Announcement

Collapse
No announcement yet.

Traynor DynaGain 30 Bad Cap?

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • Traynor DynaGain 30 Bad Cap?

    I'm trying to fix this Traynor DG30D solid state amp for a buddy. Service manual and schematic attached. The high gain channel has a weak output and the normal channel has no output. With a 1Khz test signal I scoped pin 1 on U1:A and saw a nice sine wave. However on the other side of C9 there is no signal. So far I've been able to avoid pulling the PCB out of the chassis. So in the interest of trying to determine whether C9 is the problem I was going to try connecting another .33uF cap across C9. I also thought about simply running a jumper across C9. I'm a little hesitant around the solid state circuits though so I wanted to check with you guys first. Can I do what I described without any danger of frying anything?

    B.

    Service Manual dg10-dg15-dg15r-dg30.pdf Click image for larger version

Name:	Schematic Input Zoom.jpg
Views:	1
Size:	216.1 KB
ID:	873930

  • #2
    I would do a resistance measurement on the far side of C9 to ground; so that if it looks like is not a short, sure, pop in another .33 or .47uf [the latter are more common] to see if it has gone open. As long as it is ground referenced a jumper will not damage anything but it would change the low frequency response quite a bit.

    Comment


    • #3
      Originally posted by kscience View Post
      I would do a resistance measurement on the far side of C9 to ground
      Sorry, I'm not making sense of this. C9 is not connected to ground. What resistance reading would you expect?

      Originally posted by kscience View Post
      so that if it looks like is not a short, sure, pop in another .33 or .47uf [the latter are more common] to see if it has gone open. As long as it is ground referenced a jumper will not damage anything but it would change the low frequency response quite a bit.
      The term "referenced" (or simply "reference") has always confused me. It seems ambiguous, like maybe a term that is used in more than one way and the experienced folks understand what is meant by it given the context. Can you please explain what you mean by "ground referenced"?

      Comment


      • #4
        Imagine I am standing on the floor, the top of my head is 6 feet above the floor. I am 6 feet tall with reference to the floor. If I am standing in a hot air balloon at 1000 feet, the top of my head is 1006 feet with respect to the ground, but still 6 feet with respect to the bottom of the balloon car.

        Most voltage readings are taken with respect to ground. That means your black probe of the meter is grounded. But sometimes we want to know voltage between two specific points, neither being ground, so we say what is the voltage at point X with respect to point Y?

        Reference means frame of reference.

        If you have signal up to C9, but not after, then either C9 is open, or the other side of C9 is shorted to ground. SO if you measure resistance to ground on the right end of C9, we hope to see it not shorted to ground.
        Education is what you're left with after you have forgotten what you have learned.

        Comment


        • #5
          Originally posted by Enzo View Post
          Imagine I am standing on the floor, the top of my head is 6 feet above the floor. I am 6 feet tall with reference to the floor. If I am standing in a hot air balloon at 1000 feet, the top of my head is 1006 feet with respect to the ground, but still 6 feet with respect to the bottom of the balloon car.

          Most voltage readings are taken with respect to ground. That means your black probe of the meter is grounded. But sometimes we want to know voltage between two specific points, neither being ground, so we say what is the voltage at point X with respect to point Y?

          Reference means frame of reference.
          Yes, I understand that much, but it doesn't seem to apply to this comment, as far as I can tell.

          "As long as it is ground referenced a jumper will not damage anything but it would change the low frequency response quite a bit."

          In what scenario would a jumper be "ground referenced" and in what scenario would it not be?

          Originally posted by Enzo View Post
          If you have signal up to C9, but not after, then either C9 is open, or the other side of C9 is shorted to ground. SO if you measure resistance to ground on the right end of C9, we hope to see it not shorted to ground.
          Ok, makes sense, thanks.

          Comment


          • #6
            Ground referenced = connected to ground.
            Education is what you're left with after you have forgotten what you have learned.

            Comment


            • #7
              Originally posted by bobloblaws View Post
              Yes, I understand that much, but it doesn't seem to apply to this comment, as far as I can tell.

              "As long as it is ground referenced a jumper will not damage anything but it would change the low frequency response quite a bit."

              In what scenario would a jumper be "ground referenced" and in what scenario would it not be?



              Ok, makes sense, thanks.
              You suggested running a jumper across C9. my point is that unlike discrete transistors, which have dc voltage on them that requires capacitive coupling, op amps are often at ground level [0vdc] output, and do not require ac coupling. So the jumper would not damage it. but it will certainly pass lower frequencies than the .33uf!

              Comment


              • #8
                Your initial thought of jumping over C9 is valid. The Op Amp ahead of C9 is running off +/- 15VDC. That stage is ground referenced via 470k to the (+) Input terminal, so we would expect the output terminal to be very close to 0VDC. If C9 is open, the jumper would restore signal to the next stages. If, instead, the other side of C9 is somehow shorted (not likely), then you'd get no output. Try it and see. There isn't any potential harm in doing that, though HOW you make that jumper connection needs to be accurate, so you don't inflict harm by misconnection.
                Logic is an organized way of going wrong with confidence

                Comment


                • #9
                  Originally posted by nevetslab View Post
                  If C9 is open, the jumper would restore signal to the next stages. If, instead, the other side of C9 is somehow shorted (not likely), then you'd get no output.
                  The odd thing is that I am getting some output from the hi gain channel. If C9 was in fact open I'd expect no output at all.

                  So since my inclination to try jumping across C9 has been validated, it leads me to wonder what the purpose of that cap is in the circuit. Can you enlighten me?


                  Originally posted by nevetslab View Post
                  That stage is ground referenced via 470k to the (+) Input terminal, so we would expect the output terminal to be very close to 0VDC.
                  Interesting. Forgive my lack of education, but why does it follow that the output terminal would be very close to 0VDC. In what alternative scenario would it be otherwise?
                  Last edited by bobloblaws; 08-27-2018, 08:48 PM.

                  Comment


                  • #10
                    Originally posted by Enzo View Post
                    Ground referenced = connected to ground.
                    Hmmm, OK, but not necessarily directly connected, right? Nevetslab wrote "That stage is ground referenced via 470k to the (+) Input terminal". So when kscience wrote "as long as it is ground referenced a jumper will not damage anything", are we referring to the same ground reference?

                    If a given point in a circuit is considered ground referenced by virtue of the fact it is connected via some component (resistor only?), wouldn't virtually every point in a circuit be ground referenced in some fashion?

                    Comment


                    • #11
                      the purpose of that cap. looks like to me, is to insure there is never any dc voltage on the fet channel switch associated with the volume control. It may be that the fet is shorted out, or it is on all the time; but we were going to see if there was any signal out of the op amp itself, per OP.

                      I apologize, I speak aspergers and so it may not be clear what I mean. I do attempt to be clear.

                      For fet switches you Never want dc voltage across or on them; its noisy as hell.

                      The reason you need the .33uf cap? to limit low frequencies not needed in music, eg the frequencies associated with muting the strings with your palm, in the hundreds of millisecond range. we cannot hear them but the speakers hate single digit hz signals.
                      Last edited by kscience; 08-27-2018, 09:48 PM.

                      Comment


                      • #12
                        My bad- i should have said since the output of the op amp is at or near 0 volts [or ground-ish] it would not damage things after it, like say a discrete transistor collector Not at 0 volts [though still ground referenced, so we can see how many volts the collector sits at at idle.

                        Grounded = connected to ground. Ground referenced =/= grounded; it means X volts from ground; with op amps in split supplies, it means very close to ground dc wise at idle.

                        Again, I struggle with english. I am not trying to be an ass or confuse you.

                        Comment


                        • #13
                          Originally posted by bobloblaws View Post
                          The odd thing is that I am getting some output from the hi gain channel. If C9 was in fact open I'd expect no output at all.
                          The drive channel has a gain stage of 100, then of 22, considering the 10k/10k divider. You are getting something out of it? with a gain of 2200 the input doesn't have to be much, imho. If there is no blockage, you ought to be getting a lot out. I begin to suspect the fets in series and/or shunts.

                          So since my inclination to try jumping across C9 has been validated, it leads me to wonder what the purpose of that cap is in the circuit. Can you enlighten me?
                          i forgot about reply with quote. i answered just a moment ago that one.

                          been years since i had to use bb codes :O



                          Interesting. Forgive my lack of education, but why does it follow that the output terminal would be very close to 0VDC. In what alternative scenario would it be otherwise?[/QUOTE]

                          Comment


                          • #14
                            Op amps will tend towards 0VDC on the output if biased up from ground reference, without getting into the fine details of differences between the non-inverting input and inverting input impedances. The jumpering of C9 is a temporary solution to see if you get signal past that stage. I hadn't thought about the FET stages and how they would behave with mV levels of DCV from the op amp. We're still looking for gross solutions at this point.
                            Logic is an organized way of going wrong with confidence

                            Comment


                            • #15
                              Hey, thanks for all the help guys, theoretical and otherwise. Lots to chew on. Anyway, when I went to measure some voltages etc. (in the interest of what we've been discussing) I discovered that the problem was a cold solder joint on C9 that I had missed when I initially chopstick tested. A slight nudge of the cap and my test signal came blasting through the speaker. Thanks again.

                              Comment

                              Working...
                              X