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Old 23rd March 2010, 11:05 AM   #1
franke is offline franke  United Kingdom
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Default LED power source

Hi,

I got a Bulgin illuminated switch for my amp project and I'm not sure how I should power it.
BULGIN|MP0045/1E2BL012|SWITCH, FLUSH LATCHING, BLUE ILLUM | Farnell United Kingdom

It has separate terminals for the LED and it wants 12 V DC.

I rather not use a driver circuit which means I would have to buy/build another circuit just for LED.

I think I have found 2 ways of doing it so please let me know which one you think is better.

1. Reducing voltage from my rectifier. I get 25 V out, so that means using a pretty beefy 0.5W resistor. ( I did some online calculations and it suggested a 680 Ohm resistor ).

2. Using a mobile charger or something. I just tested with a 5V one, the LED is not that bright at 5V but it looks nice. I could just put one of those chargers in the case and power it from the mains right after the powerswitch. That means using a block connector. Mains +> Transformer + charger for LED.

Are either of these options horrible?

FYI I am building a LM3886 amp from chipamp.com.
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Old 23rd March 2010, 11:32 AM   #2
AndrewT is offline AndrewT  Scotland
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both are horrible.
Try to find out what range of currents are acceptable for the LED.
Use that information to determine the value of the current limiting resistor from the 25Vdc supply.
Then determine the power dissipation in the resistor and select appropriately.

Maybe a car battery (12V) and a 1r0 resistor to supply the LED pins will tell you the preset current of the LED.
I suspect the switch has a series combination of LED + resistor inside.
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Last edited by AndrewT; 23rd March 2010 at 11:36 AM.
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Old 23rd March 2010, 10:23 PM   #3
franke is offline franke  United Kingdom
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Quote:
Originally Posted by AndrewT View Post
both are horrible.
Try to find out what range of currents are acceptable for the LED.
Use that information to determine the value of the current limiting resistor from the 25Vdc supply.
Then determine the power dissipation in the resistor and select appropriately.

Maybe a car battery (12V) and a 1r0 resistor to supply the LED pins will tell you the preset current of the LED.
I suspect the switch has a series combination of LED + resistor inside.
I used a 12V adapter that I found in the office and measured 8 mA over the LED. I'm not sure how I would power it in my amp case though. I can't use the 25v from my rectifier cause that would mean dividing the voltage that goes to the amp.
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Old 24th March 2010, 06:46 AM   #4
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13/8=1.625Kohms
Add a resistor in series of nearby value, higher than this.

Don't worry, nothing will happen to your amp.
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Old 24th March 2010, 08:22 AM   #5
Mooly is offline Mooly  United Kingdom
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Just wire it across one the DC rails with an appropriate resistor.

What voltage is the amp running on ? You mention "25v after rectifier" Do you mean a -/+25 volt supply.
If so wire from ground to either one of the rails (observing correct polarity for LED) and include a series resistor that gives around 12 volts across the LED (which as Andrew states includes an internal resistor in the switch). You mention 8 ma... sound about right by the way... so that means assuming 25 volts you need R=V/I which is 25/0.008 which is around 3k. But there is already a resistor "giving" 12 volts dropped internal to the switch so your series one needs to be around 1.5K Going higher in value will dim the LED, lower and you exceed the nominal rating. Wattage required... 0.25 watt is fine.

I have ignored the LED foward voltage in all this... it makes so little difference at that supply voltage.

It's the simplest thing possible to do
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Old 24th March 2010, 02:13 PM   #6
franke is offline franke  United Kingdom
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Quote:
Originally Posted by Mooly View Post
Just wire it across one the DC rails with an appropriate resistor.

What voltage is the amp running on ? You mention "25v after rectifier" Do you mean a -/+25 volt supply.
If so wire from ground to either one of the rails (observing correct polarity for LED) and include a series resistor that gives around 12 volts across the LED (which as Andrew states includes an internal resistor in the switch). You mention 8 ma... sound about right by the way... so that means assuming 25 volts you need R=V/I which is 25/0.008 which is around 3k. But there is already a resistor "giving" 12 volts dropped internal to the switch so your series one needs to be around 1.5K Going higher in value will dim the LED, lower and you exceed the nominal rating. Wattage required... 0.25 watt is fine.

I have ignored the LED foward voltage in all this... it makes so little difference at that supply voltage.

It's the simplest thing possible to do
Yes, I get +25V and -25V out , here is the schematic for the powesupply:

Click the image to open in full size.

The two amp boards each connect to V+, PG+, V- and PG-.
So would i connect V+ and PG+ to the led? With a resistor in series of the V+?
And would that cause problems with one of the amp boards? It would affect the voltage to one of the amp boards, right?

You can see what it would look like here:

http://www.briangt.com/gallery/album...roto_amp_3.jpg
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Old 24th March 2010, 03:03 PM   #7
franke is offline franke  United Kingdom
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Sorry for the double post, it's frustrating that I can't edit my post.

Is this correct?

Click the image to open in full size.
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Old 24th March 2010, 04:51 PM   #8
AndrewT is offline AndrewT  Scotland
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yes.
Check the voltage drop across the 1k6 resistor and the voltage drop across the LED pins, after you have wired it up.
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Old 24th March 2010, 05:19 PM   #9
Mooly is offline Mooly  United Kingdom
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That's it.
And get the LED polarity right... don't guess thinking if it doesn't work one way it will the other... LED's do have a limited reverse voltage... and it would be a shame to damage it.
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Old 25th March 2010, 11:33 AM   #10
franke is offline franke  United Kingdom
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Thanks guys! I haven't actually assembled the amp yet, all the boards are ready but I have to finish of tha case and put everything in it. I'll post some pics later.
How will it affect that channel/board? Does it steal a little current or does it do more? I do not know much about electrical calculations ...


The polarity is easy, the LED is marked with + and - on the switch so even me should do fine there.
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