Connect with us

TA7317P output protection

Discussion in 'Electronic Repair' started by John Smith, Feb 25, 2012.

  1. John Smith

    John Smith Guest

    Well wadda you know.

    I straightened the pins and put it in backwards.

    At power on there was a click and the readings now are.

    1. 0.23
    2. 0.32
    3. 0.64
    4. 0
    5. 0.82
    6. 3.7
    7. 2.57
    8. 1.44
    9. 3.35

    For socket pins 1 - 9. The device is definitely in backwards.

    I rechecked pin 5. It's +0.82 not -0.82, strange.

    Pins 7 and 8 seem strange too.

    Perhaps it's still not really a TA7317P

    Old guy
  2. Franc Zabkar

    Franc Zabkar Guest

    It sure looks that way. That said, I can't see any economic incentive
    to counterfeit a part that I can buy for AU$2.75 in single quantities.

    Maybe you should bite the bullet and buy it as a spare part from
    Panasonic or Yamaha?

    BTW, the more I examine the circuit, the more errors and anomalies I
    find. For example, AFAICS there is a 50V capacitor (C409) in the 40V
    phantom power supply that would be operating at 49V. I have a feeling
    that the correct HV supplies should be +/-55V, not 75V.

    There is also a PNP transistor (Q405) in the -15V supply that is drawn
    as an NPN, and a 2SD transistor (Q401) in the 40V supply is listed as
    a 2SC in the parts list.

    My overall impression is that it's a shoddy product.

    - Franc Zabkar
  3. Franc Zabkar

    Franc Zabkar Guest

    Maybe the chip was damaged by your first attempt?

    I think pin #8 is probably OK. AFAICS, Q7 (diode) and Q10 should both
    contribute around 0.6V when the relay is closed.

    I would think that pin#6 should be a lot closer to 0V.

    I also don't understand why the voltage on pin #3 is so high.

    - Franc Zabkar
  4. Franc Zabkar

    Franc Zabkar Guest

    When the IC was installed backwards, pins 4 and 6, and pins 3 and 7,
    would have been swapped. This means that the +75V supply would have
    been applied to the IC's ground pin via the relay coil. The IC's pin
    #7 would have been grounded via a 3K resistor (R367). This means that
    Q19's base-emitter junction would have been reverse biased by the +75V
    supply, thereby destroying it. Your measurements would suggest that
    Q19's B-E junction is now open, which would be consistent with the
    expected damage.

    Furthermore, the reading at pin #5 (Q20) would suggest that its
    connection to the IC's ground (pin #4) has been opened. AFAICS, this
    is also to be expected.

    - Franc Zabkar
  5. Franc Zabkar

    Franc Zabkar Guest

    If you are wary about the polarity of your ICs, I would use a 9V
    battery and a 1.8K series resistor to test the IC. I would connect the
    battery's negative terminal to pin #5. This pin is consistent,
    irrespective of polarity. I would then connect the battery's positive
    terminal, via the resistor, to pins 1 and 9 in turn. I expect that you
    should see approximately 1.4V at pin #1 and 3.2V at pin #9.

    - Franc Zabkar
  6. John Smith

    John Smith Guest

    I have two more unused parts.
    I'll try another unused one in backwards on Monday.
    Yes I noticed that.
    I don't either.

    I do agree it's a shoddy product but I won't get my pocket money if I can't
    fix it.

    Old guy
  7. John Smith

    John Smith Guest

    Thanks Franc, I'll take your advice and do that on Monday with the two
    unused parts I have.

    Old guy
  8. Franc Zabkar

    Franc Zabkar Guest

    I should have said that the above voltages would be with respect to
    pin #4. Add about 0.8V when measuring with respect to pin #5.

    BTW, I suspect that the 3K resistor on pin #3 would now be open. That
    would explain the higher than expected voltage on pin #3.

    - Franc Zabkar
  9. N_Cook

    N_Cook Guest

    Are you using the left hand side of the printing as your index for pin "1"
    or the chamfer or notch in the body ?
  10. John Smith

    John Smith Guest

    Both are at the same end.
    I even googled up a drawing of SIP9 to make certain that the chamfer end
    should be pin 1.

    It's not a Toshiba one but it looks just like the one on this page I found
    with google images.
    Watch out for link wrap.
  11. N_Cook

    N_Cook Guest

    I looked in my tub of salvaged and sorted on 731.... and did not find one ,
    surprisingly ( a Matshusta AN7317 9 pinner but dual amp , not protector . I
    could have done some "diode" tests.
    Its just no the sort of thing to pirate , there are other TA73** devices of
    9 pin SIL so perhaps a cock-up in the labelling section. Now removing 3055
    marking off TO3s and marking them 2N3773 or whatever , would be the actions
    of a pirater
  12. josephkk

    josephkk Guest

    It begins to sound like somebody "hot rodded" the amplifier. And did a
    shoddy job of it.
  13. John Smith

    John Smith Guest

    None of the resistors look distressed and R367 measures 2.85K with no chip

    I was going to try the battery test but first I used a resistance meter
    between pin 4 and pins 2 and 3. I was expecting these two pins to look the
    same but they didn't no matter which pin I assumed to be pin 1.

    With nothing to lose I fitted an unused chip backwards.

    At power on there was a click and:

    1. 0.22V
    2. 0.32V
    3. 0.6V
    4. 0V
    5. 0.82V
    6. 3.49V
    7. 2.38V
    8. 1.35V
    9. 3.16V

    R366 (15K) has 2.53V at the thermistor end.

    R369 is 28mV at the end not connected to the chip.

    My next move was to connect a 100K resistor between the rail end of R372 and
    This took pin 2 up to 3.11V but the relay didn't turn off.

    My conclusion is that whatever I bought in SIP9 packages marked TA7317P
    can't be TA7317P

    This could have happened accidentally, the manufacturer may have packaged
    the wrong device.

    There is no doubt that the devices are marked TA7317P but I happened to
    accidentaly type TA7137P into Google today and found that this is also SIP9.

    I think the only thing I can do to make progress is to get more TA7317P from
    a different source and perhaps tell the vendor that I don't think they
    supplied TA7317P despite the markings.

    Old guy.
  14. Franc Zabkar

    Franc Zabkar Guest

    Nice idea.
    That means that the current through R366 is ...

    (3.16V - 2.53V) / 15K = 42uA

    The current through R367 is ...

    0.6V / 2.85K = 210uA

    The current through R365 is ...

    (3.16V - 0.6V) / 180K = 14uA

    Therefore pin 3 of the IC must be sourcing 154uA (= 210 - 42 - 14).
    This is inconsistent with the equivalent circuit in the IC datasheet.

    Also, the fact that pin #5 is at a higher potential than ground would
    suggest that the IC has an internal open circuit between pin #4 and

    Assuming capacitor C316 is not leaking, then pin #7 of the IC must be
    sourcing 72uA (= 2.38V / 33K). This is also inconsistent with the
    equivalent circuit.
    It sure seems that way. :-(

    - Franc Zabkar
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day