Testing Current of AC-to-DC converter module without load

[koklimabc]

Hai all,
I got 1 240 AC to to output 12 volt 1 Amp switching power supply / converter module.

When I plug 10 A connector (Digital meter) to positive output DC of module and negative/ground (Digital Module) to negative output DC of module, then my converter module was suddenly burned (without load).

I'd no problem to test it out the current liked that from DC(battery) as direct source of voltage. Why?

Of cause I know the current can be tested by input positive(DC) to 10 A of amp meter and negative (Amp meter) to positive of component be connected, and ground componnet back to converter for safety in purpose.

 

[Alec_t]

If your meter was switched to its current-measuring mode and you connected it directly to the power supply module then you created a short circuit (a massive load presented by the internal shunt resistor in the meter) across the supply. Either the meter or the module was going to let out the magic smoke!
What you did is NOT the way to test the module's current capability. You should have used a dummy load.

 

[koklimabc]

If your meter was switched to its current-measuring mode and you connected it directly to the power supply module then you created a short circuit (a massive load presented by the internal shunt resistor in the meter) across the supply. Either the meter or the module was going to let out the magic smoke!
What you did is NOT the way to test the module's current capability. You should have used a dummy load.

Yes correct. Can you provide the way to buy a dummy load or simply build a new one?

 

[crutschow]

Can you provide the way to buy a dummy load or simply build a new one?

To provide a 1A load at 12V requires a 12Ω power resistor.
The resistor will dissipate 12W so you should get a 12Ω, 20W resistor.
If its metal-case resistor with mounting tabs, then it will need to be on a heat-sink.
If it's a ceramic/cement type than it doesn't need a sink.

 

[ivak245]

Incandescent light globes also make cheap loads , you can work out the current from the wattage, then check with an ammeter (in series with the load of course!).

 

[koklimabc]

To provide a 1A load at 12V requires a 12Ω power resistor.
The resistor will dissipate 12W so you should get a 12Ω, 20W resistor.
If its metal-case resistor with mounting tabs, then it will need to be on a heat-sink.
If it's a ceramic/cement type than it doesn't need a sink.

nice, TQ