Using an LED to replace a burnt out 5mm diameter filament bulb

[ledoux3]

I am attempting to replace an old bunt out filament bulb in a piece of HIFI gear (an older Harman Kardon Tuner). The original bulb is 5m in diameter. I have found a 12 volt DC LED that is somewhat close to the measured 14 volt DC power supplied to this device with a 5mm diameter. The LED has a power consumption rating of 20 milliamps at its rated 12 volts. The power supplied to this device 14 volts/ 2.6 amps. Can someone suggest a means for both reducing the voltage to 12 V DC while also reducing the amperage coming from the power supply? The amperage suppled seems to be 130 times what is required. Any help is greatly appreciated.

 

[sagor1]

The bulb (LED) will only pull what current it needs, no more...
Based on the 20ma at 12V, that means the internal resistance of the LED is 600 ohms. If you add another 100 ohm resistor to the 14V feed, you will keep the current consumption to 20ma. Any small 100 ohm resistor will do. There will be a 2V voltage drop across the resistor, leaving 12V on the LED.

 

[ledoux3]

The bulb (LED) will only pull what current it needs, no more...
Based on the 20ma at 12V, that means the internal resistance of the LED is 600 ohms. If you add another 100 ohm resistor to the 14V feed, you will keep the current consumption to 20ma. Any small 100 ohm resistor will do. There will be a 2V voltage drop across the resistor, leaving 12V on the LED.

Is there an issue with amperage coming from the supply to the LED? Will the 2.6 amps affect the life of the LED that only requires 0.020 amps?

 

[ledoux3]

The bulb (LED) will only pull what current it needs, no more...
Based on the 20ma at 12V, that means the internal resistance of the LED is 600 ohms. If you add another 100 ohm resistor to the 14V feed, you will keep the current consumption to 20ma. Any small 100 ohm resistor will do. There will be a 2V voltage drop across the resistor, leaving 12V on the LED.

Thank you. I have it in my head now.