1 loop + tuner out w/leds

[loki]

Hi guys, what i need to build is something similar to this:
http://www.loooper.com./1LtmWhite.JPG
that is, a box with just one bypassable loop and a tuner out, with 2 LEDs and DC plug... does anybody know how to wire exactly the pieces and if i need resistors?
all i know (from that website info) is that it uses 3PDT switches and no buffers/transformers...
maybe there's already a schematic around for something similar to this box....

thanks
bye!

[bioroids]

If you dont mind to use electronic switching (FET) I have a schem and pcb for a loop box I made. You can select between 2 guitars, send them to the loop (or not) or send them to the tuner (muting the amp output).

It works pretty well, and you may customize it if you need.

Drop me a line and I'll send it to your email.

Luck

Miguel

[loki]

here's what i actually need:

bass/guitar  --->filters--->distorsions---> box input----> box output---> amp

loop send ---> chorus----> phaser----> delay----> loop return

---> tuner out

so that the loop signal blends with the straight signal when it's switched on
and leaves the straight signal only when it's switched off.
and everything's mute when tuner switch is pressed.

Anyway bioroids here's my address: funkyass@fastwebnet.it maybe i can take a look to your loop box and see if it can suit me (even if i don't know much about FET switching)
thanks

[TheBigMan]

These diagrams are what you need.  The loop diagram is owned by Andi Allan of monkeyfx.co.uk and the tuner mute diagram is owned by Andreas Moller of stinkfoot.se.





A 1K resistor will be fine for the LEDs, but if you want to be more precise Jack Orman has a great online calculator for LEDs at muzique.com.

[smoguzbenjamin]

1k... that'll drain a lot of current. Think of it this way (ohm's law!):

Vtotal = Vbattery - Vled = 9 - 1.2 = 7.8
Vled is the forward voltage (and thus the voltage drop) of the LED. So we're gonna use a 1k resistor:

I = U/R = 7.8/1000 = 0.0078A or 7.8mA
That'll be pretty bright. Try using 8k:

7.8/8000 = 0.000975A or .975mA
That'll suffice for a high-brightness LED. If it's dull you can lower the resistor, but I use 8k to preserve my batteries.

[loki]

thank you very much for the schematics, since i'm powering it with a DC jack there should be bo problem even if it's a 1k resistor....
the way i understand it, i should ground the tuner out jack to the tuner out 3PDT switch and the other four jacks + DC jack to the loop 3PDT switch, right?

[TheBigMan]

I just said 1K as a kind of generic size.  I don't use batteries much anyway, but if I did I would use larger resistors.  With the LEDs I use 1K gives a good brightness.

If you have a 10K pot lying around you can tweak it to find what brightness level suits you best, but 1K will be fine with most LEDs.

[smoguzbenjamin]

Absolutely Just one word of advice, use a +/- 700 ohm resistor in series with the pot when you're experimenting for brightness, it'll save you from frying the LED if you make a mistake

since i'm powering it with a DC jack there should be bo problem even if it's a 1k resistor....

This is true, but I like keeping current consumption down. That way you can use more pedals on one adapter. A couple of my builds have 1mA current consumption for the circuit and 7mA for the darn LED. not that 6mA makes a lot of difference with a 250mA adaptor but it's stil 6 mA you could have used for another circuit or two