Adding two LEDs to a TS/mono cable stomp box controlling a relay?

[Oates83]

Hello!

So I've got a little project where I have a 6V relay that I want to control (A1-A2) through a footswitch.
The footswitch is located in a separate stompbox connected through a regular TS/Mono jack guitar cable.
Pretty straight forward so far, the problem comes in when I want to add two LEDs.
The setup right now is:
SLEEVE connected to the input of the footswitch
Output 1 = LED1 with a resistor (on the anode) with enough resistance to drop the voltage below what's required to operate the relay coil (and it therefore stays connected between COM and NC)         
Output 2 = LED2 with no resistor and enough voltage to operate the relay coil (COM and NO connected)

Output 1 and 2 comes together at the TIP
See picture to make it more clear

As some of you probably figured out immediately, LED2 is shining bright as the sun and it probably won't last for long,
and when it breaks down it will sever the output2 connection.

So my question is:
Is there any way to connect the LED2 (with a resistor to lower the LED voltage) parallel to a simple wire between
Output2 and TIP and still use a TS jack instead of a TRS?
Basically I want the LED2 to have a reasonable voltage without loosing so much voltage that the relay can't be "switched on".
Whether it's done  through some kind of parallel wiring or some other method is secondary.

I already thought of a TRS cable or maybe a battery inside the stomp box, or using just one LED. I just want to see if there are any actual wiring/component alternatives to these.

[antonis]

Hi & Welcome..

IMHO, your query could be utilized with more items, like a voltage comparator..
(perhaps driving a series transistor for relay current demand..)

[FiveseveN]

Welcome!
Yes, the proper way to do it is to separate the control signal (which can also drive your LEDs) from the current driving the relay. It can be as simple as one transistor and a couple more resistors, or you can go for a dedicated relay driver IC, a flip-flop, microcontroller, wireless, sky's the limit. And don't forget your flyback diode!

[Oates83]

Welcome!
Yes, the proper way to do it is to separate the control signal (which can also drive your LEDs) from the current driving the relay. It can be as simple as one transistor and a couple more resistors, or you can go for a dedicated relay driver IC, a flip-flop, microcontroller, wireless, sky's the limit. And don't forget your flyback diode!

Alright, so the solution with the transistor sounded like the most easy. I would like to point out that I'm a total beginner to all of this,
I pretty much understand the most basic function of a transistor and that's about it, but I did my best to google it so here's what I came up with
(I switched the output's places because it was easier to draw, not that it matters):



Is it right and if so It's supposed to be an NPN transistor, right? Do I need to add any resistors or something?

Thanks for taking your time with me!

[antonis]

https://www.electronics-tutorials.ws/blog/relay-switch-circuit.html

[antonis]

On a second thought : 

[niektb]

Welcome!
Yes, the proper way to do it is to separate the control signal (which can also drive your LEDs) from the current driving the relay. It can be as simple as one transistor and a couple more resistors, or you can go for a dedicated relay driver IC, a flip-flop, microcontroller, wireless, sky's the limit. And don't forget your flyback diode!

Alright, so the solution with the transistor sounded like the most easy. I would like to point out that I'm a total beginner to all of this,
I pretty much understand the most basic function of a transistor and that's about it, but I did my best to google it so here's what I came up with
(I switched the output's places because it was easier to draw, not that it matters):



Is it right and if so It's supposed to be an NPN transistor, right? Do I need to add any resistors or something?

Thanks for taking your time with me!

I believe that in this circuit, LED1 will never turn on. The current flowing into the base of the transistor (which should have been an PNP I think) is quite low.

@Antonis: Isn't LED2 always on in your diagram?

[antonis]

@Antonis: Isn't LED2 always on in your diagram?

I believe, No..

As OP stated, LED1 + current limiting resistor leave no voltage headroom for relay and/or LED2  to switch ON..

P.S.
Everything here depend on relay coil resistance..

[Oates83]

@niektb: I based this on a youtube video I found, except I removed R1 and LED1 since I figured the relay
is gonna "drive the current" instead. But the relay is on the emitter side, don't know if that makes a difference.
I'm just pulling shit out my ass right now but you can always experiment, right
https://www.youtube.com/watch?v=8ZEQEV-Stkc

I also found this circuit in the link from Antonis, which is pretty much the same principle, right?



@Antonis: Iether I'm a dumbass or you have missed an important point.
The relay is IN ANOTHER BOX. the only connection between the other box and the stomp are the TIP and the SLEEVE of the guitar cable.
The Relay is on the other side of the TIP-point.

BTW the relay coil resistance is 55ohm
https://uk.farnell.com/finder/40-61-9-006-0000/relay-spdt-6vdc-16a/dp/1169164
Its connected with a finder 95.05 relay socket 

[antonis]

@Antonis: Iether I'm a dumbass or you have missed an important point.
The relay is IN ANOTHER BOX. the only connection between the other box and the stomp are the TIP and the SLEEVE of the guitar cable.
The Relay is on the other side of the TIP-point.

As far as I know, "point" is a dimensionless item..
(whithout sides..)

P.S.
Driving a relay by BJT Emitter can't be obtained with same Vcc level for both Collector & Base..
(Bias source needs to be about 700mV + I_Base*R_Bias higher than Vcc.. - assuming  V_CESat = 0V)

You better utilize Collector relay driver..

[PRR]

Why do you need two LEDs??

If the relay isn't ON, it must be OFF.

...The relay is IN ANOTHER BOX. ....the TIP and the SLEEVE of the guitar cable.................

So draw it that way.

If it is on a tip-SLEEVE cable, electrical sanity suggests the sleeve be COMMON. Else other metallic things brushing the jack may actuate the relay unexpectedly. (And much over 6V, perhaps shockingly.)

With simple switching, 6V supply, 6V relay, you can not have any spare voltage in the box to work an LED.

The "get it done!!" path seems to be a small leak to work an LED when relay is OFF, a dead short in the box to put all power to the relay. If the relay is DT, the ON/OFF is reversable: you can make the effect (and LED) be ON when the relay is off.

[Oates83]

Why do you need two LEDs??

If the relay isn't ON, it must be OFF.

It's not to indicate the relay status, it's to indicate which channel is selected. The relay is used to switch amp channels.
And it's not absolutely necessary with two leds, I'm just exploring the possibilities.

So draw it that way.

I DID draw it that way. Notice the dashed lines in the first drawing and a text saying "separate box"?
I also very clarely wrote so in my first post:
"The footswitch is located in a separate stompbox connected through a regular TS/Mono jack guitar cable."
I didn't care to draw it again since I figured it was already established.

If it is on a tip-SLEEVE cable, electrical sanity suggests the sleeve be COMMON. Else other metallic things brushing the jack may actuate the relay unexpectedly. (And much over 6V, perhaps shockingly.)

Yes, I know Sleeve should be common. I just wrote them wrong and went with it since I figured it had nothing to do with my question and therefore was not an important issue, sorry.

With simple switching, 6V supply, 6V relay, you can not have any spare voltage in the box to work an LED.

It can operate between 4,4 and 9 Volts, if that makes any difference.

[antonis]

Your 6V relay needs about 110mA (55R coil resistance according to your statement) which is about 5 times higher than 5mm LED max current..
(10 times bigger than nominal current)..
Even if LED voltage drop is only 1.5V (leaving 4.5V for relay marginal operation) it can't be set in series with relay..
In case of current limiting resistor (to set LED current down to 15mA say), relay simply won't operate..

[Oates83]

Alright, so I got a breadboard and tested away.
I decided to change the voltage from 6V to 9V I didn't see any other good way to do it.

When I did this I actually managed to make it work by putting two resistors in parallel, one in series with the led and the other one straight to the relay,
but the voltages were not very good, around 2,1V on the LED and 3,66V on the Relay. The specified minimum for the relay is 4,4v so it probably wouldn't be very stable
So I decided to go with the transistor solution anyway.

This is what I ended up with:



I'm not quite sure why there wasn't any need for resistors in combination to LED1, but the voltages was as follows:
LED1 (when active): 1,97V
LED2 (when active): 1,97V
Relay (RL2a): 6,05V (LED1 active) = ON
                   0,97V (LED2 active) = OFF
Transistor:
K-B=1,97V (Same as LED1, naturally)
K-E=2,63V
B-E=0,65V

Both the voltages of the LEDs and Relay are ideal!
I let the LED1 path stay active for 2 hours to make sure it wasn't just a fluke and also tried it with the Amplifier (Bugera 333)
It all stayed stable.

Thanks for the tips!

[antonis]

I'm not quite sure why there wasn't any need for resistors in combination to LED1,

RL2a coil resistance serves as LED1 current limiting resistor..

P.S.
Can't figure out the use of transistor wired as diode of LED1+V_BE voltage drop..
(in the mean of it drives nothing..)

[Oates83]

Can't figure out the use of transistor wired as diode of LED1+V_BE voltage drop..
(in the mean of it drives nothing..)

I'm not sure this is what you meant but,
LED1 and 2 are simply indicators to see what Distortion-channel is selected on the Amplifier.

The Bugera 333 channels are activated by letting the current go through resistors of different sizes and some gimmick in the amp
detects the voltage drop, current change or whatever and changes to the right channel.
Clean is 100R
Crunch is 1K
Lead is 3,3K

LED1 (Red) means the Lead channel is selected (Relay coil RL2a is "active" so the NO-channel is connected and the Amp footswitch current goes through the 3,3K resistor)
LED2 (Yellow) means the Crunch channel is selected (Relay coil RL2a doesn't get enough juice so it stays "inactive" so the NC-channel is connected and the Amp footswitch current goes through the 1K resistor)


The whole point with the Transistor in the first place is to "split" the current so both the LED1 and the Relay coil could get an ideal voltage/current to work correctly.