Playing with Christmas lamp TRIAC, I got this question. Suppose some transistor is saturated, if gate current is >= 1ma, and I want to control it with 5V Arduino (or STM\PIC\etc) port. So I need to add Rb around 5K into base. Hovewer, if my transistor collector connected to 400V mains, and transistor fully opened, does base got connected to 400V too (via Rb=5K, making current around 80ma), destroing my Arduno, board, house and so on? Or I missing something?
So I did some reasearches, and found this answer:
The reason, why current does not flow from collector into base when transistor is opened, is in physical properties of the transistor. Since emiter is doped more heavily then collector, and base are made very thin, electrons are flying directly from emiter into collector, and only around 1% recombines in base. So no reverse base current exist.
Putting too much voltage on the collector of the transistor will cause the reversed biased B-C junction will break down with an avalanche of carriers. Then the current will only be limited by the Rb resistor. It will flow into the protection diodes of the Arduino output pin. Now, what will melt first? If it is the collector or base bond wire, then perhaps the transistor will stop working and your Arduino will be OK. Also, as the wire melts, and the voltage is high, an arc may form, blowing the whole transistor to pieces.
If it's the Arduino chip, it will perhaps first melt and then the bond wires from the pin to the chip will open up. Then the transistor C-E connection will take the current and it will melt until the emitter bond wire blows up. In this case, both the Arduino and the transistor will go.
It is the bond wires to the semiconductors that will act as fuses here, but it hard to tell which will blow first. Also, as they melt, arcs are likely (depending on the voltages involved). In the presence of arcs, lots of heat and pressure will be generated.
Why don't you try and report back to us on what happens?