#1
Okay im willing to add a killswitch to my frankenstrat, mostly to use it as a way of cutting the guitar signal while waiting in between songs.

Okay you would say this is pretty avarage until this: I want to add a led that will be turned on when the killswitch is deactivated, and turned off when the switch is activated. I know ill need a battery, but Ive got one.

My question is, can someone make me a diagram for this please?

another thing:


this is a diagram showing a killswitch circuit in a strat.

My question is, how do I know which of the three connection goes to the volume ground and which one goes to the volume connection that goes to the output jack?

here is my switch:



okay so this is all ive got to do this circuit:
9v battery, switch, leds, copper cable.


I just want to have 2 leds, both red, am i going to need any resistors?


Please answer my questions, this is hopefully gonna help me understand a lot more about circuits..


Thanks a lot

Martin
#2
you'll want a DPDT switch. You've got a SPDT, there.

see how theres one 'row' of 3 lugs?
you need 2 rows.. one to direct the signal where you want it to go, and the other for the LED.
The switch you have looks like this (looking at the bottom):
[a]

[c]

You need:
[a] [d]
[e]
[c] [f]
(I'll refer to each lug by the letter I labeled it in the little diagram in the rest of my reply. saying is easier than 'the middle lug ont he left side of your switch')


once you've got that, wire the killswitch like you usually would (like it shows in the diagram. Middle lug of volume pot to , [c] to ground).

Now, solder the positive lead of the battery clip to [e]. solder one end of a resistor to [d], then solder the other end to the + leg of your LED. - of the LED to + of next LED, ect.
The - leg of the last LED goes to the negative lead on the battery clip.

and ta-da, the light should come on when the signal isn't killed.
I'm not sure what value of resistor you should use with 2 LEDs... ..depends on the rating of your LEDs, but I haven't really
#3
you'll want a DPDT switch. You've got a SPDT, there.

see how theres one 'row' of 3 lugs?
you need 2 rows.. one to direct the signal where you want it to go, and the other for the LED.
The switch you have looks like this (looking at the bottom):
[a]

[c]

You need:
[a] [d]
[e]
[c] [f]
(I'll refer to each lug by the letter I labeled it in the little diagram in the rest of my reply. saying is easier than 'the middle lug ont he left side of your switch')


once you've got that, wire the killswitch like you usually would (like it shows in the diagram. Middle lug of volume pot to , [c] to ground).

Now, solder the positive lead of the battery clip to [e]. solder one end of a resistor to [d], then solder the other end to the + leg of your LED. - of the LED to + of next LED, ect.
The - leg of the last LED goes to the negative lead on the battery clip.

and ta-da, the light should come on when the signal isn't killed.
I'm not sure what value of resistor you should use with 2 LEDs... ..depends on the rating of your LEDs, but I haven't really

thakns a lot

I used http://ledcalc.com/ to calculate the resistor for two red leds and I need 250 ohms.

Im gonna get the new switch, it should look like this

[x][x]
[x][x]
[x][x]

which one is a,b,c,d,e?

again thanks a lot.

ps: internet explorer was screwing with me this is the third time I try to post this


Edit: I dont have a multimiter or so, to measure polarities
Last edited by divinorum69 at Nov 6, 2008,
#4
I'm not sure what you mean by "which one is a, b, c, d, e, f?".
they're not really assigned a, b, c, d, e, and f, I just labelled them that for reference.

For the polarities, the battery'll just say on it, and the LEDs have a few ways of identifying + and -

The negative leg (cathode) is shorter than the positive (anode)
The edge of the LED has a flat spot on the - side.
and if you can see inside the LED, you'll notice that theres 2 little bits inside. the larger one is on the negative side.
#5
Quote by james4
I'm not sure what you mean by "which one is a, b, c, d, e, f?".
they're not really assigned a, b, c, d, e, and f, I just labelled them that for reference.

For the polarities, the battery'll just say on it, and the LEDs have a few ways of identifying + and -

The negative leg (cathode) is shorter than the positive (anode)
The edge of the LED has a flat spot on the - side.
and if you can see inside the LED, you'll notice that theres 2 little bits inside. the larger one is on the negative side.


I know what you meant with A, B , C , D, E. What im asking is to which ones do I solder for the circuit to be activated when the switch is looking upwards?

I just bought a battery clip so I dont have to solder wires into the battery.

Edit: Nevermind Now I understand

I did a diagram, is it ok?


Cheers
Last edited by divinorum69 at Nov 6, 2008,
#6
That would work fine. But you could substitute two 100R resistors for one 200R.
Gear:
Schecter Hellraiser Deluxe
Boss DS-1
Crate GTD65

GAS List:
Mesa Boogie Dual Rectifier Roadster
#7
1 - your current calculation is WRONG.
you need to subtract the voltage drop across the LEDs from the supply voltage.
then divide that by the resistance.

each LED will have about 2 volts dropped across it.

9v - 4v = 5v

5V / 200ohms = 25 mA


2 - the current in those LEDs is waaaaaaay too high.
you only need a couple of mA to run an LED.
use a larger resistor.

3 - a typical alkaline 9v battery will supply about 625 mAh.
at 25 mA, you'll only get 25 hrs of use from the battery.

4 - the switch could be as simple as a SPDT, if you connected the pole of the switch to ground.

5 - it would be wise to use a stereo jack, so that the battery would be disconnected when the guitar is unplugged. this will prevent accidental battery drain while the guitar is not being used.
Meadows
Quote by Jackal58
I release my inner liberal every morning when I take a shit.
Quote by SK8RDUDE411
I wont be like those jerks who dedicate their beliefs to logic and reaosn.
#9
Quote by SomeoneYouKnew
1 - your current calculation is WRONG.
you need to subtract the voltage drop across the LEDs from the supply voltage.
then divide that by the resistance.

each LED will have about 2 volts dropped across it.

9v - 4v = 5v

5V / 200ohms = 25 mA


2 - the current in those LEDs is waaaaaaay too high.
you only need a couple of mA to run an LED.
use a larger resistor.

3 - a typical alkaline 9v battery will supply about 625 mAh.
at 25 mA, you'll only get 25 hrs of use from the battery.

4 - the switch could be as simple as a SPDT, if you connected the pole of the switch to ground.

5 - it would be wise to use a stereo jack, so that the battery would be disconnected when the guitar is unplugged. this will prevent accidental battery drain while the guitar is not being used.


I was replying this and my IEXPLORER CRASHED. I took a screenshot



EDIT: SOME PICTURES





as you can see its pretty unneat, but thats the character this guitar has
Last edited by divinorum69 at Nov 8, 2008,
#10
the position you have it in is really going to piss you off when turning the vol and tone knob around it up and down.

Also do you realise that its likely to just buzz like hell if you turn the killswitch off
#11
Quote by divinorum69
I was replying this and my IEXPLORER CRASHED. I took a screenshot
you're fortunate you were wrong. 45 mA will destroy most LEDs.

you should consider increasing the size of the series resistance.
your battery life is only 1/10 of what it could be.
Meadows
Quote by Jackal58
I release my inner liberal every morning when I take a shit.
Quote by SK8RDUDE411
I wont be like those jerks who dedicate their beliefs to logic and reaosn.
#12
Quote by guitarcam123
the position you have it in is really going to piss you off when turning the vol and tone knob around it up and down.

Also do you realise that its likely to just buzz like hell if you turn the killswitch off

Im not gonna go buckethead with the killswitch. IM gonna use it to turn the signal down while waiting between songs. Why should it buzz?

Quote by SomeoneYouKnew
you're fortunate you were wrong. 45 mA will destroy most LEDs.

you should consider increasing the size of the series resistance.
your battery life is only 1/10 of what it could be.


sorry I need to study more My physics. If I increase resistance would I have the same light?