How can I keep a boss PSU from frying an led

[breather-resistor]

Here is the deal.

I want to make a footswtich for a fender amp that lights up to notify me when verb and vib are on/off.  However, for several reasons, I want to be able to power the pedal with a boss style power input (from a voodoo labs power supply) which is 9v 200ma to my knowledge.  Is there a simple way I can distribute the voltage and current so that I can use it to power the leds??? If so, it would help if you would show me where would be the best place to introduce a switch into such a circuit.

thank you so much

love,

breather-resistor

[GibsonGM]

Well, if I understand you correctly breather, you want an on/off indicator for the thing? Easy.  Put it in the V+ side of the line.  Make sure you use the resistor.  1K is safe to start with, and can be lowered to make the LED brighter, but usually not to much more than 330 ohms...

Try a DPDT switch (or footswitch since it seems you will have this on the floor)...one set of poles controls the circuit, and the other connects the LED to ground to complete the circuit and light it up.  Like this:





Hope that does the trick...

[breather-resistor]

Ok that makes sense,but:

the power is ONLY for the leds, the rest of the circuit is passive because it is nothing more than switches.

[GibsonGM]

ok, that's more clear...

Just run the 9V to center pin of one side of the DPDT switch (no jumper!).  Use the other side for switching whatever.  The DPDT is 2 separate switches, so you can use them independently.   If you need a DPDT to switch channels or whatever else, you'll have to go to a 3PDT to include an LED, just like a stomp box.
Make sense?   

[breather-resistor]

Completely,

But wouldn't 9v and 200ma cook a LED???

[fixr1984]

The 1k is to limit the current so you dont fry it.

[GibsonGM]

The 200mA in a PSU is just what's Available...each circuit draws its own current based on Ohm's Law (current in amperes = voltage in volts / resistance in ohms).  So, 9v /  1,000 = .009 amps, or 9 milliamps...most LED's can operate in the, say, 10-20mA region.  This can get complicated with complex effects, but for what you're doing, you are ok right now.  Use the 1k ;o)  Check it out, put the 1K resistor & LED in series across the power supply wires; it will be fine.  As long as you have the cathode pointed toward ground that is, lol.


I'm adding LEDs to a pedal right now, got the R up to 3.3K for 9v/3,300=.0027 mA.  Using larger resistors = less current used (saves batteries), but a dimmer LED.  In your case, there's no battery, but keeping the LED from frying is the idea.   Read up on Ohms Law, Google it, then repeat it over & over til it makes sense...#1 concept of electronics.

If you were running high-draw effects after the LED and it was in series like it will be here, THAT could pose a problem and would require more study.  Series vs. parallel and all that.   Give it a shot, let us know how you make out )

[breather-resistor]

oops!

one more thing.  since it is a two function footswitch with two leds from the one power supply can I connect them both and use a 500ohm resistor?

[sfx1999]

Only if they are both running at the same time. Otherwise, they could share a 1K resistor.

Anyway, you can find the needed resistor like this:

R = (Vsupply - Vdrop)/I

R is resistor
Vsupply is the supply voltage (9V in this case)
Vdrop is the LED's voltage drop
I is the current you want the LED to use (it doesn't need to be the LED's max; it should be lower)

[breather-resistor]

Thanks, it was the formula that I really need.  The one thing thats left is how can I make it work if they may or may not be on at the same time depending on how it is switched.  Will one LED be naturally brighter than both of them at once?

[sfx1999]

Give both their own resistor. Calculate the value for both.

One LED may be brighter if the colors are different.