Hi, I'm trying to connect a battery pack with two wires to a power switch and the switch to two devices. The power switch is single pole single throw. It has two large terminals and two smaller ones which I guess are for the Led inside. The large are labeled 1 and 2. The small 3 and A. How would I wire them up?

 

Get out your ohmmeter and test for continuity when the switch is closed

or tell us the model number of the switch and show us a photo.

 

My guess is the small ones are for the light, and if LED the A may be the Anode?
It could require a resistor?
Max.

 

Get out your ohmmeter and test for continuity when the switch is closed

or tell us the model number of the switch and show us a photo.

Labeled as an EDK switch 240VAC 3A 125VAC 5A. Still looking for the model num

 

Here ..
http://www.jameco.com/Jameco/Products/ProdDS/2182397.pdf
looks like you have a SC-WL01A-03BB
datasheet here.. last page has the SC switches..

The smaller terminals are just the ends of a LED.. So as such it requires a resistor inline with the supply to limit the current to 10-20mA.

A switch is typically inserted "inline" on the "hot" lead of any system. So just cut the + wire going from your battery to the devices and put one end of the cut wire to one large terminal and the other cut end to the other side. If you want the LED then you put the + lead of your battery to a resistor..resistor to one terminal for the LEDs and the - lead of your battery to the other terminal.. resistor value must be calculated using the voltage of your battery.

oh and "technically" thats a bad choice of switches as its only UL approved for switching AC..not DC (battery).

 

Looks to be EDK SC-WL01A-0xBB.
x is for LED color.
If x is 1, LED is orange, Forward Voltage is 2.2 V, Forward Current 20 mA, needs resistor: R=(Supplied Voltage-2.2 volts)/(20 mA).

x is 2, LED is green, Forward Voltage is 2.1 V, Forward Current 20 mA.

x is 3, LED is red, Forward Voltage is 1.7 V, Forward Current 20 mA.

Maximum Forward Current is 40 mA.

 

Try this...

 

Here ..
http://www.jameco.com/Jameco/Products/ProdDS/2182397.pdf
looks like you have a SC-WL01A-03BB
datasheet here.. last page has the SC switches..

The smaller terminals are just the ends of a LED.. So as such it requires a resistor inline with the supply to limit the current to 10-20mA.

A switch is typically inserted "inline" on the "hot" lead of any system. So just cut the + wire going from your battery to the devices and put one end of the cut wire to one large terminal and the other cut end to the other side. If you want the LED then you put the + lead of your battery to a resistor..resistor to one terminal for the LEDs and the - lead of your battery to the other terminal.. resistor value must be calculated using the voltage of your battery.

oh and "technically" thats a bad choice of switches as its only UL approved for switching AC..not DC (battery).

Is it any danger to the electronics?

 

How bout this with the switch led integrated in? With 6V 8AA battery pack?

 

Is their any risk in using this switch for DC even though its only AC tested?

 

how much current do you intend to switch with it?
Typically if that were to be approved for DC it would only be something like .5A (500ma) max DC.

Like I said "technically" its a no-no to use any device/component in any method that its not "listed" for.

 

how much current do you intend to switch with it?
Typically if that were to be approved for DC it would only be something like .5A (500ma) max DC.

Like I said "technically" its a no-no to use any device/component in any method that its not "listed" for.

Usually much less but transient spikes somewhat above 2A.