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Powering a footswitch LED from the amp PS - will this work?

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  • Powering a footswitch LED from the amp PS - will this work?

    Hi All,

    I am trying to get the DC for the footswitch LED from the main PS rail.

    I've done a rough schematic of what I'm thinking of trying. My question is - will this work?

    Click image for larger version

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  • #2
    Why do you have to do it from the B+ rail? A half way rectifier from the filaments supply is easier and safer.

    Comment


    • #3
      Not really. First, it is just an LED, so why all the stages of filter caps and resistors and the zener? My first choice would not be running it off B+ anayway, but even if we do, say 400v of B+, and 10ma of LED current, we need then a 40k resistor at 7-10 watts. That is a non-standard value, so 47k or 39k would suffice. The one resistor and the LED.

      And why did you regulate to 5v? If it is a plain LED, then it will naturally have a voltage drop across itself of 1-3v, and the zener would never do anything. If you are using some sort of LED indicator assembly witnh internal resistor, well, that isn't the way to go about it.

      If I needed some voltage, I'd look lower. Is your amp output cathode biased? If so, tap into the voltage at the cathodes of the power tubes. That is a much lower voltage to start with.

      No cathode bias? Then you must have fixed bias, and so a bias supply. Depending on the type power tube, that supply might be anything from 10-12v to 60v. It would be negative, but the LED doesn;t care, just turn it around and use that negative supply. Again, much lower to start than B+.

      But more importantly, what circuit are you controlling? Show us that. If you are indicating channel change for example, if you use relays, then the relay coil current can run the LEDs too, no power supply required beyond the one the relays use.
      Education is what you're left with after you have forgotten what you have learned.

      Comment


      • #4
        1) that Led is upside down so it won't light

        2) Leds are current driven, not voltage driven, and drop between 1.9 and 2.5 V , so that 5V Zener will do nothing

        3) *how * are you going to send amp voltage/current to the footswitch Led?
        You'll need an extra wire besides those used for actual ... er ... footswitching.

        4) as is, Led will always stay ON, showing no indication of switch position, it will just help you find the footswitch in a large, dark stage; in fact it's an extra amplifier pilot light, nothing else.
        Juan Manuel Fahey

        Comment


        • #5
          Another way to go might be a dedicated SMPS
          As an example the Meanwell APV-12-12 puts out 12v/1A on an input of 90-264 VAC or 127-370 VDC
          its 3" x 1.5" x 1.15" and costs $6 and probably won't add hum.

          APV-12-12 Mean Well | Mouser

          you could even run an LED light show up top with one of these $3 units
          Click image for larger version

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          and a cheap RGB strip

          Comment


          • #6
            First up, thank you to all for the replies, you people are amazingly helpful!

            Secondly, I must apologise for the lack of detail and explanation in my initial post. It was late at night (here) and I fired it off in haste.

            So to flesh out the situation… The RC filtering network in the schematic is the PS for a valve amp, each of the RC powering a node.
            The footswitch for the amp is fed via a multicore (4 conductors and an earth) shielded microphone cable. The footswitch is for Reverb, Tremolo, and a tone stack bypass. For each effect switching, the effect is either contacted to ground, or disconnected from ground. No relays etc.

            The LEDs will indicate whether each effect is on or off, with the LED switching using another pole on each of the effect switches. To run power for these LEDs I plan to use the 4th wire in the multicore cable, and the return ground of the shield. But as the cable has connections to the signal path running through it, I thought that the LED PS should be very well filtered DC.

            The Zener diode is there to provide a dual purpose, to act as an AC bypass to ground to filter the DC, and also to set the voltage of the DC running to the footswitch. 5V here is nominal, I just thought that this would be more voltage than the LEDs will need, but a safe low figure. As stated above, as soon as one of the footswitch LEDs is turned on, the voltage at this node will drop down to that across the LED.

            The reason i thought of using the B+ network to generate the LED DC was twofold. Physically the end of the PS chain is exactly where I need it to be to run to the footswitch (trivial I guess!). The other reason is that as all of the footswitch uses the same ground return, I wanted to keep it all on the same star ground node to avoid hum.

            I'm at work currently, and break is ending, so I'll respond to the great suggestions ^ properly this evening.

            Comment


            • #7
              OK, home from work, dinner cooked and family fed, dog walked, now can reply

              First I remembered another reason I was proposing the circuit at top - the large value dropping resistor and zener would act as a voltage bleeder when the amp was turned off, at least down to 5V.

              Gregg, I hadn't thought of a half wave rectified heater supply. I will think about it now, thanks for the suggestion.

              Enzo, as I mentioned above, I'm sorry I forgot to include the extra info to help explain the schematic, particularly WRT the string of RC. These are the RC/dropping nodes for the rest of the amp PS.

              This amp is cathode biased, at least partially. The bias supply has a bias resistor connected to the output valves cathodes, then via a zener diode to ground. So the bias voltage is partially "fixed" in voltage (zener), and partially resistive. The cathodes are also connected to a false CT (via 100 ohm resistors) on the heater supply.
              At the top of the zener, the voltage should be perfectly AC bypassed to ground, and so should offer a well filtered DC supply, at the zener diode voltage, in this case 7.5V. How can I use this as a PS for the footswitch LEDs?
              If I connect a high value resistor here, then connect to the LEDs, that might work well. As long as the resistor prevents the voltage at the top of the zener from dropping when the LED is activated, which would change the bias voltage.

              Re. the effects circuits that the footswitch is controlling, they are all passive, think Fender blackface Rev and Trem switching.

              JM Fahey, it really was a rough and rushed schematic, I didn't even look at the semiconductor polarities, if the zener was correct that was just luck
              The zener purpose is so that this resistor/zener circuit can act as a PS bleeder, even when the footswitch LED are disconnected/switched off.
              As I wrote above, the footswitch lead has 4 conductors plus shield, leaving one conductor for the LED DC supply, and sharing the shield as a ground return with the effect switching.

              tedmich, I don't even know what these SMPS are! Talk about a luddite, I've mostly been sticking to the old technology, the most advanced parts I use ATM are diodes I'll read up, thanks!

              Comment


              • #8
                Ok.

                1) Forget about using the pedal or *any* externally accesible connection to bleed +V .
                That means there is a contact connected to +300/450V through a power resistor which can source quite a few mA ... did I say Electric Chair?

                Do not trust the Zener to keep accessible voltage to +5V ; the minute it opens/solder cracks/loses ground/whatever you have full +V there.

                Just add the bleeding resistors inside and all will be fine.

                2) full wave rectifying a center tapped 6.3V winding will give you ~4.5 or 5V at any current you want (we are speaking driving Leds here) and be real safe.

                3) Passive switching may have 2 problems, only actual testing (and some old fashioned luck) will tell:
                a) long cables have capacitance and may kill highs.
                Not applicable for tremolo and acceptable, sort of, for Reverb, but actual guitar signal may be dulled by such capacitance
                b) did I mention hum?
                Audio carrying cables should be well shielded.

                4) I suggest using a Led/LDR optocoupler for actual switching on the amp itself (signal never leaves the chassis) , and putting internal Led and pedal mounted Led in series, with an approppriate current limiting resistor, then when grounding that string , both Leds light, one "acting" and one "showing".
                Or relays, etc.
                Juan Manuel Fahey

                Comment


                • #9
                  Thanks again for your detailed response J M Fahey. You have definitely talked me out of my original plan, that is unless there is large, lucrative, and non-litigious market for guitar amps that can double as electric chairs

                  I have used this passive switching method with previous builds, and I couldn't discern any difference in hum levels or HF when the footswitch plug was either connected or disconnected from the amp. I was initially concerned about this, so I did test this to my satisfaction. However these previous builds did not have LED indicators in the footswitch, and no LED PS running through the cable. This could well be a source of extra hum injection, either from any ripple coupling, or else from grounding issues (earth loops etc.)

                  However I greatly appreciate your suggestions re. using LED/LDR for the switching, and I may well investigate that in the future, as well as using the heater supply to obtain the LED DC.

                  What I have been following up is the suggestion (thanks Enzo) to use the output valve cathode voltage as a DC source. I wrote this a couple of days ago -

                  "This amp is cathode biased, at least partially. The bias supply has a bias resistor connected to the output valves cathodes, then via a zener diode to ground. So the bias voltage is partially "fixed" in voltage (zener), and partially resistive. The cathodes are also connected to a false CT (via 100 ohm resistors) on the heater supply.
                  At the top of the zener, the voltage should be perfectly AC bypassed to ground, and so should offer a well filtered DC supply, at the zener diode voltage, in this case 7.5V. How can I use this as a PS for the footswitch LEDs?
                  If I connect a high value resistor here, then connect to the LEDs, that might work well. As long as the resistor prevents the voltage at the top of the zener from dropping when the LED is activated, which would change the bias voltage."

                  I tried it, using ever diminishing values of dropping resistor. I got down to 4.7K ohm, which fully lit the trial LED, and made no measurable difference to the bias voltage at the top of the bias zener. I haven't tried any smaller values of dropping resistor yet. But for now using a 4.7K gives me a usable and safe DC PS, with only the addition of 1 resistor to the amp!

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