ShortScale.org

portugalwillie wrote:Okay. I've been learning about diodes, capacitors, and resistors. Most of what the Prof is talking about is over my head but no worries, I'm slowly catching on.

My soldering skills are pretty decent. They're not great but I've rewired several guitars (including Jags!) so I have a decent amount of practice, I'm just trying to figure out the best way to lay out all of my parts. Right now I'm getting a little stuck on the points that are not connected to anything. For example on the top circuit I see a capacitor followed by a resistor then the plus sign. Obviously resistors have two ends, so do I just leave one end not connected to anything? If that is that case, should I cut the end off so it doesn't come in contact with anything on accident? And all of the grounds, what should I be grounding them to?

I know these are super noob questions but I am one!

And just out of curiosity, the part of the circuit with the LED is only connected to the other parts of the circuit through the ground, I assume they all need to be grounded to the same place but does that part of the circuit actually effect the tone, or is all of that just for the LED?

See the VR by C11? All points marked VR connect to there. So the ends of R3 and R4 would connect to the positive side of C11.
Same exact thing with all the grounds. They all have to connect together.

For the LED, you need to look online for true bypass switching diagrams. R16 is the Drop Down Resistor which controls how bright your LED is and the LED pad would connect to the positive leg of the LED. The other leg of the LED would connect to the switch. Hence looking online for bypass switching diagrams.

Would this work?

No, you wouldn't want +9v going to the switch. You would want ground there. I mean, I guess you could wire up the LED that way but I just don't like putting 9v on the switch for some reason.

I would do it this way:
True Bypass Switching, LED, Grouded Input

I found this . . . and the person seemed to know what they're talking about. Second opinion?

Bypass Explanation

Yes, I've seen that guys stuff and he is very knowledgeable. Following that will work. By the way, it's the same as in the link I gave you.

Yeah don't put 9V on the switch. Use the scheme Bill linked. It is the one I use and it is the best.

Cool, thanks guys.

today I should be actually drawing a layout for everything and putting it on a bread board.

Annnd

Bill, where would recommend buying 3 pdt switches. I wouldn't mind buying 5 or 6 but not if they're selling for $10 a piece. Is there a place that sells them cheaper?

pedalpartsplus.com

Awesome!

Also, I can't find the integrated circuit JRC 45580 anywhere? I assume the IC1A just means integrated circuit but if I can't find that exact one, is there another I can use?

Seriously, thanks for the help. I know I'm asking questions about everything but I'm just trying to learn everything I can so when I finally start building, it may actually work.

Any dual Op-amp will work really.

This may become your new friend: Octopart

JRC4558

Edit: Actually the jrc4558 octopart search really didn't come up with anything usable.

This is better: RC4558

Sweet. Bookparked octopart and pedal parts. Thanks for the info

You can also sub in lots of low cost IC dual op-amps in place of that if you want to experiment:

TL072
TL082

etc.

Okay, I'm into the soldering part now, but I'm still a little mystified by the power socket I bought. There are two tags, one longer than the other, and then a metal bit in the centre. Now on the other power socket that was supplied with the kit, there were three tags, one of which appeared to be connected directly to the centre bit of metal. On this one, neither tag appears to be directly connected on the back of the socket, but I can see on the inside that the shorter tag is connected to the middle bit (I think).

Now with the black/negative wires, do I solder them directly to the centre bit of the plug, or the shorter tag? And for the red wires, are they all soldered to the longer tag?

I ask because the red wires were soldered to two different tags in the original schematic, so it's thrown me a little bit. Is it actually possible to simply have all of the red wires soldered to a single tag, and have they split them between two to make it simpler for noobs like me? Or do they genuinely need to not touch?

Apart from the question of where and how to wire up the power socket, I've soldered everything else up, plugged it in, and...

NOTHING.

Because one of the switches was on, and there was nothing in that loop. Switched it off, and BANG. Works. Tried the loops with a simple patch across each one, and still received glorious rez-o-matic goodness from my Batwing through everything.

Dead chuffed, I am. Proper fucking chuffed. To mintballs. I feel like I've done something worthwhile (yes, I know, it's just a bypass loop).

There's plenty wrong with it; my soldering is wank, the footswitches were drilled so close together it looks like a pisstake (there's an inch between them), power socket isn't wired up, etc, but I'm proud of myself for having followed a piss-simple plan and actually managed to create something.

One question: Why does my soldering iron look like it's on its last legs? The tip has gone all knobbly, and now that it's cooled down, I've noticed it's all wobbly too. It looks like the nib has etched grooves into itself from being stood in my ghetto-ass homemade stand I made this morning (triangle of wood with a coiled metal coathanger) for some bizarre reason. This is the only time I've used it. Looks like I'm going to have to do a bit more practice. And maybe not use that stand. And probably buy a new soldering iron.

And another thing that just struck me: If I'm leaving the battery clip out then is the red wire from the input jack going to the power supply still necessary?

No, and really that wire should be black. You also didn't have to use a stereo jack for the input either.

Wiring scheme looks weird.

If I'm using a battery clip, I sold the black wire of the battery clip to the large tag (centre negative, ground) of the DC jack, and the red one to the switching power tag (bottom one if the ground tag is on the top), and then the power to the circuit goes from the right lug. This way when a DC jack is plugged in the battery is disconnected from the circuit. Bugger knows why they've connected the 9V output of the battery to the ground loop on that diagram.

What you need to do:

1. Show me the DC jack again so I can ID the ground and 9V lugs for you.
2. Solder a wire from the ground lug to one of the jack grounds, and then connect all ground connections to this ground lug on the jack (easier to solder many things here), either directly or via other jack grounds (hopefully that makes sense)
3. Wire from the 9V lug goes to the two LED anodes. Their cathodes go to ground via current limiting resistors (and the 3PDTs switching them to ground)

Your soldering iron looks like shit because:

1. It's cheap
2. You're probably not using a sponge to clean off gunk before tinning and post soldering
3. It's too hot?
4. It's cheap.

If you're going to solder a few things get one of these:
http://www.circuitspecialists.eu/csi-de ... -p-11.html

Yeah Mike's right. Looking at the diagram shit isn't the way I do things either.

The only thing that is odd/different about that schem is that they are using the input to switch the positive of the battery on/off when the cable in inserted into it. Normally you would see the negative wire of the battery connected to the ring of the input jack and the positive wire of the battery on the DC jack.