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Want to increase brightness of LED headlight

Discussion in 'LEDs and Optoelectronics' started by seanspotatobusiness, Dec 13, 2016.

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  1. seanspotatobusiness

    seanspotatobusiness

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    Sep 11, 2012
    I have one of these magnifiers and they're sort of almost useable at 2.5x but the light isn't very bright at all. I attached my bench power supply and determined that at 3.6 V a current of 20 mA would flow which is my guess at what the LEDs are rated for (maybe it's not; all I know for sure is that they're cheap, white and 5 mm). At 3 V it's 7 mA and at 2.7 V (batteries only a bit discharged) it's 2 mA!! I want to use a little wotzit to increase the current to 20 mA but I have a question regarding which wotzit to use.

    It needs to be tiny to fit into this battery compartment and the current contenders are:

    QX5252F available in TO-94 form
    YX8018 available in TO-94 form
    ZXSC380 available in SOT23 form
    Alf available in pog form

    The ZXSC380 has the advantage of a teeny tiny package whereas I think the QX5252F and YX8018 have a leg I could somehow connect to prevent overdischarging NiMh batteries. Basically a different solution is needed for alkaline versus rechargeable batteries, I think.

    Could anyone tell me alternative ICs or how to connect the TO-94 parts to not discharge past 1.6 V (0.8 V per cell)?
     
  2. Audioguru

    Audioguru

    3,040
    678
    Sep 24, 2016
    The low current solar garden light ICs are made for a single cell 1.25V input, not two cells that you have.
    The ZXSC380 has a wide input voltage range with a mean output current of about 18mA which might be fine, but the peak current is 65mA with a 2.8V fully charged battery that might zap the cheap LEDs.
    Do you have space for the inductor?

    Ni-MH cells do not care if they are discharged very low. Rechargeable Lithium cells demand to be disconnected.

    Wait a minute. Why are you using low voltage rechargeable battery cells? You should simply use Energizer L92 (Ultimate Lithium) disposable AAA battery cells that are more than 1.5V for the first half of their life and are above 1.4V for the second half.
     
  3. seanspotatobusiness

    seanspotatobusiness

    193
    4
    Sep 11, 2012
    I'm using cheap alkalines but I want to use rechargeables because then I don't need to keep remembering to switch it off when I don't need it, Fancy Energizer batteries sound relatively expensive but I already have a few rechargeable AAAs that aren't in use.

    I think I can use an SMD inductor. I have access to a microscope to be able to solder the parts together,

    Are you sure NiMH aren't damaged by discharge below 0.8 V? Many sources in my first couple pages of Google results suggest they would be.
     
  4. Audioguru

    Audioguru

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    Sep 24, 2016
    A SMD inductor might have a low maximum current then it will saturate.

    The Ni-MH Battery Applications Manual from Energizer shows that a cell provides little power when discharged below about 0.9V per cell. The cell is damaged only when it is the lowest capacity one in series with a mismatched higher capacity one and the discharge continues until the stronger cell reverse charges the weaker cell.

    Google has lies from people who think they know stuff.
     
  5. seanspotatobusiness

    seanspotatobusiness

    193
    4
    Sep 11, 2012
    Thanks! I'll be sure to check the inductor specifications for saturation currents; there's one series I'm thinking of using which have currents in excess of an amp.

    It occured to me that the two LEDs must be in parallel in which case they might be able to take 40 mA total and not just 20 mA so that's cool. I wish the current-voltage curves of the ZXSC380 were flatter nonetheless.

    Graph: http://imgur.com/QIf18jQ.png
    [​IMG]
     
  6. Audioguru

    Audioguru

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    678
    Sep 24, 2016
    The current curves in the datasheet are for the peak current, but our vision sees the average current that is much lower because the LEDs are pulsed on and off. That is why a solar garden light is rather dim, the pulses are PWM dimming.
     
  7. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    This is the circuit you need:

    [​IMG]
     
  8. Audioguru

    Audioguru

    3,040
    678
    Sep 24, 2016
    The headlight has only two rechargeable AAA cells for a maximum of 2.8V that drops to about 1.6V during their discharge so a voltage boosting circuit is needed.
     
  9. cjdelphi

    cjdelphi

    1,096
    104
    Oct 26, 2011
    If the other models are similar to the qx5252f, you can feed up to (i think) 4 or 5v, i'm actually feeding 2nimh batteries in series supplying it to the qx5252f

    Why? Because i then perfboarded a circuit with a 3-5uf home wound torroid which results in a 300ma output, but it was pulling .7amp/2.8v! So i switched for a 10uf inductor at 2.8v pulling 80ma (LED side) on a 1w (300ma rated) LED

    It's illuminating the toilet closet nicely it's 4 or 5x brighter than 1 (5mm) led[/QUOTE]
     
  10. Audioguru

    Audioguru

    3,040
    678
    Sep 24, 2016
    The circuit converts a low voltage to a higher voltage and the power must not change except to make up for loss of efficiency. If the LEDs need 3.6V at 80mA but the battery is only 2.4V then the battery current is 3.6/2.4= 1.5 times higher plus about 20% more for losses which is a total battery current of 144mA. This is probably the momentary peak current, the average current that you see is less.
     
  11. seanspotatobusiness

    seanspotatobusiness

    193
    4
    Sep 11, 2012
    Oh I see, thanks. The frequencies seem really high so am I right in thinking a small ceramic capacitor would smooth out the peaks so the LEDs wouldn't see large currents? Also I think I would need to use two ZXSC380 chips in parallel to provide the current I want to both LEDs? Could they share a single inductor? Presumably they would get half the microHenrys each?


    Does that boost voltage somehow? Or is it a low-battery cut-off?


    The only datasheet I can find for the QX5252 is this: http://www.mikrocontroller.net/attachment/158139/QX5252.pdf

    The LEDs need 4.1 V to get 40 mA of current to them. That's 164 mW. I'm providing 2.4 V (dropping to ~1.6 V) .

    Inductor needed at start = 2 x 2.4 V x 10E-6 / 0.164 W = 29 uH. Is that calculation correct?


    Components are pretty cheap on AliExpress so I think I could try both approaches (in about two months when the components arrive!).
     
  12. Audioguru

    Audioguru

    3,040
    678
    Sep 24, 2016
    No and no. If you add a smoothing capacitor then the IC will need extra output current to charge it.
    You cannot parallel ICs because there is no way to sync their oscillators.
    Use a separate circuit for each LED if one IC cannot drive enough current for two parallel LEDs.

    It has a 6V battery made from two 3V Lithium coin cells and feeds a constant current of 47mA until the battery voltage becomes so low that the LED does not light anymore.

    How do you know if your LEDs will survive 40mA?

    I did not check your calculation because the IC is Chinese.

    I think Alliexpress is Chinese and has bad reviews.
     
  13. seanspotatobusiness

    seanspotatobusiness

    193
    4
    Sep 11, 2012
    I have the QX5252F IC and it's in a circuit where it increases the current to the LEDs up to 29 mA depending on the inductor used but there seems to be a lot of current bypassing the LEDs, going through the QX5252F. Is that normal? My inductors are cheap, low power, moulded axial ones. Are they appropriate or do I need lower resistance ones?

    33 uH (0.9 Ω) = 27 mA through LEDs and 140 mA total consumed
    68 uH (2.0 Ω) = 22 mA through LEDs and 160 mA total consumed
    68 uH (1.1 Ω) = 29 mA through LEDs and 150 mA total consumed (higher power/lower resistance inductor)
    100 uH (2.8 Ω) = 15 mA through LEDs and 160 mA total consumed

    Image not showing for some reason: http://imgur.com/ml4OtBM.png

    [​IMG]
     
  14. Audioguru

    Audioguru

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    678
    Sep 24, 2016
    The datasheet for the QX5252 IC shows only one rechargeable cell battery but you have two cells in series.
    How are you measuring the output current? It is pulses that a meter cannot see.
    Of course the input current is higher than the output current, the input is a low voltage and the output is a higher voltage and the powers must be similar (plus inefficiency).
    I think you should try better and larger inductors.
     
  15. seanspotatobusiness

    seanspotatobusiness

    193
    4
    Sep 11, 2012
    Thanks. I'll try to find larger inductors to test.

    I forgot to include the capacitor in my diagram which I think helps stabilise the current measured (updated diagram attached). The current is measured immediately before the LED. Do I need to consider the frequency of the circuit to choose a suitable capacitor? What I have is some kind of ceramic capacitor that says:
    104
    0M0

    It looks like this and I thought it was 104 uF but maybe I should find a bigger one.
    [​IMG]
     

    Attached Files:

  16. Audioguru

    Audioguru

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    Sep 24, 2016
    You should not put a capacitor in that switching circuit (the capacitor is a dead short to the high frequency) because it might burn out the Mosfet switch. You show a film capacitor, not a ceramic one. Its value is 10 followed by 4 zeros which is 100000 pF which is 100nF which is 0.1μF.

    On my solar garden lights I removed the white LED and replaced it with a colors-changing LED in series with a Schottky diode. The colors-changing LED has a ceramic capacitor in parallel but the diode isolates the capacitor from the switch.

    Your current meter cannot measure the high frequency pulses so it reads low. The pulses are on and off over and over which has a high short-duration current, but a fairly low average current.
     
  17. seanspotatobusiness

    seanspotatobusiness

    193
    4
    Sep 11, 2012
    Thanks! I didn't think about the high frequency draining the capacitor. Is there nothing else I can do to smooth the current in lieu of a capacitor?

    I might be able to use an oscilloscope at my local Hacklab/Maker space to measure the current.
     
  18. Audioguru

    Audioguru

    3,040
    678
    Sep 24, 2016
    An oscilloscope looking at the voltage pulses in a resistor that is in series with the LED shows pulses, not the average current that creates the light that you see You do not see the peak light of the high frequency pulses, the low duty cycle dims the light (that is how a dimmer works). The resistor reduces the output current anyway.
     
  19. seanspotatobusiness

    seanspotatobusiness

    193
    4
    Sep 11, 2012
    Is the current reading on the power supply affected by the frequency? I put a capacitor across the PSU and there was no change in the current displayed. I think I'm gonna use a 68 uH inductor since the LEDs are then plenty bright and the power supply reports a 40-50 mA consumption which is presumably fine. Thanks for your help. I'm pretty much all the way there now.
     
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