Her is my relay schematic

[Tdale]

Or actually, the whole foot switch schem.

The way it's drawn now, each led is supposed to light up when the relay-coil circuit is broken, because that's when OD/Boost is activated.

When the coil circuits are closed, the led's turn off and the amp is in Clean/NoBoost mode.

The red lines are those who leads back to the "-" of the bridge. The circuit is not grounded to the chassis, to prevent noise.

Does it seem to be a working circuit?

Tommy

[Tdale]

I'd really appreciate any comments on this schematic... especially if it won't work....

Tommy

[Luthierwnc]

The LED switches should be on the same throws as the relays -- not on the far side. You want them to light up when you ground the coils. You need diodes in reverse parallel with the coils and a .01 cap wouldn't hurt there either. I'd put two 1000uf caps on either side of the regulator rather than the 2200 on the input. sh

[Tdale]

I forgot to draw the diodes, they are already in place.

As for the diodes, I want them to light up when the coils aren't grounded. The reason is that when the coils are not grounded, the OD and Boost circuits are active, and that's when I want the led's to light up.

Tommy

[tictac]

Here's an easier way to do what you want. Your way you need the expensive DPDT footswitches, this way you can use the less expensive SPST footswitches.

Also notice there are only 3 wires comming out of the amp to the footswitch. Now you can use a 3 conductor XLR connector or even a stereo jack (with an isolated ground)

Notice the cap after the regulator too.

[Tdale]

Yes, that's a way to do it, although I already got the DTDP foot switches..

But I can't see how this circuit will turn on the leds when the relay is off? I mean, the relay is off when the foot switch is off, and when the foot switch is off, how can the leds light up?

Tommy

[Luthierwnc]

The relay coil is on when the footswitch is on. It works just like a light switch. Diagram one shows a lightbulb on a wall switch. Electrical potential on the hot wire is waiting for a flow of electrons from ground. The resistance of the filament causes heat which lights up.

Diagram two shows the same thing but with the switch on the ground side rather than the hot. Not recommended from an electrical code standpoint but it would work the same.

Diagram three shows the relay coil instead of a light bulb. Electrons flow from ground through the coil, the resistance from which creates an electro-magnet that pulls the leaf on the switch to the normally-open position. Some of the little solid-state relays use a different method but the result is the same. When the current stops, the leaf springs back to the normally-closed position.

The last one shows a flyback diode on the coil and an LED like TicTac and I use so you don't need a separate hot. There is reason not to use it though and it may be quieter.

Think of a doorbell buzzer. When you push the button (and for as long as you hold the button in), the circuit is completed and the bell rings. When you let your finger off, the circuit collapses and the noise stops. The reason the buzzer or bell rings is because with AC current, the field is created and collapses 60 times a second (50 times where you live). The clapper jerks each time and hits the bell. We smooth out the line by converting it to direct current so it holds the connection.

Trust us, it works!

Skip

[Tdale]

I understand how this works, but I think maybe you misunderstood me.

I want the led to light up when the foot switch is off. In other words:

When there is no current flowing through the coil, I want the led to light up.

When the foot switch is pressed, and current flows through the relay coil to ground, I want the led to be turned off.

Are you saying that your modified circuit will do that? I'm having a hard time understanding how that can be, since the coil and led is in seried, through the same switch.

The way I read your schematic, is that when the foot switch is on, current flows through the coil and the led to ground, and that way both the relay and led are active at the same time, and that's the oposite of what I want...

Tommy

[Luthierwnc]

Nope. If you want them to light when the boost and od are off, the switch in your schematic will work fine. sh

[Tdale]

Just to clarify.. The OD/Boost circuits are activated when the relay is in it's "normally closed" position (no current through the coil".

That's why I want the leds to light when the relays are off.

Tommy

[tictac]

Yes that right you want the OD to be one when the relay is turned off so the OD is going to be on a N.C. contact the clean on a N.O. contact.

Look at the schem. When the front panel or footswitch is open the relay is off which means DC voltage is going thru the coil and lighting the LED. The LED and the resistor are limiting the current flow and preventing the coil from energizing. This is OD or Boost mode.

When you close a switch the both the and the LED coil is are grounded. The LED shuts off, current flows, coil is energized now your in clean or boost off mode.

The beauty of it all is the simplicity.

TT

[tictac]

One more thing. If you've already got DPST footswitchs you can use one to turn on both relays at the same time. I have my footswitch wired with 2 switches. One for the regular OD/Clean switching the other one activates the boost and OD at the same time.

Great for HRM amps since you want the boost to be active when your in OD mode anyway. This way you do need to step on 2 seperate switches to get the HRM going.

I never use the boost in clean mode anyway so this is the setup I like. You could have 3 footswtches if you wanted to. One for clean/OD, one for Boost and one clean/OD+Boost.
TT

[Tdale]

Hmmm still don't get it.

First of all. Does "closed switch" mean that it's leading or not leading? I get a little confused, probably because of the language.

To me, a closed switch is conducting and an open switch does not conduct, sinc it's open.. But it looks like you mean the oposite?

Anyway. When one of the foot switches is pressed, so that the switch is not conducting, there is no contact between the +5V and ground. How can anything light up then?

I get that the resistor and led in series, will lower the voltage across the coil, so that it won't switch, and the leds will light, when the switch is conducting...

But when you break the circuit, using the foot switch, there isn't current flowing through, either the coil nor the resistor/led...

Is it me that's just stupid enough to not see how this should work...?

Tommy

[tictac]

Of course you don't see how it works, because the way it is drawn it won't work! Sorry bout that, here's the correct drawing

TT

[Tdale]

Phew! I'll call and cancel my appointment with the shrink!

Now it makes sense. The only concern I might have, is how the relay likes the constant voltage on the coil, both high and low? could it be damaged by this in some way?

Tommy