ZXSC300 0.8v-4.5V 700mA 3W LED Driver -help with components

Discussion in 'The Projects Forum' started by smilem, Oct 14, 2010.

  1. smilem

    Thread Starter Active Member

    Jul 23, 2008
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    Hi, I'm trying to make LED driver.

    My LED is 3W 700mA max
    AVAGO TECHNOLOGIES|ASMT-JW32-NVV01|LED, 3W, COOL-WHITE, MINIMUM 100LM | Farnell LT

    Power source is 1.5V batteries x3 = 4.5Volts.

    I would like to use ZXSC300 because it works up to 0.8V and I don't want to waste any power from my batteries :)

    Here is datasheet for ZXCS300:
    http://www.diodes.com/datasheets/ZXSC300.pdf

    I tried to use ZXSC300 Feed Forward calculator here:
    http://www.diodes.com/_files/calculators/ZXSC300_Calculator.xls

    However I'm unsure about some values so I attach my screen shot.
    Please help to calculate and select parts for my project.

    Here is a list I made for screenshot calculation:
    U1 ZXSC300 | SOT23-5 Elfa 73-298-35
    Q1 FMMT617 transistor NPN 15 V | SOT23 Elfa 71-029-33
    D1 Schottky diode DO-214AA 1A 15V SMB Elfa 70-217-02
    R Sense Resistor SMD up to 20 W 12 mΩ 0.5 W ±1 % Elfa 60-314-39
    R1 WELWYN - PCF0805R-17K4BT1 - RESISTOR, 0805, 17K4 Farnell 1353223
    Offset resistor Resistor 97,6ohm 1% 0,6W Elfa 60-711-38
    RT NTC resistor SMD 0603 4.7 kΩ Elfa 60-304-56
    L WUERTH ELEKTRONIK CHOKE, 1045 SIZE, 1.1UH Farnell 1800264
    C1 KEMET CAPACITOR, 0603, 2.2UF, X5R Farnell 1108322
     
  2. smilem

    Thread Starter Active Member

    Jul 23, 2008
    132
    0
    Any comments, anyone ?
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    Your LED has a Vf of 3.5v @ 700mA current. However, you're trying to use a boost converter with 4.5v in. The Schottky diode will drop around 0.4v across itself, so you have:
    4.5v - (3.5v+0.4v) = 4.5-3.9 = 0.6v too much on the input side.

    You'll toast the LED, the Schottky, or the inductor - or maybe all three, as there will be NO current regulation until the input falls below 3.9v.

    Vin must be <= (Vf_LED+Vf_D1), so use two 1.5v in series, or four in series-parallel.

    You should really use a capacitor across your batteries, as otherwise the peak current draws will drain your batteries in half the time.
     
  4. DickCappels

    Moderator

    Aug 21, 2008
    2,653
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    There two possible ways around the hazard that Sgt. Wookie pointed out (the average voltage at the output is enough to kill the LED and other parts as well) is to connect the LED across the inductor.

    The easiest is to use a lower voltage battery. Such as a single 1.5 volt cell or a 3 volt battery.

    If you really must use a 4.5 volt battery, connect the negative end of C1 and the cathode of LED1 to the battery.

    All related Wookieisms heartily endorsed.
     
  5. smilem

    Thread Starter Active Member

    Jul 23, 2008
    132
    0
    Thank you for your calculations.

    I think the LED is really a 3.8V as this is what specified at the datasheet.

    4.5v - (3.8v+0.4v) = 4.5-4.2 = 0.2v is it still too much?
    Battery voltage drops when you connect a load so I think it will drop a little.


    I thought the voltage would self regulate like when using LM338T constant current regulator, you set the current you need for LED and voltage self regulates.
    I always thought LEDs are more current than voltage devices, if current is OK the voltage will flow as specified by forward voltage.

    I could use a Schottky diode that drops more voltage but whole Idea was to drop as less as possible, to keep the circuit working longer.

    I will use simple 3x AAA battery, I'm modifying china made LED torch, so doubt I can use single cell only.
    I changed LED voltage to 3.8V in calculator data sheet and had to change the Offset resistor to 54Ohm, the peak current of 682mA vs. 567mA at 65 degrees does not look promising as it's 115mA difference. I couldn't tweak the calculator better, perhaps someone could take a look at it.
     
    Last edited: Oct 15, 2010
  6. smilem

    Thread Starter Active Member

    Jul 23, 2008
    132
    0
    Very good idea, I found a schematic that illustrates this here:

    [​IMG]

    You can read article in russian:
    http://www.pitaemled.biz/driver/zxsc300/5-pitanie-svetodiodov-s-pomoshhyu-zxsc300.html

    However I need to be able to power from 0.8V to 4.5V and because I now connect LED differently I can't use the manufactures calculator. Could you help me to determine components that i would run as close to 0.8V to 4.5V as possible? There is no temperature compensation in this schematic, is not required?
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    You should read this Application Note: http://www.diodes.com/zetex/_pdfs/3.0/appnotes/apps/an65.pdf

    Your inductor is much too small at 1.1uH. You'll probably need more like 33uH to 50uH; otherwise the current will decay much too rapidly, requiring an extremely high frequency to drive. Otherwise, your inductor will saturate in the blink of an eye, and you will fry the transistor.

    If you drain your batteries excessively, you will likely wind up with leakage. If you tried to use NiMH or NiCD batteries, they'd be deeply discharged.

    If you want to use 4.5v in, you will need a different configuration; a buck-boost, which will be more complex than a simple buck regulator.
     
  8. smilem

    Thread Starter Active Member

    Jul 23, 2008
    132
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    At some china site I found these screenshots:
    They tested ZXCS300 and AMC7135
    LED was LUMILEDS LXHL-PW09
    Forward voltage 3.9V Peak Forward Current:1000mA

    They damaged the LED running at more than 1360mA, the LED data sheet shows values for 1400mA but the schematic did work. I'm puzzled how did it work, it's the same schematic you said would fry the LED if voltage is more than 3V. But if you look at the table LED with Vf 3.9V worked until 5.4V

    I tried to get AMC7135 but to no avail, seems only available in china.
     
  9. smilem

    Thread Starter Active Member

    Jul 23, 2008
    132
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    So I built the thing with ZXCS310 chip, and 22uH inductor. What can be done to increase current?
    I did order 11mΩ resistor, but if I can't reach the needed amperage?

    C1 is 0.1uF
    C2 is 10uF

    Transistor is FMMT617

    I tested from 1AAA to 4AAA batteries.

    3AAA is 4.3V at LED voltage is 3.26V current 451mA
    4AAA is 5.8V at LED voltage is 3.5V current 500mA
    Need I remind you my LED is rated at ~700mA ~3.6-3.8V
    My current set resitor is 16mΩ, tried to connect 2 in parallel but then circuit does not work at all since pdf specifies lowest value 10mΩ
     
    Last edited: Oct 30, 2010
  10. smilem

    Thread Starter Active Member

    Jul 23, 2008
    132
    0
    I tried 11mΩ resistor but it did not help.

    Anyone?
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    AAA batteries are too anemic for your application. You'll need at least AA batteries, or go to LiPo. Otherwise, the battery internal resistance will limit your current too much.

    You need to use a capacitor across your batteries to extend their lives. That will effectively cut their internal resistance in half or more, depending on the duty cycle of the IC.
     
  12. smilem

    Thread Starter Active Member

    Jul 23, 2008
    132
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    I have 0.1Uf do I need larger cap? I tried AA batteries, result is the same.
    I even tried Nokia 3.7 Li-ion battery, still result is the same 400-450mA max

    My diode is https://www1.elfa.se/data1/wwwroot/assets/datasheets/07022940.pdf
    Can it be the problem?
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    The drain will be pretty heavy. I forget offhand what your switcher's frequency is. Try 330uF or higher.
     
  14. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
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    Sgt, from the datasheet it looks like the circuit regulates for peak current through the LED, so it should not fry it with overvoltage.
     
  15. smilem

    Thread Starter Active Member

    Jul 23, 2008
    132
    0
    Thanks for you reply,
    I tried to connect 470uf electrolytic cap as you say to the battery. No changes. Then I added another 470uf electrolytic cap, but still no changes.

    Current will not rise :confused: i think kubeek is correct.
    Voltage is kept within boundaries, tried 4AA still the voltage was 3.5V at the LED.
     
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