# basic questions about LED in connection with batteries

Discussion in 'Electronic Basics' started by Roland Grichnik, Dec 2, 2003.

1. ### Roland GrichnikGuest

Hi there!

As I'm not very much into electronics up to now, I decided to post
this message.
I'm trying to modify my electrical guitar a bit, so I wanted to build
in some diffuse LEDs, 4 LEDs at each pick-up (3 singlecoil pickups--->
12 LEDs) to indicate which pickup is selected at the moment.
The 3 pickups can be selected with a 5-way-switch, where the 2. and 4.
position of the switch mix two of the pickups into one audio-signal.
I want all 4 LEDs of one pickup to shine if it's the only one
activated and for the mixed positions I only want the inner LEDs to
shine (there should always be 4 shining LEDs).

Now to my real question:
-Which specifications should the LEDs match?
-What about the Resistor (sorry, I don't now the english word, but the
symbol is R)? How much Ohm do I need?
-How long will the batteries last (I think 2 mignon-batteries would be
ok to put them into my guitar)
-Could i also build in a dimmer to dim the LEDs or is it not possible
with them?
-Is there a way of building a sound-steered dimmer, so the LEDs change
their brightness due to the music?

Everybody, who could help me, please message, thank you very much,

newbie, Roland

2. ### John PopelishGuest

The brightness of LEDs is proportional to the current that passes
through them. The voltage drop across them is roughly constant (over
the usable brightness range of current. The voltage drop is different
for different colors. Red leds drop about 2 volts, so you calculate
the resistor to use up the excess voltage while the desired current is
passed.

For instance, if you used a red led and a 3 volt battery, you would
need a resistor that uses up about 1 volt while it passes some current
less than the maximum rated current of the LED. Many LEDs rae rated
for .02 amperes (20 milliamps) current. Lets say you decide that .01
amp gives enough brightness. Resistance (ohms) is the ratio of
voltage across to current through a resistor. So 1 volt across, with
..01 ampere through implies 100 ohms of resistance. If you were using
a 9 volt battery, instead, you would need a resistor that drops 7
volts (leaving the remaining 2 volts to be dropped across the LED)
while passing .01 ampere, or 7/.01=700 ohms, approximately.

3. ### Roger JohanssonGuest

The big question is how to tell the leds to turn on and off.
The 5-way switch is already used to switch the guitar signal, so you
would need a way to detect when it is in a certain position.

The best solution is probably to replace the 5-way switch with 3 small
2-pole switches, one for each pic-up, where you can use 1 pole to
switch the sound signal and the other to control the leds.

Do you like the way the 5-way switch works? Have you gotten used to
have it working as it is working? Then you probably would not want to
exchange it for 3 separate switches. Then the problem becomes
complicated.

If it is okay to use 3 switches instead of the 5-way switch you have
to find a type which is nice to work with mechanically, and install
them in the slot where the 5-way is now, or some other place.

Then you can connect the leds to the un-used contacts on the new
switches and get a visual indication of which pic-ups are in use.

Your idea that fewer leds per pic-up should be lighted when more
pic-ups are used can be realized, but it will complicate the circuits
a little.
It is a lot easier to use one (or two) leds for each pic-up, and
accept that when you use all three pic-ups you will have 3 (or 6) leds
on.

There are advantages in replacing the 5-way switch, because they often
fail after some years, but it will change the second-hand value of the
guitar if you rebuild it with 3 separate switches instead.

Each led needs a resistor in series with it, to set the current to a
suitable value. But you need a switch to turn it on and off, and that
is the problem. The existing 5-way switch does not have any free
contacts you can use, that is why you need to replace it with other
switches.

Instead of 3 separate 2-way switches you could use a rotary 5-way
2-pole switch, if that sounds like a better alternative.
It would work more like the old switch.

Or maybe you can find a 5-way which fits like the old switch but has 2
poles. (Probably an expensive solution.)

4. ### Roland GrichnikGuest

hmm, first of all thx for the good ideas, I think I'll build in 3
2-pole-switches, it might look quite cool too, and as it's just my old
guitar I decided to use for modifications.
I also sprayed the pickguard and other things, besides the 5-
way-switch doesn't work perfectly any more, so it won't be a problem
^^

ok, so that's the new plan:

-build in 3 2-pole-switches, each one connecting one of the pickups
with the tone- and volume-knobs at one pole and the LEDs that belong
to it with the batteries at the other pole

I think that's right for now, isn't it?

ok, then problem no. 2:
how do I connect the LEDs?
do I have to put them in a series or do I have to connect them
parallel? which factor is important for the batteries' life? the LEDs
current or the voltage? and which factor denotes how much "capacity" a
battery has?

thx, roland grichnik

5. ### Rich GriseGuest

Well, I'm not too good with ASCII art, so I'll try to describe it in
words. Each switch has two sections, call them "A" and "B". For this
app, you only need to use two contacts on each switch for the LEDs
and the other two for the pickups. If the switch has 6 contacts, then
just ignore the two outside ones (the ones on the "off" side.)

Call the switch terminals 1 and 2.

The pickup lead goes to 1A, and 2A goes to the volume control.
The battery goes to 1B of all three switches, and 2A will go to
the resistor, then to the LED, then back to the negative of the
battery.

lessee ....

[pickup] [pickup] [pickup]
| | |
batt. pos.--------|----+-----------|----+-----------|----+
-|- = NO CONNECT (CROSS)
| | | | | |
-+- = CONNECT
1A 1B 1A 1B 1A 1B
SW1 SW2 SW3
2A 2B 2A 2B 2A 2B
| | | | | |
| | | | | |
| | | | | |
vol. | vol. | vol. |
| | |
[R] [R] [R]
| | |
| | |
LED LED LED
| | |
| | |
batt. neg.-------------+----------------+----------------+

R = (Vbatt - Vled) / Iled.

Observe polarity, and you should be good to go!

Have Fun!
Rich

6. ### Roger JohanssonGuest

Rich Grise gave you a nice schematic for that so I won't go into that
so much.
In a 2-pole switch you have two completely separate switches, often
with 3 contacts for each, but you need just two of them for the led.

Switches with 3 contacts are used to direct the current this way or
that way, but we often use just two of the contacts if we only want to
turn the current on or off.

Keep the sound circuit totally separated from the led circuit.
Use half of each switch to connect, or not connect, the pickup to the
volume and tone pots, the other half to turn a led on and off.

I recommend using just one led for each pickup, if the battery gives 3
volts, because the battery will last longer, but if you have a big
battery or you can change the battery often you can use more leds.
If you put components in series you can only turn them all on or odd,
because the current either goes through them or all, or it doesn't.

(If you have a 3 volt battery and want to use two leds for each pickup
you have to put a resistor in series with each led and then put these
two led-resistor combinations in parallell with each other, and that
circuit is series with a switch and the battery.)
A led uses a certain voltage but you decide how much current to allow
it to use, if you give it too much it will be destroyed.
You use a resistor in series with the led to set the current.

With a 3 volt battery (or 2*1.5volt) you can try with a 1kOhm resistor
to begin with, and lower it to 500Ohm or lower to see how much light
you get.
The battery has a capacity, for example 200 mAH, which means it can
deliver 200mA for 1 Hour, or 20mA for 10 Hours, or 2mA for 100 Hours,
etc..

A led can use 20mA maximum, but you should choose a lower value to
save the battery, something like 1-5mA could be suitable.

Try putting a resistor, a led, a switch and a battery in series before
you actually build it into the guitar, and try out a suitable value of
the resistor.
Start with a high value resistor and go lower until you get as much
light from it as you think you need.

You have not told us the battery voltage, I think, so we cannot give
exact numbers here, but you can use a little math to find out
approximately what resistor values you should try.

You need at least 3 volt from the battery, if the battery is 1.5volt
you need two batteries.

If the battery is 6 volts you can use 2 leds in series, plus resistor,
for each indicator.
We need to use a significant part of the voltage, say at least 25%,
for the resistor to regulate the current.

Some trial and error here will only cost you a few burnt-out leds and
they are cheap

If you turn the led backwards it will not work, and may get destroyed,
they are very sensitive to backward voltages, some leds get destroyed
by only 4 Volts backwards.