Electronic – use an inductive load in an ATX-to-bench supply project

The ground terminal is almost certainly not connected to the power supply output terminals. To measure the output voltage of the power supply, you must turn the current limit knob up somewhat, and measure the voltage between the black and red output jacks.

With the current control turned to minimum, the supply probably won't output any voltage, as the current is limited to zero.

To test the current limit, set your meter to a current scale, and connect to the red and black terminals (again, the yellow ground terminal is not connected to the power supply output - it is just there to give you convenient access to the AC power ground). Start with the current control fully counterclockwise (minimum), turn the voltage pots up somewhat, then slowly turn the current limit knob clockwise - you should see increasing current on your meter, which should match the current indication on the power supply.

Each of the 3 power rails (3.3V, 5V, +12V) are connected to a fuse, then to a binding post, while the negative binding posts go to ground.

This sounds like the problem. Keep in mind, voltages are differences in potential between two points. There's nothing special about ground, it's just an arbitrary point which we pick. It's 0V because the difference of something with itself is 0.

You can call anything you want "ground". That's why this circuit works. If you call the -12V output as "ground", then everything else is 12V higher. That's including what was previously called "ground": now it's 12V, because it's 12V more than what you are now calling ground.

Now consider what you've done:

^{simulate this circuit – Schematic created using CircuitLab}

The "ground" of the power supply connects all of the voltages supplied together (connections labeled B and C). The output voltages are relative to this. Notice how the -12V (V4) makes a negative voltage because it's positive side is attached to "ground".

Then, you attached the negative binding posts of all the supplies together. Largely this is redundant: you are duplicating connections B and C. But you are also adding connection A.

See the problem? You've shorted out V4. A wire has ideally zero resistance. By Ohm's law, the current that will flow is:

$$ \frac{12V}{0\Omega} = $$

In reality, the wires used to make this connection actually have some very small resistance, and you get a whole ton of current. This far exceeded the current the voltage regulator can handle and the smoke got out.