LED lighting on open circuit

Kwagmire Wagner · February 06, 2016, 07:10:24 PM

Hey folks.

I am wiring a custom channel selector/effects loop and have run into a problem in the test phase and wanted to run it by you for your thoughts.

First, the circuit: When the circuit is "open", the overdrive channel is selected (LED lit). When closed, the clean channel is selected (LED not lit). Now, when you look at the pic below, I realize the circuit is never really open. The switch lugs are bridged with the LED however because the LED gives resistance the electricity flows through the switch and doesn't light the LED when the switch lugs are connected via the throw. Using an LED with match voltage sent from the amp down the line (orange 5mm @ 3.2V) I constructed the following switch:

It seems to work however, when testing it out with a couple of AAs powering the circuit, the wire which I was using to connect to the negative terminal of the batteries heated up when the circuit was closed (LED not lit). Clearly, that is resistance making it heat up.

Question 1: Do I need to place a resistor between the switch lug and the negative terminal of the battery OR will this not matter in my channel switch pedal as I would image the power the amp is sending is AC?

Question 2: I tested the voltage on the negative side of the LED when it is lit and it has only dropped to 1.3V meaning a forward voltage of 1.9V. I'm not sure if this is enough of a drop to cause the channel to switch. How do I drop that if it isn't?

Jdansti · February 07, 2016, 02:47:10 AM

Aren't you just shorting the power supply (battery) when you close the switch, and that's what's causing the wire to get hot? The LED doesn't light up because the path of least resistance is through the switch contacts. This isn't such a big deal when using a battery because the battery has limited capacity to deliver current. But if you try this with a power supply, your going to fry something.

Am I missing something?

bluebunny · February 07, 2016, 11:07:02 AM

My thoughts exactly, John. @KW: You close that switch and you short your battery. I hope you have shares in Duracell!

blackieNYC · February 07, 2016, 11:29:38 AM

It would be very convenient at times if the opening of a switch could turn on an LED But think of the inverse: that would mean closing the switch turns it off, and that means your switch must connect that current to something else. If you really need to do this, have a look at the logic circuits in The Tonegod Wicked Switches

slacker · February 07, 2016, 12:05:57 PM

Some amp channel switches are wired like this, my old Marshall was. The amp has a resistor or something else to limit the current to the switch and it detects whether the switch is closed, the voltage across it is zero, or open the voltage across it is the forward voltage of the LED.
We'd need to know what the rest of circuitry is to know whether or not this will work. If it's not designed to work like this then you could blow the LED up or damage whatever the 3.2 volts is coming from.

Kwagmire Wagner · February 07, 2016, 12:20:19 PM

Hmm.

So here's the deal as I understand it with my foot switch and keep in mind, the batteries were just there to provide the same voltage as the footswitch line from my amp. I've confirmed this with a voltmeter on a normal patch cable plugged into the amp footswitch input and by measuring the voltage between the tip and sleeve of the cord.

3.2V of phantom power travel down the footswitch line to the pedal. When the sleeve and tip are connected and the circuit closed, the amp is on the clean channel. When they aren't connected the amp switches to the overdrive channel. The SPDT in my schematic does the same thing does it not? The LED only lights up when the SPDT's throw is to the left side and doesn't provide a connection with less resistance than the LED offers. In essence, the circuit is always closed but the resulting voltage drops enough after flowing through the LED to cause the channel to switch to the overdrive channel. I imagine the voltage trips a relay in the amp and a lack thereof does the opposite. I have manually completed the circuit on a patch cable plugged into the footswitch input and it has not heated up.

So, with that in mind, is it not plausible that wiring a pedal that way would work and would seem to negate shorting of the power source? I suppose the easy alternative is simply to have the LED only light when the circuit is closed and on the clean channel and not when it is on the overdrive channel and open however this tends to be the opposite of footswitches work.

PRR · February 07, 2016, 03:15:46 PM

> Do I need to place a resistor between....

YES.

An LED has very-little resistance and a nearly fixed voltage. If you don't have some added resistance in the circuit, it WILL suck way too much current, burn itself up or its power supply or drain all your batteries.

(* Exception which violates the rules but is now very common. LED makers don't mind extra resistance in the LED. They make LEDs for *specific* voltages: 3V battery, 12V supply. They "work" without added resistors. But they also go dim when a hair under-voltage and burn-up a hair over-voltage. This is usually a Bad Plan except in specific flashlights or appliances.)

> match voltage sent from the amp down the line

What is "match voltage"? What voltage, how much current available, is there any added resistance?

If you built this shorted-LED affair with a 9V battery, a 1K resistor, and a 2V red LED, it would "work". Without the switch, there would be 9V-2V= 7V across the 1K, 7mA of current to the LED. With switch shorting the LED, the full 9V flows through resistor and switch contacts, 9mA. Yes, "off consumption" is larger than "on consumption".
7mA or 9mA is a significant but not dangerous current for a 9V battery.

It will work fine with 9V and your 3.2V LED. The LED current will be less than 6mA, which should still be plenty of light.

If this "match voltage" is some other voltage or resistance, that's an important detail you need to say.

slacker · February 07, 2016, 03:35:34 PM

Quote from: Kwagmire Wagner on February 07, 2016, 12:20:19 PM
So, with that in mind, is it not plausible that wiring a pedal that way would work...

Yes it might work fine but without seeing the circuit that's providing the 3.2 volts we don't know for sure. What is the amp? Have you tried looking for a schematic for it or for the original footswitch.

Kwagmire Wagner · February 07, 2016, 03:46:50 PM

I'm replacing the Blackstar FS-4 footswitch

slacker · February 07, 2016, 04:07:39 PM

There's a thread here http://www.diystompboxes.com/smfforum/index.php?topic=102217.0 where someone is talking about an FS-1 footswitch and that is wired the same as what you've done, so assuming the FS-4 is the same it should work.

Kwagmire Wagner · February 07, 2016, 04:21:14 PM

Yeah, just went and grabbed my footswitch and tested the actual sleeve and tip with an ohmmeter and there is no detectable resistance when it is in the closed position. Thanks Slacker.

Kwagmire Wagner · February 07, 2016, 04:29:41 PM

And having opened up the switch there is a sum total of one led and one SPDT in it. Wow... talk about budget.

Jdansti · February 07, 2016, 04:59:49 PM

That foot switch works with your amp because there is other circuitry inside the amp that the foot switch connects to. In other words, instead of putting all of the switching circuit in the foot switch enclosure, the manufacturer only put the components that were absolutely necessary for remote operation-that would be the switch and LED. The rest of the circuit is inside the amp.

It sounds like you are building a custom selector/loop that is totally independent of the amp. Correct? If so, you can't use just the switch and LED as your means of switching. If the new box is to only be used with your Blackstar amp, then you're probably OK.

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LED lighting on open circuit