short circuit with two power supplies

Question

I have a radio controlled power switch (one that you plug in between the wall and a lamp to switch it on and off with a remote), that has an atmel microcontroller on it. I connected the radio switch to an atmel programmer (Gnd, Vcc(3.3v), Miso, Mosi, Clk, Rst). The atmel programmer is connected and powered via USB.

I accidentally left the programmer connected when I plugged in the radio switch and both the switch and the programmer blew up.

Now I wonder why.

The radio switch's power supply is a resistor followed by a 4 diodes (rectifier?) followed by a capacitor (filter?) followed by a VIPer12A switcher (provides lower voltage?).

I assume that somehow there was a voltage difference between Gnd of my USB port and Gnd of the radio switch, but how can that be?

Answer 1

Power supplies are (usually) voltage sources, approximating ideal voltage sources, so they have a very low impedance looking in. A nonideal voltage source can be approximated as a resistance in series with an ideal voltage source. If we connect two such sources in parallel, we get the following situation, which reduces to a simpler equivalent circuit:

^{simulate this circuit – Schematic created using CircuitLab}

So if \$V_1\$ and \$V_2\$ are not precisely the same, a nonzero difference voltage \$V_1-V_2\$ is applied across \$R_1 + R_2\$. Since \$R_1\$ and \$R_2\$ are very small (ideally zero), even a small \$V_1-V_2\$ can cause a large current to flow.

If \$V_1 \neq V_2\$ and \$R_1 = R_2 = 0\$, then there is a pure short circuit: the circuit cannot be solved for current flow due to division by zero in \$I = V/R\$.