40Vdc Power on LED help.

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I am making a Class D amplifier that I am powering with a external 40Vdc power supply. I want to put in a power on LED, but the LED resistor calculator that I used said, "With a supply voltage of more than 24V, you'll dissipate excessive heat in a current-limiting resistor." Is there a better way to do a power LED using this higher voltage?
 

Less than 1Watt power dissipation. You CAN use led with a resistor.
The power supply is stabilized? If yes, use 2000Ohm resistor 1Watt in series with the LED.
If not, you should use a bigger resistor or a zener to help you limit the voltage before it comes to led.zener to help
 
Yes, it is a regulated power supply. Thank you.
 

Do you want the LED to be an indicator that there is power, or do you want it to be extremely bright and blind people?
An LED is pretty bright when it has a current of 20mA. Its voltage is from 2V for a red one to 3.5V for other colors and white. Then the current-limiting resistor has a voltage of about 37.25V and it dissipates 37.25V X 20mA= 0.745W. A 1W resistor should not be enclosed unless there is lots of space. If you reduce the current to 5mA maybe it will be bright enough and the resistor can be 1/2W.
 
It depends om your requirements for package size , qty and Luminous Intensity requirements.

10 mcd very dim,
100 mcd visible short range,
1000 mcd very bright,
10,000 mcd (10Cd) blinding at arm's length @20mA
.... but at 2mA or 10% , 1 Cd is very bright and uses only 2mA x38V = 76mW for the series drop resistor.
 

Some very bright LEDs cheat and focus the light beam into a very narrow beam to make it bright like a spotlight. You probably want your indicator light to shine all around with a wide viewing angle.
 

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