DC On/Off Switch Help

[jhayx7]

I have built a little stir plate using a computer fan with a rare earth mag attached to it. It is fairly simple setup with a 12V DC power supply connected to a on/off rocker switch then to a potentiometer to control the speed of the fan. It is working fine but the little rocker switch has an LED that lights up in the on position. Problem is, the LED is not lighting up when it is in the on position. The switch has a three prong connector in which the negative supply goes to the middle prong and then one of the outer prongs goes to the potentiometer. Here is the article I used to build it to show more detail:

http://www.thebitterbrewer.com/viewtopic.php?f=1&t=64

Since I am clueless about how electronics work, does anyone have a suggestion on how to get the led to light up when it is in the on position? I assume that I need to wire up the unused prong but not sure...

 

[Demon-Xanth]

One of the prongs goes to the LED, one goes to the power source, the other to the powered equipment.

 

[jhayx7]

Ok, so I need another negative supply going to the LED prong?

 

[Demon-Xanth]

No, you connect the LED prong to ground (through a resistor) so that when you apply power to the common lead the LED turns on.

 

[Paperdoc]

The problem is that the LED needs a certain voltage to light up (often about 5 v dc), but you are only giving it a low voltage from the pot designed to keep the fan motor running slowly. The LED supply needs to be independent of the pot.

To do this you need a toggle switch that is called Double Pole, Single Throw - DPST. It is really two independent sets of switch contacts operated at one time. For one set you connect the variable voltage supplied from the pot, and the output terminal goes to the motor. (The motor's other terminal is connected back to the original power supply's Ground.) On the other switch contacts you connect the original full voltage from BEFORE the pot, and feed it to the LED. (Likewise, the other leg of the LED comes back to power supply Ground.) Now, there is still an adjustment to be made on the LED leg of the circuit. Your full supply is 12 vdc, but the LED only wants maybe 5 vdc, and it will "consume" a current of some milliamps (perhaps 10 mA) given that supply. So you need in series with the LED (between the switch output terminal and the LED) a voltage-dropping resistor that drops from 12 vdc to 5 vdc when the current through it is the LED's operating current. IF the LED specs say it uses 10 mA with a 5 vdc supply, then the resistor value will be given by:

R = (12-5) x 1000 / 10 - NOTE that I used the current in MILLIAMPS, and that's where the "1000" factor comes from.

The value with those numbers plugged in is 700 ohms. In terms of power rating of the resistor, it will be dissipating 7v x 10/1000 amps = 0.07 watts, so a ¼ watt resistor is surely big enough.

Look up the spec's for your LED, then plug in the voltage and current ratings for it and calculate your own correct resistor value. Anything close will do the job. Doing it this way will give your LED exactly the right voltage independent of the motor speed pot setting, but it will turn on and off as the switch is operated.