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.
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.
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.
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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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.
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
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
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:
An externally hosted image should be here but it was not working when we last tested it.
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/albums/lm3886amp/proto_amp_3.jpg
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.
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.
My best guess is that the calculation is something like that of a 1st order high pass filter fora treble unit, so when you go below fn of the circuit the resistance will go up and since you have a fixed frequence in the supply it should be possiple to calculate something.
Maybe some clever heads here can come up with the math
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 ...
It draws current... it doesn't steal it from anything.
In real terms it's half of a quarter of nothing
If you are a perfectionist... and this will make Zero difference... I see that you have a 2K2 across each rail anyway. So on the rail you add the LED too (which can be either the plus or minus) remove the 2K2 on that rail (the LED supply one) to even things up. Easy
It draws current... it doesn't steal it from anything.
In real terms it's half of a quarter of nothing
If you are a perfectionist... and this will make Zero difference... I see that you have a 2K2 across each rail anyway. So on the rail you add the LED too (which can be either the plus or minus) remove the 2K2 on that rail (the LED supply one) to even things up. Easy
I think I'm beginning to understand. The current the LED will use is so small so it doesn't make a difference, unless you are a purist. There's lots of current going into the amp anyway.
thanks
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