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132 Posts
Discussion Starter #1
OK electrical gurus.

I have recently installed a manual isolation switch in my dash so I can totally isolate the duel battery by switching the solenoid off.

The switch I used has an LED indicator which is very bright even during the day.

Can I install a resistor or something on the earth wire that I have grounded for the switch LED to reduce the voltage flow and therefore the LED brightness?

I would like to about halve the brightness so if a resistor would work what rating should I use?

If a resistor is not an option is there any other simple way to do it?


645 Posts
Simple answer, use a permanant marker and blacken the LED. :) If you can't see it, scratch some off.

Seriously, though. Just to clarify, and I don't mean to sound condescending, just letting you know, Voltage has potential. Current (amps) is flow.

LED's are diodes (Light Emitting Diode) and have a breakdown voltage of around 1.7V. In most 12V situations I think it is a 220ohm or 220K ohm resistor that is used for current limiting, so as not to blow the LED. If the voltage reaching the diode is less than 1.7V it will turn off. Theoretically a Diode is either on or off, but I have seen them dim and brighten depending on current.

You could try installing a potentiometer (or variable resistor for those not familiar with electronics) that will allow you to adjust the resistance. That way, during the day you can have it brighter and at night you can make it dimmer.

Or use the Pot to set a comfortable level, take it out, measure the resistance and buy a resistor in that area of value.

Or just buy a pack of various value resistors and try each one by one. Resistors are a dime a dozen these days.

Without extensive calculations, and measurements it can be difficult to pick a value and in this instance trial an error may yield quicker results.


Just an example on the calculations,

Calculating the Required Resistor

The calculation used to find the value of the series resistor we need to know the diode forward voltage and current and its connections. This information can be obtained from the package if you purchased the LED.

In this example it is 2 volts and 20mA (0.02 amps).

The cathode lead is the one nearest a "flat" on the body.

Since the voltage across the diode (LED) is 2 volts and the battery voltage is 12 volts, then the voltage across the resistor is 12-2 = 10 volts.

The diode is in series with the resistor, so the current through then both is the same, 0.02 amps.

We now know the voltage across, and the current through the resistor.

From Ohm's Law we can now calculate the value of the resistor.

Resistance = Volts divided by Amps = V/I = 10/0.02 =500 ohms.

Since this is not a standard value we can use a 470 or 560 ohm resistor as this application is not critical of values.

For a 4 volt (Blue or White LED) the formula would be:

Resistance = Volts divided by Amps = V/I = 8/0.02 =400 ohms.

The common 390 or 470 ohm resistor can be used for these.

Another thing, to avoid loosing too much voltage drop across the LED and causing it to turn off, is to put a resistor in Parallel to the LED, that way the voltage drop would be less affected by the value of resistor, mind you it does have an affect, but needless to say, with a resistor in parallel will sink some of the current, thereby reducing the current in the LED.

Calculating resistance in parallel uses fractions and I hate fractions! :)

1,555 Posts
What he said ^^^ with one addition. The LED acts as a constant voltage source within certain current ranges, therefore you can reduce its brightness by decreasing the current flowing through it. the LED is probably drawing somewhere between 20 and 50 mA. This means that the inbuilt resistor is probably around the 240R to 600R mark. Resistors are worth a few cents each, so to half the current draw and reduce the brightness go buy say a 220R, 390R and 560R, maybe even an 820R, 1/4 or 1/2 watt resistor from Jaycar and try each one between the ground wire and ground until you get the brightness you like.
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