7
\$\begingroup\$

I was experimenting with ways to cut power to my board and so I added an NPN transistor on the low side of an optocoupler (Q202.) I noticed my voltage regulator was oscillating from 3.3V to GND every few seconds and then the transistor caught on fire. I am very confused as to why this would cause a catch on fire. After the transistor burned up I removed it and noticed the voltage regulator was stabilized.

fire transistor

\$\endgroup\$
1
  • \$\begingroup\$ What is the ign voltage and why the opto coupler? You only need to limit and clamp the input. With many microcontrollers you can have the interrupt wake the cpu and capture the time on the one port pin. \$\endgroup\$
    – Kartman
    May 16 at 13:32

1 Answer 1

Reset to default
13
\$\begingroup\$

No base resistors to limit current. BJTs aren't like MOSFETs. The base-emitter junction on BJT transistors is a diode and you are straight up connecting a voltage source across it.

The PN in your NPN transistors is the diode of the base-emitter junction.

UPDATE:

Why did only Q202 catch on fire and not Q203?

No two diodes are identical due to manufacturing tolerances so when in parallel the one with the lowest voltage drop will conduct and cause the others to have insufficient voltage to fully conduct thus acting like a short-circuit across the others and hogging all the current. That's why paralleling diodes isn't good practice. And semiconductors tend to fail short so the one that flamed up was probably still conducting as a short. If it failed opened I expect Q203 to have followed it.

\$\endgroup\$
4
  • \$\begingroup\$ Why did only Q202 catch on fire and not Q203? \$\endgroup\$
    – Feynman137
    May 16 at 4:22
  • 4
    \$\begingroup\$ @Feynman137 No two diodes are identical due to manufacturing tolerances so when in parallel the one with the lowest voltage drop will conduct and cause the others to have insufficient voltage to fully conduct thus acting like a short-circuit across the others and hogging all the current. That's why paralleling diodes isn't good practice. And semiconductors tend to fail short so the one that flamed up was probably still conducting as a short. If it failed opened I expect Q203 to have followed it. \$\endgroup\$
    – DKNguyen
    May 16 at 4:23
  • 1
    \$\begingroup\$ I see now I feel like an idiot. I usually put a 1k resistor at the base for this transistor and base voltage. I just totally forgot \$\endgroup\$
    – Feynman137
    May 16 at 4:25
  • 4
    \$\begingroup\$ I have this mental image of you standing over the circuit screaming at the flaming BJTs "Conduct! Conduct like you never have before!" \$\endgroup\$
    – DKNguyen
    May 16 at 4:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.