I have a circuit where a LED has its cathode connected to the collector of my C8050 (or SS8050) NPN transistor, with the emitter grounded. Even though the base is left floating, the current is still flowing freely to the emitter. Applying any current/voltage to the base doesn't change anything.

Here's the circuit: Circuit

Is this how it's supposed to work? Why? What can I do to use this transistor as a switch correctly?

(EDIT: I tried the same circuit setup with two different C8050 transistors, with the same result)

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    \$\begingroup\$ Does connecting base to emitter change anything? What power supply are you using i.e. voltage and polarity. \$\endgroup\$ – Andy aka Jun 11 '15 at 16:26
  • \$\begingroup\$ No, connecting B to E doesn't change anything, I've checked both current and voltage. I'm using the Arduino's 5V output pin as LED's power supply (with a 1K resistor in series connected to it) \$\endgroup\$ – James Game Jun 11 '15 at 16:33
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    \$\begingroup\$ A floating base is something you should generally avoid. A floating gate, in a mos, is even worse. But you say you apply "any current/voltage" to the base, that strucks me as unlikely. Please add a diagram of your circuit and expand a bit on that "any". \$\endgroup\$ – Vladimir Cravero Jun 11 '15 at 16:33
  • \$\begingroup\$ How freely is "freely". And what is your LED supply voltage? Note that the data sheet defines BVce as 2 mA at 25 volts. If you are using a high-efficiency LED and high LED voltage, you might be getting enough current to visibly light the LED, and still be within specs. \$\endgroup\$ – WhatRoughBeast Jun 11 '15 at 16:48
  • \$\begingroup\$ I added the photo including the circuit. By 'any' I mean 3.3V and a few voltages around 2V and 1V, also currents ranging from 0.5mA to 50mA \$\endgroup\$ – James Game Jun 11 '15 at 16:51

8050 transistors are made with the Japanese and the American style pinouts- it is not a JEDEC or JIS registered part.

Looks like the one you happen to have has the Japanese pinout.


  • \$\begingroup\$ @Spehro How did you ever figure that out??? That would have never cross my mind \$\endgroup\$ – Kvegaoro Jun 11 '15 at 20:48
  • \$\begingroup\$ @Kvegaoro Experience buying transistors in Asia. ;-) I have both varieties in my lab. Fortunately the SMT versions are standardized. \$\endgroup\$ – Spehro Pefhany Jun 11 '15 at 20:49
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    \$\begingroup\$ @SpehroPefhany Nothing beats experience!!! \$\endgroup\$ – Kvegaoro Jun 11 '15 at 20:52

No, that's not how it should work; with the base floating the transistor should not conduct.

My first guess would be that you connected it wrong. It's important to note that different transistors have different pinouts, make sure you refer to the datasheet for the C8050, and not some other transistor.

If you would mistake the base for the collector then the transistor will always conduct, whether you connect the emitter at the other end, or the collector. (On their own, base-collector and base-emitter junctions act as diodes.)

  • \$\begingroup\$ I also thought I connected it wrong, but the datasheet says otherwise. The current also flows from Collector to Base as well. My only guess here is that the datasheet is wrong \$\endgroup\$ – James Game Jun 11 '15 at 16:45
  • \$\begingroup\$ In an NPN transistor the collector-base junction would be reverse-polarized when you connect the collector to a higher voltage. There shouldn't be any current. I think either your transistor is broken, or it's not a C8050 (since you're sure you have the datasheet for that one). \$\endgroup\$ – Joris Groosman Jun 11 '15 at 16:50
  • \$\begingroup\$ The datasheet says that pins 1 2 3 are respectively Emitter Base Collector, but while connecting 1 2 3 as Emitter Collector Base the transistor behaves as expected. It's a very unusual setup, and it doesn't match the datasheet \$\endgroup\$ – James Game Jun 11 '15 at 16:58

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