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.
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
On Fri, 2 Mar 2012 15:30:36 -0500, "John Smith" put finger to keyboard and composed:
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
-- Please remove one 'i' from my address when replying by email.
On Fri, 2 Mar 2012 16:42:59 -0500, "John Smith" put finger to keyboard and composed:
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
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On Fri, 2 Mar 2012 16:42:59 -0500, "John Smith" put finger to keyboard and composed:
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
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On Sat, 03 Mar 2012 10:00:29 +1100, Franc Zabkar put finger to keyboard and composed:
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
-- Please remove one 'i' from my address when replying by email.
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
Thanks Franc, I'll take your advice and do that on Monday with the two unused parts I have.
Old guy
On Fri, 2 Mar 2012 18:52:18 -0500, "John Smith" put finger to keyboard and composed:
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
-- Please remove one 'i' from my address when replying by email.
Are you using the left hand side of the printing as your index for pin "1" or the chamfer or notch in the body ?
...
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.
"1"
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 c*ck-up in the labelling section. Now removing 3055 marking off TO3s and marking them 2N3773 or whatever , would be the actions of a pirater
It begins to sound like somebody "hot rodded" the amplifier. And did a shoddy job of it.
None of the resistors look distressed and R367 measures 2.85K with no chip fitted.
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:
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 pin2. 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.
On Mon, 5 Mar 2012 18:44:57 -0500, "John Smith" put finger to keyboard and composed:
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 = 210uAThe 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 Q20.
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
-- Please remove one 'i' from my address when replying by email.
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