How to "Dim" a lamp.

[pfenlon]

I have fitted a push button starter unit, this has an integral light. Problem is that the lamp is very very bright.

How can I reduce the luminosity please. 

[Lebro]

I think we need a bit more information about it please.

Bob

[pfenlon]

The bulb or LED is inaccessible in the unit Bob, thought about a resistor on the +v side but as to value, I haven't a clue.

Have yet to fit the bolt "Cover" caps too. Hope this helps.

 

Edited February 26 by pfenlon

[RogerH]

Hi Pete,

If access to the bulb terminals is too difficult how about a coat of red paint on the red cover.

 

 

Roger

[Ian Vincent]

My understanding is that if it’s an LED, which I would have thought likely, you need to reduce the voltage. At which point you need input from one of the electronic gurus on this site.

 Rgds Ian

[RogerH]

59 minutes ago, Ian Vincent said:

My understanding is that if it’s an LED, which I would have thought likely, you need to reduce the voltage. At which point you need input from one of the electronic gurus on this site.

 Rgds Ian

That all depends on access to the LED pins.

Roger

[RobH]

If the switch is like this one 

https://www.merlinmotorsport.co.uk/p/push-button-starter-switch-rated-at-50-amps-ge338r

there are three terminals, 1 Power, 2 Output, and 3 Earth for the LED. 

Dimming the led can probably  be done by putting a resistor between pin 3 and earth instead of a direct connection. The value depends on the LED used but I would start by trying a 2.2K Ohm 1/4 watt resistor and seeing what happens.  If it is too dim, go to a lower value, or if still too bright increase it.  

[You can determine the resistance needed more scientifically by using a multimeter on its DC milliamps range to measure the current in the earth wire, which will tell you the approximate value of the internal resistor.    ( LED elements work on about 3V so a resistor which sets the current must drop another 9 volts for a 12v system. )

 R in kilOhms = 9/mA - ie if you read 5mA then the internal resistor is 9/5=1.8k, so putting another resistor of the same value in series will halve the LED current. ]    

 

[Lebro]

What he said 

Bob

[pfenlon]

1 hour ago, RobH said:

If the switch is like this one 

https://www.merlinmotorsport.co.uk/p/push-button-starter-switch-rated-at-50-amps-ge338r

there are three terminals, 1 Power, 2 Output, and 3 Earth for the LED. 

Dimming the led can probably  be done by putting a resistor between pin 3 and earth instead of a direct connection. The value depends on the LED used but I would start by trying a 2.2K Ohm 1/4 watt resistor and seeing what happens.  If it is too dim, go to a lower value, or if still too bright increase it.  

[You can determine the resistance needed more scientifically by using a multimeter on its DC milliamps range to measure the current in the earth wire, which will tell you the approximate value of the internal resistor.    ( LED elements work on about 3V so a resistor which sets the current must drop another 9 volts for a 12v system. )

 R in kilOhms = 9/mA - ie if you read 5mA then the internal resistor is 9/5=1.8k, so putting another resistor of the same value in series will halve the LED current. ]    

 

Its the same switch Rob, many thanks for the info, off to RS components tomorrow.  What a forum!

[pfenlon]

Lamp now at a level of luminosity that isn't all pervading,  2 X 2K resistors in series did the job very well,  Thank you Rob H.

[tthomson]

On 2/26/2024 at 3:56 PM, Ian Vincent said:

My understanding is that if it’s an LED, which I would have thought likely, you need to reduce the voltage. At which point you need input from one of the electronic gurus on this site.

 Rgds Ian

The voltage across the LED remains the same when a series resistor is added, but the current through the LED changes, and hence so does the brightness. (The LED has a minimum voltage below which it will not operate.)

I know it is pedantry, but it all helps with understanding electrickery.

Tony ;-}