What's the problem with this.

[QSQCaito]

Well, i think if it's not done like this, there might be a problem with it, which my whole experience(:p) does not let me guess. If you know it, i'd  fully apreciate you tell it to me

Excuse me for knowing too little But maybe u cant send a signal to the output:P
Any way of avoiding that, just let it flow one side, like with a diode?

I fully apreciate your answers and don't worry, im about to get the art of electronics first and 2nd edition so i know a bit, and maybe, who knows, help somebody as u may help me

cya!

Diego Andrés Cao

PS: excuse my english, im from Argentina

[Peter Snowberg]

When you are using a SPDT switch to bypass an effect, you need to wire it so that the output jack is switched between the input jack and the effect output. The input jack is always wired to the effect input so it is desirable to have a high impedance input on that effect. If it does not have a high impedance input, you can always add one using a JFET buffer stage.

Try this:


Don't worry about what you know at this time. The important thing is that you are learning. If you have any questions, the community here will be glad to help answer them.

Best wishes for your projects! 

PS: your English is very good..... and it's much better than my Spanish.    We have several members from Argentina here.

[QSQCaito]

The input jack is always wired to the effect input so it is desirable to have a high impedance input on that effect. If it does not have a high impedance input, you can always add one using a JFET buffer stage.

What does a high impedance input means?what would happen if i don't have a high impedance input?, adding one, will change sound quality?

making it this way, you're always using youre battery supply right?

Thanks a lot for helping me, and for the complimments too I'll soon post more of my great drawings haah, with some more doubt remaining, by now, ive donde one fx including the pcb, but im looking forward to know what im doing so i can expand and do better things.

Thanks a lot!!
Diego Andrés cao

[QSQCaito]

Is this ok???


Once more little question, is there anyway to have bypass without using a foot dpdt? as you know, some amps have the option of connecting an external pedal to command the distorsion or whatever effect is it, because here only foot dpdt is imported, and i was going to make small racks with an effect each, with a normal dpdt to toggle it on and off, but i was told that i could also add that bypass pedal i was talking of..

Once again and again , thanks a lot for your help, and i hope you get my ideas, now its time for me to go to sleep almost 2 am here..

Cya

Diego Andrés Cao


PS im seeing the pic and theres a trace that goes from the (+)   to ground that is  not intended, a tiny one

bye, thanks.

[QSQCaito]

Does any1 knows how to build footswitch, lik the ones that go on the back of the amps to control effects, thanks!!!


cya thnks
andrés cao

[zpyder]

QSQCaito-

a lot of amps (and also some more complex like multifx pedals) have inputs for extra stuff.  Like switching from clean to distorted channels on an amp, or switching patches on a loop sampler... I believe this is what you're referring to.  If so, it is as easy as pie and cheap to make your own switchers for them (don't buy the $40 ones!!!!) All they are (usually) is a simple MOMENTARY switch which connects (momentarily) the two conductors of a 1/4 instrument cable.  All you need to do (I've done this several times) is get a SPST Momentary switch, and hook a two wires coming out of a 1/4" instrument cable to the two different switch lugs.  Put a 1/4 plug on the other end of the wire and plug it into your amp/multifx pedal.  When you press the switch, the two wires are momentarily connected, creating a closed circuit and tellign the amp/pedal to change.  It's probably the easiest build you could do.  I like to put a SPST switch in a box with a mono 1/4 jack on the side, and just use patch cables to connect it to the amp/pedal.

To test out if your amp works this way, just plug a regular 1/4 patch cable into the back of of your amp.  Then take any ol wire and touch the two parts of the plug at the other end of the cable together.  If your amp switches modes, then this will work for you.

Hope this helps,

cheers
zpyder

[QSQCaito]

Hope this helps,

I didn't understood, or understood poorly what you where trying to say, but it indeed helped me to develop my own(it probably already exists.. but i designed from scratch by myself, really) complicated system of bypassing, using a normal dpdt(not footstomp, which i can't get inhere) and a spst footstomp(which i  can acquire)..

Context: im about to create several effects, all mounted in small racks, so they'll be controlled by normal dpdt, but in case i need to control them with my foot, i've got an alternative..

Look at this awful .bmp




Here's the explanation. The foot-bypass only works when the dpdt is in the "down" position. When the stereo jack is unplugged, the #3(as theyre labeled with light blue in the stereo jack), that is in contact wi the other #3, touchs #2. That connects the input signal to the FX's input. So nothing is really happening, the effect is working normally.

But when you connect a stereo plug, the #3 and the#2 stop having contact. Now depending on the spst position, you con have either a bypass, or the effect. If it's in the upper position, once again, #3 does contact with #2, but if it's in its lower position, #2 contacts #1, making it bypassed, directly, from the input to the output...

Well finally i think it's a messy drawing, with a messy explanation.. But a great solution(i hope)..

Tell me if you find any flaws please, i think no buffer or anything be needed, input is not always connected, so i don't think i will have this problem "The input jack is always wired to the effect input so it is desirable to have a high impedance input on that effect. If it does not have a high impedance input, you can always add one using a JFET buffer stage."

Well that's it for now.. hope it helps for me and many others

thank you a lot for your help

Cya

Diego Andrés Cao

PS Excuse my english.

EDIT:  I added labels to the stereo jacks, and did some modifications to the text, was wrong written

[Seljer]

If you're going to have your effects mounted remotely somewhere in a rack use relays or some other such method ( geofex.com has a couple of articles on various system using CMOS switches for bypass) that will allow you to switch it remotely without the audio signal ever leaving the rack

[R.G.]

Welcome to the DIY effects community, QsQCaito.

Just like you're reading the Art of Electronics, there are some things that exist that you can read that will answer a lot of your questions.

Go to http://www.geofex.com and look for "The Technology of Bypasses". Click on the link for "The Technology of..." in the upper left hand corner of the entry page.

To bypass an effect, you need to connect the input jack to the effect input and the output jack to the effect output when the effect is engaged; when the effect is bypassed, you have to connect the input jack to the output jack, as well as disconnecting the effect output from the output jack.

There are several ways to do this. Some effects leave the input of the effect attached to the input jack when the effect is bypassed. This is OK if the effect input does not load down the input signal since it always stays attached, and a bad idea if the effect is a low impedance and loads down the signal.

You can find out about impedance at GEOFEX (http://www.geofex.com) in the "circuit sweepings" section in the article "What is impedance?"

For remotely switched bypasses, again GEOFEX has several articles for remote switching.

[QSQCaito]

Mm.. Thanks for the answers..

It seems that my system failed, but i still can't figure out why. Seems time to go to RG's web to look for some better remote switching.

Thanks a lot for the answers, cya!

Diego Andrés Cao

PS What happens if when bypassing the FX i leave the output of fx connected to the output, if there's no input connected to the fx, are there noises coming out anyway?

Or i cant send a signal to the output, can i avoid that with any kind of diode or something?

[R.G.]

It's tied up in that impedances business.

First, to get used to the idea, impedance is like resistance if we're only talking about DC. A high resistance means that you have to use lots of volts to get a little current to flow. A low resistance means that you can get large currents with only a small voltage.

Impedance is the AC circuit version of that. A low impedance lets lots of current flow for a few volts. A high impedance needs lots of volts to make little currents flow.

A high impedance input lets only tiny currents in, even for large input voltages. It does not "eat" much electrical current. So we can put signals into a high impedance input and not load down the source of the signal.

A high output impedance output acts like there is a big resistor in series with it. It cannot supply much current - the resistance in series with the output drops all of the voltage. A high impedance output is easy to load down.

A low impedance input can eat a lot of current - you have to have lots of current flowing into a low impedance input to get much voltage to show there.

A low impedance output can supply a lot of current. So it has the same output voltage even if loaded.

So what we usually want in the effects and audio world is an output impedance that is at least ten times lower impedance than the load that it drives. That means that the low impedance source can put out lots of current, and the high impedance input can be driven with little current. So almost all of the signal - which is almost always a voltage - shows up at the input.

The reverse of that, a high impedance output and a low impedance input is a recipe for big signal losses because of the loading.

With that as background:
A guitar is about a 8K to 18K source impedance. Effects impedances vary, but 10K to 100K is common. Amps have high input impedances so they can be easily driven by a guitar.

If you leave an effect output connected to the signal wires and the effect input open. the effect output tries to make the signal be zero - dead quiet. It does this through its 10K to 100K output impedance. The signal wire is trying to get signal to the amp, but the effect output is trying to make it be quiet, so the effect output eats a lot of the signal, and the signal to the amp is much smaller. This is why it is a bad idea to leave the effect output connected. You cannot use diodes to do this because a diode conducts one way, and signals are both ways. You only cause distortion.

[QSQCaito]

wow.. that was cool hard to understand but finally got it, thanks RG!, and all of the rest too.

Im actually reading your text: "A Remote Indicating Effects Bypass System", to see how that can help me, by now seems great, but i have some doubts, things that i can't clear.

I will be asking soon, if it does not bothers.


Thank you a lot

Bye bye

Diego Andrés Cao

[QSQCaito]

Here come my doubts , as i believe i can't psot the image because its copyrighted, ill levae the link,http://www.geofex.com/Article_Folders/rmtswtch/rmtsw3.gif

Ok, here i come How can i know which should be the ressistance values if i use a 9v relay?
The "ground" would go to the negative of the battery?
The +On/-Off  is a switch between the negative of the battery and the circuit?

The remote switches, NEED to have those ic's, because i can't get them in here. Can't it be just an on/off switch? can a led be added'?

Well, thank you all from advance. what i would do, would be place the relays in the mini-racks, and have an optional output to control the FX remotely..

Thank you all again, and i hope my question doesnt bother 

Bye bye!!

Diego Andrés Cao

PS Excuse my english

[QSQCaito]

Forget the last post, seems like i got to understand the circuit a bit more.

Correct me if im wrong, + and - of the battery are always connected. When you send a (+)signal from the +on/-off, Q1 acts, letting the (-) current flow, right?

Now the question is, which should be the voltage of the +on/-off, could it be 9v, or is it too much?
And, what whould be the resistance values if i use a 9v power supply and relay?

Aannd last but not least, can the +on/-off signal be remotely controled with a single on off switch, can a led be added to it? (i can't get those chips ca... in here, neithr the small switches, etc)

Thanks a lottt, hope it doesn't bother too much questioning:P and hopes it helps many others too..


Bye bye!!

Diego Andrés Cao