hi everybody,
I am using my old phone charger as a dc source (rated at 5V 500ma).
ok, what would happen if I short the dc source, would it cause the transformer to go in flames or break? or would it simply make a 500ma go through the wires, assuming the wire can hold it?
If you are wondering about the purpose, I am experimenting with EM, and would like to make an electro-magnet using the phone charger.
thank you.

 

Hello,

It depends on how the charger is protected.
There are some types that will limit the current.
Others will blow a (poly) fuse.
Some will get very hot and sometimes even catch fire.

Bertus

 

If your "short" was to actually have 10 ohms of resistance. (a long coil of small AWG wire- consult tables for size/current/resistance specs.)

Then only 500 mA of current would flow through the wire.

 

In most cases there's an internal fuse in the primary circuit of the transformer - if it has one. On rare occasions you can take the out covering layers off the transformer, find it and replace it but in most cases it's either buried or the actual sall gauge of the wire is the self-fusing part.

A lot of the more modern supplies convert ACC, drive a simple oscillator at a high frequency into a small toroid then rectify the output with some sort of opto feedback back to the oscillator drive. Take for example the tiny USB output supplies that are used with Apple I-Pods, they're barely larger than a simple wall plug by itself. There would be some sort of protection in there on the primary side, probably a tiny SMD picofuse.