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?
Best Answer
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\$.