I want to light a LED when 2.1mm plug is inserted. Got any idea's?

[jimmybjj]

I just built the AMZ power supply and it includes 6 output 2.1mm jacks. I would like to put a LED on each output that lights when a plug is inserted. Being a novice i not really sure on how to go about doing this. The only thing i can think of is the opposite (turning off the led) of what i want using the battery tab on the jack. Anybody have any ideas?

[jaki54321]

I just built the AMZ power supply and it includes 6 output 2.1mm jacks. I would like to put a LED on each output that lights when a plug is inserted. Being a novice i not really sure on how to go about doing this. The only thing i can think of is the opposite (turning off the led) of what i want using the battery tab on the jack. Anybody have any ideas?

Im a newbie as well but I think wat you can do is hook each LED to the main input power jack(on the positive or "Long" leg), and hook the led(Small/Negative Leg) to the negative terminal on each power jack, so when the plug is inserted to send power to the device, the ground will be completed. But I do not know for sure, you could always test it.



Look at the bottom jack, the one facing toward the rails. Lets say this is the input jack for the entire power supply. Wat you could do is hook each LED up to the positive rail(the one that says board +), but make sure you have a resistor for the LED because if ur inputting 18 Volts you will destroy it in less then the blink of an eye 

Maybe a 4.7K resistor will work ok, depending on the specifications on your LED. Each LED will probably need one unless its possible to hook them all to 1 resistor, but I wont encourage you because i have no clue xD but ill have to test it myself.

Once you hooked the LED's to the positive power jack, hook the negative leg of each LED(the small leg) to the - terminal(The one that says Board - and battery -). So once you plug in each of your plugs to the power supply, the ground will be completed, thus the LED will light up.

I cannot confirm this because I have not worked with this power supply before, but it is worth a try, it wont damage anything. But test only 1 LED. Just get a resistor that can drop the power so the LED wont blow. I used a 4.7K and it worked ok. 2.2 works for 12 volt so it might work for 18. Id jus try 4.7K 1/2 or 1/4 watt. Only test 1 LED before putting the rest on there, you dont wana jus wire all the LED's up and then realize it doesnt work correctly.

Good luck, if u need help lemme know. I hope this works out for u, i might jus test it myself with my spare LED's

[jimmybjj]

I just built the AMZ power supply and it includes 6 output 2.1mm jacks. I would like to put a LED on each output that lights when a plug is inserted. Being a novice i not really sure on how to go about doing this. The only thing i can think of is the opposite (turning off the led) of what i want using the battery tab on the jack. Anybody have any ideas?

Im a newbie as well but I think wat you can do is hook each LED to the main input power jack(on the positive or "Long" leg), and hook the led(Small/Negative Leg) to the negative terminal on each power jack, so when the plug is inserted to send power to the device, the ground will be completed. But I do not know for sure, you could always test it.



Look at the bottom jack, the one facing toward the rails. Lets say this is the input jack for the entire power supply. Wat you could do is hook each LED up to the positive rail(the one that says board +), but make sure you have a resistor for the LED because if ur inputting 18 Volts you will destroy it in less then the blink of an eye 

Maybe a 4.7K resistor will work ok, depending on the specifications on your LED. Each LED will probably need one unless its possible to hook them all to 1 resistor, but I wont encourage you because i have no clue xD but ill have to test it myself.

Once you hooked the LED's to the positive power jack, hook the negative leg of each LED(the small leg) to the - terminal(The one that says Board - and battery -). So once you plug in each of your plugs to the power supply, the ground will be completed, thus the LED will light up.

I cannot confirm this because I have not worked with this power supply before, but it is worth a try, it wont damage anything. But test only 1 LED. Just get a resistor that can drop the power so the LED wont blow. I used a 4.7K and it worked ok. 2.2 works for 12 volt so it might work for 18. Id jus try 4.7K 1/2 or 1/4 watt. Only test 1 LED before putting the rest on there, you dont wana jus wire all the LED's up and then realize it doesnt work correctly.

Good luck, if u need help lemme know. I hope this works out for u, i might jus test it myself with my spare LED's

Thanks for your input, but "ground will be complete" regardless of there is a plug or not. The result would be a led that is always on. Thanks again.

[earthtonesaudio]

Did you use this type of jack (3 terminals, two of which are normally closed contacts that open when a plug is inserted)?

If so then all you need is the LED and resistor.  You have two contacts which are closed when no plug is present.  Put the LED across them, and a resistor to the other power terminal.  With no plug the current flows through the switch, not the LED, and you just have a resistor across the power supply.  When you insert a plug, the switch that's in parallel with the LED opens and now current has to flow through the LED+resistor combo.

Now the only things left to work out are the polarities.


If you used a two-terminal, non-switching jack, it's gonna be more complicated.  Now the simplest option is to put an LED in series with the power supply, but this option also sucks in terms of performance.
At the other end of the performance spectrum would be to use a high-side current monitor, wired to light the led when anything but an open circuit is connected to the jack.
In between are a lot of possibilities with transistors, diodes, and resistors.

In my opinion, the easiest route would be to install switching jacks if you haven't already and use the first option above.

[jimmybjj]

Did you use this type of jack (3 terminals, two of which are normally closed contacts that open when a plug is inserted)?

If so then all you need is the LED and resistor.  You have two contacts which are closed when no plug is present.  Put the LED across them, and a resistor to the other power terminal.  With no plug the current flows through the switch, not the LED, and you just have a resistor across the power supply.  When you insert a plug, the switch that's in parallel with the LED opens and now current has to flow through the LED+resistor combo.

Now the only things left to work out are the polarities.


If you used a two-terminal, non-switching jack, it's gonna be more complicated.  Now the simplest option is to put an LED in series with the power supply, but this option also sucks in terms of performance.
At the other end of the performance spectrum would be to use a high-side current monitor, wired to light the led when anything but an open circuit is connected to the jack.
In between are a lot of possibilities with transistors, diodes, and resistors.

In my opinion, the easiest route would be to install switching jacks if you haven't already and use the first option above.

Thank you very much, that worked. I'm still trying to wrap my head around why. I guess when the switch if closed the led doesn't have a reference to ground and there for no current flowing thru it. Is that right?

[jaki54321]

I just built the AMZ power supply and it includes 6 output 2.1mm jacks. I would like to put a LED on each output that lights when a plug is inserted. Being a novice i not really sure on how to go about doing this. The only thing i can think of is the opposite (turning off the led) of what i want using the battery tab on the jack. Anybody have any ideas?

Im a newbie as well but I think wat you can do is hook each LED to the main input power jack(on the positive or "Long" leg), and hook the led(Small/Negative Leg) to the negative terminal on each power jack, so when the plug is inserted to send power to the device, the ground will be completed. But I do not know for sure, you could always test it.



Look at the bottom jack, the one facing toward the rails. Lets say this is the input jack for the entire power supply. Wat you could do is hook each LED up to the positive rail(the one that says board +), but make sure you have a resistor for the LED because if ur inputting 18 Volts you will destroy it in less then the blink of an eye 

Maybe a 4.7K resistor will work ok, depending on the specifications on your LED. Each LED will probably need one unless its possible to hook them all to 1 resistor, but I wont encourage you because i have no clue xD but ill have to test it myself.

Once you hooked the LED's to the positive power jack, hook the negative leg of each LED(the small leg) to the - terminal(The one that says Board - and battery -). So once you plug in each of your plugs to the power supply, the ground will be completed, thus the LED will light up.

I cannot confirm this because I have not worked with this power supply before, but it is worth a try, it wont damage anything. But test only 1 LED. Just get a resistor that can drop the power so the LED wont blow. I used a 4.7K and it worked ok. 2.2 works for 12 volt so it might work for 18. Id jus try 4.7K 1/2 or 1/4 watt. Only test 1 LED before putting the rest on there, you dont wana jus wire all the LED's up and then realize it doesnt work correctly.

Good luck, if u need help lemme know. I hope this works out for u, i might jus test it myself with my spare LED's

Thanks for your input, but "ground will be complete" regardless of there is a plug or not. The result would be a led that is always on. Thanks again.

ya ur right i tested it out. sorry i couldnt be of more help, hope u get it workin 

[pleiades]

When the switch is closed, both sides of the LED are held at ground.  No voltage difference across the LED means it is not forward biased and doesn't conduct, no current flow through it, no light.

Less rigorously, closing the switch gives the current an direct path to ground which bypasses the LED (current prefers the easiest path), so no current flows through it and it doesn't light up.

[jimmybjj]

When the switch is closed, both sides of the LED are held at ground.  No voltage difference across the LED means it is not forward biased and doesn't conduct, no current flow through it, no light.

Less rigorously, closing the switch gives the current an direct path to ground which bypasses the LED (current prefers the easiest path), so no current flows through it and it doesn't light up.

Thanks for the education  I have the fundamental's and I understand the logic behind most of it but it seems that I just need a lot of guidance in the application. I seem to constantly go "Why didn't i think of that" or "I knew that, why didn't i think of that"

Thanks again.

[Sanguinicus]

First post here. Firstly, I want to apologise for bumping such an old topic.

I want to do this same thing for my R.G Keen 'Spyder' style power supply.

I don't understand the explanations given, is it possible to get a layout of how everything is connected? I can't wrap my head around how it's done. Also, in addition to the layout, is it possible to get a schematic view of it?

Thanks a lot.

EDIT: Here is what I want to achieve. Instead of doing this at the effect, I want to do this at the power supply. Instead of having a bunch of unused leads on the power supply, i want to insert cables as needed and have an LED light up upon insertion of the power lead. And will this only work when an effect is plugged in at the other end?

[earthtonesaudio]

If you just want a LED lit when the plug is inserted, see above.  If you want it lit only when there's an effect connected to the other end, that's a little more tricky.  You need to sense current being drawn by the effect (through a very low voltage drop) and use that current as a signal to switch on the LED.  I would do it with a comparator and resistor, but there are specialty chips made for the job.

9V from supply --------- current sense resistor --------- to jack
                           |                                           |
                   comparator input(-)                  comparator input(+)
                                                  \        /
                                           comparator output to the point labeled "somewhere" (from your drawing)

Ground-------------------------------------------------------- to jack