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!
 
REMUS <suk@itspambot.com> wrote:
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.
Click to expand...
http://www.maplin.co.uk/

You may get a better response over on sci.electronics.basics.
Take the forward voltage of all the LEDs, say it's 2V * 3 = 6V.
Now, 6V/.02A (typical maxiumum current) = 300 ohms.
So, you need a 300 ohm resistor, and 3 LEDs in parallel connected up to the
0V battery.
 
"Ian Stirling" <root@mauve.demon.co.uk> wrote in message
news:416aa90e$0$44845$ed2619ec@ptn-nntp-reader02.plus.net...
REMUS <suk@itspambot.com> wrote:
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.

http://www.maplin.co.uk/

You may get a better response over on sci.electronics.basics.
Take the forward voltage of all the LEDs, say it's 2V * 3 = 6V.
Now, 6V/.02A (typical maxiumum current) = 300 ohms.
So, you need a 300 ohm resistor, and 3 LEDs in parallel connected up to
the
0V battery.
Click to expand...

Oops!! Wrong answer...
If you are using three 2-volt LED's and a 9-volt battery, the calculation
would be:
9 - (3 * 2) = 3V
3V/0.02A = 150 ohms

If the LED's drop 6 volts, that leaves 3 volts to drop across the resistor.

--
Dave M
MasonDG44 at comcast dot net (Just subsitute the appropriate characters in
the address)

Never take a laxative and a sleeping pill at the same time!!
 
[crossposted to sci.electronics.design,sci.electronics.basics, with
followups-to the same.]

On Monday 11 October 2004 08:38 am, Ian Stirling did deign to grace us with
the following:

REMUS <suk@itspambot.com> wrote:
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.

http://www.maplin.co.uk/

You may get a better response over on sci.electronics.basics.
Take the forward voltage of all the LEDs, say it's 2V * 3 = 6V.
Now, 6V/.02A (typical maxiumum current) = 300 ohms.
So, you need a 300 ohm resistor, and 3 LEDs in parallel connected up to
the 0V battery.
Click to expand...
If you're going to answer the guy, at least give accurate answers.

In the first place, 6V LED voltage from 9V (the battery) leaves 3V
for the dropping resistor. 3V/.02A = 150 ohm. You put this, three
LEDs, the battery, and the switch all in series.

- + K K K
+-[9V Batt]--[sw]--[150R]--[led]--[led]--[led]--+
| |
+-----------------------------------------------+

where the - and + by the battery are the polarity, the +'s in the
wire just mean "bend here", and the K's are the LED cathodes.

FWIW, a 9V "transistor" battery in this app will last about one gig.

Have Fun!
Rich
 
I think i'll go with Rich's set up, it looks easier!
http://www.maplin.co.uk/Module.aspx?ModuleNo=35736&TabID=1&source=15&WorldID=&doy=11m10
I found these, they are described as having low current and power
consumption, so which resistor would I need to buy, to run 3 of these?

Thanks very much for the advice so far!
 
REMUS <suk@itspambot.com> wrote:
I think i'll go with Rich's set up, it looks easier!
http://www.maplin.co.uk/Module.aspx?ModuleNo=35736&TabID=1&source=15&WorldID=&doy=11m10
I found these, they are described as having low current and power
consumption, so which resistor would I need to buy, to run 3 of these?

Thanks very much for the advice so far!
Click to expand...
Something like UK19V might be more appropriate.
You get more light out of ultrabright LEDs when run at low current than
low current LEDs.
Do you own a multimeter, one would be handy for working out exact
voltages and currents you want to use.
If the LEDs are too bright, you can just increase the value of the
resistor.
(get several, maybe 150 ohm, 300 ohm, 600 ohm, ...)
 
Rich Grise wrote:

[crossposted to sci.electronics.design,sci.electronics.basics, with
followups-to the same.]

On Monday 11 October 2004 08:38 am, Ian Stirling did deign to grace us with
the following:


REMUS <suk@itspambot.com> wrote:

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.

http://www.maplin.co.uk/

You may get a better response over on sci.electronics.basics.
Take the forward voltage of all the LEDs, say it's 2V * 3 = 6V.
Now, 6V/.02A (typical maxiumum current) = 300 ohms.
So, you need a 300 ohm resistor, and 3 LEDs in parallel connected up to
the 0V battery.


If you're going to answer the guy, at least give accurate answers.

In the first place, 6V LED voltage from 9V (the battery) leaves 3V
for the dropping resistor. 3V/.02A = 150 ohm. You put this, three
LEDs, the battery, and the switch all in series.

- + K K K
+-[9V Batt]--[sw]--[150R]--[led]--[led]--[led]--+
| |
+-----------------------------------------------+

where the - and + by the battery are the polarity, the +'s in the
wire just mean "bend here", and the K's are the LED cathodes.

FWIW, a 9V "transistor" battery in this app will last about one gig.

Have Fun!
Rich

I liked the "0V" battery part.
Click to expand...

A "9V" battery has 6 cells; for NiCd and dry cells the recommendation
from battery manufacturers is to plan on 1.0 to 0.9V per cell at full
discharge (this is just past the "knee" on the NiCd curve, and about 3/4
of the way down the dry cell discharge curve). So for a 9V battery you
should design for 5.4V, or expect the thing to die before you've really
wrung everything out of it that you can.

Now, having burbled on about all that, I don't know if you'll get better
battery life with your circuits, or with four LED's in series parallel
running 15mA in each leg -- but it may be a good thing to try both.
Here's the other circuit to try:

- + K K
+-[9V Batt]--[sw]-+-[270R]--[led]--[led]-+
| | |
| | K K |
| +-[270R]--[led]--[led]-+
| |
+----------------------------------------+


The _best_ thing for battery life would be a little switcher, but that
would be the _worst_ thing for a circuit that you could just whip up and
toss into your ax. If it were me I'd just try Rich's circuit and mine,
use the one that works best, and forget about the fancy stuff.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
 
On Monday 11 October 2004 12:19 pm, Ian Stirling did deign to grace us with
the following:

REMUS <suk@itspambot.com> wrote:
I think i'll go with Rich's set up, it looks easier!

http://www.maplin.co.uk/Module.aspx?ModuleNo=35736&TabID=1&source=15&WorldID=&doy=11m10
I found these, they are described as having low current and power
consumption, so which resistor would I need to buy, to run 3 of these?

Thanks very much for the advice so far!



Something like UK19V might be more appropriate.
You get more light out of ultrabright LEDs when run at low current than
low current LEDs.
Do you own a multimeter, one would be handy for working out exact
voltages and currents you want to use.
If the LEDs are too bright, you can just increase the value of the
resistor.
(get several, maybe 150 ohm, 300 ohm, 600 ohm, ...)
Click to expand...
I can't tell from the catalog page what the brightness is supposed to
be, or what the forward voltage (Vf) is at 3 mA. But if you ass-u-me
that it's 2V like the others, then you're still looking to drop 3V, and
3V at 3mA is 1K. E = I * R, R = I / E, I = E / R - that's Ohm's law
in a nutshell.

Cheers!
Rich
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top