need help with this ec ballast

Discussion in 'General Electronics Chat' started by paachu802, Jul 12, 2014.

  1. paachu802

    Thread Starter New Member

    Feb 5, 2014
    27
    2
    this circuit have non polarised 5 caps 3 green and 2 blue

    green ones
    104J - (2)
    223F - (1)

    blue ones(disc type)-behind transistors
    471,1kv - (2)

    1.how to check these caps
    2.how to check diac
    3.there is 1w resitor at input (color- yellow,violet,gold,gold,black) what is its value..DMM show 7.6 ohms


    checked all resistors,diodes and transistor--ok

    plz help
     
    Last edited by a moderator: Jul 13, 2014
  2. MrCarlos

    Active Member

    Jan 2, 2010
    400
    134
    Hello paachu802

    Surely your DMM, with you measured the resistance (yellow, violet, gold, gold, black), does not have the function to measure capacitors.

    To measure a capacitor a "capacitance meter ' is required.
    There are some very cheap.

    To simply try that are not short or open, you can use any of the procedures that look at the following link:
    Asking to Google.com by:
    -How to measure capacitance with DMM-
    https://www.google.com.mx/?gfe_rd=cr&ei=huHBU83BMKqR8QeD74GwDw&gws_rd=ssl#q=how+to+measure+capacitance+with+dmm

    A method for measuring capacitance with your DMM might be:
    Since we know the formula to calculate the capacitive reactance (Xc):
    Xc = 1 / 2 x p x F x C.

    We also need to know the input impedance of your DMM.
    Okay:
    We will assume that the input impedance of Your DMM is of 10 Megohms.
    This is the input impedance for almost all DMM. you can consult the manual of your DMM.
    We require also a sinusoidal AC signal generator.

    Connecting all in series:
    The generator, the capacitor and your DMM on AC Volts function, on an adequate scale.
    We will get a reading on the DMM.

    Calculating the Xc of the capacitor at the frequency set in the generator, we can get the value of the capacitance.
    Xc = 1 / 2 x p x F x C.
    But we do not know the C value.

    well:
    But we have a reading on the DMM.
    let's assume that we adjust the generator to 100V @ 1kHz. and reading at the DMM is 50.
    we can easily deduce that Xc is equal to the input impedance of the DMM 10MW. . . Right ?

    We have deduced the voltage drop across the capacitor because:
    VC = VG – RDMM.
    Where:
    VC, Voltage drop across the capacitor.
    VG, Generator Voltage.
    RDMM, Voltage reading on the DMM.

    The data we know are:
    Z = 10 Megohms input impedance of the DMM.
    V = 100 The voltage we set the generator.
    F = 1 kHz. the frequency at which we set the generator.
    R = Reading on the DMM.

    Also know that in a series circuit the same current flows through all components.
    So IDMM = IC.
    The current flowing through the DMM equals the current flowing through the capacitor.
    so if:
    IDMM = RDMM / Z = (50 / 10'000, 000 = 0.000005 Amp. 5 uAmp)
    The IC (Current flowing in the capacitor) would also like:
    IC = VC / XC. Again, do not know the XC value.
    Well then:
    XC = VC / IC, We Know that IC = IDMM.
    thus:
    XC = 50 / 0.000005 = 10 Megohms @ 1KHz.

    Now we know the XC value, so you can solve C in the formula below.
    ! Make a try, this is not very difficult.
    Xc = 1 / 2 x p x F x C.

    Regarding the Diac device.
    Briefly, what makes this device is getting short to a certain level of voltage at its terminals.
    You can try putting a resistor (10K) in series with the device (Diac) and applying a DC voltage higher and higher.
    You can use your DMM connected in series, with the previous circuit or in parallel with the resistor.
    In series, your DMM in function of measuring current.
    In parallel, your DMM in function to measure voltage.

    Surely this Diac has an identification number.
    Proceed with caution when desolder. with that number you can find the data sheets.
     
  3. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    The resistor yel/viol/gold is 4.7 ohm.

    I can see a cooked resistor just to the left of the top transistor.

    The fact the other matching resistor seems ok usually indicates the top transistor is shorted, and that cooked the top resistor. i would check the diode next to it too.

    As a general fault finding system; unsolder and lift ONE leg free of the board for all resistors and diodes, then test resistance of the resistors, and text the diodes on multimeter diode test.

    And you probably need to replace that top transistor, or both.
     
  4. paachu802

    Thread Starter New Member

    Feb 5, 2014
    27
    2
    again checked components......removed 1w 4.7Ω at input with 3.3Ω as later not available......resistors seems to be ok....transistors ,diodes ok......only caps not checked...i think caps are problem....but it is dangerous to measure voltage in mains (PLZ DON'T DO IT--ballast is dc-ac inverter) i just connected DMM across rectifier o/p cap 10μF,450V......measured around 367V...:eek:...NOW STOPPED EVERYTHING...how to repair this item???...everytime replace each component faulty & reconnect to a tube???...
     
  5. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    Please show a better photo of the PCB, I would like to see that burnt resistor (1 ohm?) more clearly.

    Also please show the bottom of the PCB. That resistor has a cooked leg that may be a dry joint, and seeing the bottom of the PCB also lets us understand the circuit better.
     
  6. alfacliff

    Well-Known Member

    Dec 13, 2013
    2,449
    428
    if your getting 367 volts output, why do you think it is broken?
     
  7. paachu802

    Thread Starter New Member

    Feb 5, 2014
    27
    2
    i mentioned across rectifier o/p...:cool:..not choke o/p
     
  8. paachu802

    Thread Starter New Member

    Feb 5, 2014
    27
    2
    here it is...not so clear..although resistor tip(@ board) seems to be burnt resistor shows same value and also other components when checked
     
  9. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
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    Thanks for more photos. :)

    But they are still not very good.

    That resistor seems ok in the new photos, but I can see some heat damage on the leg of C3 green poly cap and the diode next to it (compound photo, bottom left).

    It looks like the thing has had a hard life?

    I would be lifting one leg of every resistor and every diode, and testing them well. And then replace the two transistors.

    Of maybe you should cut your losses and just buy a new one?
     
  10. paachu802

    Thread Starter New Member

    Feb 5, 2014
    27
    2
    i checked that diode no problem with diode..cap i do not check... i will replace it...transistor both ok(tested 2 times)...only thing i do not check in this ckt are caps...i do not know ..it need capacitance meter that i don't have......can anyone explain working of the circuit ...working of each section...how much voltage.../any website
     
  11. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    Looking at your photos again, there looks to be some nasty corrosion of two PCB copper tracks. One at the far right of the PCB (dark areas on track) and one might be even worse, far left side.

    Measure these tracks with your multimeter on ohms setting. Check other tracks with a magnifier. You can scratch off the green lacquer coating and inspect the copper directly in bad areas.

    For a schematic, it is a standard type push-pull 2 transistor inverter. They are simple circuits and usually easy to repair unless it fried the transformer.
     
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