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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? schematic 1

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  • \$\begingroup\$ where in the world do you have 400V mains? \$\endgroup\$ – jsotola Jan 2 '18 at 6:33
  • \$\begingroup\$ Mains is AC. Your circuit needs DC to function. To switch AC from a microcomputer circuit, one does use a solid state relay. They are cheap and safe to use. \$\endgroup\$ – Janka Jan 2 '18 at 6:36
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    \$\begingroup\$ you are missing common sense. do not play around with mains powered equipment unless you know what you are doing. what you describe can easily kill you. \$\endgroup\$ – jsotola Jan 2 '18 at 6:36
  • \$\begingroup\$ Yeah, I rounded numbers a bit. In Russia we have 220AC in line, so after full bridge there is 310V, not 400. But it does not change question anyway. Also I'm grown adult, and understand risk, thank you. \$\endgroup\$ – Keroronsk Jan 2 '18 at 8:26
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    \$\begingroup\$ One thing you are NOT missing : there is the potential when interfacing to mains for components to fail and distribute high voltages to circuitry that cannot cope with it. Either pay for proper isolation barriers between mains and LV circuitry (such as transformers) or spend several years learning about the failure mechanisms and how to design for safety despite them. \$\endgroup\$ – Brian Drummond Jan 2 '18 at 11:07
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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.

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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?

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  • \$\begingroup\$ Well, I was trying to build a TRIAC controlled Christmas Tree lamp and my Arduino was burned (it was powered with cheap RC chain from mains). And I thought maybe it was because of this "reverse base current". But it seems like real reason was in bad TRIAC/wiring etc since I made another lamp with same schematic and it working flawessly. \$\endgroup\$ – Keroronsk Jul 22 '19 at 6:10
  • \$\begingroup\$ Glad to heat it works now. \$\endgroup\$ – user69795 Jul 22 '19 at 19:44

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