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Author Topic: Adding LEDs to switches, etc  (Read 8442 times)
expanoncolin
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« on: December 08, 2004, 09:25:33 AM »

As per requested in Mocker's help me thread...

LEDs light up when you send current through them.  As a general rule, 1-3mA works well for most LEDs.  The voltage is determined by the supply voltage and the voltage drop across the LED-each LED loses voltage when you light it up, basically.  Most LEDs have a 2v voltage drop, sometimes 4v for bright LEDs.  In terms of current, I tend to light LEDs with about 2mA of current... more current means brighter LED.  If Vs is the source voltage, Vd is the voltage drop, and I is the current...
(Vs-Vd)/I=R
R=(9-2)/I
R is about 3500, or 3k3 or 4k7
Current is a matter of preference.  These calculations were done using a 9v battery.  If you don't have a 9v on hand or want to use something else, measure the DC voltage at certain points on the board, or from the supply, and redo the calculations.

Now that you have your LED lit, you need a way to switch it on.  If you want it to be coupled with a switch, you need a new pull and throw.  So, if you have a SPDT switch and want an LED, you need to use a DPDT switch instead.  The difference is a whole "column" of throw, pull, throw.  If you know which way you want the switch to be switched to turn on the LED, you can do one of two things-wire the resistor that creates the LEDs current to the pull (center lug) and the throw to the positive side (longer wire) of the LED with the negative side to ground or you could wire ground to the pull (center lug) and the throw to the negative side with the positive side always connected to V+.  You are either giving the LED current or allowing the current to flow, either way works... some people think that if you use V+ then you will get popping.  If you want to wire an LED for either direction, just wire a second LED to the other throw.

Hope this all makes sense.  Probably more info than anyone needed-if there are any errors let me know.

-Colin
« Last Edit: April 13, 2006, 06:27:28 PM by expanoncolin » Logged

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dnny
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« Reply #1 on: July 14, 2005, 05:09:25 PM »

hi here is different method for adding LED if you don´t use switches you can hook led whit NPN transistor - this is good way to make LED indicators to show any rate of signal.
Bear in mind that the LEDs will eat current and kill your batteries faster though.
but they look so cool :)

every time the point A gets signal the LED is lit - use any general type NPN transistor. the resistor values you need to figure out (read the post above)


or if you have SPST switch and NPN


signal goes from point A to B this is whit DPDT


or if you have one signal to two points and you want two LEDs


dnny
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expanoncolin
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« Reply #2 on: April 13, 2006, 06:27:10 PM »

Brilliant diagrams, dnny... this thread's been stickied.

-Colin
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specialcowboy
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« Reply #3 on: June 27, 2006, 09:30:17 AM »

How do I figure out I? I'm doing lights on a frequency analyzer it's got the 40v wall wart so would that make (40-3)/I=R?   All of my circuit bending is lodged deeply in the anti-theory/breakin stuff by accident/weirdo art school.
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expanoncolin
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« Reply #4 on: June 27, 2006, 10:00:33 AM »

Quote from: "specialcowboy"
How do I figure out I? I'm doing lights on a frequency analyzer it's got the 40v wall wart so would that make (40-3)/I=R?   All of my circuit bending is lodged deeply in the anti-theory/breakin stuff by accident/weirdo art school.

Chances are there's a lower voltage in there somewhere.  Use your multimeter to find something like 12v DC and wire a 4k7 to that.

-Colin
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paper&tape
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« Reply #5 on: October 29, 2008, 09:01:47 AM »

This answer to this is probably very simple, but I thought I would try to clarify anyway...

I want to setup an SPST with LED as an On/Off (power) switch using Dnny's schematic for the SPST w/ NPN transistor.  My question is regarding "A" on the schematic.  For and On/Off (power) switch "A" would be the power source, correct?  The power source also feeds into the resistor, then into the LED.  It seems like this could create a problem...kinda like a feedback loop.  Do the resistors negate this problem?  Does this problem even actually exist?

Keep in mind, this is my first project and I am new to all of this, so my understanding could be completely off base.

Thanks,

Micah
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expanoncolin
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« Reply #6 on: October 29, 2008, 09:05:49 AM »

Quote from: "paper&tape"
This answer to this is probably very simple, but I thought I would try to clarify anyway...

I want to setup an SPST with LED as an On/Off (power) switch using Dnny's schematic for the SPST w/ NPN transistor.  My question is regarding "A" on the schematic.  For and On/Off (power) switch "A" would be the power source, correct?  The power source also feeds into the resistor, then into the LED.  It seems like this could create a problem...kinda like a feedback loop.  Do the resistors negate this problem?  Does this problem even actually exist?

Keep in mind, this is my first project and I am new to all of this, so my understanding could be completely off base.

Thanks,

Micah

Welcome Micah.  Because you're wiring a power switch, just wire V+ on one side and the resistor to led to ground on the other side.  Just pretend like you're switching the LED on and off, except on the side that isn't connected to V+, connect that to the rest of the circuit (which is receiving power).  The LED won't get power unless the rest of the circuit is, so it will only get power when the circuit isn't.

-Colin
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paper&tape
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« Reply #7 on: October 29, 2008, 09:53:58 AM »

Thanks Colin.  I think I've got it.  Once my resistors arrive (hopefully today) I'll give it a whirl.
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BrettNephew
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« Reply #8 on: February 28, 2010, 09:56:49 AM »

Ok so this is a noob question about LED lighting.  I have an amp I'm building (project updates in the general DIY section)  and I have a LED pilot lamp.  My heater primaries are 6.3 volts at 2.5A..   So If I assume a 2v drop across the LED, and I want around .003 amps,
I should be able to do E=I*R  (with the voltage drop calculation you showed there) and come up 4.3/.003 = 1433  Ohms. 

So I guess now i have a question of wiring this.  I hope its not considered trolling to post a layout/schematic..

So, if anyone looks,  I'm running my heater wires off the 269EX up to the Legs of the LED, the + which will have a 1500K resistor on it, (side note.. should I use a high wattage resistor?)  and then back down to the board there, where they join on opposite ends of R2 and R3, and then split off and go heat my tubes. I would need to have the wires that leave the LED and head to the board, join up before the resistor wouldn't I? As not to change my amperage delivered to the board/heaters?

Anyway, I know that's a big question, but you guys seem to have the LED stuff figured out, so I thought I'd ask! Smiley   Thanks. 

-Brett

* AX84 P1 Schematic.pdf (248.5 KB - downloaded 183 times.)
* AX84 P1 Chassis Layout - Ham.pdf (122.73 KB - downloaded 119 times.)
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expanoncolin
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« Reply #9 on: February 28, 2010, 02:27:01 PM »

Don't really follow your post, but yes, if you wire a 1500k resistor (or 2k2 or 1k) from the 6.3V supply, you should be fine.

-Colin
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