I am only guessing here.


  19. #19
    Silver Moderator
    SilverFox's Avatar
    Join Date
    Jan 2003
    Location
    Bellingham WA
    Posts
    11,641

    Default Re: Electrical help....

    Hello Mudman cj,

    With Li-Ion chemistry, you simply clamp the voltage at 4.2 volts and terminate the charge when the charge rate drops to a low value.

    With NiMh and NiCd, you can use peak voltage, negative delta voltage, peak temperature, rate of change in temperature, and/or time.

    Tom
    Behind every Great man there's always a woman rolling her eyes...

    Most batteries don't die - they are tortured to near death, then murdered...

  20. #20
    Enlightened
    Join Date
    Feb 2007
    Location
    Finland, Oulu, beyond 65N
    Posts
    97

    Default Re: Electrical help....

    One option would be to use only one nimh cell and step-up converter(plus led) ripped of from this http://www.dealextreme.com/details.dx/sku.1120

    Would make it also easier to generate power, no need to turn that motor at 11000rpm anymore, 3000-4000rpm should be enough. I would seriously consider this option if you must use that kind of like "toy" motor.

    Charge termination based on temperature, when battery feels hot user should stop cranking

  21. #21
    Flashaholic* mudman cj's Avatar
    Join Date
    Dec 2005
    Location
    Where corn and pigs are grown unimpeded by trees
    Posts
    1,827

    Default Re: Electrical help....

    I found that Li-ion charging circuit I was looking for, but the input requires 12V. Sorry, no dice. Here it is anyway.


  22. #22
    Silver Moderator
    SilverFox's Avatar
    Join Date
    Jan 2003
    Location
    Bellingham WA
    Posts
    11,641

    Default Re: Electrical help....

    Hello Mzzj,

    Your solution would simplify things...

    Tom
    Behind every Great man there's always a woman rolling her eyes...

    Most batteries don't die - they are tortured to near death, then murdered...

  23. #23

    Default Re: Electrical help....

    If i were to use the three AA NiCd rated at 1200mAh, would i still need to limit my incoming current so as not to "overcharge" the batteries? Also, if my generator produces a voltage range of 4.5v-6v, will that be too much voltage for the 3.6v (3x 1.2v) batteries to handle? or is the incoming current the only thing I should be worried about?

  24. #24
    Silver Moderator
    SilverFox's Avatar
    Join Date
    Jan 2003
    Location
    Bellingham WA
    Posts
    11,641

    Default Re: Electrical help....

    Hello Goucho,

    NiCd chemistry is tough. It will probably handle anything you can throw at it. You can keep an eye on cell temperature, and if the cells start to heat up (above 120-140 F) stop charging them.

    Tom
    Behind every Great man there's always a woman rolling her eyes...

    Most batteries don't die - they are tortured to near death, then murdered...

  25. #25

    Default Re: Electrical help....

    Quote Originally Posted by SilverFox
    Hello Goucho,

    You may want to consider running 3 AA NiCd 1000 - 1200 mAh cells. If you can manage to get 3 amps charging current from your set up, it would only take around 22 minutes to go to a full charge, and 5 - 6 minutes at 3 amps charging would probably give you your 20 minute run time with a 350 mA draw.

    Tom
    Thanks for all your help in the matter. I am just wondering how you went about getting these numbers... If you could explain (including any equations you used) that would be extremely helpful. I'm just trying to gain a better understanding! Also, how would I know what charging current i would be getting without physically testing it? Finally, wouldn't I be limited to the 1.36A given from the max output of the generator? Thanks again!

  26. #26
    Silver Moderator
    SilverFox's Avatar
    Join Date
    Jan 2003
    Location
    Bellingham WA
    Posts
    11,641

    Default Re: Electrical help....

    Hello Goucho,

    I was looking at your stall current of 6.9 amps and thinking you may be able to get around 50% of that.

    Lets look at some numbers. I am going to round things off so we have even numbers, but you can tweak them however you see fit.

    First of all, yes, you will have to do some testing to see how well things work out.

    Your LED current is 350 mA. By the time you add a circuit or resistor, you will most likely be drawing around 500 mA from the battery pack. You want a 20 minute run time, so you are looking at around 170 mAh of capacity from the battery.

    Let's look at a charge rate of 1000 mA. Charging has losses, so you will probably need to generate around 200 mAh to get 170 mAh out. 200/1000 = 0.2 hours of charge time, or 12 minutes.

    If you wanted to charge the 1200 mAh battery up from empty, you would need to generate around 1440 mAh of capacity. If you are capable of 1000 mA, it would take you around 86 minutes to do a full charge.

    Tom
    Behind every Great man there's always a woman rolling her eyes...

    Most batteries don't die - they are tortured to near death, then murdered...

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •