Wah heel bypass Idea

[Paul.Ruby]

If you want to make the 555 based board work, do this. Leave out the red-X components. Wire the reed switch as the green lines between pins 3 and 8 of the 555 IC socket. BUT, polarity is backwards, so, do not connect anything to "IN" on the module; Connect the input jack to the input of the wah AND to the OUT on the relay module; Connect the output of the wah to the "JACK IN" on the module. JACK OUT on the module is still the output. Of course... Programming is more fun. I would use the intelligent module and program it.

Also leave out Q1 and R4, assuming there is no LED. If there is an LED, keep in mind polarity is backwards now.

 

[Paul.Ruby]

 

[RobotZombie23]

there is no need to reverse the in out wires. I think your way would be for a normally open switch I'm a using normally closed switch.

I would like to throw that ne555 back in there and make it delay the bypass... but that's for another time...

It's working wired like this: except I can't get the led to work but thats ok


Thank you everyone

 

[Stickman393]

If he's using the non-latching simple relay bypass, it uses an NE555, not a microcontroller.

Using a will be easier. Basically, you want the reed switch to energize the relay coil, which is component K1.1 on the below schematic.

The basic idea here is to route vref (should be about 4.5Vdc) through the reed switch (FS in the diagram) to the coil of the relay.

The simplest way to achieve this is to remove the NE555 and replace R3 with a jumper.

Since your switch is Normally Closed this is the order of operations:

Normal state (closed)= wah pedal is in use (heel magnet is not "switching" the reed switch)
Vref is able to pass FS
K1.1 is Energized
Relay common is connected to Normally Open Contacts (this is your effect send and return on the relay board)

Energized Reed switch (open) = Wah pedal is at rest, heel down position. (Magnet is "switching" the reed switch)
Vref cannot pass FS
K1.1 is de-energized
Relay common is connected to normally closed contacts (this is your bypass route)

Now...this has some drawbacks. If you quickly cycle between engaging and disengaging the reed switch your contacts can flutter. There's a chance of burning up the coil if done in rapid succession...at least, that's been my experience with much larger devices. Not sure how much it matters in this sort of use case.

Someone who's really clever could probably steer ya towards a way to utilize the NE555 to give you a short time delay (nanoseconds-milliseconds) between the reed switch triggering and the relay contacts switching. I cant say I'm familiar enough with the NE555 to know how to accomplish something like that offhand.

 

[Stickman393]

I don't recommend removing C1, honestly. That guy will give you a little buffer between when the switch is opened and when the relay contacts open, yeah?

Otherwise the aforementioned fluttering issue could be a lot worse. The value could be potentially be reduced if ya find that it takes too long for the relay to revert to normal position.

 

[Stickman393]

YO! Another thing:

That relay has a 4.5Vdc coil. VCC is 9volts. DO NOT USE VCC. You want to use Vref. That will be 4.5VDC.

Trust me. You'll burn up your relay coil like what like woah.

 

[RobotZombie23]

You want to use Vref.

I tried hooking up to lug 6 and 2 of where the ne555 used to be for vref. Didn't work. I'm just leaving it the way it is. If it ain't broke don't fix it. If it does break I'll just have to figure something else out.

 

[Stickman393]

That last sentence is true.

Just sayin'. Ya probably got a good 8.4Vdc acting on that coil after the schottkey. A good 186% of the rated voltage. It's not a matter of if, it's a matter of when.

 

[RobotZombie23]

That last sentence is true.

Just sayin'. Ya probably got a good 8.4Vdc acting on that coil after the schottkey. A good 186% of the rated voltage. It's not a matter of if, it's a matter of when.

Can I just throw a resistor in there to bring the voltage down?

 

[Feral Feline]

I'd throw in two resistors to form a voltage divider, one to ground the other supplying whatever's within the spec the relay requires.

I'm not as learned about relays as Stickman and others, though.

 

[Stickman393]

Can I just throw a resistor in there to bring the voltage down?

Possibly! Though I'd recommend trying to make use of the 4.5vdc source on the board already.

This is how I'd do this:

Instead of jumping across pin 8 to 3 with your reed switch, jump across 6 to 3 with the reed switch. Put R3 and C1 back where they were. Those will help ya keep those contacts from bouncing around.

Also: to fix your LED: remove LED from the relay board. Solder the cathode to a current limiting resistor of some kind (4.7k is a pretty standard value). Bring the other end of that resistor to the SW pad.

Then, bring your anode to VCC. Pin 8 will do.

And viola: your Q1 is now behaving like a switch. When current flows from the base to the emitter, it will allow current to flow from the collector to the emitter.

Right now, that is also happening, but you don't have any potential at the collector. Your LED is connected to ground at the cathode and nothing or also ground depending on the state of Q1.

 

[Paul.Ruby]

The original circuit is is driving the relay from about 7.5V out of pin 3 of the 555, not Vref, but the specified relay is rated at 4.5V. You can add a 130 ohm resistor in series with the relay to drive it from 9V. lift one side of D2 and add 130ohm in series with it. There is no Vref supply voltage on the board. It's only a reference voltage from 10k resistors and won't drive the relay. Alternatively, replace the 78L09 with 78L05 to reduce Vcc instead of adding a resistor.

 

[Paul.Ruby]

And this is the LED wiring needed. (This is my original picture showing the reversed polarity, so ignore that part.)

 

[Stickman393]

A'ite, that makes sense. Thanks for that explanation Paulie.

Paul sounds like he's got a better grip than I do regarding the functionality of the circuit. Imma defer to him. My reasoning was based on a less holistic look at the circuit: I see exactly what I was missing now.

*Edit*. One little question though: R4 and Q1 are crossed out. This isn't gonna work: with Q1 gone, pad SW is left floating. There's nothing to connect the LED cathode to ground: The LED won't light.

Shouldn't he leave R4 and Q1 in place?

 

[Paul.Ruby]

A'ite, that makes sense. Thanks for that explanation Paulie.

Paul sounds like he's got a better grip than I do regarding the functionality of the circuit. Imma defer to him. My reasoning was based on a less holistic look at the circuit: I see exactly what I was missing now.

*Edit*. One little question though: R4 and Q1 are crossed out. This isn't gonna work: with Q1 gone, pad SW is left floating. The LED won't light.

Shouldn't he leave R4 and Q1 in place?

That's right. But I lost my original drawing in a PC crash and didn't feel like redoing it from scratch. But, yes, add back in R4 and Q1 for the LED.

 

[Stickman393]

That's right. But I lost my original drawing in a PC crash and didn't feel like redoing it from scratch. But, yes, add back in R4 and Q1 for the LED.

Coo coo. Whelp, ya see that @RobotZombie23? We got a long time member (me) leaning from the folks on this board. It's a never-ending journey. It's human. Good to roll with it.

Except I am not human, and this is not supposed to happen to me. The world will feel my wrath when my retribution comes for that time I was wrong about a relay.

 

[Paul.Ruby]

Oh hey... Visio did an auto save... Here's the corrected full circuit... Fixed it for normally closed reed switch, the LED driver and added a cap to ensure no relay bounce when switching. (EDIT: 100uf might be too big. Depends on the switching voltages of the relay. If you can feel the delay, then reduce to 10uf.)

 

[Paul.Ruby]

Or... If you want to add a resistor instead of swapping the 78L09... (EDIT: 100uf might be too big. Depends on the switching voltages of the relay. If you can feel the delay, then reduce to 10uf.)

 

[RobotZombie23]

After making a Cream Pie and going to Valhalla. (Both builds went very smooth btw) I tried to resurrect my Sad Zombie wah. After burning up the solder points to the ne555, ruining my reed switch,and I'm pretty sure I destroyed the relay. It clicks on then immediately off when I give it 4.5v to pin 1. I think I'm going to build something from scratch on a breadboard and eventually a prototype pcb. I'll have to get a new relay. probably use a 78L05 and I want to use a micro switch instead of the reed switch.

When it was working as wired above thanks to @Paul.Ruby it was amazing except it had a pop that almost sounded like a digital beep when the relay went on/off. How would I prevent this pop? I'm going to try get a schematic together. I'm trying to figure out what I need to do this from scratch. any tips?

Just saw this in similar threads

 

[Stickman393]

Feel ya. No worries, I've burnt up more boards than I care to admit.

A micro switch is one idea: Dunlop uses tactile switches on their auto-on bass crybaby 105Q.

Another idea, though, with no moving parts and a similar magnet-based principle: A hall switch. Little solid state device that looks like a transistor and controls current based on proximity to a source of magnetism. The T-Rex wah pedals use this kind of setup.

The blip you hear is likely one of two things: relay bounce, or DC offset between the relay terminals.


DC offset:

The tearjerker has a 2.2M pull down resistor tied to its input and ground, as well as a DC-blocking cap before Q1. The chances of that being the source of DC-offset are low.

The other option is whatever you're feeding into the pedal has a slight DC offset, compared to the input of the pedal. Typically, this will come from a leaky DC-blocking cap on the pedal that you're using immediately before the one making noise. I imagine something like that could occur with active pickups as well, if that's what you're using.

If you're using passive pickups and the wah is the first pedal in series, either the tearjerker's C2 is *extremely* leaky, or (more likely) you've got a bit of relay bounce.

IF I'VE GOT THIS RIGHT (which is entirely questionable), it would help to run a capacitor in parallel to the relay coil. That's C1 in the relay documentation.

Granted: this is based off my understanding of how AC relays work (which is that they are inductive loads that produce a magnetic field that attracts a small lever that carries the signal between two different sets of contact points. Which set is currently connected depends on if the coil is magnetized or not)

That inductive load has a proportionally larger inrush current to its normal operating state, and that inrush of current can lead to a momentary voltage sag across the terminals.

A capacitor in parallel with this will also see a proportionally larger inrush current, but it will ramp up the voltage as it charges and keep the voltage constant across the relay coil.

The "bounce" is the result of that voltage sag when the relay coil is first energized. It magnetized, voltage sag for a picosecond and it loses some of its magnetism, and then it magnetizes again.