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  • Roland Super Cube 100 LED

    I have a Roland Super Cube 100 (not bass version) with an power indicator LED that isn't lighting up. The circuit is foreign to me and is different than what you see in the schematic for the bass version. I'm uploading images of the bass version schematic and my own sketch of the circuit in question including voltages. All the surrounding components that are pictured check out as in spec on the DMM. In my experience these small red LEDs typically run with around a 2V voltage drop and I tested this one outside of the amp and it checked out as such and lit up normally. I tried adding different values of resistance at different points in the circuit to try and adjust the voltages on either side of the LED hoping that if I could modify it so that the voltage drop across the LED was in the 2V neighborhood it would work as it did in my experiment outside of the amp. But any change in voltage was negligible with the resistors I tried (e.g. 1K, 100K) and the voltage drop on the LED remained the same.

    I need guidance please!

    ..and thanks,

    - B



    Click image for larger version

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    Click image for larger version

Name:	LED_And_Power_Bass_Version.jpg
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  • #2
    First off, this is a commercial unit, and one presumes it worked at some point. Likely Roland did not ship products with non-functioning power lights. My point is that redesigning the circuit is not the approach, we just need to fix it.

    Is there anything else wrong? I mean the power light is dark, so does that mean it also won't amplify? Or is the LED the ONLY problem?

    LEDs MAKE the 2v (varies with color of LED) by their nature, we don't light them by putting two voltages on the ends that are 2v different. HAven't you got one of the diodes in your drawing backwards? Zeners do not zene in forward bias. They break down predictably in reverse bias. As drawn the current freely flows through the zener.

    So the bass version we see is different, but do you have similar power supplies? In other words do you have roughly 54v main power rails? And they are made from a roughly 38 - 0 - 38 VAC center tapped power transformer? If so, and you want to change something, why not just steal the LED circuit from the bass version? A resistor, a diode, and an LED.
    Education is what you're left with after you have forgotten what you have learned.

    Comment


    • #3
      Originally posted by Enzo View Post
      First off, this is a commercial unit, and one presumes it worked at some point. Likely Roland did not ship products with non-functioning power lights. My point is that redesigning the circuit is not the approach, we just need to fix it.

      Is there anything else wrong? I mean the power light is dark, so does that mean it also won't amplify? Or is the LED the ONLY problem?

      LEDs MAKE the 2v (varies with color of LED) by their nature, we don't light them by putting two voltages on the ends that are 2v different. HAven't you got one of the diodes in your drawing backwards? Zeners do not zene in forward bias. They break down predictably in reverse bias. As drawn the current freely flows through the zener.

      So the bass version we see is different, but do you have similar power supplies? In other words do you have roughly 54v main power rails? And they are made from a roughly 38 - 0 - 38 VAC center tapped power transformer? If so, and you want to change something, why not just steal the LED circuit from the bass version? A resistor, a diode, and an LED.

      I had a feeling you might be burning the midnight oil :-)

      Point taken about redesigning the circuit, I guess I was just experimenting hoping I could learn something at the same time. I have to admit I don't fully comprehend the relationship between the AC and DC voltages observed and/or the effects of the components in that circuit. We use capacitors to block DC from passing to subsequent gain stages (coupling caps). In my diagram there is 122VAC at the power switch end of the 1nF cap. On the LED side of that cap I measure 55VAC and 85VDC. I would have expected 0VDC. That's just the tip of the iceberg as far as how confusing it is to me. So borrowing from the other diagram makes sense but I'd like to learn the concept of this particular design and why it isn't working and why I'm seeing the voltages that I am.

      Regarding the zener, I jumped to a conclusion based on the appearance of it. In fact there is a symbol drawn on the component side of the PCB showing it is a regular ol' diode. Here's my corrected diagram:


      Click image for larger version

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      Edit - Yes, the power light is the only problem.

      Comment


      • #4
        Originally posted by bobloblaws View Post
        Here's my corrected diagram:


        [ATTACH=CONFIG]56104[/ATTACH]
        Your diagram is incorrect

        - LED diode is not galvanic separated from AC main voltage.
        - If the AC socket turns 180 degrees and no phase coming to the on/off switch, LED diode will not illuminate.
        - When the fuse burns LED diode still lights.

        Isn't it easier to replace in kit R47, D8 and D9.
        Who does not know and knows that he does not know - teach him Confucius)
        Who knows and does not know that he knows - wake him Confucius)

        Comment


        • #5
          Originally posted by vintagekiki View Post
          Your diagram is incorrect

          - LED diode is not galvanic separated from AC main voltage.
          - If the AC socket turns 180 degrees and no phase coming to the on/off switch, LED diode will not illuminate.
          - When the fuse burns LED diode still lights.

          Isn't it easier to replace in kit R47, D8 and D9.
          There's a chance I may have overlooked something, but I went over the circuit pretty thoroughly. Where would you suggest I look specifically for where you think it is incorrect? Or are you just saying it wasn't designed correctly?

          - LED diode is not galvanic separated from AC main voltage.

          Do you mean the LED should be isolated via the power transformer as in the bass version schematic? You would think so, but not the case with this amp.

          - If the AC socket turns 180 degrees and no phase coming to the on/off switch, LED diode will not illuminate.

          In what circumstance would the AC socket turn 180 degrees? You mean an wall outlet that was incorrectly wired up?

          Isn't it easier to replace in kit R47, D8 and D9.
          Maybe, but if this is the way they did it I'd like to figure out why it is not working given that the LED itself seems fine, for my own education.

          Comment


          • #6
            The cap is a reactance at AC and passes AC to the circuit. The diodes rectify the AC into DC, though it is unfiltered. LED lighting current need not be filtered, why waste a cap?

            Just idly looking at it, it looks like one or both diodes could be open.
            Education is what you're left with after you have forgotten what you have learned.

            Comment


            • #7
              All which is incorrect writes in post # 4. Let me repeat.
              There's galvanic connection AC main voltage with chassis.
              When the fuse burns LED diode still lights.
              What is the problem to replace R47, D8 and D9?

              1)
              LEDs constantly illuminate regardless without position of on/off switch
              Last edited by vintagekiki; 11-29-2019, 10:49 AM. Reason: 1)
              Who does not know and knows that he does not know - teach him Confucius)
              Who knows and does not know that he knows - wake him Confucius)

              Comment


              • #8
                Originally posted by Enzo View Post
                The diodes rectify the AC into DC
                OK, so in my question I was referring to the DC voltage at the on/off switch end of the LED. But do I need to be thinking in terms of electron flow as opposed to "conventional current"? Because my first thought is to question how the voltage measured at that point has been rectified since it has only "seen" one terminal of the first diode (the LED). Further, I can't grok how there could be 85VDC there when a complete half cycle of 120VAC is only 60 volts.

                Originally posted by Enzo View Post
                Just idly looking at it, it looks like one or both diodes could be open
                They appear to be OK on both the diode check and resistance check on my DMM.

                Comment


                • #9
                  Originally posted by bobloblaws View Post
                  OK, so in my question I was referring to the DC voltage at the on/off switch end of the LED. But do I need to be thinking in terms of electron flow as opposed to "conventional current"? Because my first thought is to question how the voltage measured at that point has been rectified since it has only "seen" one terminal of the first diode (the LED). Further, I can't grok how there could be 85VDC there when a complete half cycle of 120VAC is only 60 volts.


                  They appear to be OK on both the diode check and resistance check on my DMM.
                  Dear colleague constantly questions, little theory will not bother anyone.
                  Why thinking when it's simpler replace R47, D8 and D9?

                  May be cold soldered in the LED connection.

                  1)
                  Question
                  Either D8 and D9 are accidentally inverse soldered or one of the terminals (R47, D8, D9) is interrupted.
                  Last edited by vintagekiki; 11-29-2019, 03:59 PM. Reason: 1)
                  Who does not know and knows that he does not know - teach him Confucius)
                  Who knows and does not know that he knows - wake him Confucius)

                  Comment


                  • #10
                    Here is a crop from the Cube 100.

                    Cube-100 Pwr S.pdf

                    Super duper easy.
                    A diode, a resistor & an LED to ground.

                    Comment


                    • #11
                      Originally posted by Jazz P Bass View Post
                      Here is a crop from the Cube 100.

                      [ATTACH]56114[/ATTACH]

                      Super duper easy.
                      A diode, a resistor & an LED to ground.
                      Thanks Jazz. I may end up going that route. I didn't want to give up on the existing circuit without a fight though.

                      Comment


                      • #12
                        Originally posted by bobloblaws View Post
                        I didn't want to give up on the existing circuit without a fight though.
                        Please explain difference between these two attachments:

                        https://music-electronics-forum.com/attachment.php?attachmentid=56103&d=1575005854

                        https://music-electronics-forum.com/attachment.php?attachmentid=56114&d=1575047278

                        And what did you learn from fight?
                        Who does not know and knows that he does not know - teach him Confucius)
                        Who knows and does not know that he knows - wake him Confucius)

                        Comment


                        • #13
                          Originally posted by vintagekiki View Post
                          My point is I see the situation as a challenge and I wasn't prepared to immediately throw in the towel without figuring out what is going on with the original circuit. I obviously don't have the same knowledge as most on here. I'm trying to learn as much as I can. Enzo suggested borrowing the configuration from the bass cube in post #2 and I acknowledged that possibility in my reply to him.

                          Comment


                          • #14
                            Both circuits have AC source, 2 diodes, and 3.9K resistor in series. (cap steps down AC in one circuit, transformer does it in other)
                            Diodes are biased reversed in one circuit, but should not matter as long as source is AC, and both diodes in circuit biased same way.

                            What does not look right is that your drawn in DC voltages are positive instead of negative, and show the LED diode turned off.
                            "Everything is better with a tube. I have a customer with an all-tube pacemaker. His heartbeat is steady, reassuring and dependable, not like a modern heartbeat. And if it goes wrong he can fix it himself. You can't do that with SMD." - Mick Bailey

                            Comment


                            • #15
                              Originally posted by g1 View Post
                              What does not look right is that your drawn in DC voltages are positive instead of negative, and show the LED diode turned off.
                              Please clarify for me. For the LED to be lit should the voltage be negative on both sides? (in this configuration)

                              Comment

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