# Trying to get the most out of 48V 500mA switch

I am sorry, I don't have a lot of experience with electricity, but I'm going to give this a shot. What I'd like to do is utilize a switch, that is part of a large device. When the device turns on, the switch is bridged. I'd like the switch to power a light, as the device turns on. However the switch is limited to 48V 500mA, and I can't change it because it is within the components of the machine.

After looking on the forum I found that 48V 500mA is equivalent to 12V 2A, but I also read it wouldn't be smart to have higher or lower volts running through the switch than it is rated for.

I have uploaded a picture above with a way I thought to create the circuit.

120V 20A comes out of the wall to a 48V 500mA power adaptor. The power would then be consistent with the switch and a simple circuit could be made. I couldn't find a 48V 500mA adaptor online so I was wondering about having a higher voltage run through the light, then through a resistor (to lower it for the switch rating), into the switch, and back to the adaptor.

Does anyone know of a 48V 500mA power adapter out there I could buy? Would the circuit work this way, or would I need a resistor, then an amplifier on the other side of the switch?

• It's not clear what the 48V switch is for. Is that the main power switch to the machine, or is that switch somehow tripped (closed) when the machine turns on? Commented Oct 21, 2013 at 21:57
• Take a step back and ignoring any internal components tell us what you are trying to achieve - as if you were telling a non-technical person - like... when power comes on I want a lamp to light up or dancing girls etc... (eg) Commented Oct 21, 2013 at 21:57

Short answer: No, you can't use a resistor to reduce the voltage seen by the switch in your circuit.

It's the voltage difference across the component's terminals that matters, not the voltage at the component.

In your circuit, when the switch is closed the voltage across it is (close to) zero:

When the switch is opened, it will see the full voltage of the source across its terminals, regardless of the resistance of the rest of the circuit:

The purpose of a switch is to extinguish the arc that forms whenever a current-carrying circuit is interrupted. If the voltage and/or current exceeds the rating of the switch, the arc will not be interrupted quickly enough which will cause damage to the contacts of the switch, and eventually cause it to fail. In extreme cases (hundreds of volts and hundreds of amps) the arc will not be interrupted at all and the switch will fail catastrophically (think sparks, fire, molten metal etc).

If you want to operate your light bulb at a higher voltage or current than the switch is rated for you will need to use an interposing relay, like this:

A switch rated at 48V and 500mA would work fine with a lower voltage or a lower current. Those are maximum ratings. Also, the wall outlet may supply up to 20A at 120V, but likely supplies less than 20A. The machine will draw the amount of current that it needs; if it works on, say, 10A, then that's what will flow out of the outlet. The voltage would still be 120V.

If the existing 48V switch is not currently electrically connected to anything, but closes when the machine starts, then you could use a much smaller power supply for a light if you just want an indication that the machine is on. You may or may not need a resistor, depending on the indicator. For a little LED, you'd need a current limiting resistor.

If you're after illumination, like a light bulb, then an even simpler approach is to use a device sometimes found in a woodshop. It has two wall outlets. Into one, you plug the machine, and into the other, you plug something else that should turn on when the machine does. In a woodshop, it's the sawdust vacuum. It could also be a lamp or some other electrical device. Here's one: http://www.rockler.com/product.cfm?page=17351&site=ROCKLER