LED circuit?

R

REMUS

Guest
Hi i'm new to the group, and i'm not really to awesome when it comes to
electronics, so I thought I would ask people who really know!

Heres the situation: I play bass in a thrash/rock-punk band and I want to
customise my bass a little. And my idea is to place red LED's under the
scratch plate which I will raise up with washers/nut(s) to give a glow
eminateing from underneath on dark gig's. Ideally I suppose 4 or 5 LED's on
a 9v battery would be about as much as I could fit under there.

I dont know how to make a circuit with a switch and LED's or where to buy
them in the UK.

A freind mentioned I could get a sort of fibre optic wire, that he has on
his PC for cosmetic effect but I guess that runs off 12v?

Cheers for any advice + sorry for being so clueless!
 
That should be a fairly easy circuit to create. Basically string the 4 or 5
LED's in series and stick a resistor in with them to reduce the current. I
don't know where in the UK you can buy LEDs, but any regular electronics
store would sell them. If there's nothing around, I'm sure you could do a
search for LEDs in UK to find a place you can order them online.

The most important thing is to make sure you keep the current below the
specified value of the LED. When you buy the LED, it should tell you the
maximum current it can accept. Also you need to understand that LEDs are
diodes, so they act as voltage dropping devices. So, for example, if you
have a 9V battery and an LED that drops the voltage 2V and has a max current
of 0.02A (20mA), you could use the equation V = I*R to get:

(9V - 2V) = .02A * R

Solve for R to get 350. Thus, you need a 350 Ohm resistor. Simarily, if
you had 4 LEDs, it would be 2V*4 = 8V, so you'd have:

(9V - 8V) = .02A * R

and you'd get R = 50 Ohms. Make sure that the total of the voltage drops
across the LEDs is not greater than the battery voltage. In this case, 5
LEDs would have been too much for a 9V battery (2V * 5 LEDs = 10V, which is
more than the battery can give). But I'm only using example numbers, so
they could change depending on which LEDs you buy.


"REMUS" <suk@itspambot.com> wrote in message
news:ckdteg$efl$1@newsg4.svr.pol.co.uk...
Hi i'm new to the group, and i'm not really to awesome when it comes to
electronics, so I thought I would ask people who really know!

Heres the situation: I play bass in a thrash/rock-punk band and I want to
customise my bass a little. And my idea is to place red LED's under the
scratch plate which I will raise up with washers/nut(s) to give a glow
eminateing from underneath on dark gig's. Ideally I suppose 4 or 5 LED's
on
a 9v battery would be about as much as I could fit under there.

I dont know how to make a circuit with a switch and LED's or where to buy
them in the UK.

A freind mentioned I could get a sort of fibre optic wire, that he has on
his PC for cosmetic effect but I guess that runs off 12v?

Cheers for any advice + sorry for being so clueless!
Click to expand...
 
"Damn Dan" <a@a.com> wrote in message
news:ZCIcd.83162$Lo6.21314@fed1read03...
That should be a fairly easy circuit to create. Basically string the
4 or 5
LED's in series and stick a resistor in with them to reduce the
current. I
don't know where in the UK you can buy LEDs, but any regular
electronics
store would sell them. If there's nothing around, I'm sure you could
do a
search for LEDs in UK to find a place you can order them online.

The most important thing is to make sure you keep the current below
the
specified value of the LED. When you buy the LED, it should tell you
the
maximum current it can accept. Also you need to understand that LEDs
are
diodes, so they act as voltage dropping devices. So, for example, if
you
have a 9V battery and an LED that drops the voltage 2V and has a max
current
of 0.02A (20mA), you could use the equation V = I*R to get:

(9V - 2V) = .02A * R

Solve for R to get 350. Thus, you need a 350 Ohm resistor. Simarily,
if
you had 4 LEDs, it would be 2V*4 = 8V, so you'd have:

(9V - 8V) = .02A * R

and you'd get R = 50 Ohms. Make sure that the total of the voltage
drops
across the LEDs is not greater than the battery voltage. In this
case, 5
LEDs would have been too much for a 9V battery (2V * 5 LEDs = 10V,
which is
more than the battery can give). But I'm only using example numbers,
so
they could change depending on which LEDs you buy.
Click to expand...
Although they might run off 9V, 4 2V LEDs would go dim very rapidly as
soon as the battery lost a half volt or 1V as it discharged. So it's
not a good idea to use that many off a 9V battery.

Use flashing LEDs or use a flasher with LEDs to get much longer battery
life and more visual effect.

Suggest you don't top-post, or you may get flamed for it. :)


"REMUS" <suk@itspambot.com> wrote in message
news:ckdteg$efl$1@newsg4.svr.pol.co.uk...
Hi i'm new to the group, and i'm not really to awesome when it comes
to
electronics, so I thought I would ask people who really know!

Heres the situation: I play bass in a thrash/rock-punk band and I
want to
customise my bass a little. And my idea is to place red LED's under
the
scratch plate which I will raise up with washers/nut(s) to give a
glow
eminateing from underneath on dark gig's. Ideally I suppose 4 or 5
LED's
on
a 9v battery would be about as much as I could fit under there.

I dont know how to make a circuit with a switch and LED's or where
to buy
them in the UK.

A freind mentioned I could get a sort of fibre optic wire, that he
has on
his PC for cosmetic effect but I guess that runs off 12v?

Cheers for any advice + sorry for being so clueless!
Click to expand...
 
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