Using three TIP3055 transistors for RGB LED switching. Collectors are tied to cases. They need to be heat sunk and everything's already mounted and assembled and I don't have time to get rubber isolators.

Surprisingly it seems to work if I tie the collectors to ground and put the LED loads between the emitters and +12v. Running just a few LEDs right now to test the Arduino circuit, but I'll have around 4-5 amps of current through each transistor when finished. Is this safe or am I going to have a meltdown when I ramp up the current load?

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    \$\begingroup\$ Schematic where? \$\endgroup\$ Aug 14, 2014 at 19:28
  • \$\begingroup\$ A biploar transistor with emitter and collector swapped is still a transistor in terms of the junctions it has... but not a very good one as the geometric optimization is backwards. If it works without underperforming/overheating, it probably means your chosen part was overspec'd to begin with. \$\endgroup\$ Aug 14, 2014 at 19:36
  • \$\begingroup\$ @IgnacioVazquez-Abrams it scarcely matters. The question really hinges on device properties, the arrangement is trivial from the wording. \$\endgroup\$ Aug 14, 2014 at 19:36
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    \$\begingroup\$ @ChrisStratton: Train them early, train them often. \$\endgroup\$ Aug 14, 2014 at 19:40
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    \$\begingroup\$ @IgnacioVazquez-Abrams - no. What has general utility about the question is the fundamental 'can I put it in backwards' part - circuit diagrams would only distract from that. But a diagram of how a transistor is actually fabricated could be a great part of an answer. \$\endgroup\$ Aug 14, 2014 at 19:43

1 Answer 1


If you have the following:


simulate this circuit – Schematic created using CircuitLab

Yes it will work. This operating region of your BJT is called "Reverse active region". Now is it a good idea? Well... it will work but you will get crappy performance, since the current gain will be much lower. Also, the component may overheat (that region is rarely used and your device is not designed to be operated in that region). It is more a side effect of BJT's internal than a real expected usage. Usually, it is used only as failsafe in some circuits so you don't have to put diodes to discharge another part of the circuit or to do very clever circuits with very little component count. There is no point in you doing this, really.

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    \$\begingroup\$ While technically equivalent, this is a really confusing way to draw the circuit. It would be much better if you could talk not about the data sheet property, but how the collector and emitter differ. \$\endgroup\$ Aug 14, 2014 at 19:45
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    \$\begingroup\$ I'd like to, really. But I'm no semiconductor specialist at all. I understand how it works and side effects, but I'm very likely to provide a wrong explanation, so I prefer providing none to providing a bad one :) Sorry. But feel free to add another more detailed answer or complete mine if you feel like it. \$\endgroup\$
    – Mishyoshi
    Aug 14, 2014 at 19:51
  • \$\begingroup\$ Its good you put up what the operating region is called though. It is a good lead for anybody to go research should an answer explaining what Chris stated not come about. \$\endgroup\$
    – Funkyguy
    Aug 14, 2014 at 19:54
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    \$\begingroup\$ Actually, this is well outside the specs for the transistor. According to the datasheet, the maximum emitter-base voltage is just 7.0 volts. \$\endgroup\$
    – Dave Tweed
    Aug 14, 2014 at 20:20
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    \$\begingroup\$ @Mishyoshi - when you deactivate the transistor all the supply voltage will be reverse biasing the emitter-base region and then it will likely fail. This technique works for low voltages and low powers. \$\endgroup\$
    – Andy aka
    Aug 15, 2014 at 8:34

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