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Repairing Grundig cd-player, transformer identification problem

Discussion in 'Electronic Repair' started by [email protected], Apr 25, 2007.

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  1. Guest

    Hello!

    I have a over 10-year old Grundig model CDC477 (5 cd changer) with a
    dead transformer and i need to replace it. The problem is that I don't
    know the exact mA and V ratings for it. I tried to contact Grundig and
    AMC about it but they haven't answered.

    The transformer is made by AMC and it has the following markings:
    1st line: AMC-logo and "95" "30"
    2nd line: "99.2139302FM"
    3rd line: "3104 113 70491"

    The input voltage is 230V. The transformer has two outputs which each
    have 3 wires. The first has white, green, white. The second has blue,
    red, red.

    These wires go to some sort of circuit with mainly electrolytic
    capasitors (16V and 50V types) and diodes. From this circuit leaves 7
    wires which are marked to be +10V, -10V, +3V9, GND, -27V, 4V8~ and
    4V8~

    Those 4V8 AC lines come straight through the circuit I described
    above, so one of the transformer outputs is 4V8 AC. But what might be
    the other output?

    This really sounds like a puzzle, but thanks in advance ;-)
     
  2. Franc Zabkar

    Franc Zabkar Guest

    Probably a YYWW date code.
    That looks like a Philips part number.

    - Franc Zabkar
     
  3. Gary Tait

    Gary Tait Guest

    wrote in @u32g2000prd.googlegroups.com:
    I am guessing the unit has a VFD, which the 4V8~ would be the filament,
    the -27 the drive, and a very small current. Depending, the 3v9 could be
    a logic frequency.

    I would think there would be`separate supplies for the motors, logic,
    and audio circuitry.

    Outward guess, one winding is the centre tpped 4v8, the CT to ground.
    The -27v derived from the other winding with a doubler, the CT making
    10V or so, either way.

    I think the 10V supplies might be 500ma each, the 3v9 maybe 1A, if it
    does the main system logic.
     
  4. N Cook

    N Cook Guest

    If the transformer is an ex-transformer, desolder it. Excavate under the
    wrapping of the coils and you may find a thermal fuse. Bridge that to a
    working but probably unsafe one, and power up to determine the voltages. If
    no such fuse continue breaking it up counting turns/weighing and gauging the
    secondary wires and apply the fomulae.
     
  5. N Cook

    N Cook Guest

    see my post above on a blown primary simulation for determining unknown
    secondary ratings
     
  6. Guest

    Thanks for your tips everyone! I'll perform the transformer autopsy
    asap.
     
  7. Guest

    The mains circuit on the device (before the transformer) has a fuse
    that's rated 250V / 80mA.

    The circuit I described in my first posting has a three-terminal low
    current negative voltage regulator MC79L24ACP. The spec sheet for this
    indicates typical output voltage of -24V.

    The second transformer output is going throught diode 1N4002GI and a
    22kOhm resistor (connected in parallel) to a electrolytic capacitor
    rated at 16V.


    This might be a n00b question, but is it safe to assume that the
    second transformer output is 27V and under 80mA?
     
  8. Jim Poore

    Jim Poore Guest

    The primary current is under 80ma probably close to 20ma if it a
    fast-blo fuse up to 40ma if is a slo-blo type. The seconday current
    would be higher up to perhaps 500-1000ma

    Jim
     
  9. N Cook

    N Cook Guest

    I experimented with a few more transformers, simulating a blown primary.
    Added to my tips file off URL below

    Blown primary, how to determine unknown secondary voltages of a mains
    transformer
    Simulating a multi-secondary transformer using a known
    good one but not using the primary, to get some data.
    I used a variac supply near the bottom of its range at 18volts and a 25 ohm,
    20W dropper to feed 50Hz (UK)ac into a secondary. Assuming you have a
    reasonable idea of the voltage of one 'unknown' secondary.
    The transformer I used first was a high grade enclosed Gardners, 1.3Kg but
    only 15W combined outputs,
    240V (UK) with marked 2 separate secondaries of
    6.3V, 0.6A and a 150-0-150 at 25mA.
    With 3.43V ac on one '6.3V' secondary there was open circuit 3.40 on the
    other isolated '6.3' and 161.4V end-to-end on the '150-0-150' and incidently
    116.4 on the primary.
    Then loading with different resistors
    100K, 161.4 drops to 159.1
    5.8K on 161.4 drops to 55.8, 3.43 input drops to 1.64
    swapping to 5.8K on 3.4 , no change
    1K on 161.4 to 12.1 and 3.43 to 0.771
    swap to 1K on 3.4 , drops to 3.39
    270 ohm , 161.4 to 3.34V
    270 on 3.40, drops to 3.37
    56 ohm on 161.4 to .704 and 3.43 to .54V
    56 on 3.4 , drops to 3.28 and 3.43 to 3.42
    8.2 ohm on 3.4 , drops to 2.55 and 3.43 drops to 2.99V
    A bit more generalised.
    Noting that for one secondary for this test transformer was rating 300V,
    25mA then V/I of 12K and the 6.3V, 0.6 secondary of 10.5 ohm.
    Doing as before powering a 6.3V secondary to 3.43V and '300V' was 161.4V
    then loading it until the voltage ratio was 80 per cent that is 161.4V down
    to 101.5V and 3.43 falling to 2.69V so 101.5/2.69 = .8 then that R is 12K.
    So for similar transformer construction and high V, low I then find that
    value of R for 80% then if V is known then current rating is V/R.
    Doing the same for the low V,high I one then for R=10.5 ohm then
    corresponding ratio drops from 1:1 ie ==3.43:3.4 down to 3.03/3.43 is 88%
    for high current , low voltage.
    So for similar transformer construction and high I, low V then find that
    value of R for 88% then if V is known then current rating is V/R.
    Other clues would be the gauge of the wires if they can be seen and the
    overall size and weight giving an idea of the overall power rating.
    Resistance checks would show which are more likely high V or high I.
    Second test with a more basic Albion make, .8Kg, 20W open
    construction 245V primary, 2 secondaries 17V,1A and 6.3V,.6A.
    Again putting current into the lowest secondary giving 5.59V on
    '6.3' and 14.52 on '17' (185.8V on 'primary')
    6.3/.6 wire was 24thou diameter and 17V,1A wire was 27 thou diameter.
    17/1 = 17 ohm. This time loading the 17V secondary with
    17 ohm meant the ratio had dropped 69 per cent (15.52/5.59 to 2.778/1.544 )
    Usually you would get some idea of one rectified V from max or min, by
    capacitor ratings or a regulator voltage etc.
    Valve radios would have one secondary connected to the
    heaters so usually 6.3V. A vacuum fluorescent
    display is likely to have a feed in the range only 2 to 5V
    Toroidal transformer 2x 120V to 2x 15V,2A, .75Kg and 2A wires
    33 thou diameter. Characteristic R = 15/2 = 7.5 ohm.
    Critical ratio in this case was 82 per cent with 7.5 ohm.
    15.27 input on '15' giving 15.26 on the othe rand 108V on
    one of the primaries.
    With 7.5 ohm 15.27 i/p drops to 3.91 and 15.26 drops to 3.19.
    For a large toroidal 500W 2x 35V, 7.1A , weight 4.8 kg
    Secondary wires consist of 2 paralled 56 thou diameter wires
    per secondary.
    Characteristic R= 35/7.1 = 5 ohm.
    With 15.16 on one 'secondary' 15.1 on the other and 49.8V on
    a 'primary'
    15.16 dropped to 2.59V and 15.1 dropped to 2.46
    so characteristic ratio is 95 percent for this transformer.
     
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